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Advanced Engineering Mathematics by Erwin Kreyszig. Check also for solution manual

Advanced Engineering Mathematics by Erwin Kreyszig. Check also for solution manual

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## Advanced.engineering.mathematics.10th.edition.bdDocument Transcript

• Systems of Units. Some Important Conversion FactorsThe most important systems of units are shown in the table below. The mks system is also known asthe International System of Units (abbreviated SI), and the abbreviations sec (instead of s),gm (instead of g), and nt (instead of N) are also used.System of units Length Mass Time Forcecgs system centimeter (cm) gram (g) second (s) dynemks system meter (m) kilogram (kg) second (s) newton (nt)Engineering system foot (ft) slug second (s) pound (lb)1 inch (in.) ϭ 2.540000 cm 1 foot (ft) ϭ 12 in. ϭ 30.480000 cm1 yard (yd) ϭ 3 ft ϭ 91.440000 cm 1 statute mile (mi) ϭ 5280 ft ϭ 1.609344 km1 nautical mile ϭ 6080 ft ϭ 1.853184 km1 acre ϭ 4840 yd2ϭ 4046.8564 m21 mi2ϭ 640 acres ϭ 2.5899881 km21 fluid ounce ϭ 1/128 U.S. gallon ϭ 231/128 in.3ϭ 29.573730 cm31 U.S. gallon ϭ 4 quarts (liq) ϭ 8 pints (liq) ϭ 128 fl oz ϭ 3785.4118 cm31 British Imperial and Canadian gallon ϭ 1.200949 U.S. gallons ϭ 4546.087 cm31 slug ϭ 14.59390 kg1 pound (lb) ϭ 4.448444 nt 1 newton (nt) ϭ 105dynes1 British thermal unit (Btu) ϭ 1054.35 joules 1 joule ϭ 107ergs1 calorie (cal) ϭ 4.1840 joules1 kilowatt-hour (kWh) ϭ 3414.4 Btu ϭ 3.6 • 106joules1 horsepower (hp) ϭ 2542.48 Btu/h ϭ 178.298 cal/sec ϭ 0.74570 kW1 kilowatt (kW) ϭ 1000 watts ϭ 3414.43 Btu/h ϭ 238.662 cal/s°F ϭ °C • 1.8 ϩ 32 1° ϭ 60Ј ϭ 3600Љ ϭ 0.017453293 radianFor further details see, for example, D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics. 9th ed., Hoboken,N. J: Wiley, 2011. See also AN American National Standard, ASTM/IEEE Standard Metric Practice, Institute of Electrical andElectronics Engineers, Inc. (IEEE), 445 Hoes Lane, Piscataway, N. J. 08854, website at www.ieee.org.fendpaper.qxd 11/4/10 12:05 PM Page 2
• Integration͵uvЈ dx ϭ uv Ϫ ͵ uЈv dx (by parts)͵xndx ϭ ϩ c (n Ϫ1)͵ dx ϭ ln ͉x͉ ϩ c͵eaxdx ϭ eaxϩ c͵sin x dx ϭ Ϫcos x ϩ c͵cos x dx ϭ sin x ϩ c͵tan x dx ϭ Ϫln ͉cos x͉ ϩ c͵cot x dx ϭ ln ͉sin x͉ ϩ c͵sec x dx ϭ ln ͉sec x ϩ tan x͉ ϩ c͵csc x dx ϭ ln ͉csc x Ϫ cot x͉ ϩ c͵ ϭ arctan ϩ c͵ ϭ arcsin ϩ c͵ ϭ arcsinh ϩ c͵ ϭ arccosh ϩ c͵sin2x dx ϭ 1_2 x Ϫ 1_4 sin 2x ϩ c͵cos2x dx ϭ 1_2 x ϩ 1_4 sin 2x ϩ c͵tan2x dx ϭ tan x Ϫ x ϩ c͵cot2x dx ϭ Ϫcot x Ϫ x ϩ c͵ln x dx ϭ x ln x Ϫ x ϩ c͵eaxsin bx dxϭ (a sin bx Ϫ b cos bx) ϩ c͵eaxcos bx dxϭ (a cos bx ϩ b sin bx) ϩ ceaxa2ϩ b2eaxa2ϩ b2xᎏadxᎏᎏ͙x2ෆ Ϫෆ aෆ2ෆxᎏadxᎏᎏ͙x2ෆ ϩෆ aෆ2ෆxᎏadxᎏᎏ͙aෆ2ෆϪෆ xෆ2ෆxᎏa1ᎏadxᎏx2ϩ a21a1xxnϩ1n ϩ 1Differentiation(cu)Ј ϭ cuЈ (c constant)(u ϩ v)Ј ϭ uЈ ϩ vЈ(uv)Ј ϭ uЈv ϩ uvЈ( )Јϭϭ • (Chain rule)(xn)Ј ϭ nxn؊1(ex)Ј ϭ ex(eax)Ј ϭ aeax(ax)Ј ϭ axln a(sin x)Ј ϭ cos x(cos x)Ј ϭ Ϫsin x(tan x)Ј ϭ sec2x(cot x)Ј ϭ Ϫcsc2x(sinh x)Ј ϭ cosh x(cosh x)Ј ϭ sinh x(ln x)Ј ϭ(loga x)Ј ϭ(arcsin x)Ј ϭ(arccos x)Ј ϭ Ϫ(arctan x)Ј ϭ(arccot x)Ј ϭ Ϫ1ᎏ1 ϩ x21ᎏ1 ϩ x21ᎏᎏ͙1ෆ Ϫෆ xෆ2ෆ1ᎏᎏ͙1ෆ Ϫෆ xෆ2ෆloga eᎏx1ᎏxdyᎏdxduᎏdyduᎏdxuЈv Ϫ uvЈᎏᎏv2uᎏvfendpaper.qxd 11/4/10 12:05 PM Page 3
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• ADVANCEDENGINEERINGMATHEMATICSffirs.qxd 11/8/10 3:50 PM Page iii
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• 10T H E D I T I O NADVANCEDENGINEERINGMATHEMATICSERWIN KREYSZIGProfessor of MathematicsOhio State UniversityColumbus, OhioIn collaboration withHERBERT KREYSZIGNew York, New YorkEDWARD J. NORMINTONAssociate Professor of MathematicsCarleton UniversityOttawa, OntarioJOHN WILEY & SONS, INC.ffirs.qxd 11/8/10 3:50 PM Page v
• PUBLISHER Laurie RosatonePROJECT EDITOR Shannon CorlissMARKETING MANAGER Jonathan CottrellCONTENT MANAGER Lucille BuonocorePRODUCTION EDITOR Barbara RussielloMEDIA EDITOR Melissa EdwardsMEDIA PRODUCTION SPECIALIST Lisa SabatiniTEXT AND COVER DESIGN Madelyn LesurePHOTO RESEARCHER Sheena GoldsteinCOVER PHOTO © Denis Jr. Tangney/iStockphotoCover photo shows the Zakim Bunker Hill Memorial Bridge inBoston, MA.This book was set in Times Roman. The book was composed by PreMedia Global, and printed and bound byRR Donnelley & Sons Company, Jefferson City, MO. The cover was printed by RR Donnelley & Sons Company,Jefferson City, MO.This book is printed on acid free paper.Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for morethan 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company isbuilt on a foundation of principles that include responsibility to the communities we serve and where we live andwork. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social,economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact,paper specifications and procurement, ethical conduct within our business and among our vendors, and communityand charitable support. For more information, please visit our website: www.wiley.com/go/citizenship.Copyright © 2011, 2006, 1999 by John Wiley & Sons, Inc. All rights reserved. No part of this publication maybe reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical,photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 UnitedStates Copyright Act, without either the prior written permission of the Publisher, or authorization through paymentof the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA01923 (Web site: www.copyright.com). Requests to the Publisher for permission should be addressed to thePermissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011,fax (201) 748-6008, or online at: www.wiley.com/go/permissions.Evaluation copies are provided to qualified academics and professionals for review purposes only, for use intheir courses during the next academic year. These copies are licensed and may not be sold or transferred to athird party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructionsand a free of charge return shipping label are available at: www.wiley.com/go/returnlabel. Outside of the UnitedStates, please contact your local representative.ISBN 978-0-470-45836-5Printed in the United States of America10 9 8 7 6 5 4 3 2 1ϱffirs.qxd 11/4/10 10:50 AM Page vi
• P R E F A C ESee also http://www.wiley.com/college/kreyszigPurpose and Structure of the BookThis book provides a comprehensive, thorough, and up-to-date treatment of engineeringmathematics. It is intended to introduce students of engineering, physics, mathematics,computer science, and related fields to those areas of applied mathematics that are mostrelevant for solving practical problems. A course in elementary calculus is the soleprerequisite. (However, a concise refresher of basic calculus for the student is includedon the inside cover and in Appendix 3.)The subject matter is arranged into seven parts as follows:A. Ordinary Differential Equations (ODEs) in Chapters 1–6B. Linear Algebra. Vector Calculus. See Chapters 7–10C. Fourier Analysis. Partial Differential Equations (PDEs). See Chapters 11 and 12D. Complex Analysis in Chapters 13–18E. Numeric Analysis in Chapters 19–21F. Optimization, Graphs in Chapters 22 and 23G. Probability, Statistics in Chapters 24 and 25.These are followed by five appendices: 1. References, 2. Answers to Odd-NumberedProblems, 3. Auxiliary Materials (see also inside covers of book), 4. Additional Proofs,5. Table of Functions. This is shown in a block diagram on the next page.The parts of the book are kept independent. In addition, individual chapters are kept asindependent as possible. (If so needed, any prerequisites—to the level of individualsections of prior chapters—are clearly stated at the opening of each chapter.) We give theinstructor maximum flexibility in selecting the material and tailoring it to his or herneed. The book has helped to pave the way for the present development of engineeringmathematics. This new edition will prepare the student for the current tasks and the futureby a modern approach to the areas listed above. We provide the material and learningtools for the students to get a good foundation of engineering mathematics that will helpthem in their careers and in further studies.General Features of the Book Include:• Simplicity of examples to make the book teachable—why choose complicatedexamples when simple ones are as instructive or even better?• Independence of parts and blocks of chapters to provide flexibility in tailoringcourses to specific needs.• Self-contained presentation, except for a few clearly marked places where a proofwould exceed the level of the book and a reference is given instead.• Gradual increase in difficulty of material with no jumps or gaps to ensure anenjoyable teaching and learning experience.• Modern standard notation to help students with other courses, modern books, andjournals in mathematics, engineering, statistics, physics, computer science, and others.Furthermore, we designed the book to be a single, self-contained, authoritative, andconvenient source for studying and teaching applied mathematics, eliminating the needfor time-consuming searches on the Internet or time-consuming trips to the library to geta particular reference book.viifpref.qxd 11/8/10 3:16 PM Page vii
• viii PrefaceGUIDES AND MANUALSMaple Computer GuideMathematica Computer GuideStudent Solutions Manualand Study GuideInstructor’s ManualPART AChaps. 1–6Ordinary Differential Equations (ODEs)Chaps. 1–4Basic MaterialChap. 5 Chap. 6Series Solutions Laplace TransformsPART BChaps. 7–10Linear Algebra. Vector CalculusChap. 7 Chap. 9Matrices, Vector DifferentialLinear Systems CalculusChap. 8 Chap. 10Eigenvalue Problems Vector Integral CalculusPARTS AND CHAPTERS OF THE BOOKPART CChaps. 11–12Fourier Analysis. Partial DifferentialEquations (PDEs)Chap. 11Fourier AnalysisChap. 12Partial Differential EquationsPART DChaps. 13–18Complex Analysis,Potential TheoryChaps. 13–17Basic MaterialChap. 18Potential TheoryPART EChaps. 19–21Numeric AnalysisChap. 19 Chap. 20 Chap. 21Numerics in Numeric Numerics forGeneral Linear Algebra ODEs and PDEsPART FChaps. 22–23Optimization, GraphsChap. 22 Chap. 23Linear Programming Graphs, OptimizationPART GChaps. 24–25Probability, StatisticsChap. 24Data Analysis. Probability TheoryChap. 25Mathematical Statisticsfpref.qxd 11/8/10 3:16 PM Page viii
• Four Underlying Themes of the BookThe driving force in engineering mathematics is the rapid growth of technology and thesciences. New areas—often drawing from several disciplines—come into existence.Electric cars, solar energy, wind energy, green manufacturing, nanotechnology, riskmanagement, biotechnology, biomedical engineering, computer vision, robotics, spacetravel, communication systems, green logistics, transportation systems, financialengineering, economics, and many other areas are advancing rapidly. What does this meanfor engineering mathematics? The engineer has to take a problem from any diverse areaand be able to model it. This leads to the first of four underlying themes of the book.1. Modeling is the process in engineering, physics, computer science, biology,chemistry, environmental science, economics, and other fields whereby a physical situationor some other observation is translated into a mathematical model. This mathematicalmodel could be a system of differential equations, such as in population control (Sec. 4.5),a probabilistic model (Chap. 24), such as in risk management, a linear programmingproblem (Secs. 22.2–22.4) in minimizing environmental damage due to pollutants, afinancial problem of valuing a bond leading to an algebraic equation that has to be solvedby Newton’s method (Sec. 19.2), and many others.The next step is solving the mathematical problem obtained by one of the manytechniques covered in Advanced Engineering Mathematics.The third step is interpreting the mathematical result in physical or other terms tosee what it means in practice and any implications.Finally, we may have to make a decision that may be of an industrial nature orrecommend a public policy. For example, the population control model may implythe policy to stop fishing for 3 years. Or the valuation of the bond may lead to arecommendation to buy. The variety is endless, but the underlying mathematics issurprisingly powerful and able to provide advice leading to the achievement of goalstoward the betterment of society, for example, by recommending wise policiesconcerning global warming, better allocation of resources in a manufacturing process,or making statistical decisions (such as in Sec. 25.4 whether a drug is effective in treatinga disease).While we cannot predict what the future holds, we do know that the student has topractice modeling by being given problems from many different applications as is donein this book. We teach modeling from scratch, right in Sec. 1.1, and give many examplesin Sec. 1.3, and continue to reinforce the modeling process throughout the book.2. Judicious use of powerful software for numerics (listed in the beginning of Part E)and statistics (Part G) is of growing importance. Projects in engineering and industrialcompanies may involve large problems of modeling very complex systems with hundredsof thousands of equations or even more. They require the use of such software. However,our policy has always been to leave it up to the instructor to determine the degree of use ofcomputers, from none or little use to extensive use. More on this below.3. The beauty of engineering mathematics. Engineering mathematics relies onrelatively few basic concepts and involves powerful unifying principles. We point themout whenever they are clearly visible, such as in Sec. 4.1 where we “grow” a mixingproblem from one tank to two tanks and a circuit problem from one circuit to two circuits,thereby also increasing the number of ODEs from one ODE to two ODEs. This is anexample of an attractive mathematical model because the “growth” in the problem isreflected by an “increase” in ODEs.Preface ixfpref.qxd 11/8/10 3:16 PM Page ix
• 4. To clearly identify the conceptual structure of subject matters. For example,complex analysis (in Part D) is a field that is not monolithic in structure but was formedby three distinct schools of mathematics. Each gave a different approach, which we clearlymark. The first approach is solving complex integrals by Cauchy’s integral formula (Chaps.13 and 14), the second approach is to use the Laurent series and solve complex integralsby residue integration (Chaps. 15 and 16), and finally we use a geometric approach ofconformal mapping to solve boundary value problems (Chaps. 17 and 18). Learning theconceptual structure and terminology of the different areas of engineering mathematics isvery important for three reasons:a. It allows the student to identify a new problem and put it into the right group ofproblems. The areas of engineering mathematics are growing but most often retain theirconceptual structure.b. The student can absorb new information more rapidly by being able to fit it into theconceptual structure.c. Knowledge of the conceptual structure and terminology is also important when usingthe Internet to search for mathematical information. Since the search proceeds by puttingin key words (i.e., terms) into the search engine, the student has to remember the importantconcepts (or be able to look them up in the book) that identify the application and areaof engineering mathematics.Big Changes in This EditionProblem Sets ChangedThe problem sets have been revised and rebalanced with some problem sets having moreproblems and some less, reflecting changes in engineering mathematics. There is a greateremphasis on modeling. Now there are also problems on the discrete Fourier transform(in Sec. 11.9).Series Solutions of ODEs, Special Functions and Fourier Analysis ReorganizedChap. 5, on series solutions of ODEs and special functions, has been shortened. Chap. 11on Fourier Analysis now contains Sturm–Liouville problems, orthogonal functions, andorthogonal eigenfunction expansions (Secs. 11.5, 11.6), where they fit better conceptually(rather than in Chap. 5), being extensions of Fourier’s idea of using orthogonal functions.Openings of Parts and Chapters Rewritten As Well As Parts of SectionsIn order to give the student a better idea of the structure of the material (see UnderlyingTheme 4 above), we have entirely rewritten the openings of parts and chapters.Furthermore, large parts or individual paragraphs of sections have been rewritten or newsentences inserted into the text. This should give the students a better intuitiveunderstanding of the material (see Theme 3 above), let them draw conclusions on theirown, and be able to tackle more advanced material. Overall, we feel that the book hasbecome more detailed and leisurely written.Student Solutions Manual and Study Guide EnlargedUpon the explicit request of the users, the answers provided are more detailed andcomplete. More explanations are given on how to learn the material effectively by pointingout what is most important.More Historical Footnotes, Some EnlargedHistorical footnotes are there to show the student that many people from different countriesworking in different professions, such as surveyors, researchers in industry, etc., contributed54321x Prefacefpref.qxd 11/8/10 3:16 PM Page x
• to the field of engineering mathematics. It should encourage the students to be creative intheir own interests and careers and perhaps also to make contributions to engineeringmathematics.Further Changes and New Features• Parts of Chap. 1 on first-order ODEs are rewritten. More emphasis on modeling, alsonew block diagram explaining this concept in Sec. 1.1. Early introduction of Euler’smethod in Sec. 1.2 to familiarize student with basic numerics. More examples ofseparable ODEs in Sec. 1.3.• For Chap. 2, on second-order ODEs, note the following changes: For ease of reading,the first part of Sec. 2.4, which deals with setting up the mass-spring system, hasbeen rewritten; also some rewriting in Sec. 2.5 on the Euler–Cauchy equation.• Substantially shortened Chap. 5, Series Solutions of ODEs. Special Functions:combined Secs. 5.1 and 5.2 into one section called “Power Series Method,” shortenedmaterial in Sec. 5.4 Bessel’s Equation (of the first kind), removed Sec. 5.7(Sturm–Liouville Problems) and Sec. 5.8 (Orthogonal Eigenfunction Expansions) andmoved material into Chap. 11 (see “Major Changes” above).• New equivalent definition of basis (Sec. 7.4).• In Sec. 7.9, completely new part on composition of linear transformations withtwo new examples. Also, more detailed explanation of the role of axioms, inconnection with the definition of vector space.• New table of orientation (opening of Chap. 8 “Linear Algebra: Matrix EigenvalueProblems”) where eigenvalue problems occur in the book. More intuitive explanationof what an eigenvalue is at the begining of Sec. 8.1.• Better definition of cross product (in vector differential calculus) by properlyidentifying the degenerate case (in Sec. 9.3).• Chap. 11 on Fourier Analysis extensively rearranged: Secs. 11.2 and 11.3combined into one section (Sec. 11.2), old Sec. 11.4 on complex Fourier Seriesremoved and new Secs. 11.5 (Sturm–Liouville Problems) and 11.6 (OrthogonalSeries) put in (see “Major Changes” above). New problems (new!) in problem set11.9 on discrete Fourier transform.• New section 12.5 on modeling heat flow from a body in space by setting up the heatequation. Modeling PDEs is more difficult so we separated the modeling processfrom the solving process (in Sec. 12.6).• Introduction to Numerics rewritten for greater clarity and better presentation; newExample 1 on how to round a number. Sec. 19.3 on interpolation shortened byremoving the less important central difference formula and giving a reference instead.• Large new footnote with historical details in Sec. 22.3, honoring George Dantzig,the inventor of the simplex method.• Traveling salesman problem now described better as a “difficult” problem, typicalof combinatorial optimization (in Sec. 23.2). More careful explanation on how tocompute the capacity of a cut set in Sec. 23.6 (Flows on Networks).• In Chap. 24, material on data representation and characterization restructured interms of five examples and enlarged to include empirical rule on distribution ofPreface xifpref.qxd 11/8/10 3:16 PM Page xi
• xii Prefacedata, outliers, and the z-score (Sec. 24.1). Furthermore, new example on encription(Sec. 24.4).• Lists of software for numerics (Part E) and statistics (Part G) updated.• References in Appendix 1 updated to include new editions and some references towebsites.Use of ComputersThe presentation in this book is adaptable to various degrees of use of software,Computer Algebra Systems (CAS’s), or programmable graphic calculators, rangingfrom no use, very little use, medium use, to intensive use of such technology. The choiceof how much computer content the course should have is left up to the instructor, therebyexhibiting our philosophy of maximum flexibility and adaptability. And, no matter whatthe instructor decides, there will be no gaps or jumps in the text or problem set. Someproblems are clearly designed as routine and drill exercises and should be solved byhand (paper and pencil, or typing on your computer). Other problems require morethinking and can also be solved without computers. Then there are problems where thecomputer can give the student a hand. And finally, the book has CAS projects, CASproblems and CAS experiments, which do require a computer, and show its power insolving problems that are difficult or impossible to access otherwise. Here our goal isto combine intelligent computer use with high-quality mathematics. The computerinvites visualization, experimentation, and independent discovery work. In summary,the high degree of flexibility of computer use for the book is possible since there areplenty of problems to choose from and the CAS problems can be omitted if desired.Note that information on software (what is available and where to order it) is at thebeginning of Part E on Numeric Analysis and Part G on Probability and Statistics. SinceMaple and Mathematica are popular Computer Algebra Systems, there are two computerguides available that are specifically tailored to Advanced Engineering Mathematics:E. Kreyszig and E.J. Norminton, Maple Computer Guide, 10th Edition and MathematicaComputer Guide, 10th Edition. Their use is completely optional as the text in the book iswritten without the guides in mind.Suggestions for Courses: A Four-Semester SequenceThe material, when taken in sequence, is suitable for four consecutive semester courses,meeting 3 to 4 hours a week:1st Semester ODEs (Chaps. 1–5 or 1–6)2nd Semester Linear Algebra. Vector Analysis (Chaps. 7–10)3rd Semester Complex Analysis (Chaps. 13–18)4th Semester Numeric Methods (Chaps. 19–21)Suggestions for Independent One-Semester CoursesThe book is also suitable for various independent one-semester courses meeting 3 hoursa week. For instance,Introduction to ODEs (Chaps. 1–2, 21.1)Laplace Transforms (Chap. 6)Matrices and Linear Systems (Chaps. 7–8)fpref.qxd 11/8/10 3:16 PM Page xii
• Vector Algebra and Calculus (Chaps. 9–10)Fourier Series and PDEs (Chaps. 11–12, Secs. 21.4–21.7)Introduction to Complex Analysis (Chaps. 13–17)Numeric Analysis (Chaps. 19, 21)Numeric Linear Algebra (Chap. 20)Optimization (Chaps. 22–23)Graphs and Combinatorial Optimization (Chap. 23)Probability and Statistics (Chaps. 24–25)AcknowledgmentsWe are indebted to former teachers, colleagues, and students who helped us directly orindirectly in preparing this book, in particular this new edition. We profited greatly fromdiscussions with engineers, physicists, mathematicians, computer scientists, and others,and from their written comments. We would like to mention in particular ProfessorsY. A. Antipov, R. Belinski, S. L. Campbell, R. Carr, P. L. Chambré, Isabel F. Cruz,Z. Davis, D. Dicker, L. D. Drager, D. Ellis, W. Fox, A. Goriely, R. B. Guenther,J. B. Handley, N. Harbertson, A. Hassen, V. W. Howe, H. Kuhn, K. Millet, J. D. Moore,W. D. Munroe, A. Nadim, B. S. Ng, J. N. Ong, P. J. Pritchard, W. O. Ray, L. F. Shampine,H. L. Smith, Roberto Tamassia, A. L. Villone, H. J. Weiss, A. Wilansky, Neil M. Wigley,and L. Ying; Maria E. and Jorge A. Miranda, JD, all from the United States; ProfessorsWayne H. Enright, Francis. L. Lemire, James J. Little, David G. Lowe, Gerry McPhail,Theodore S. Norvell, and R. Vaillancourt; Jeff Seiler and David Stanley, all from Canada;and Professor Eugen Eichhorn, Gisela Heckler, Dr. Gunnar Schroeder, and WiltrudStiefenhofer from Europe. Furthermore, we would like to thank Professors JohnB. Donaldson, Bruce C. N. Greenwald, Jonathan L. Gross, Morris B. Holbrook, JohnR. Kender, and Bernd Schmitt; and Nicholaiv Villalobos, all from Columbia University,New York; as well as Dr. Pearl Chang, Chris Gee, Mike Hale, Joshua Jayasingh, MD,David Kahr, Mike Lee, R. Richard Royce, Elaine Schattner, MD, Raheel Siddiqui, RobertSullivan, MD, Nancy Veit, and Ana M. Kreyszig, JD, all from New York City. We wouldalso like to gratefully acknowledge the use of facilities at Carleton University, Ottawa,and Columbia University, New York.Furthermore we wish to thank John Wiley and Sons, in particular Publisher LaurieRosatone, Editor Shannon Corliss, Production Editor Barbara Russiello, Media EditorMelissa Edwards, Text and Cover Designer Madelyn Lesure, and Photo Editor SheenaGoldstein for their great care and dedication in preparing this edition. In the same vein,we would also like to thank Beatrice Ruberto, copy editor and proofreader, WordCo, forthe Index, and Joyce Franzen of PreMedia and those of PreMedia Global who typeset thisedition.Suggestions of many readers worldwide were evaluated in preparing this edition.Further comments and suggestions for improving the book will be gratefully received.KREYSZIGPreface xiiifpref.qxd 11/8/10 8:51 PM Page xiii
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• xvC O N T E N T SP A R T A Ordinary Differential Equations (ODEs) 1CHAPTER 1 First-Order ODEs 21.1 Basic Concepts. Modeling 21.2 Geometric Meaning of yЈ ϭ ƒ(x, y). Direction Fields, Euler’s Method 91.3 Separable ODEs. Modeling 121.4 Exact ODEs. Integrating Factors 201.5 Linear ODEs. Bernoulli Equation. Population Dynamics 271.6 Orthogonal Trajectories. Optional 361.7 Existence and Uniqueness of Solutions for Initial Value Problems 38Chapter 1 Review Questions and Problems 43Summary of Chapter 1 44CHAPTER 2 Second-Order Linear ODEs 462.1 Homogeneous Linear ODEs of Second Order 462.2 Homogeneous Linear ODEs with Constant Coefficients 532.3 Differential Operators. Optional 602.4 Modeling of Free Oscillations of a Mass–Spring System 622.5 Euler–Cauchy Equations 712.6 Existence and Uniqueness of Solutions. Wronskian 742.7 Nonhomogeneous ODEs 792.8 Modeling: Forced Oscillations. Resonance 852.9 Modeling: Electric Circuits 932.10 Solution by Variation of Parameters 99Chapter 2 Review Questions and Problems 102Summary of Chapter 2 103CHAPTER 3 Higher Order Linear ODEs 1053.1 Homogeneous Linear ODEs 1053.2 Homogeneous Linear ODEs with Constant Coefficients 1113.3 Nonhomogeneous Linear ODEs 116Chapter 3 Review Questions and Problems 122Summary of Chapter 3 123CHAPTER 4 Systems of ODEs. Phase Plane. Qualitative Methods 1244.0 For Reference: Basics of Matrices and Vectors 1244.1 Systems of ODEs as Models in Engineering Applications 1304.2 Basic Theory of Systems of ODEs. Wronskian 1374.3 Constant-Coefficient Systems. Phase Plane Method 1404.4 Criteria for Critical Points. Stability 1484.5 Qualitative Methods for Nonlinear Systems 1524.6 Nonhomogeneous Linear Systems of ODEs 160Chapter 4 Review Questions and Problems 164Summary of Chapter 4 165CHAPTER 5 Series Solutions of ODEs. Special Functions 1675.1 Power Series Method 1675.2 Legendre’s Equation. Legendre Polynomials Pn(x) 175ftoc.qxd 11/4/10 11:48 AM Page xv
• 5.3 Extended Power Series Method: Frobenius Method 1805.4 Bessel’s Equation. Bessel Functions J␯(x) 1875.5 Bessel Functions of the Y␯(x). General Solution 196Chapter 5 Review Questions and Problems 200Summary of Chapter 5 201CHAPTER 6 Laplace Transforms 2036.1 Laplace Transform. Linearity. First Shifting Theorem (s-Shifting) 2046.2 Transforms of Derivatives and Integrals. ODEs 2116.3 Unit Step Function (Heaviside Function).Second Shifting Theorem (t-Shifting) 2176.4 Short Impulses. Dirac’s Delta Function. Partial Fractions 2256.5 Convolution. Integral Equations 2326.6 Differentiation and Integration of Transforms.ODEs with Variable Coefficients 2386.7 Systems of ODEs 2426.8 Laplace Transform: General Formulas 2486.9 Table of Laplace Transforms 249Chapter 6 Review Questions and Problems 251Summary of Chapter 6 253P A R T B Linear Algebra. Vector Calculus 255CHAPTER 7 Linear Algebra: Matrices, Vectors, Determinants.Linear Systems 2567.1 Matrices, Vectors: Addition and Scalar Multiplication 2577.2 Matrix Multiplication 2637.3 Linear Systems of Equations. Gauss Elimination 2727.4 Linear Independence. Rank of a Matrix. Vector Space 2827.5 Solutions of Linear Systems: Existence, Uniqueness 2887.6 For Reference: Second- and Third-Order Determinants 2917.7 Determinants. Cramer’s Rule 2937.8 Inverse of a Matrix. Gauss–Jordan Elimination 3017.9 Vector Spaces, Inner Product Spaces. Linear Transformations. Optional 309Chapter 7 Review Questions and Problems 318Summary of Chapter 7 320CHAPTER 8 Linear Algebra: Matrix Eigenvalue Problems 3228.1 The Matrix Eigenvalue Problem.Determining Eigenvalues and Eigenvectors 3238.2 Some Applications of Eigenvalue Problems 3298.3 Symmetric, Skew-Symmetric, and Orthogonal Matrices 3348.4 Eigenbases. Diagonalization. Quadratic Forms 3398.5 Complex Matrices and Forms. Optional 346Chapter 8 Review Questions and Problems 352Summary of Chapter 8 353xvi Contentsftoc.qxd 11/4/10 11:48 AM Page xvi
• CHAPTER 9 Vector Differential Calculus. Grad, Div, Curl 3549.1 Vectors in 2-Space and 3-Space 3549.2 Inner Product (Dot Product) 3619.3 Vector Product (Cross Product) 3689.4 Vector and Scalar Functions and Their Fields. Vector Calculus: Derivatives 3759.5 Curves. Arc Length. Curvature. Torsion 3819.6 Calculus Review: Functions of Several Variables. Optional 3929.7 Gradient of a Scalar Field. Directional Derivative 3959.8 Divergence of a Vector Field 4029.9 Curl of a Vector Field 406Chapter 9 Review Questions and Problems 409Summary of Chapter 9 410CHAPTER 10 Vector Integral Calculus. Integral Theorems 41310.1 Line Integrals 41310.2 Path Independence of Line Integrals 41910.3 Calculus Review: Double Integrals. Optional 42610.4 Green’s Theorem in the Plane 43310.5 Surfaces for Surface Integrals 43910.6 Surface Integrals 44310.7 Triple Integrals. Divergence Theorem of Gauss 45210.8 Further Applications of the Divergence Theorem 45810.9 Stokes’s Theorem 463Chapter 10 Review Questions and Problems 469Summary of Chapter 10 470P A R T C Fourier Analysis. Partial Differential Equations (PDEs) 473CHAPTER 11 Fourier Analysis 47411.1 Fourier Series 47411.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions 48311.3 Forced Oscillations 49211.4 Approximation by Trigonometric Polynomials 49511.5 Sturm–Liouville Problems. Orthogonal Functions 49811.6 Orthogonal Series. Generalized Fourier Series 50411.7 Fourier Integral 51011.8 Fourier Cosine and Sine Transforms 51811.9 Fourier Transform. Discrete and Fast Fourier Transforms 52211.10 Tables of Transforms 534Chapter 11 Review Questions and Problems 537Summary of Chapter 11 538CHAPTER 12 Partial Differential Equations (PDEs) 54012.1 Basic Concepts of PDEs 54012.2 Modeling: Vibrating String, Wave Equation 54312.3 Solution by Separating Variables. Use of Fourier Series 54512.4 D’Alembert’s Solution of the Wave Equation. Characteristics 55312.5 Modeling: Heat Flow from a Body in Space. Heat Equation 557Contents xviiftoc.qxd 11/4/10 11:48 AM Page xvii
• 12.6 Heat Equation: Solution by Fourier Series.Steady Two-Dimensional Heat Problems. Dirichlet Problem 55812.7 Heat Equation: Modeling Very Long Bars.Solution by Fourier Integrals and Transforms 56812.8 Modeling: Membrane, Two-Dimensional Wave Equation 57512.9 Rectangular Membrane. Double Fourier Series 57712.10 Laplacian in Polar Coordinates. Circular Membrane. Fourier–Bessel Series 58512.11 Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential 59312.12 Solution of PDEs by Laplace Transforms 600Chapter 12 Review Questions and Problems 603Summary of Chapter 12 604P A R T D Complex Analysis 607CHAPTER 13 Complex Numbers and Functions.Complex Differentiation 60813.1 Complex Numbers and Their Geometric Representation 60813.2 Polar Form of Complex Numbers. Powers and Roots 61313.3 Derivative. Analytic Function 61913.4 Cauchy–Riemann Equations. Laplace’s Equation 62513.5 Exponential Function 63013.6 Trigonometric and Hyperbolic Functions. Euler’s Formula 63313.7 Logarithm. General Power. Principal Value 636Chapter 13 Review Questions and Problems 641Summary of Chapter 13 641CHAPTER 14 Complex Integration 64314.1 Line Integral in the Complex Plane 64314.2 Cauchy’s Integral Theorem 65214.3 Cauchy’s Integral Formula 66014.4 Derivatives of Analytic Functions 664Chapter 14 Review Questions and Problems 668Summary of Chapter 14 669CHAPTER 15 Power Series, Taylor Series 67115.1 Sequences, Series, Convergence Tests 67115.2 Power Series 68015.3 Functions Given by Power Series 68515.4 Taylor and Maclaurin Series 69015.5 Uniform Convergence. Optional 698Chapter 15 Review Questions and Problems 706Summary of Chapter 15 706CHAPTER 16 Laurent Series. Residue Integration 70816.1 Laurent Series 70816.2 Singularities and Zeros. Infinity 71516.3 Residue Integration Method 71916.4 Residue Integration of Real Integrals 725Chapter 16 Review Questions and Problems 733Summary of Chapter 16 734xviii Contentsftoc.qxd 11/4/10 11:48 AM Page xviii
• CHAPTER 17 Conformal Mapping 73617.1 Geometry of Analytic Functions: Conformal Mapping 73717.2 Linear Fractional Transformations (Möbius Transformations) 74217.3 Special Linear Fractional Transformations 74617.4 Conformal Mapping by Other Functions 75017.5 Riemann Surfaces. Optional 754Chapter 17 Review Questions and Problems 756Summary of Chapter 17 757CHAPTER 18 Complex Analysis and Potential Theory 75818.1 Electrostatic Fields 75918.2 Use of Conformal Mapping. Modeling 76318.3 Heat Problems 76718.4 Fluid Flow 77118.5 Poisson’s Integral Formula for Potentials 77718.6 General Properties of Harmonic Functions.Uniqueness Theorem for the Dirichlet Problem 781Chapter 18 Review Questions and Problems 785Summary of Chapter 18 786P A R T E Numeric Analysis 787Software 788CHAPTER 19 Numerics in General 79019.1 Introduction 79019.2 Solution of Equations by Iteration 79819.3 Interpolation 80819.4 Spline Interpolation 82019.5 Numeric Integration and Differentiation 827Chapter 19 Review Questions and Problems 841Summary of Chapter 19 842CHAPTER 20 Numeric Linear Algebra 84420.1 Linear Systems: Gauss Elimination 84420.2 Linear Systems: LU-Factorization, Matrix Inversion 85220.3 Linear Systems: Solution by Iteration 85820.4 Linear Systems: Ill-Conditioning, Norms 86420.5 Least Squares Method 87220.6 Matrix Eigenvalue Problems: Introduction 87620.7 Inclusion of Matrix Eigenvalues 87920.8 Power Method for Eigenvalues 88520.9 Tridiagonalization and QR-Factorization 888Chapter 20 Review Questions and Problems 896Summary of Chapter 20 898CHAPTER 21 Numerics for ODEs and PDEs 90021.1 Methods for First-Order ODEs 90121.2 Multistep Methods 91121.3 Methods for Systems and Higher Order ODEs 915Contents xixftoc.qxd 11/4/10 11:48 AM Page xix
• 21.4 Methods for Elliptic PDEs 92221.5 Neumann and Mixed Problems. Irregular Boundary 93121.6 Methods for Parabolic PDEs 93621.7 Method for Hyperbolic PDEs 942Chapter 21 Review Questions and Problems 945Summary of Chapter 21 946P A R T F Optimization, Graphs 949CHAPTER 22 Unconstrained Optimization. Linear Programming 95022.1 Basic Concepts. Unconstrained Optimization: Method of Steepest Descent 95122.2 Linear Programming 95422.3 Simplex Method 95822.4 Simplex Method: Difficulties 962Chapter 22 Review Questions and Problems 968Summary of Chapter 22 969CHAPTER 23 Graphs. Combinatorial Optimization 97023.1 Graphs and Digraphs 97023.2 Shortest Path Problems. Complexity 97523.3 Bellman’s Principle. Dijkstra’s Algorithm 98023.4 Shortest Spanning Trees: Greedy Algorithm 98423.5 Shortest Spanning Trees: Prim’s Algorithm 98823.6 Flows in Networks 99123.7 Maximum Flow: Ford–Fulkerson Algorithm 99823.8 Bipartite Graphs. Assignment Problems 1001Chapter 23 Review Questions and Problems 1006Summary of Chapter 23 1007P A R T G Probability, Statistics 1009Software 1009CHAPTER 24 Data Analysis. Probability Theory 101124.1 Data Representation. Average. Spread 101124.2 Experiments, Outcomes, Events 101524.3 Probability 101824.4 Permutations and Combinations 102424.5 Random Variables. Probability Distributions 102924.6 Mean and Variance of a Distribution 103524.7 Binomial, Poisson, and Hypergeometric Distributions 103924.8 Normal Distribution 104524.9 Distributions of Several Random Variables 1051Chapter 24 Review Questions and Problems 1060Summary of Chapter 24 1060CHAPTER 25 Mathematical Statistics 106325.1 Introduction. Random Sampling 106325.2 Point Estimation of Parameters 106525.3 Confidence Intervals 1068xx Contentsftoc.qxd 11/4/10 11:48 AM Page xx
• 25.4 Testing Hypotheses. Decisions 107725.5 Quality Control 108725.6 Acceptance Sampling 109225.7 Goodness of Fit. ␹2-Test 109625.8 Nonparametric Tests 110025.9 Regression. Fitting Straight Lines. Correlation 1103Chapter 25 Review Questions and Problems 1111Summary of Chapter 25 1112APPENDIX 1 References A1APPENDIX 2 Answers to Odd-Numbered Problems A4APPENDIX 3 Auxiliary Material A63A3.1 Formulas for Special Functions A63A3.2 Partial Derivatives A69A3.3 Sequences and Series A72A3.4 Grad, Div, Curl, ٌ2in Curvilinear Coordinates A74APPENDIX 4 Additional Proofs A77APPENDIX 5 Tables A97INDEX I1PHOTO CREDITS P1Contents xxiftoc.qxd 11/4/10 11:48 AM Page xxi
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• CHAPTER 1 First-Order ODEsCHAPTER 2 Second-Order Linear ODEsCHAPTER 3 Higher Order Linear ODEsCHAPTER 4 Systems of ODEs. Phase Plane. Qualitative MethodsCHAPTER 5 Series Solutions of ODEs. Special FunctionsCHAPTER 6 Laplace TransformsMany physical laws and relations can be expressed mathematically in the form of differentialequations. Thus it is natural that this book opens with the study of differential equations andtheir solutions. Indeed, many engineering problems appear as differential equations.The main objectives of Part A are twofold: the study of ordinary differential equationsand their most important methods for solving them and the study of modeling.Ordinary differential equations (ODEs) are differential equations that depend on a singlevariable. The more difficult study of partial differential equations (PDEs), that is,differential equations that depend on several variables, is covered in Part C.Modeling is a crucial general process in engineering, physics, computer science, biology,medicine, environmental science, chemistry, economics, and other fields that translates aphysical situation or some other observations into a “mathematical model.” Numerousexamples from engineering (e.g., mixing problem), physics (e.g., Newton’s law of cooling),biology (e.g., Gompertz model), chemistry (e.g., radiocarbon dating), environmental science(e.g., population control), etc. shall be given, whereby this process is explained in detail,that is, how to set up the problems correctly in terms of differential equations.For those interested in solving ODEs numerically on the computer, look at Secs. 21.1–21.3of Chapter 21 of Part F, that is, numeric methods for ODEs. These sections are keptindependent by design of the other sections on numerics. This allows for the study ofnumerics for ODEs directly after Chap. 1 or 2.1P A R T AOrdinaryDifferentialEquations (ODEs)c01.qxd 7/30/10 8:14 PM Page 1
• 2C H A P T E R 1First-Order ODEsChapter 1 begins the study of ordinary differential equations (ODEs) by deriving them fromphysical or other problems (modeling), solving them by standard mathematical methods,and interpreting solutions and their graphs in terms of a given problem. The simplest ODEsto be discussed are ODEs of the first order because they involve only the first derivativeof the unknown function and no higher derivatives. These unknown functions will usuallybe denoted by or when the independent variable denotes time t. The chapter endswith a study of the existence and uniqueness of solutions of ODEs in Sec. 1.7.Understanding the basics of ODEs requires solving problems by hand (paper and pencil,or typing on your computer, but first without the aid of a CAS). In doing so, you willgain an important conceptual understanding and feel for the basic terms, such as ODEs,direction field, and initial value problem. If you wish, you can use your Computer AlgebraSystem (CAS) for checking solutions.COMMENT. Numerics for first-order ODEs can be studied immediately after thischapter. See Secs. 21.1–21.2, which are independent of other sections on numerics.Prerequisite: Integral calculus.Sections that may be omitted in a shorter course: 1.6, 1.7.References and Answers to Problems: App. 1 Part A, and App. 2.1.1 Basic Concepts. ModelingIf we want to solve an engineering problem (usually of a physical nature), we firsthave to formulate the problem as a mathematical expression in terms of variables,functions, and equations. Such an expression is known as a mathematical model of thegiven problem. The process of setting up a model, solving it mathematically, andinterpreting the result in physical or other terms is called mathematical modeling or,briefly, modeling.Modeling needs experience, which we shall gain by discussing various examples andproblems. (Your computer may often help you in solving but rarely in setting up models.)Now many physical concepts, such as velocity and acceleration, are derivatives. Hencea model is very often an equation containing derivatives of an unknown function. Sucha model is called a differential equation. Of course, we then want to find a solution (afunction that satisfies the equation), explore its properties, graph it, find values of it, andinterpret it in physical terms so that we can understand the behavior of the physical systemin our given problem. However, before we can turn to methods of solution, we must firstdefine some basic concepts needed throughout this chapter.y1t2y1x2PhysicalSystemPhysicalInterpretationMathematicalModelMathematicalSolutionFig. 1. Modeling,solving, interpretingc01.qxd 7/30/10 8:14 PM Page 2
• An ordinary differential equation (ODE) is an equation that contains one or severalderivatives of an unknown function, which we usually call (or sometimes if theindependent variable is time t). The equation may also contain y itself, known functionsof x (or t), and constants. For example,(1)(2)(3) yryt Ϫ 32 yr2ϭ 0ys ϩ 9y ϭ e؊2xyr ϭ cos xy(t)y(x)SEC. 1.1 Basic Concepts. Modeling 3hOutflowing water(Sec. 1.3)Water level hh′ = –kVibrating masson a spring(Secs. 2.4, 2.8)Displacement yymmy″ + ky = 0(Sec. 1.1)Falling stoney″ = g = const.yBeats of a vibratingsystem(Sec. 4.5)Lotka–Volterrapredator–prey model(Sec. 4.5)PendulumLθ″ + g sin θ = 0L(Sec. 1.2)Parachutistmv′ = mg – bv2Velocityvθ(Sec. 3.3)Deformation of a beamEIyiv= f(x)(k)θ(Sec. 2.9)Current I in anRLC circuitLI″ + RI′ + I = E′hCLERyty1Cy′ = ky1y2– ly2y′ = ay1– by1y212(Sec. 2.8)y″ + w02y = cos wt, w0≈ wω ω ω ωFig. 2. Some applications of differential equationsc01.qxd 7/30/10 8:14 PM Page 3
• are ordinary differential equations (ODEs). Here, as in calculus, denotes ,etc. The term ordinary distinguishes them from partial differentialequations (PDEs), which involve partial derivatives of an unknown function of twoor more variables. For instance, a PDE with unknown function u of two variables xand y isPDEs have important engineering applications, but they are more complicated than ODEs;they will be considered in Chap. 12.An ODE is said to be of order n if the nth derivative of the unknown function y is thehighest derivative of y in the equation. The concept of order gives a useful classificationinto ODEs of first order, second order, and so on. Thus, (1) is of first order, (2) of secondorder, and (3) of third order.In this chapter we shall consider first-order ODEs. Such equations contain only thefirst derivative and may contain y and any given functions of x. Hence we can writethem as(4)or often in the formThis is called the explicit form, in contrast to the implicit form (4). For instance, the implicitODE (where ) can be written explicitly asConcept of SolutionA functionis called a solution of a given ODE (4) on some open interval if isdefined and differentiable throughout the interval and is such that the equation becomesan identity if y and are replaced with h and , respectively. The curve (the graph) ofh is called a solution curve.Here, open interval means that the endpoints a and b are not regarded aspoints belonging to the interval. Also, includes infinite intervals(the real line) as special cases.E X A M P L E 1 Verification of SolutionVerify that (c an arbitrary constant) is a solution of the ODE for all Indeed, differentiateto get Multiply this by x, obtaining thus, the given ODE. ᭿xyr ϭ Ϫy,xyr ϭ Ϫc>x;yr ϭ Ϫc>x2.y ϭ c>xx 0.xyr ϭ Ϫyy ϭ c>xa Ͻ x Ͻ ϱ, Ϫϱ Ͻ x Ͻ ϱϪϱ Ͻ x Ͻ b,a Ͻ x Ͻ ba Ͻ x Ͻ bhryrh(x)a Ͻ x Ͻ by ϭ h(x)yr ϭ 4x3y2.x 0x؊3yr Ϫ 4y2ϭ 0yr ϭ f(x, y).F(x, y, yr) ϭ 0yr02u0x2ϩ02u0y2ϭ 0.ys ϭ d2y>dx2,dy>dxyr4 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:14 PM Page 4
• E X A M P L E 2 Solution by Calculus. Solution CurvesThe ODE can be solved directly by integration on both sides. Indeed, using calculus,we obtain where c is an arbitrary constant. This is a family of solutions. Each valueof c, for instance, 2.75 or 0 or gives one of these curves. Figure 3 shows some of them, for᭿Ϫ1, 0, 1, 2, 3, 4.c ϭ Ϫ3, Ϫ2,Ϫ8,y ϭ ͐cos x dx ϭ sin x ϩ c,yr ϭ dy>dx ϭ cos xSEC. 1.1 Basic Concepts. Modeling 5yx0–42ππ–π42–2Fig. 3. Solutions of the ODE yr ϭ cos xy ϭ sin x ϩ c00.51.01.52.52.00 2 4 6 8 10 12 14 tyFig. 4B. Solutions ofin Example 3 (exponential decay)yr ϭ Ϫ0.2y0102030400 2 4 6 8 10 12 14 tyFig. 4A. Solutions ofin Example 3 (exponential growth)yr ϭ 0.2yE X A M P L E 3 (A) Exponential Growth. (B) Exponential DecayFrom calculus we know that has the derivativeHence y is a solution of (Fig. 4A). This ODE is of the form With positive-constant k it canmodel exponential growth, for instance, of colonies of bacteria or populations of animals. It also applies tohumans for small populations in a large country (e.g., the United States in early times) and is then known asMalthus’s law.1We shall say more about this topic in Sec. 1.5.(B) Similarly, (with a minus on the right) has the solution (Fig. 4B) modelingexponential decay, as, for instance, of a radioactive substance (see Example 5). ᭿y ϭ ce؊0.2t,yr ϭ Ϫ0.2yr ϭ ky.yr ϭ 0.2yyr ϭdydtϭ 0.2e0.2tϭ 0.2y.y ϭ ce0.2t1Named after the English pioneer in classic economics, THOMAS ROBERT MALTHUS (1766–1834).c01.qxd 7/30/10 8:14 PM Page 5
• We see that each ODE in these examples has a solution that contains an arbitraryconstant c. Such a solution containing an arbitrary constant c is called a general solutionof the ODE.(We shall see that c is sometimes not completely arbitrary but must be restricted to someinterval to avoid complex expressions in the solution.)We shall develop methods that will give general solutions uniquely (perhaps except fornotation). Hence we shall say the general solution of a given ODE (instead of a generalsolution).Geometrically, the general solution of an ODE is a family of infinitely many solutioncurves, one for each value of the constant c. If we choose a specific c (e.g., or 0or ) we obtain what is called a particular solution of the ODE. A particular solutiondoes not contain any arbitrary constants.In most cases, general solutions exist, and every solution not containing an arbitraryconstant is obtained as a particular solution by assigning a suitable value to c. Exceptionsto these rules occur but are of minor interest in applications; see Prob. 16 in ProblemSet 1.1.Initial Value ProblemIn most cases the unique solution of a given problem, hence a particular solution, isobtained from a general solution by an initial condition with given valuesand , that is used to determine a value of the arbitrary constant c. Geometricallythis condition means that the solution curve should pass through the pointin the xy-plane. An ODE, together with an initial condition, is called an initial valueproblem. Thus, if the ODE is explicit, the initial value problem is ofthe form(5)E X A M P L E 4 Initial Value ProblemSolve the initial value problemSolution. The general solution is ; see Example 3. From this solution and the initial conditionwe obtain Hence the initial value problem has the solution . This is aparticular solution.More on ModelingThe general importance of modeling to the engineer and physicist was emphasized at thebeginning of this section. We shall now consider a basic physical problem that will showthe details of the typical steps of modeling. Step 1: the transition from the physical situation(the physical system) to its mathematical formulation (its mathematical model); Step 2:the solution by a mathematical method; and Step 3: the physical interpretation of the result.This may be the easiest way to obtain a first idea of the nature and purpose of differentialequations and their applications. Realize at the outset that your computer (your CAS)may perhaps give you a hand in Step 2, but Steps 1 and 3 are basically your work.᭿y(x) ϭ 5.7e3xy(0) ϭ ce0ϭ c ϭ 5.7.y(x) ϭ ce3xy(0) ϭ 5.7.yr ϭdydxϭ 3y,y(x0) ϭ y0.yr ϭ f(x, y),yr ϭ f(x, y),(x0, y0)y0x0y(x0) ϭ y0,Ϫ2.01c ϭ 6.456 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:14 PM Page 6
• And Step 2 requires a solid knowledge and good understanding of solution methodsavailable to you—you have to choose the method for your work by hand or by thecomputer. Keep this in mind, and always check computer results for errors (which mayarise, for instance, from false inputs).E X A M P L E 5 Radioactivity. Exponential DecayGiven an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.Physical Information. Experiments show that at each instant a radioactive substance decomposes—and is thusdecaying in time—proportional to the amount of substance present.Step 1. Setting up a mathematical model of the physical process. Denote by the amount of substance stillpresent at any time t. By the physical law, the time rate of change is proportional to . Thisgives the first-order ODE(6)where the constant k is positive, so that, because of the minus, we do get decay (as in [B] of Example 3).The value of k is known from experiments for various radioactive substances (e.g.,approximately, for radium ).Now the given initial amount is 0.5 g, and we can call the corresponding instant Then we have theinitial condition This is the instant at which our observation of the process begins. It motivatesthe term initial condition (which, however, is also used when the independent variable is not time or whenwe choose a t other than ). Hence the mathematical model of the physical process is the initial valueproblem(7)Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decayand has the general solution (with arbitrary constant c but definite given k)(8)We now determine c by using the initial condition. Since from (8), this gives Hencethe particular solution governing our process is (cf. Fig. 5)(9)Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!)that your solution (9) satisfies (7) as well asStep 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts fromthe correct initial amount and decreases with time because k is positive. The limit of y as is zero. ᭿t : ϱdydtϭ Ϫ0.5ke؊ktϭ Ϫk ؒ 0.5e؊ktϭ Ϫky, y(0) ϭ 0.5e0ϭ 0.5.y(0) ϭ 0.5:(k Ͼ 0).y(t) ϭ 0.5e؊kty(0) ϭ c ϭ 0.5.y(0) ϭ cy(t) ϭ ce؊kt.dydtϭ Ϫky, y(0) ϭ 0.5.t ϭ 0y(0) ϭ 0.5.t ϭ 0.22688Rak ϭ 1.4 ؒ 10؊11sec؊1,dydtϭ Ϫkyy(t)yr(t) ϭ dy>dty(t)SEC. 1.1 Basic Concepts. Modeling 70.10.20.30.40.50y0 0.5 1.5 2 2.5 31 tFig. 5. Radioactivity (Exponential decay,with as an example)k ϭ 1.5y ϭ 0.5eϪkt,c01.qxd 7/30/10 8:14 PM Page 7
• 8 CHAP. 1 First-Order ODEs1–8 CALCULUSSolve the ODE by integration or by remembering adifferentiation formula.1.2.3.4.5.6.7.8.9–15 VERIFICATION. INITIAL VALUEPROBLEM (IVP)(a) Verify that y is a solution of the ODE. (b) Determinefrom y the particular solution of the IVP. (c) Graph thesolution of the IVP.9.10.11.12.13.14.15. Find two constant solutions of the ODE in Prob. 13 byinspection.16. Singular solution. An ODE may sometimes have anadditional solution that cannot be obtained from thegeneral solution and is then called a singular solution.The ODE is of this kind. Showby differentiation and substitution that it has thegeneral solution and the singular solution. Explain Fig. 6.y ϭ x2>4y ϭ cx Ϫ c2yr2Ϫ xyr ϩ y ϭ 0yr tan x ϭ 2y Ϫ 8, y ϭ c sin2x ϩ 4, y(12 p) ϭ 0yr ϭ y Ϫ y2, y ϭ11 ϩ ce؊x , y(0) ϭ 0.25yyr ϭ 4x, y2Ϫ 4x2ϭ c (y Ͼ 0), y(1) ϭ 4yr ϭ y ϩ ex, y ϭ (x ϩ c)ex, y(0) ϭ 12yr ϩ 5xy ϭ 0, y ϭ ce؊2.5x2, y(0) ϭ pyr ϩ 4y ϭ 1.4, y ϭ ce؊4xϩ 0.35, y(0) ϭ 2yt ϭ e؊0.2xyr ϭ cosh 5.13xys ϭ Ϫyyr ϭ 4e؊xcos xyr ϭ Ϫ1.5yyr ϭ yyr ϩ xe؊x2>2ϭ 0yr ϩ 2 sin 2px ϭ 017–20 MODELING, APPLICATIONSThese problems will give you a first impression of modeling.Many more problems on modeling follow throughout thischapter.17. Half-life. The half-life measures exponential decay.It is the time in which half of the given amount ofradioactive substance will disappear. What is the half-life of (in years) in Example 5?18. Half-life. Radium has a half-life of about3.6 days.(a) Given 1 gram, how much will still be present after1 day?(b) After 1 year?19. Free fall. In dropping a stone or an iron ball, airresistance is practically negligible. Experimentsshow that the acceleration of the motion is constant(equal to called theacceleration of gravity). Model this as an ODE for, the distance fallen as a function of time t. If themotion starts at time from rest (i.e., with velocity), show that you obtain the familiar law offree fall20. Exponential decay. Subsonic flight. The efficiencyof the engines of subsonic airplanes depends on airpressure and is usually maximum near ft.Find the air pressure at this height. Physicalinformation. The rate of change is proportionalto the pressure. At ft it is half its valueat sea level. Hint. Remember from calculusthat if then Can you seewithout calculation that the answer should be closeto ?y0>4yr ϭ kekxϭ ky.y ϭ ekx,y0 ϭ y(0)18,000yr(x)y(x)35,000y ϭ 12 gt2.v ϭ yr ϭ 0t ϭ 0y(t)g ϭ 9.80 m>sec2ϭ 32 ft>sec2,22488Ra22688RaP R O B L E M S E T 1 . 1–4 42yx213–4–5–2–3–2–1Fig. 6. Particular solutions and singularsolution in Problem 16c01.qxd 7/30/10 8:15 PM Page 8
• 1.2 Geometric Meaning ofDirection Fields, Euler’s MethodA first-order ODE(1)has a simple geometric interpretation. From calculus you know that the derivative ofis the slope of . Hence a solution curve of (1) that passes through a pointmust have, at that point, the slope equal to the value of f at that point; that is,Using this fact, we can develop graphic or numeric methods for obtaining approximatesolutions of ODEs (1). This will lead to a better conceptual understanding of an ODE (1).Moreover, such methods are of practical importance since many ODEs have complicatedsolution formulas or no solution formulas at all, whereby numeric methods are needed.Graphic Method of Direction Fields. Practical Example Illustrated in Fig. 7. Wecan show directions of solution curves of a given ODE (1) by drawing short straight-linesegments (lineal elements) in the xy-plane. This gives a direction field (or slope field)into which you can then fit (approximate) solution curves. This may reveal typicalproperties of the whole family of solutions.Figure 7 shows a direction field for the ODE(2)obtained by a CAS (Computer Algebra System) and some approximate solution curvesfitted in.yr ϭ y ϩ xyr(x0) ϭ f(x0, y0).yr(x0)(x0, y0)y(x)y(x)yr(x)yr ϭ f(x, y)yr ϭ f(x, y).SEC. 1.2 Geometric Meaning of yЈ ϭ ƒ(x, y). Direction Fields, Euler’s Method 9120.5 1–0.5–1–1.5–2–1–2yxFig. 7. Direction field of with three approximate solutioncurves passing through (0, 1), (0, 0), (0, ), respectivelyϪ1yr ϭ y ϩ x,c01.qxd 7/30/10 8:15 PM Page 9
• If you have no CAS, first draw a few level curves const of , then parallellineal elements along each such curve (which is also called an isocline, meaning a curveof equal inclination), and finally draw approximation curves fit to the lineal elements.We shall now illustrate how numeric methods work by applying the simplest numericmethod, that is Euler’s method, to an initial value problem involving ODE (2). First wegive a brief description of Euler’s method.Numeric Method by EulerGiven an ODE (1) and an initial value Euler’s method yields approximatesolution values at equidistant x-values namely,(Fig. 8), etc.In general,where the step h equals, e.g., 0.1 or 0.2 (as in Table 1.1) or a smaller value for greateraccuracy.yn ϭ ynϪ1 ϩ hf(xnϪ1, ynϪ1)y2 ϭ y1 ϩ hf(x1, y1)y1 ϭ y0 ϩ hf(x0, y0)x0, x1 ϭ x0 ϩ h, x2 ϭ x0 ϩ 2h, Á ,y(x0) ϭ y0,f(x, y)f(x, y) ϭ10 CHAP. 1 First-Order ODEsyxx0x1y0y1y(x1)Solution curveError of y1hf(x0, y0)hFig. 8. First Euler step, showing a solution curve, its tangent at ( ),step h and increment in the formula for y1hf(x0, y0)x0, y0Table 1.1 shows the computation of steps with step for the ODE (2) andinitial condition corresponding to the middle curve in the direction field. Weshall solve the ODE exactly in Sec. 1.5. For the time being, verify that the initial valueproblem has the solution . The solution curve and the values in Table 1.1are shown in Fig. 9. These values are rather inaccurate. The errors are shownin Table 1.1 as well as in Fig. 9. Decreasing h would improve the values, but would soonrequire an impractical amount of computation. Much better methods of a similar naturewill be discussed in Sec. 21.1.y(xn) Ϫ yny ϭ exϪ x Ϫ 1y(0) ϭ 0,h ϭ 0.2n ϭ 5c01.qxd 7/30/10 8:15 PM Page 10
• Table 1.1. Euler method for forwith step h ‫؍‬ 0.2x ‫؍‬ 0, Á , 1.0yr ‫؍‬ y ؉ x, y(0) ‫؍‬ 0SEC. 1.2 Geometric Meaning of yЈ ϭ ƒ(x, y). Direction Fields, Euler’s Method 110.70.50.30.10 0.2 0.4 0.6 0.8 1 xyFig. 9. Euler method: Approximate values in Table 1.1 and solution curven Error0 0.0 0.000 0.000 0.0001 0.2 0.000 0.021 0.0212 0.4 0.04 0.092 0.0523 0.6 0.128 0.222 0.0944 0.8 0.274 0.426 0.1525 1.0 0.488 0.718 0.230y(xn)ynxn1–8 DIRECTION FIELDS, SOLUTION CURVESGraph a direction field (by a CAS or by hand). In the fieldgraph several solution curves by hand, particularly thosepassing through the given points .1.2.3.4.5.6.7.8.9–10 ACCURACY OF DIRECTION FIELDSDirection fields are very useful because they can give youan impression of all solutions without solving the ODE,which may be difficult or even impossible. To get a feel forthe accuracy of the method, graph a field, sketch solutioncurves in it, and compare them with the exact solutions.9.10. (Sol. )11. Autonomous ODE. This means an ODE not showingx (the independent variable) explicitly. (The ODEs inProbs. 6 and 10 are autonomous.) What will the levelcurves const (also called isoclines curvesϭf(x, y) ϭ1y ϩ 52 x ϭ cyr ϭ Ϫ5y1>2yr ϭ cos pxyr ϭ Ϫ2xy, (0, 12), (0, 1), (0, 2)yr ϭ ey>x, (2, 2), (3, 3)yr ϭ sin2y, (0, Ϫ0.4), (0, 1)yr ϭ x Ϫ 1>y, (1, 12)yr ϭ 2y Ϫ y2, (0, 0), (0, 1), (0, 2), (0, 3)yr ϭ 1 Ϫ y2, (0, 0), (2, 12)yyr ϩ 4x ϭ 0, (1, 1), (0, 2)yr ϭ 1 ϩ y2, (14 p, 1)(x, y)of equal inclination) of an autonomous ODE look like?Give reason.12–15 MOTIONSModel the motion of a body B on a straight line withvelocity as given, being the distance of B from a pointat time t. Graph a direction field of the model (theODE). In the field sketch the solution curve satisfying thegiven initial condition.12. Product of velocity times distance constant, equal to 2,13.14. Square of the distance plus square of the velocity equalto 1, initial distance15. Parachutist. Two forces act on a parachutist, theattraction by the earth mg (m mass of person plusequipment, the acceleration of gravity)and the air resistance, assumed to be proportional to thesquare of the velocity v(t). Using Newton’s second lawofmotion(mass acceleration resultantoftheforces),set up a model (an ODE for v(t)). Graph a direction field(choosing m and the constant of proportionality equal to 1).Assume that the parachute opens when vGraph the corresponding solution in the field. What is thelimiting velocity? Would the parachute still be sufficientif the air resistance were only proportional to v(t)?ϭ 10 m>sec.ϭϫg ϭ 9.8 m>sec2ϭ1>12Distance ϭ Velocity ϫ Time, y(1) ϭ 1y(0) ϭ 2.y ϭ 0y(t)P R O B L E M S E T 1 . 2c01.qxd 7/30/10 8:15 PM Page 11
• 1.3 Separable ODEs. ModelingMany practically useful ODEs can be reduced to the form(1)by purely algebraic manipulations. Then we can integrate on both sides with respect to x,obtaining(2)On the left we can switch to y as the variable of integration. By calculus, , so that(3)If f and g are continuous functions, the integrals in (3) exist, and by evaluating them weobtain a general solution of (1). This method of solving ODEs is called the method ofseparating variables, and (1) is called a separable equation, because in (3) the variablesare now separated: x appears only on the right and y only on the left.E X A M P L E 1 Separable ODEThe ODE is separable because it can be writtenBy integration, or .It is very important to introduce the constant of integration immediately when the integration is performed.If we wrote then and then introduced c, we would have obtained whichis not a solution (when ). Verify this. ᭿c 0y ϭ tan x ϩ c,y ϭ tan x,arctan y ϭ x,y ϭ tan (x ϩ c)arctan y ϭ x ϩ cdy1 ϩ y2ϭ dx.yr ϭ 1 ϩ y2Ύg(y) dy ϭ Ύf(x) dx ϩ c.yrdx ϭ dyΎg(y) yrdx ϭ Ύf(x) dx ϩ c.g(y) yr ϭ f(x)12 CHAP. 1 First-Order ODEs16. CAS PROJECT. Direction Fields. Discuss directionfields as follows.(a) Graph portions of the direction field of the ODE (2)(see Fig. 7), for instance,Explain what you have gained by this enlargement ofthe portion of the field.(b) Using implicit differentiation, find an ODE withthe general solution Graph itsdirection field. Does the field give the impressionthat the solution curves may be semi-ellipses? Can youdo similar work for circles? Hyperbolas? Parabolas?Other curves?(c) Make a conjecture about the solutions offrom the direction field.(d) Graph the direction field of and somesolutions of your choice. How do they behave? Whydo they decrease for ?y Ͼ 0yr ϭ Ϫ12 yyr ϭ Ϫx>yx2ϩ 9y2ϭ c (y Ͼ 0).Ϫ5 Ϲ x Ϲ 2, Ϫ1 Ϲ y Ϲ 5.17–20 EULER’S METHODThis is the simplest method to explain numerically solvingan ODE, more precisely, an initial value problem (IVP).(More accurate methods based on the same principle areexplained in Sec. 21.1.) Using the method, to get a feel fornumerics as well as for the nature of IVPs, solve the IVPnumerically with a PC or a calculator, 10 steps. Graph thecomputed values and the solution curve on the samecoordinate axes.17.18.19.Sol.20.Sol. y ϭ 1>(1 ϩ x)5yr ϭ Ϫ5x4y2, y(0) ϭ 1, h ϭ 0.2y ϭ x Ϫ tanh xyr ϭ (y Ϫ x)2, y(0) ϭ 0, h ϭ 0.1yr ϭ y, y(0) ϭ 1, h ϭ 0.01yr ϭ y, y(0) ϭ 1, h ϭ 0.1c01.qxd 7/30/10 8:15 PM Page 12
• E X A M P L E 2 Separable ODEThe ODE is separable; we obtainE X A M P L E 3 Initial Value Problem (IVP). Bell-Shaped CurveSolveSolution. By separation and integration,This is the general solution. From it and the initial condition, Hence the IVP has thesolution This is a particular solution, representing a bell-shaped curve (Fig. 10). ᭿y ϭ 1.8e؊x2.y(0) ϭ ce0ϭ c ϭ 1.8.dyyϭ Ϫ2x dx, ln y ϭ Ϫx2ϩ cෂ, y ϭ ce؊x2.yr ϭ Ϫ2xy, y(0) ϭ 1.8.᭿By integration, Ϫy؊1ϭ Ϫ(x ϩ 2)e؊xϩ c, y ϭ1(x ϩ 2)eϪxϪ c.y؊2dy ϭ (x ϩ 1)e؊xdx.yr ϭ (x ϩ 1)e؊xy2SEC. 1.3 Separable ODEs. Modeling 13110–1–2 2 xyFig. 10. Solution in Example 3 (bell-shaped curve)ModelingThe importance of modeling was emphasized in Sec. 1.1, and separable equations yieldvarious useful models. Let us discuss this in terms of some typical examples.E X A M P L E 4 Radiocarbon Dating2In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found inthe ice of the Oetztal Alps (hence the name “Oetzi”) in Southern Tyrolia near the Austrian–Italian border, causeda scientific sensation. When did Oetzi approximately live and die if the ratio of carbon to carbon inthis mummy is 52.5% of that of a living organism?Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon (maderadioactive by cosmic rays) to ordinary carbon is constant. When an organism dies, its absorption ofby breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactivecarbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of , whichis 5715 years (CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page 11–52,line 9).Solution. Modeling. Radioactive decay is governed by the ODE (see Sec. 1.1, Example 5). Byseparation and integration (where t is time and is the initial ratio of to )(y0 ϭ ec).y ϭ y0 ektln ƒ yƒ ϭ kt ϩ c,dyyϭ k dt,126C146Cy0yr ϭ ky146C146C126C146C126C146C2Method by WILLARD FRANK LIBBY (1908–1980), American chemist, who was awarded for this workthe 1960 Nobel Prize in chemistry.c01.qxd 7/30/10 8:15 PM Page 13
• Next we use the half-life to determine k. When , half of the original substance is still present. Thus,Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),Answer: About 5300 years ago.Other methods show that radiocarbon dating values are usually too small. According to recent research, this isdue to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing.E X A M P L E 5 Mixing ProblemMixing problems occur quite frequently in chemical industry. We explain here how to solve the basic modelinvolving a single tank. The tank in Fig. 11 contains 1000 gal of water in which initially 100 lb of salt is dissolved.Brine runs in at a rate of 10 gal min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank iskept uniform by stirring. Brine runs out at 10 gal min. Find the amount of salt in the tank at any time t.Solution. Step 1. Setting up a model. Let denote the amount of salt in the tank at time t. Its time rateof change isBalance law.5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This isof the total brine content in the tank, hence 0.01 of the salt content , that is, 0.01 . Thus themodel is the ODE(4)Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on bothsides givesInitially the tank contains 100 lb of salt. Hence is the initial condition that will give the uniquesolution. Substituting and in the last equation gives HenceHence the amount of salt in the tank at time t is(5)This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically thatshould increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE?The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35)or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flowrates (in and out) may be different and known only very roughly. ᭿y(t)y(t) ϭ 5000 Ϫ 4900e؊0.01t.c ϭ Ϫ4900.100 Ϫ 5000 ϭ ce0ϭ c.t ϭ 0y ϭ 100y(0) ϭ 100y Ϫ 5000 ϭ ce؊0.01t.ln ƒy Ϫ 5000ƒ ϭ Ϫ0.01t ϩ c*,dyy Ϫ 5000ϭ Ϫ0.01 dt,yr ϭ 50 Ϫ 0.01y ϭ Ϫ0.01(y Ϫ 5000).y(t)y(t)(ϭ 1%)10>1000 ϭ 0.01yr ϭ Salt inflow rate Ϫ Salt outflow ratey(t)>>᭿t ϭln 0.525Ϫ0.0001213ϭ 5312.ektϭ e؊0.0001213tϭ 0.525,k ϭln 0.5Hϭ Ϫ0.6935715ϭ Ϫ0.0001213.ekHϭ 0.5,y0ekHϭ 0.5y0,t ϭ HH ϭ 571514 CHAP. 1 First-Order ODEs100200030001000500040001000 300200 400 500Salt content y(t)tTankTankyFig. 11. Mixing problem in Example 5c01.qxd 7/30/10 8:15 PM Page 14
• E X A M P L E 6 Heating an Office Building (Newton’s Law of Cooling3)Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heatingis shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M.was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. Whatwas the temperature inside the building when the heat was turned on at 6 A.M.?Physical information. Experiments show that the time rate of change of the temperature T of a body B (whichconducts heat well, for example, as a copper ball does) is proportional to the difference between T and thetemperature of the surrounding medium (Newton’s law of cooling).Solution. Step 1. Setting up a model. Let be the temperature inside the building and TA the outsidetemperature (assumed to be constant in Newton’s law). Then by Newton’s law,(6)Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if amodel seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information.To see how good a model is, the engineer will collect experimental data and compare them with calculationsfrom the model.Step 2. General solution. We cannot solve (6) because we do not know TA, just that it varied between 50°Fand 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. Wesolve (6) with the unknown function TA replaced with the average of the two known values, or 45°F. For physicalreasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M.For constant (or any other constant value) the ODE (6) is separable. Separation, integration, andtaking exponents gives the general solutionStep 3. Particular solution. We choose 10 P.M. to be Then the given initial condition is andyields a particular solution, call it . By substitution,Step 4. Determination of k. We use where is 2 A.M. Solving algebraically for k and insertingk into gives (Fig. 12)Tp(t) ϭ 45 ϩ 25e؊0.056t.k ϭ 14 ln 0.8 ϭ Ϫ0.056,e4kϭ 0.8,Tp(4) ϭ 45 ϩ 25e4kϭ 65,Tp(t)t ϭ 4T(4) ϭ 65,Tp(t) ϭ 45 ϩ 25ekt.c ϭ 70 Ϫ 45 ϭ 25,T(0) ϭ 45 ϩ ce0ϭ 70,TpT(0) ϭ 70t ϭ 0.(c ϭ ec*).T(t) ϭ 45 ϩ cektln ƒ T Ϫ 45 ƒ ϭ kt ϩ c*,dTT Ϫ 45ϭ k dt,TA ϭ 45dTdtϭ k(T Ϫ TA).T(t)SEC. 1.3 Separable ODEs. Modeling 156264687060y2 4 6 80 t666165Fig. 12. Particular solution (temperature) in Example 63Sir ISAAC NEWTON (1642–1727), great English physicist and mathematician, became a professor atCambridge in 1669 and Master of the Mint in 1699. He and the German mathematician and philosopherGOTTFRIED WILHELM LEIBNIZ (1646–1716) invented (independently) the differential and integral calculus.Newton discovered many basic physical laws and created the method of investigating physical problems bymeans of calculus. His Philosophiae naturalis principia mathematica (Mathematical Principles of NaturalPhilosophy, 1687) contains the development of classical mechanics. His work is of greatest importance to bothmathematics and physics.c01.qxd 7/30/10 8:15 PM Page 15
• Step 5. Answer and interpretation. 6 A.M. is (namely, 8 hours after 10 P.M.), andHence the temperature in the building dropped 9°F, a result that looks reasonable.E X A M P L E 7 Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law)This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from acylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank atany time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when thehole is opened is 2.25 m. When will the tank be empty?Physical information. Under the influence of gravity the outflowing water has velocity(7) (Torricelli’s law4),where is the height of the water above the hole at time t, and is theacceleration of gravity at the surface of the earth.Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in water level to theoutflow. The volume of the outflow during a short time is(A Area of hole).must equal the change of the volume of the water in the tank. Now(B Cross-sectional area of tank)where is the decrease of the height of the water. The minus sign appears because the volume ofthe water in the tank decreases. Equating and givesWe now express v according to Torricelli’s law and then let (the length of the time interval considered)approach 0—this is a standard way of obtaining an ODE as a model. That is, we haveand by letting we obtain the ODE,where This is our model, a first-order ODE.Step 2. General solution. Our ODE is separable. is constant. Separation and integration givesandDividing by 2 and squaring gives . Insertingyields the general solutionh(t) ϭ (c Ϫ 0.000332t)2.13.28A>B ϭ 13.28 ؒ 0.52p>1002p ϭ 0.000332h ϭ (c Ϫ 13.28At>B)221h ϭ c* Ϫ 26.56ABt.dh1hϭ Ϫ26.56ABdtA>B26.56 ϭ 0.60022 ؒ 980.dhdtϭ Ϫ26.56AB1h¢t : 0¢h¢tϭ ϪABv ϭ ϪAB0.60012gh(t)¢tϪB ¢h ϭ Av ¢t.¢V*¢Vh(t)¢h (Ͼ 0)ϭ¢V* ϭ ϪB ¢h¢V*¢Vϭ¢V ϭ Av ¢t¢t¢Vh(t)g ϭ 980 cm>sec2ϭ 32.17 ft>sec2h(t)v(t) ϭ 0.60022gh(t)᭿Tp(8) ϭ 45 ϩ 25e؊0.056 ؒ 8ϭ 613°F4.t ϭ 816 CHAP. 1 First-Order ODEs4EVANGELISTA TORRICELLI (1608–1647), Italian physicist, pupil and successor of GALILEO GALILEI(1564–1642) at Florence. The “contraction factor” 0.600 was introduced by J. C. BORDA in 1766 because thestream has a smaller cross section than the area of the hole.c01.qxd 7/30/10 8:15 PM Page 16
• Step 3. Particular solution. The initial height (the initial condition) is cm. Substitution ofand gives from the general solution and thus the particular solution (Fig. 13)Step 4. Tank empty. if [hours].Here you see distinctly the importance of the choice of units—we have been working with the cgs system,in which time is measured in seconds! We usedStep 5. Checking. Check the result. ᭿g ϭ 980 cm>sec2.t ϭ 15.00>0.000332 ϭ 45,181 c sec d ϭ 12.6hp(t) ϭ 0hp(t) ϭ (15.00 Ϫ 0.000332t)2.c2ϭ 225, c ϭ 15.00h ϭ 225t ϭ 0h(0) ϭ 225SEC. 1.3 Separable ODEs. Modeling 172.25 m2.00 mh(t)OutflowingwaterWater levelat time tht250200150100500100000 30000 50000Tank Water level h(t) in tankFig. 13. Example 7. Outflow from a cylindrical tank (“leaking tank”).Torricelli’s lawExtended Method: Reduction to Separable FormCertain nonseparable ODEs can be made separable by transformations that introduce fory a new unknown function. We discuss this technique for a class of ODEs of practicalimportance, namely, for equations(8)Here, f is any (differentiable) function of , such as sin , , and so on. (Suchan ODE is sometimes called a homogeneous ODE, a term we shall not use but reservefor a more important purpose in Sec. 1.5.)The form of such an ODE suggests that we set ; thus,(9) and by product differentiationSubstitution into then gives or . We see thatif , this can be separated:(10)duf(u) Ϫ uϭdxx.f(u) Ϫ u 0urx ϭ f(u) Ϫ uurx ϩ u ϭ f(u)yr ϭ f(y>x)yr ϭ urx ϩ u.y ϭ uxy>x ϭ u(y>x)4(y>x)y>xyr ϭ f ayxb.c01.qxd 7/30/10 8:15 PM Page 17
• E X A M P L E 8 Reduction to Separable FormSolveSolution. To get the usual explicit form, divide the given equation by 2xy,Now substitute y and from (9) and then simplify by subtracting u on both sides,You see that in the last equation you can now separate the variables,By integration,Take exponents on both sides to get or . Multiply the last equation by toobtain (Fig. 14)ThusThis general solution represents a family of circles passing through the origin with centers on the x-axis. ᭿ax Ϫc2b2ϩ y2ϭc24.x2ϩ y2ϭ cx.x21 ϩ (y>x)2ϭ c>x1 ϩ u2ϭ c>xln (1 ϩ u2) ϭ Ϫln ƒxƒ ϩ c* ϭ ln `1x` ϩ c*.2u du1 ϩ u2ϭ Ϫdxx.urx ϭ Ϫu2Ϫ12uϭϪu2Ϫ 12u.urx ϩ u ϭu2Ϫ12u,yryr ϭy2Ϫ x22xyϭy2xϪx2y.2xyyr ϭ y2Ϫ x2.18 CHAP. 1 First-Order ODEs4–4yx–4–8 4 82–2Fig. 14. General solution (family of circles) in Example 81. CAUTION! Constant of integration. Why is itimportant to introduce the constant of integrationimmediately when you integrate?2–10 GENERAL SOLUTIONFind a general solution. Show the steps of derivation. Checkyour answer by substitution.2.3.4.5.6.7.8.9.10. xyr ϭ x ϩ y (Set y>x ϭ u)xyr ϭ y2ϩ y (Set y>x ϭ u)yr ϭ (y ϩ 4x)2(Set y ϩ 4x ϭ v)xyr ϭ y ϩ 2x3sin2yx(Set y>x ϭ u)yr ϭ e2xϪ1y2yyr ϩ 36x ϭ 0yr sin 2px ϭ py cos 2pxyr ϭ sec2yy3yr ϩ x3ϭ 011–17 INITIAL VALUE PROBLEMS (IVPS)Solve the IVP. Show the steps of derivation, beginning withthe general solution.11.12.13.14.15.16.(Set )17.18. Particular solution. Introduce limits of integration in(3) such that y obtained from (3) satisfies the initialcondition y(x0) ϭ y0.(Set y>x ϭ u)xyr ϭ y ϩ 3x4cos2(y>x), y(1) ϭ 0v ϭ x ϩ y Ϫ 2yr ϭ (x ϩ y Ϫ 2)2, y(0) ϭ 2yr ϭ Ϫ4x>y, y(2) ϭ 3dr>dt ϭ Ϫ2tr, r(0) ϭ r0yrcosh2x ϭ sin2y, y(0) ϭ 12 pyr ϭ 1 ϩ 4y2, y(1) ϭ 0xyr ϩ y ϭ 0, y(4) ϭ 6P R O B L E M S E T 1 . 3c01.qxd 7/30/10 8:15 PM Page 18
• 19–36 MODELING, APPLICATIONS19. Exponential growth. If the growth rate of the numberof bacteria at any time t is proportional to the numberpresent at t and doubles in 1 week, how many bacteriacan be expected after 2 weeks? After 4 weeks?20. Another population model.(a) If the birth rate and death rate of the number ofbacteria are proportional to the number of bacteriapresent, what is the population as a function of time.(b) What is the limiting situation for increasing time?Interpret it.21. Radiocarbon dating. What should be the content(in percent of ) of a fossilized tree that is claimed tobe 3000 years old? (See Example 4.)22. Linear accelerators are used in physics foraccelerating charged particles. Suppose that an alphaparticle enters an accelerator and undergoes a constantacceleration that increases the speed of the particlefrom to sec. Find theacceleration a and the distance traveled during thatperiod of sec.23. Boyle–Mariotte’s law for ideal gases.5Experimentsshow for a gas at low pressure p (and constanttemperature) the rate of change of the volumeequals . Solve the model.24. Mixing problem. A tank contains 400 gal of brinein which 100 lb of salt are dissolved. Fresh water runsinto the tank at a rate of The mixture, keptpractically uniform by stirring, runs out at the samerate. How much salt will there be in the tank at theend of 1 hour?25. Newton’s law of cooling. A thermometer, reading5°C, is brought into a room whose temperature is 22°C.One minute later the thermometer reading is 12°C.How long does it take until the reading is practically22°C, say, 21.9°C?26. Gompertz growth in tumors. The Gompertz modelis , where is the mass oftumor cells at time t. The model agrees well withclinical observations. The declining growth rate withincreasing corresponds to the fact that cells inthe interior of a tumor may die because of insufficientoxygen and nutrients. Use the ODE to discuss thegrowth and decline of solutions (tumors) and to findconstant solutions. Then solve the ODE.27. Dryer. If a wet sheet in a dryer loses its moisture ata rate proportional to its moisture content, and if itloses half of its moisture during the first 10 min ofy Ͼ 1y(t)yr ϭ ϪAy ln y (A Ͼ 0)2 gal>min.ϪV>pV(p)10؊3104m>sec in 10؊3103m>secy0146CSEC. 1.3 Separable ODEs. Modeling 19drying, when will it be practically dry, say, when willit have lost 99% of its moisture? First guess, thencalculate.28. Estimation. Could you see, practically without calcu-lation, that the answer in Prob. 27 must lie between60 and 70 min? Explain.29. Alibi? Jack, arrested when leaving a bar, claims thathe has been inside for at least half an hour (whichwould provide him with an alibi). The police checkthe water temperature of his car (parked near theentrance of the bar) at the instant of arrest and again30 min later, obtaining the values 190°F and 110°F,respectively. Do these results give Jack an alibi?(Solve by inspection.)30. Rocket. A rocket is shot straight up from the earth,with a net acceleration ( acceleration by the rocketengine minus gravitational pullback) ofduring the initial stage of flight until the engine cut outat sec. How high will it go, air resistanceneglected?31. Solution curves of Show that any(nonvertical) straight line through the origin of thexy-plane intersects all these curves of a given ODE atthe same angle.32. Friction. If a body slides on a surface, it experiencesfriction F (a force against the direction of motion).Experiments show that (Coulomb’s6law ofkinetic friction without lubrication), where N is thenormal force (force that holds the two surfaces together;see Fig. 15) and the constant of proportionality iscalled the coefficient of kinetic friction. In Fig. 15assume that the body weighs 45 nt (about 10 lb; seefront cover for conversion). (correspondingto steel on steel), the slide is 10 m long, theinitial velocity is zero, and air resistance isnegligible. Find the velocity of the body at the endof the slide.a ϭ 30°,␮ ϭ 0.20␮ƒFƒ ϭ ␮ƒNƒyr ‫؍‬ g1y>x2.t ϭ 107t m>sec2ϭ5ROBERT BOYLE (1627–1691), English physicist and chemist, one of the founders of the Royal Society. EDME MARIOTTE (about1620–1684), French physicist and prior of a monastry near Dijon. They found the law experimentally in 1662 and 1676, respectively.6CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer.v(t)WNBodyαs(t)Fig. 15. Problem 32c01.qxd 7/30/10 8:15 PM Page 19
• 33. Rope. To tie a boat in a harbor, how many timesmust a rope be wound around a bollard (a verticalrough cylindrical post fixed on the ground) so that aman holding one end of the rope can resist a forceexerted by the boat 1000 times greater than the mancan exert? First guess. Experiments show that thechange of the force S in a small portion of therope is proportional to S and to the small anglein Fig. 16. Take the proportionality constant 0.15.The result should surprise you!¢␾¢S20 CHAP. 1 First-Order ODEsthis as the condition for the two families to beorthogonal (i.e., to intersect at right angles)? Do yourgraphs confirm this?(e) Sketch families of curves of your own choice andfind their ODEs. Can every family of curves be givenby an ODE?35. CAS PROJECT. Graphing Solutions. A CAS canusually graph solutions, even if they are integrals thatcannot be evaluated by the usual analytical methods ofcalculus.(a) Show this for the five initial value problems, , graphing all five curveson the same axes.(b) Graph approximate solution curves, using the firstfew terms of the Maclaurin series (obtained by term-wise integration of that of ) and compare with theexact curves.(c) Repeat the work in (a) for another ODE and initialconditions of your own choice, leading to an integralthat cannot be evaluated as indicated.36. TEAM PROJECT. Torricelli’s Law. Suppose thatthe tank in Example 7 is hemispherical, of radius R,initially full of water, and has an outlet of 5 cm2cross-sectional area at the bottom. (Make a sketch.) Setup the model for outflow. Indicate what portion ofyour work in Example 7 you can use (so that it canbecome part of the general method independent of theshape of the tank). Find the time t to empty the tank(a) for any R, (b) for Plot t as function ofR. Find the time when (a) for any R, (b) forR ϭ 1 m.h ϭ R>2R ϭ 1 m.yry(0) ϭ 0, Ϯ1, Ϯ2yr ϭ e؊x2S + ΔSΔ␾SSmallportionof ropeFig. 16. Problem 3334. TEAM PROJECT. Family of Curves. A family ofcurves can often be characterized as the generalsolution of(a) Show that for the circles with center at the originwe get(b) Graph some of the hyperbolas Find anODE for them.(c) Find an ODE for the straight lines through theorigin.(d) You will see that the product of the right sides ofthe ODEs in (a) and (c) equals Do you recognizeϪ1.xy ϭ c.yr ϭ Ϫx>y.yr ϭ f(x, y).1.4 Exact ODEs. Integrating FactorsWe recall from calculus that if a function has continuous partial derivatives, itsdifferential (also called its total differential) isFrom this it follows that if thenFor example, if , thenoryr ϭdydxϭ Ϫ1 ϩ 2xy33x2y2,du ϭ (1 ϩ 2xy3) dx ϩ 3x2y2dy ϭ 0u ϭ x ϩ x2y3ϭ cdu ϭ 0.u(x, y) ϭ c ϭ const,du ϭ0u0xdx ϩ0u0ydy.u(x, y)c01.qxd 7/30/10 8:15 PM Page 20
• an ODE that we can solve by going backward. This idea leads to a powerful solutionmethod as follows.A first-order ODE written as (use as in Sec. 1.3)(1)is called an exact differential equation if the differential formis exact, that is, this form is the differential(2)of some function . Then (1) can be writtenBy integration we immediately obtain the general solution of (1) in the form(3)This is called an implicit solution, in contrast to a solution as defined in Sec.1.1, which is also called an explicit solution, for distinction. Sometimes an implicit solutioncan be converted to explicit form. (Do this for ) If this is not possible, yourCAS may graph a figure of the contour lines (3) of the function and help you inunderstanding the solution.Comparing (1) and (2), we see that (1) is an exact differential equation if there is somefunction such that(4) (a) (b)From this we can derive a formula for checking whether (1) is exact or not, as follows.Let M and N be continuous and have continuous first partial derivatives in a region inthe xy-plane whose boundary is a closed curve without self-intersections. Then by partialdifferentiation of (4) (see App. 3.2 for notation),By the assumption of continuity the two second partial derivaties are equal. Thus(5)0M0yϭ0N0x.0N0xϭ02u0x 0y.0M0yϭ02u0y 0x,0u0yϭ N.0u0xϭ M,u(x, y)u(x, y)x2ϩ y2ϭ 1.y ϭ h(x)u(x, y) ϭ c.du ϭ 0.u(x, y)du ϭ0u0xdx ϩ0u0ydyM(x, y) dx ϩ N(x, y) dyM(x, y) dx ϩ N(x, y) dy ϭ 0dy ϭ yrdxM(x, y) ϩ N(x, y)yr ϭ 0,SEC. 1.4 Exact ODEs. Integrating Factors 21c01.qxd 7/30/10 8:15 PM Page 21
• This condition is not only necessary but also sufficient for (1) to be an exact differentialequation. (We shall prove this in Sec. 10.2 in another context. Some calculus books, forinstance, [GenRef 12], also contain a proof.)If (1) is exact, the function can be found by inspection or in the followingsystematic way. From (4a) we have by integration with respect to x(6)in this integration, y is to be regarded as a constant, and plays the role of a “constant”of integration. To determine , we derive from (6), use (4b) to get , andintegrate to get k. (See Example 1, below.)Formula (6) was obtained from (4a). Instead of (4a) we may equally well use (4b).Then, instead of (6), we first have by integration with respect to y(6*)To determine , we derive from (6*), use (4a) to get , and integrate. Weillustrate all this by the following typical examples.E X A M P L E 1 An Exact ODESolve(7)Solution. Step 1. Test for exactness. Our equation is of the form (1) withThusFrom this and (5) we see that (7) is exact.Step 2. Implicit general solution. From (6) we obtain by integration(8)To find , we differentiate this formula with respect to y and use formula (4b), obtainingHence By integration, Inserting this result into (8) and observing (3),we obtain the answeru(x, y) ϭ sin (x ϩ y) ϩ y3ϩ y2ϭ c.k ϭ y3ϩ y2ϩ c*.dk>dy ϭ 3y2ϩ 2y.0u0yϭ cos (x ϩ y) ϩdkdyϭ N ϭ 3y2ϩ 2y ϩ cos (x ϩ y).k(y)u ϭ ΎM dx ϩ k(y) ϭ Ύcos (x ϩ y) dx ϩ k(y) ϭ sin (x ϩ y) ϩ k(y).0N0xϭ Ϫsin (x ϩ y).0M0yϭ Ϫsin (x ϩ y),N ϭ 3y2ϩ 2y ϩ cos (x ϩ y).M ϭ cos (x ϩ y),cos (x ϩ y) dx ϩ (3y2ϩ 2y ϩ cos (x ϩ y)) dy ϭ 0.dl>dx0u>0xl(x)u ϭ ΎN dy ϩ l(x).dk>dydk>dy0u>0yk(y)k(y)u ϭ ΎM dx ϩ k(y);u(x, y)22 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 22
• Step 3. Checking an implicit solution. We can check by differentiating the implicit solutionimplicitly and see whether this leads to the given ODE (7):(9)This completes the check.E X A M P L E 2 An Initial Value ProblemSolve the initial value problem(10)Solution. You may verify that the given ODE is exact. We find u. For a change, let us use (6*),Fromthis, Hence Byintegration,This gives the general solution From the initial condition,Hence the answer is cos y cosh Figure 17 shows the particular solutions for(thicker curve), 1, 2, 3. Check that the answer satisfies the ODE. (Proceed as in Example 1.) Also check that theinitial condition is satisfied. ᭿c ϭ 0, 0.358x ϩ x ϭ 0.358.0.358 ϭ c.cos 2 cosh 1 ϩ 1 ϭu(x, y) ϭ cos y cosh x ϩ x ϭ c.l(x) ϭ x ϩ c*.dl>dx ϭ 1.0u>0x ϭ cos y sinh x ϩ dl>dx ϭ M ϭ cos y sinh x ϩ 1.u ϭ ϪΎsin y cosh x dy ϩ l(x) ϭ cos y cosh x ϩ l(x).y(1) ϭ 2.(cos y sinh x ϩ 1) dx Ϫ sin y cosh x dy ϭ 0,᭿du ϭ0u0xdx ϩ0u0ydy ϭ cos (x ϩ y) dx ϩ (cos (x ϩ y) ϩ 3y2ϩ 2y) dy ϭ 0.u(x, y) ϭ cSEC. 1.4 Exact ODEs. Integrating Factors 23yx0 1.0 2.0 3.00.5 1.5 2.51.02.00.51.52.5Fig. 17. Particular solutions in Example 2E X A M P L E 3 WARNING! Breakdown in the Case of NonexactnessThe equation is not exact because and so that in (5), butLet us show that in such a case the present method does not work. From (6),henceNow, should equal by (4b). However, this is impossible because can depend only on . Try(6*); it will also fail. Solve the equation by another method that we have discussed.Reduction to Exact Form. Integrating FactorsThe ODE in Example 3 is It is not exact. However, if we multiply itby , we get an exact equation [check exactness by (5)!],(11)Integration of (11) then gives the general solution y>x ϭ c ϭ const.Ϫy dx ϩ x dyx2ϭ Ϫyx2dx ϩ1xdy ϭ d ayxb ϭ 0.1>x2Ϫy dx ϩ x dy ϭ 0.᭿yk(y)N ϭ x,0u>0y0u0yϭ Ϫx ϩdkdy.u ϭ ΎM dx ϩ k(y) ϭ Ϫxy ϩ k(y),0N>0x ϭ 1.0M>0y ϭ Ϫ1N ϭ x,M ϭ ϪyϪy dx ϩ x dy ϭ 0c01.qxd 7/30/10 8:15 PM Page 23
• This example gives the idea. All we did was to multiply a given nonexact equation, say,(12)by a function F that, in general, will be a function of both x and y. The result was an equation(13)that is exact, so we can solve it as just discussed. Such a function is then calledan integrating factor of (12).E X A M P L E 4 Integrating FactorThe integrating factor in (11) is Hence in this case the exact equation (13) isSolutionThese are straight lines through the origin. (Note that is also a solution of )It is remarkable that we can readily find other integrating factors for the equation namely,and because(14)How to Find Integrating FactorsIn simpler cases we may find integrating factors by inspection or perhaps after some trials,keeping (14) in mind. In the general case, the idea is the following.For the exactness condition (5) is Hence for (13),the exactness condition is(15)By the product rule, with subscripts denoting partial derivatives, this givesIn the general case, this would be complicated and useless. So we follow the Golden Rule:If you cannot solve your problem, try to solve a simpler one—the result may be useful(and may also help you later on). Hence we look for an integrating factor depending onlyon one variable: fortunately, in many practical cases, there are such factors, as we shallsee. Thus, let Then and so that (15) becomesDividing by FQ and reshuffling terms, we have(16) where R ϭ1Qa0P0yϪ0Q0xb.1FdFdxϭ R,FPy ϭ FrQ ϩ FQx.Fx ϭ Fr ϭ dF>dx,Fy ϭ 0,F ϭ F(x).FyP ϩ FPy ϭ FxQ ϩ FQx.00y(FP) ϭ00x(FQ).FP dx ϩ FQ dy ϭ 0,0M>0y ϭ 0N>0x.M dx ϩ N dy ϭ 0᭿Ϫy dx ϩ x dyx2ϩ y2ϭ d aarctanyxb.Ϫy dx ϩ x dyxyϭ Ϫd alnxyb,Ϫy dx ϩ x dyy2ϭ d axyb,1>(x2ϩ y2),1>y2, 1>(xy),Ϫy dx ϩ x dy ϭ 0,Ϫy dx ϩ x dy ϭ 0.x ϭ 0y ϭ cxyxϭ c.FP dx ϩ FQ dy ϭϪy dx ϩ x dyx2ϭ d ayxb ϭ 0.F ϭ 1>x2.F(x, y)FP dx ϩ FQ dy ϭ 0P(x, y) dx ϩ Q(x, y) dy ϭ 0,24 CHAP. 1 First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 24
• This proves the following theorem.T H E O R E M 1 Integrating Factor F(x)If (12) is such that the right side R of (16) depends only on x, then (12) has anintegrating factor which is obtained by integrating (16) and takingexponents on both sides.(17)Similarly, if then instead of (16) we get(18) whereand we have the companionT H E O R E M 2 Integrating Factor F*(y)If (12) is such that the right side R* of (18) depends only on y, then (12) has anintegrating factor , which is obtained from (18) in the form(19)E X A M P L E 5 Application of Theorems 1 and 2. Initial Value ProblemUsing Theorem 1 or 2, find an integrating factor and solve the initial value problem(20)Solution. Step 1. Nonexactness. The exactness check fails:butStep 2. Integrating factor. General solution. Theorem 1 fails because R [the right side of (16)] depends onboth x and y.Try Theorem 2. The right side of (18) isHence (19) gives the integrating factor From this result and (20) you get the exact equation(exϩ y) dx ϩ (x Ϫ e؊y) dy ϭ 0.F*(y) ϭ e؊y.R* ϭ1Pa0Q0xϪ0P0yb ϭ1exϩyϩ yey (eyϪ exϩyϪ eyϪ yey) ϭ Ϫ1.R ϭ1Qa0P0yϪ0Q0xb ϭ1xeyϪ 1(exϩyϩ eyϩ yeyϪ ey).0Q0xϭ00x(xeyϪ 1) ϭ ey.0P0yϭ00y(exϩyϩ yey) ϭ exϩyϩ eyϩ yeyy(0) ϭ Ϫ1(exϩyϩ yey) dx ϩ (xeyϪ 1) dy ϭ 0,F*(y) ϭ expΎR*(y) dy.F* ϭ F*(y)R* ϭ1Pa0Q0xϪ0P0yb1F*dF*dyϭ R*,F* ϭ F*(y),F(x) ϭ expΎR(x) dx.F ϭ F(x),SEC. 1.4 Exact ODEs. Integrating Factors 25c01.qxd 7/30/10 8:15 PM Page 25
• Test for exactness; you will get 1 on both sides of the exactness condition. By integration, using (4a),Differentiate this with respect to y and use (4b) to getHence the general solution isSetp 3. Particular solution. The initial condition gives Hence theanswer is Figure 18 shows several particular solutions obtained as level curvesof obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast asolution into explicit form. Note the curve that (nearly) satisfies the initial condition.Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initialcondition. ᭿u(x, y) ϭ c,exϩ xy ϩ e؊yϭ 1 ϩ e ϭ 3.72.u(0, Ϫ1) ϭ 1 ϩ 0 ϩ e ϭ 3.72.y(0) ϭ Ϫ1u(x, y) ϭ exϩ xy ϩ e؊yϭ c.k ϭ e؊yϩ c*.dkdyϭ Ϫe؊y,0u0yϭ x ϩdkdyϭ N ϭ x Ϫ e؊y,u ϭ Ύ(exϩ y) dx ϭ exϩ xy ϩ k(y).26 CHAP. 1 First-Order ODEsyx0–1–2–3131 2 3–1–2–32Fig. 18. Particular solutions in Example 51–14 ODEs. INTEGRATING FACTORSTest for exactness. If exact, solve. If not, use an integratingfactor as given or obtained by inspection or by the theoremsin the text. Also, if an initial condition is given, find thecorresponding particular solution.1.2.3.4.5.6.7. 2x tan y dx ϩ sec2y dy ϭ 03(y ϩ 1) dx ϭ 2x dy, (y ϩ 1)x؊4(x2ϩ y2) dx Ϫ 2xy dy ϭ 0e3u(dr ϩ 3r du) ϭ 0sin x cos y dx ϩ cos x sin y dy ϭ 0x3dx ϩ y3dy ϭ 02xy dx ϩ x2dy ϭ 08.9.10.11. 2 cosh x cos y12.13.14.15. Exactness. Under what conditions for the constants a,b, k, l is exact? Solvethe exact ODE.(ax ϩ by) dx ϩ (kx ϩ ly) dy ϭ 0F ϭ xayb(a ϩ 1)y dx ϩ (b ϩ 1)x dy ϭ 0, y(1) ϭ 1,e؊ydx ϩ e؊x(Ϫe؊yϩ 1) dy ϭ 0, F ϭ exϩy(2xy dx ϩ dy)ex2ϭ 0, y(0) ϭ 2dx ϭ sinh x sin y dyy dx ϩ 3y ϩ tan (x ϩ y)4 dy ϭ 0, cos (x ϩ y)e2x(2 cos y dx Ϫ sin y dy) ϭ 0, y(0) ϭ 0ex(cos y dx Ϫ sin y dy) ϭ 0P R O B L E M S E T 1 . 4c01.qxd 7/30/10 8:15 PM Page 26
• 16. TEAM PROJECT. Solution by Several Methods.Show this as indicated. Compare the amount of work.(a) as an exact ODEand by separation.(b) by Theorem 2and by separation.(c) by Theorem 1 or 2 andby separation with(d) by Theorems 1 and 2 andby separation.(e) Search the text and the problems for further ODEsthat can be solved by more than one of the methodsdiscussed so far. Make a list of these ODEs. Findfurther cases of your own.17. WRITING PROJECT. Working Backward.Working backward from the solution to the problemis useful in many areas. Euler, Lagrange, and othergreat masters did it. To get additional insight intothe idea of integrating factors, start from a ofyour choice, find destroy exactness bydivision by some and see what ODE’ssolvable by integrating factors you can get. Can youproceed systematically, beginning with the simplestF(x, y)?F(x, y),du ϭ 0,u(x, y)3x2y dx ϩ 4x3dy ϭ 0v ϭ y>x.(x2ϩ y2) dx Ϫ 2xy dy ϭ 0(1 ϩ 2x) cos y dx ϩ dy>cos y ϭ 0ey(sinh x dx ϩ cosh x dy) ϭ 0SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 27yx04π2π–1–2–31233ππParticular solutions in CAS Project 1818. CAS PROJECT. Graphing Particular Solutions.Graph particular solutions of the following ODE,proceeding as explained.(21)(a) Show that (21) is not exact. Find an integratingfactor using either Theorem 1 or 2. Solve (21).(b) Solve (21) by separating variables. Is this simplerthan (a)?(c) Graph the seven particular solutions satisfying thefollowing initial conditions(see figure below).(d) Which solution of (21) do we not get in (a) or (b)?Ϯ23, Ϯ1y(p>2) ϭ Ϯ12,y(0) ϭ 1,dy Ϫ y2sin x dx ϭ 0.1.5 Linear ODEs. Bernoulli Equation.Population DynamicsLinear ODEs or ODEs that can be transformed to linear form are models of variousphenomena, for instance, in physics, biology, population dynamics, and ecology, as weshall see. A first-order ODE is said to be linear if it can be brought into the form(1)by algebra, and nonlinear if it cannot be brought into this form.The defining feature of the linear ODE (1) is that it is linear in both the unknownfunction y and its derivative whereas p and r may be any given functions ofx. If in an application the independent variable is time, we write t instead of x.If the first term is (instead of ), divide the equation by to get the standardform (1), with as the first term, which is practical.For instance, is a linear ODE, and its standard form isThe function on the right may be a force, and the solution a displacement ina motion or an electrical current or some other physical quantity. In engineering, isfrequently called the input, and is called the output or the response to the input (and,if given, to the initial condition).y(x)r(x)y(x)r(x)yr ϩ y tan x ϭ x sec x.yr cos x ϩ y sin x ϭ xyrf(x)yrf(x)yryr ϭ dy>dx,yr ϩ p(x)y ϭ r(x),c01.qxd 7/30/10 8:15 PM Page 27
• 28 CHAP. 1 First-Order ODEsHomogeneous Linear ODE. We want to solve (1) in some interval callit J, and we begin with the simpler special case that is zero for all x in J. (This issometimes written ) Then the ODE (1) becomes(2)and is called homogeneous. By separating variables and integrating we then obtainthusTaking exponents on both sides, we obtain the general solution of the homogeneousODE (2),(3)here we may also choose and obtain the trivial solution for all x in thatinterval.Nonhomogeneous Linear ODE. We now solve (1) in the case that in (1) is noteverywhere zero in the interval J considered. Then the ODE (1) is called nonhomogeneous.It turns out that in this case, (1) has a pleasant property; namely, it has an integrating factordepending only on x. We can find this factor by Theorem 1 in the previous sectionor we can proceed directly, as follows. We multiply (1) by obtainingF(x),F(x)r(x)y(x) ϭ 0c ϭ 0(c ϭ Ϯec*when y ѥ 0);y(x) ϭ ce؊͐p(x)dxln ƒy ƒ ϭ ϪΎp(x)dx ϩ c*.dyyϭ Ϫp(x)dx,yr ϩ p(x)y ϭ 0r(x) ϵ 0.r(x)a Ͻ x Ͻ b,(1*)The left side is the derivative of the product Fy ifBy separating variables, By integration, writingWith this F and Eq. (1*) becomesBy integration,Dividing by we obtain the desired solution formula(4) y(x) ϭ e؊ha Ύehr dx ϩ cb, h ϭ Ύp(x) dx.eh,ehy ϭ Ύehr dx ϩ c.ehyr ϩ hrehy ϭ ehyr ϩ (eh)ry ϭ (ehy)r ϭ reh.hr ϭ p,ln ƒFƒ ϭ h ϭ Ύp dx, thus F ϭ eh.h ϭ ͐p dx,dF>F ϭ p dx.pFy ϭ Fry, thus pF ϭ Fr.(Fy)r ϭ Fry ϩ FyrFyr ϩ pFy ϭ rF.c01.qxd 7/30/10 8:15 PM Page 28
• SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 29This reduces solving (1) to the generally simpler task of evaluating integrals. For ODEsfor which this is still difficult, you may have to use a numeric method for integrals fromSec. 19.5 or for the ODE itself from Sec. 21.1. We mention that h has nothing to do within Sec. 1.1 and that the constant of integration in h does not matter; see Prob. 2.The structure of (4) is interesting. The only quantity depending on a given initialcondition is c. Accordingly, writing (4) as a sum of two terms,(4*)we see the following:(5)E X A M P L E 1 First-Order ODE, General Solution, Initial Value ProblemSolve the initial value problemSolution. Here andFrom this we see that in (4),and the general solution of our equation isFrom this and the initial condition, thus and the solution of our initial value problemis Here 3 cos x is the response to the initial data, and is the response to theinput sin 2x.E X A M P L E 2 Electric CircuitModel the RL-circuit in Fig. 19 and solve the resulting ODE for the current A (amperes), where t istime. Assume that the circuit contains as an EMF (electromotive force) a battery of V (volts), whichis constant, a resistor of (ohms), and an inductor of H (henrys), and that the current is initiallyzero.Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm’s law) anda voltage drop across the conductor, and the sum of these two voltage drops equals the EMF(Kirchhoff’s Voltage Law, KVL).Remark. In general, KVL states that “The voltage (the electromotive force EMF) impressed on a closedloop is equal to the sum of the voltage drops across all the other elements of the loop.” For Kirchoff’s CurrentLaw (KCL) and historical information, see footnote 7 in Sec. 2.9.Solution. According to these laws the model of the RL-circuit is in standard form(6) Ir ϩRLI ϭE(t)L.LIr ϩ RI ϭ E(t),LIr ϭ L dI>dtL ϭ 0.1R ϭ 11 ⍀E ϭ 48E(t)I(t)᭿Ϫ2 cos2xy ϭ 3 cos x Ϫ 2 cos2x.c ϭ 31 ϭ c # 1 Ϫ 2 # 12;y(x) ϭ cos x a2 Ύsin x dx ϩ cb ϭ c cos x Ϫ 2 cos2x.ehr ϭ (sec x)(2 sin x cos x) ϭ 2 sin x,e؊hϭ cos x,ehϭ sec x,h ϭ Ύp dx ϭ Ύtan x dx ϭ ln ƒ sec xƒ.p ϭ tan x, r ϭ sin 2x ϭ 2 sin x cos x,y(0) ϭ 1.yr ϩ y tan x ϭ sin 2x,Total Output ϭ Response to the Input r ϩ Response to the Initial Data.y(x) ϭ e؊hΎehr dx ϩ ce؊h,h(x)c01.qxd 7/30/10 8:15 PM Page 29
• 30 CHAP. 1 First-Order ODEsWe can solve this linear ODE by (4) with obtaining the general solutionBy integration,(7)In our case, and thus,In modeling, one often gets better insight into the nature of a solution (and smaller roundoff errors) by insertinggiven numeric data only near the end. Here, the general solution (7) shows that the current approaches the limitfaster the larger is, in our case, and the approach is very fast, frombelow if or from above if If the solution is constant (48/11 A). SeeFig. 19.The initial value gives and the particular solution(8)᭿I ϭER(1 Ϫ e؊(R>L)t), thus I ϭ4811(1 Ϫ e؊110t).c ϭ ϪE>RI(0) ϭ E>R ϩ c ϭ 0,I(0) ϭ 0I(0) ϭ 48>11,I(0) Ͼ 48>11.I(0) Ͻ 48>11R>L ϭ 11>0.1 ϭ 110,R>LE>R ϭ 48>11I ϭ 4811 ϩ ce؊110t.E(t) ϭ 48>0.1 ϭ 480 ϭ const;R>L ϭ 11>0.1 ϭ 110I ϭ e؊(R>L)taELe1R>L2tR>Lϩ cb ϭERϩ ce؊(R>L)t.I ϭ e؊(R>L)ta Ύe(R>L)t E(t)Ldt ϩ cb.x ϭ t, y ϭ I, p ϭ R>L, h ϭ (R>L)t,Fig. 19. RL-circuitE X A M P L E 3 Hormone LevelAssume that the level of a certain hormone in the blood of a patient varies with time. Suppose that the time rateof change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuousremoval rate proportional to the level present. Set up a model for the hormone level in the blood and find itsgeneral solution. Find the particular solution satisfying a suitable initial condition.Solution. Step 1. Setting up a model. Let be the hormone level at time t. Then the removal rate isThe input rate is where and A is the average input rate; here to makethe input rate nonnegative. The constants A, B, K can be determined from measurements. Hence the model is thelinear ODEThe initial condition for a particular solution is with suitably chosen, for example,6:00 A.M.Step 2. General solution. In (4) we have and Hence (4) gives thegeneral solution (evaluate by integration by parts)͐eKtcos vt dtr ϭ A ϩ B cos vt.p ϭ K ϭ const, h ϭ Kt,t ϭ 0ypart(0) ϭ y0ypartyr(t) ϭ In Ϫ Out ϭ A ϩ B cos vt Ϫ Ky(t), thus yr ϩ Ky ϭ A ϩ B cos vt.A м Bv ϭ 2p>24 ϭ p>12A ϩ B cos vt,Ky(t).y(t)L = 0.1 HCircuit Current I(t)I(t)E = 48 VR = 11 ⍀0.01 0.02 0.03 0.04 0.05 t24680c01.qxd 7/30/10 8:15 PM Page 30
• SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 31Fig. 20. Particular solution in Example 3010152025100 20005tyThe last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initialcondition). The other part of is called the steady-state solution because it consists of constant and periodicterms. The entire solution is called the transient-state solution because it models the transition from rest to thesteady state. These terms are used quite generally for physical and other systems whose behavior depends on time.Step 3. Particular solution. Setting in and choosing we havethusInserting this result into we obtain the particular solutionwith the steady-state part as before. To plot we must specify values for the constants, say,and Figure 20 shows this solution. Notice that the transition period is relatively short (althoughK is small), and the curve soon looks sinusoidal; this is the response to the input᭿1 ϩ cos ( 112 pt).A ϩ B cos ( 112 pt) ϭK ϭ 0.05.A ϭ B ϭ 1ypartypart(t) ϭAKϩBK2ϩ (p>12)2aK cospt12ϩp12sinpt12b Ϫ aAKϩKBK2ϩ (p>12)2b e؊Ky(t),c ϭ ϪAKϪKBK2ϩ (p>12)2.y(0) ϭAKϩBK2ϩ (p>12)2upK ϩ c ϭ 0,y0 ϭ 0,y(t)t ϭ 0y(t)ϭAKϩBK 2ϩ (p>12)2aK cospt12ϩp12sinpt12b ϩ ce؊Kt.ϭ e؊KteKtcAKϩBK 2ϩ v2aK cos vt ϩ v sin vtbd ϩ ce؊Kty(t) ϭ e؊KtΎeKtaA ϩ B cos vtb dt ϩ ce؊KtReduction to Linear Form. Bernoulli EquationNumerous applications can be modeled by ODEs that are nonlinear but can be transformedto linear ODEs. One of the most useful ones of these is the Bernoulli equation7(9) (a any real number).yr ϩ p(x)y ϭ g(x)ya7JAKOB BERNOULLI (1654–1705), Swiss mathematician, professor at Basel, also known for his contributionto elasticity theory and mathematical probability. The method for solving Bernoulli’s equation was discovered byLeibniz in 1696. Jakob Bernoulli’s students included his nephew NIKLAUS BERNOULLI (1687–1759), whocontributed to probability theory and infinite series, and his youngest brother JOHANN BERNOULLI (1667–1748),who had profound influence on the development of calculus, became Jakob’s successor at Basel, and had amonghis students GABRIEL CRAMER (see Sec. 7.7) and LEONHARD EULER (see Sec. 2.5). His son DANIELBERNOULLI (1700–1782) is known for his basic work in fluid flow and the kinetic theory of gases.c01.qxd 7/30/10 8:15 PM Page 31
• 32 CHAP. 1 First-Order ODEs8PIERRE-FRANÇOIS VERHULST, Belgian statistician, who introduced Eq. (8) as a model for humanpopulation growth in 1838.If or Equation (9) is linear. Otherwise it is nonlinear. Then we setWe differentiate this and substitute from (9), obtainingSimplification giveswhere on the right, so that we get the linear ODE(10)For further ODEs reducible to linear form, see lnce’s classic [A11] listed in App. 1. Seealso Team Project 30 in Problem Set 1.5.E X A M P L E 4 Logistic EquationSolve the following Bernoulli equation, known as the logistic equation (or Verhulst equation8):(11)Solution. Write (11) in the form (9), that is,to see that so that Differentiate this u and substitute from (11),The last term is Hence we have obtained the linear ODEThe general solution is [by (4)]Since this gives the general solution of (11),(12) (Fig. 21)Directly from (11) we see that is also a solution. ᭿y ϵ 0 (y(t) ϭ 0 for all t)y ϭ1uϭ1ce؊Atϩ B>Au ϭ 1>y,u ϭ ce؊Atϩ B>A.ur ϩ Au ϭ B.ϪAy؊1ϭ ϪAu.ϭ B Ϫ AyϪ1.Ϫy؊2(Ay Ϫ By2)ur ϭ Ϫy؊2yr ϭyru ϭ y1؊aϭ y؊1.a ϭ 2,yr Ϫ Ay ϭ ϪBy2yr ϭ Ay Ϫ By2ur ϩ (1 Ϫ a)pu ϭ (1 Ϫ a)g.y1؊aϭ uur ϭ (1 Ϫ a)(g Ϫ py1؊a),ur ϭ (1 Ϫ a)y؊ayr ϭ (1 Ϫ a)y؊a(gyaϪ py).yru(x) ϭ 3y(x)41؊a.a ϭ 1,a ϭ 0c01.qxd 7/30/10 8:15 PM Page 32
• SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 33Fig. 21. Logistic population model. Curves (9) in Example 4 with A>B ϭ 41 2 3 4Population yTime t20= 468ABPopulation DynamicsThe logistic equation (11) plays an important role in population dynamics, a fieldthat models the evolution of populations of plants, animals, or humans over time t.If then (11) is In this case its solution (12) isand gives exponential growth, as for a small population in a large country (theUnited States in early times!). This is called Malthus’s law. (See also Example 3 inSec. 1.1.)The term in (11) is a “braking term” that prevents the population from growingwithout bound. Indeed, if we write we see that if thenso that an initially small population keeps growing as long as But ifthen and the population is decreasing as long as The limitis the same in both cases, namely, See Fig. 21.We see that in the logistic equation (11) the independent variable t does not occurexplicitly. An ODE in which t does not occur explicitly is of the form(13)and is called an autonomous ODE. Thus the logistic equation (11) is autonomous.Equation (13) has constant solutions, called equilibrium solutions or equilibriumpoints. These are determined by the zeros of because gives by(13); hence These zeros are known as critical points of (13). Anequilibrium solution is called stable if solutions close to it for some t remain closeto it for all further t. It is called unstable if solutions initially close to it do not remainclose to it as t increases. For instance, in Fig. 21 is an unstable equilibriumsolution, and is a stable one. Note that (11) has the critical points andE X A M P L E 5 Stable and Unstable Equilibrium Solutions. “Phase Line Plot”The ODE has the stable equilibrium solution and the unstable as the directionfield in Fig. 22 suggests. The values and are the zeros of the parabola in the figure.Now, since the ODE is autonomous, we can “condense” the direction field to a “phase line plot” giving andand the direction (upward or downward) of the arrows in the field, and thus giving information about thestability or instability of the equilibrium solutions. ᭿y2,y1f(y) ϭ (y Ϫ 1)(y Ϫ 2)y2y1y2 ϭ 2,y1 ϭ 1yr ϭ (y Ϫ 1)(y Ϫ 2)y ϭ A>B.y ϭ 0y ϭ 4y ϭ 0y ϭ const.yr ϭ 0f(y) ϭ 0f(y),yr ϭ f(y)yr ϭ f(t, y)A>B.y Ͼ A>B.yr Ͻ 0y Ͼ A>B,y Ͻ A>B.yr Ͼ 0,y Ͻ A>B,yr ϭ Ay31 Ϫ (B>A)y4,ϪBy2y ϭ (1>c)eAtyr ϭ dy>dt ϭ Ay.B ϭ 0,c01.qxd 7/30/10 8:15 PM Page 33
• 34 CHAP. 1 First-Order ODEsy(x)x0 2–2(a)y1 y2y1y2y1y2(b) (c)1–11.02.00.51.52.53.0yx0 2.0 2.5 3.00.5 1.0 1.51.00.51.52.0Fig. 22. Example 5. (A) Direction field. (B) “Phase line”. (C) Parabola f(y)A few further population models will be discussed in the problem set. For some moredetails of population dynamics, see C. W. Clark. Mathematical Bioeconomics: TheMathematics of Conservation 3rd ed. Hoboken, NJ, Wiley, 2010.Further applications of linear ODEs follow in the next section.1. CAUTION! Show that and2. Integration constant. Give a reason why in (4) you maychoose the constant of integration in to be zero.3–13 GENERAL SOLUTION. INITIAL VALUEPROBLEMSFind the general solution. If an initial condition is given,find also the corresponding particular solution and graph orsketch it. (Show the details of your work.)3.4.5.6.7.8.9.10.11.12.13. yr ϭ 6(y Ϫ 2.5)tanh 1.5xxyr ϩ 4y ϭ 8x4, y(1) ϭ 2yr ϭ (y Ϫ 2) cot xyr cos x ϩ (3y Ϫ 1) sec x ϭ 0, y(14p) ϭ 4>3yr ϩ y sin x ϭ ecos x, y(0) ϭ Ϫ2.5yr ϩ y tan x ϭ e؊0.01xcos x, y(0) ϭ 0xyr ϭ 2y ϩ x3exyr ϩ 2y ϭ 4 cos 2x, y(14p) ϭ 3yr ϩ ky ϭ e؊kxyr ϭ 2y Ϫ 4xyr Ϫ y ϭ 5.2͐p dxe؊ln(sec x)ϭ cos x.e؊ln xϭ 1>x (not Ϫx) 14. CAS EXPERIMENT. (a) Solve the ODEFind an initial condition for which thearbitrary constant becomes zero. Graph the resultingparticular solution, experimenting to obtain a goodfigure near(b) Generalizing (a) from to arbitrary n, solve theODE Find an initialcondition as in (a) and experiment with the graph.15–20 GENERAL PROPERTIES OF LINEAR ODEsThese properties are of practical and theoretical importancebecause they enable us to obtain new solutions from givenones. Thus in modeling, whenever possible, we prefer linearODEs over nonlinear ones, which have no similar properties.Show that nonhomogeneous linear ODEs (1) and homo-geneous linear ODEs (2) have the following properties.Illustrate each property by a calculation for two or threeequations of your choice. Give proofs.15. The sum of two solutions and of thehomogeneous equation (2) is a solution of (2), and so isa scalar multiple for any constant a. These propertiesare not true for (1)!ay1y2y1y1 ϩ y2yr Ϫ ny>x ϭ Ϫxn؊2cos (1>x).n ϭ 1x ϭ 0.Ϫx؊1cos (1>x).yr Ϫ y>x ϭP R O B L E M S E T 1 . 5c01.qxd 7/30/10 8:15 PM Page 34
• SEC. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics 3516. (that is, for all x, also written )is a solution of (2) [not of (1) if !], called thetrivial solution.17. The sum of a solution of (1) and a solution of (2) is asolution of (1).18. The difference of two solutions of (1) is a solution of (2).19. If is a solution of (1), what can you say about20. If and are solutions of andrespectively (with the same p!), whatcan you say about the sum21. Variation of parameter. Another method of obtaining(4) results from the following idea. Write (3) aswhere is the exponential function, which is a solutionof the homogeneous linear ODEReplace the arbitrary constant c in (3) with a functionu to be determined so that the resulting functionis a solution of the nonhomogeneous linear ODE22–28 NONLINEAR ODEsUsing a method of this section or separating variables, findthe general solution. If an initial condition is given, findalso the particular solution and sketch or graph it.22.23.24.25.26.27.28.29. REPORT PROJECT. Transformation of ODEs.We have transformed ODEs to separable form, to exactform, and to linear form. The purpose of suchtransformations is an extension of solution methods tolarger classes of ODEs. Describe the key idea of eachof these transformations and give three typical exam-ples of your choice for each transformation. Show eachstep (not just the transformed ODE).30. TEAM PROJECT. Riccati Equation. ClairautEquation. Singular Solution.A Riccati equation is of the form(14)A Clairaut equation is of the form(15)(a) Apply the transformation to theRiccati equation (14), where Y is a solution of (14), andobtain for u the linear ODEExplain the effect of the transformation by writing itas y ϭ Y ϩ v, v ϭ 1>u.ur ϩ (2Yg Ϫ p)u ϭ Ϫg.y ϭ Y ϩ 1>uy ϭ xyr ϩ g(yr).yr ϩ p(x)y ϭ g(x)y2ϩ h(x).2xyyr ϩ (x Ϫ 1)y2ϭ x2ex(Set y2ϭ z)yr ϭ 1>(6eyϪ 2x)y(0) ϭ 12 pyr ϭ (tan y)>(x Ϫ 1),yr ϭ 3.2y Ϫ 10y2yr ϩ y ϭ Ϫx>yyr ϩ xy ϭ xy؊1, y(0) ϭ 3yr ϩ y ϭ y2, y(0) ϭ Ϫ13yr ϩ py ϭ r.y ϭ uy*y*r ϩ py* ϭ 0.y*cy*,y1 ϩ y2?y2r ϩ py2 ϭ r2,y1r ϩ py1 ϭ r1y2y1cy1?y1r(x) 0y(x) ϵ 0y(x) ϭ 0y ϭ 0 (b) Show that is a solution of the ODEand solve thisRiccati equation, showing the details.(c) Solve the Clairaut equation asfollows. Differentiate it with respect to x, obtainingThen solve (A) and (B)separately and substitute the two solutions(a) and (b) of (A) and (B) into the given ODE. Thusobtain (a) a general solution (straight lines) and (b) aparabola for which those lines (a) are tangents (Fig. 6in Prob. Set 1.1); so (b) is the envelope of (a). Such asolution (b) that cannot be obtained from a generalsolution is called a singular solution.(d) Show that the Clairaut equation (15) has assolutions a family of straight lines anda singular solution determined by wherethat forms the envelope of that family.31–40 MODELING. FURTHER APPLICATIONS31. Newton’s law of cooling. If the temperature of a cakeis when it leaves the oven and is tenminutes later, when will it be practically equal to theroom temperature of say, when will it be32. Heating and cooling of a building. Heating andcooling of a building can be modeled by the ODEwhere is the temperature in the building attime t, the outside temperature, the temperaturewanted in the building, and P the rate of increase of Tdue to machines and people in the building, and andare (negative) constants. Solve this ODE, assumingand varying sinusoidallyover 24 hours, say, Discussthe effect of each term of the equation on the solution.33. Drug injection. Find and solve the model for druginjection into the bloodstream if, beginning at aconstant amount A g min is injected and the drug issimultaneously removed at a rate proportional to theamount of the drug present at time t.34. Epidemics. A model for the spread of contagiousdiseases is obtained by assuming that the rate of spreadis proportional to the number of contacts betweeninfected and noninfected persons, who are assumed tomove freely among each other. Set up the model. Findthe equilibrium solutions and indicate their stability orinstability. Solve the ODE. Find the limit of theproportion of infected persons as and explainwhat it means.35. Lake Erie. Lake Erie has a water volume of aboutand a flow rate (in and out) of about 175 km2450 km3t : ϱ>t ϭ 0,Ta ϭ A Ϫ C cos(2p>24)t.TaP ϭ const, Tw ϭ const,k2k1TwTaT ϭ T(t)Tr ϭ k1(T Ϫ Ta) ϩ k2(T Ϫ Tv) ϩ P,61°F?60°F,200°F300°Fs ϭ yr,gr(s) ϭ Ϫx,y ϭ cx ϩ g(c)2yr Ϫ x ϭ 0ys ϭ 0ys(2yr Ϫ x) ϭ 0.yr2Ϫ xyr ϩ y ϭ 0y ϭ Ϫx2y2Ϫ x4Ϫ x ϩ1(2x3ϩ 1)yr Ϫy ϭ Y ϭ xc01.qxd 7/30/10 10:01 PM Page 35
• 36 CHAP. 1 First-Order ODEsper year. If at some instant the lake has pollutionconcentration how long, approximately,will it take to decrease it to p 2, assuming that theinflow is much cleaner, say, it has pollutionconcentration p 4, and the mixture is uniform (anassumption that is only imperfectly true)? First guess.36. Harvesting renewable resources. Fishing. Supposethat the population of a certain kind of fish is givenby the logistic equation (11), and fish are caught at arate Hy proportional to y. Solve this so-called Schaefermodel. Find the equilibrium solutions andwhen The expression is calledthe equilibrium harvest or sustainable yield corre-sponding to H. Why?37. Harvesting. In Prob. 36 find and graph the solutionsatisfying when (for simplicity)and What is the limit? What does it mean?What if there were no fishing?38. Intermittent harvesting. In Prob. 36 assume that youfish for 3 years, then fishing is banned for the next3 years. Thereafter you start again. And so on. This iscalled intermittent harvesting. Describe qualitativelyhow the population will develop if intermitting iscontinued periodically. Find and graph the solution forthe first 9 years, assuming thatand y(0) ϭ 2.A ϭ B ϭ 1, H ϭ 0.2,H ϭ 0.2.A ϭ B ϭ 1y(0) ϭ 2Y ϭ Hy2H Ͻ A.y2 (Ͼ 0)y1y(t)>>p ϭ 0.04%,39. Extinction vs. unlimited growth. If in a populationthe death rate is proportional to the population, andthe birth rate is proportional to the chance encountersof meeting mates for reproduction, what will the modelbe? Without solving, find out what will eventuallyhappen to a small initial population. To a large one.Then solve the model.40. Air circulation. In a room containing of air,of fresh air flows in per minute, and the mixture(made practically uniform by circulating fans) isexhausted at a rate of 600 cubic feet per minute (cfm).What is the amount of fresh air at any time ifAfter what time will 90% of the air be fresh?y(0) ϭ 0?y(t)600 ft320,000 ft3y(t)Fig. 23. Fish population in Problem 380.811.21.41.61.822 4 6 80 ty1.6 Orthogonal Trajectories. OptionalAn important type of problem in physics or geometry is to find a family of curves thatintersects a given family of curves at right angles. The new curves are called orthogonaltrajectories of the given curves (and conversely). Examples are curves of equaltemperature (isotherms) and curves of heat flow, curves of equal altitude (contour lines)on a map and curves of steepest descent on that map, curves of equal potential(equipotential curves, curves of equal voltage—the ellipses in Fig. 24) and curves ofelectric force (the parabolas in Fig. 24).Here the angle of intersection between two curves is defined to be the angle betweenthe tangents of the curves at the intersection point. Orthogonal is another word forperpendicular.In many cases orthogonal trajectories can be found using ODEs. In general, if weconsider to be a given family of curves in the xy-plane, then each value ofc gives a particular curve. Since c is one parameter, such a family is called a one-parameter family of curves.In detail, let us explain this method by a family of ellipses(1) (c Ͼ 0)12 x2ϩ y2ϭ cG(x, y, c) ϭ 0c01.qxd 7/30/10 8:15 PM Page 36
• Step 2. Find an ODE for the orthogonal trajectories This ODE is(3)with the same f as in (2). Why? Well, a given curve passing through a point hasslope at that point, by (2). The trajectory through has slopeby (3). The product of these slopes is , as we see. From calculus it is known that thisis the condition for orthogonality (perpendicularity) of two straight lines (the tangents at), hence of the curve and its orthogonal trajectory at .Step 3. Solve (3) by separating variables, integrating, and taking exponents:This is the family of orthogonal trajectories, the quadratic parabolas along which electronsor other charged particles (of very small mass) would move in the electric field betweenthe black ellipses (elliptic cylinders).yෂϭ c*x2.ln ƒyෂƒ ϭ 2 ln x ϩ c,dyෂyෂ ϭ 2dxx,(x0, y0)(x0, y0)Ϫ1Ϫ1>f(x0, y0)(x0, y0)f(x0, y0)(x0, y0)yෂr ϭ Ϫ1f(x, yෂ)ϭ ϩ2yෂxyෂϭ yෂ(x).SEC. 1.6 Orthogonal Trajectories. Optional 37–6 6yx4–4Fig. 24. Electrostatic field between two ellipses (elliptic cylinders in space):Elliptic equipotential curves (equipotential surfaces) and orthogonaltrajectories (parabolas)and illustrated in Fig. 24. We assume that this family of ellipses represents electricequipotential curves between the two black ellipses (equipotential surfaces between twoelliptic cylinders in space, of which Fig. 24 shows a cross-section). We seek theorthogonal trajectories, the curves of electric force. Equation (1) is a one-parameter familywith parameter c. Each value of c corresponds to one of these ellipses.Step 1. Find an ODE for which the given family is a general solution. Of course, thisODE must no longer contain the parameter c. Differentiating (1), we haveHence the ODE of the given curves is(2) yr ϭ f(x, y) ϭ Ϫx2y.x ϩ 2yyr ϭ 0.(Ͼ 0)c01.qxd 7/30/10 8:15 PM Page 37
• 38 CHAP. 1 First-Order ODEs1–3 FAMILIES OF CURVESRepresent the given family of curves in the formand sketch some of the curves.1. All ellipses with foci and 3 on the x-axis.2. All circles with centers on the cubic parabolaand passing through the origin3. The catenaries obtained by translating the catenaryin the direction of the straight line .4–10 ORTHOGONAL TRAJECTORIES (OTs)Sketch or graph some of the given curves. Guess what theirOTs may look like. Find these OTs.4. 5.6. 7.8. 9.10.11–16 APPLICATIONS, EXTENSIONS11. Electric field. Let the electric equipotential lines(curves of constant potential) between two concentriccylinders with the z-axis in space be given by(these are circular cylinders inthe xyz-space). Using the method in the text, find theirorthogonal trajectories (the curves of electric force).12. Electric field. The lines of electric force of two oppositecharges of the same strength at and arethe circles through and . Show that thesecircles are given by . Showthat the equipotential lines (which are orthogonaltrajectories of those circles) are the circles given by(dashed in Fig. 25).(x ϩ c*)2ϩ yෂ2ϭ c*2Ϫ 1x2ϩ (y Ϫ c)2ϭ 1 ϩ c2(1, 0)(Ϫ1, 0)(1, 0)(Ϫ1, 0)u(x, y) ϭ x2ϩ y2ϭ cx2ϩ (y Ϫ c)2ϭ c2y ϭ ce؊x2y ϭ 2x ϩ cy ϭ c>x2xy ϭ cy ϭ cxy ϭ x2ϩ cy ϭ xy ϭ cosh x(0, 0).y ϭ x3Ϫ3G(x, y; c) ϭ 0P R O B L E M S E T 1 . 6Fig. 25. Electric field in Problem 1213. Temperature field. Let the isotherms (curves ofconstant temperature) in a body in the upper half-planebe given by . Find the ortho-gonal trajectories (the curves along which heat willflow in regions filled with heat-conducting material andfree of heat sources or heat sinks).14. Conic sections. Find the conditions under whichthe orthogonal trajectories of families of ellipsesare again conic sections. Illustrateyour result graphically by sketches or by using yourCAS. What happens if If15. Cauchy–Riemann equations. Show that for a familyconst the orthogonal trajectoriesconst can be obtained from the followingCauchy–Riemann equations (which are basic incomplex analysis in Chap. 13) and use them to find theorthogonal trajectories of const. (Here, sub-scripts denote partial derivatives.)16. Congruent OTs. If with f independent of y,show that the curves of the corresponding family arecongruent, and so are their OTs.yr ϭ f(x)uy ϭ Ϫvxux ϭ vy,exsin y ϭc* ϭv(x, y) ϭu(x, y) ϭ c ϭb : 0?a : 0?x2>a2ϩ y2>b2ϭ c4x2ϩ 9y2ϭ cy Ͼ 01.7 Existence and Uniqueness of Solutionsfor Initial Value ProblemsThe initial value problemhas no solution because (that is, for all x) is the only solution of the ODE.The initial value problemy(0) ϭ 1yr ϭ 2x,y(x) ϭ 0y ϭ 0y(0) ϭ 1ƒyr ƒ ϩ ƒyƒ ϭ 0,c01.qxd 7/30/10 8:15 PM Page 38
• SEC. 1.7 Existence and Uniqueness of Solutions 39Theorems that state such conditions are called existence theorems and uniquenesstheorems, respectively.Of course, for our simple examples, we need no theorems because we can solve theseexamples by inspection; however, for complicated ODEs such theorems may be ofconsiderable practical importance. Even when you are sure that your physical or othersystem behaves uniquely, occasionally your model may be oversimplified and may notgive a faithful picture of reality.T H E O R E M 1 Existence TheoremLet the right side of the ODE in the initial value problem(1)be continuous at all points in some rectangle(Fig. 26)and bounded in R; that is, there is a number K such that(2) for all in R.Then the initial value problem (1) has at least one solution . This solution existsat least for all x in the subinterval of the intervalhere, is the smaller of the two numbers a and b K.>aƒx Ϫ x0 ƒ Ͻ a;ƒx Ϫ x0 ƒ Ͻ ay(x)(x, y)ƒf(x, y) ƒ Ϲ Kƒy Ϫ y0 ƒ Ͻ bR: ƒx Ϫ x0 ƒ Ͻ a,(x, y)y(x0) ϭ y0yr ϭ f(x, y),f(x, y)has precisely one solution, namely, The initial value problemhas infinitely many solutions, namely, where c is an arbitrary constant becausefor all c.From these examples we see that an initial value problem(1)may have no solution, precisely one solution, or more than one solution. This fact leadsto the following two fundamental questions.Problem of ExistenceUnder what conditions does an initial value problem of the form (1) have at leastone solution (hence one or several solutions)?Problem of UniquenessUnder what conditions does that problem have at most one solution (hence excludingthe case that is has more than one solution)?y(x0) ϭ y0yr ϭ f(x, y),y(0) ϭ 1y ϭ 1 ϩ cx,y(0) ϭ 1xyr ϭ y Ϫ 1,y ϭ x2ϩ 1.c01.qxd 7/30/10 8:15 PM Page 39
• 40 CHAP. 1 First-Order ODEsyxy0+ bx0+ ax0– a x0y0y0– bRFig. 26. Rectangle R in the existence and uniqueness theorems(Example of Boundedness. The function is bounded (with ) in thesquare . The function is not bounded for. Explain!)T H E O R E M 2 Uniqueness TheoremLet f and its partial derivative be continuous for all in the rectangleR (Fig. 26) and bounded, say,(3) (a) (b) for all in R.Then the initial value problem (1) has at most one solution . Thus, by Theorem 1,the problem has precisely one solution. This solution exists at least for all x in thatsubinterval ƒx Ϫ x0 ƒ Ͻ a.y(x)(x, y)ƒ fy(x, y) ƒ Ϲ Mƒ f(x, y) ƒ Ϲ K,(x, y)fy ϭ 0f>0yƒx ϩ yƒ Ͻ p>2f(x, y) ϭ tan (x ϩ y)ƒxƒ Ͻ 1, ƒy ƒ Ͻ 1K ϭ 2f(x, y) ϭ x2ϩ y2Understanding These TheoremsThese two theorems take care of almost all practical cases. Theorem 1 says that ifis continuous in some region in the xy-plane containing the point , then the initialvalue problem (1) has at least one solution.Theorem 2 says that if, moreover, the partial derivative of f with respect to yexists and is continuous in that region, then (1) can have at most one solution; hence, byTheorem 1, it has precisely one solution.Read again what you have just read—these are entirely new ideas in our discussion.Proofs of these theorems are beyond the level of this book (see Ref. [A11] in App. 1);however, the following remarks and examples may help you to a good understanding ofthe theorems.Since , the condition (2) implies that that is, the slope of anysolution curve in R is at least and at most K. Hence a solution curve that passesthrough the point must lie in the colored region in Fig. 27 bounded by the linesand whose slopes are and K, respectively. Depending on the form of R, twodifferent cases may arise. In the first case, shown in Fig. 27a, we have andtherefore in the existence theorem, which then asserts that the solution exists for allx between and . In the second case, shown in Fig. 27b, we have .Therefore, and all we can conclude from the theorems is that the solutiona ϭ b>K Ͻ a,b>K Ͻ ax0 ϩ ax0 Ϫ aa ϭ ab>K м aϪKl2l1(x0, y0)ϪKy(x)ƒyr ƒ Ϲ K;yr ϭ f(x, y)0f>0y(x0, y0)f(x, y)c01.qxd 7/30/10 8:15 PM Page 40
• and take the rectangle Then , andIndeed, the solution of the problem is (see Sec. 1.3, Example 1). This solution is discontinuous at, and there is no continuous solution valid in the entire interval from which we started.The conditions in the two theorems are sufficient conditions rather than necessary ones,and can be lessened. In particular, by the mean value theorem of differential calculus wehavewhere and are assumed to be in R, and is a suitable value betweenand . From this and (3b) it follows that(4) ƒ f(x, y2) Ϫ f(x, y1)ƒ Ϲ Mƒy2 Ϫ y1 ƒ.y2y1yෂ(x, y2)(x, y1)f(x, y2) Ϫ f(x, y1) ϭ (y2 Ϫ y1)0f0y`yϭyෂ᭿ƒ xƒ Ͻ 5Ϯp>2y ϭ tan xa ϭbKϭ 0.3 Ͻ a.`0f0y` ϭ 2ƒ yƒ Ϲ M ϭ 6,ƒ f(x, y)ƒ ϭ ƒ1 ϩ y2ƒ Ϲ K ϭ 10,a ϭ 5, b ϭ 3R; ƒ xƒ Ͻ 5, ƒyƒ Ͻ 3.SEC. 1.7 Existence and Uniqueness of Solutions 41y yxy0+ bl1l2x0(a)y0y0– bRxy0+ bl1l2x0(b)y0y0– bRa a= a = aα αα αLet us illustrate our discussion with a simple example. We shall see that our choice ofa rectangle R with a large base (a long x-interval) will lead to the case in Fig. 27b.E X A M P L E 1 Choice of a RectangleConsider the initial value problemy(0) ϭ 0yr ϭ 1 ϩ y2,exists for all x between and . For larger or smaller x’s the solutioncurve may leave the rectangle R, and since nothing is assumed about f outside R, nothingcan be concluded about the solution for those larger or amaller x’s; that is, for such x’sthe solution may or may not exist—we don’t know.x0 ϩ b>Kx0 Ϫ b>KFig. 27. The condition (2) of the existence theorem. (a) First case. (b) Second casec01.qxd 7/30/10 8:15 PM Page 41
• 42 CHAP. 1 First-Order ODEs9RUDOLF LIPSCHITZ (1832–1903), German mathematician. Lipschitz and similar conditions are importantin modern theories, for instance, in partial differential equations.10EMILE PICARD (1856–1941). French mathematician, also known for his important contributions tocomplex analysis (see Sec. 16.2 for his famous theorem). Picard used his method to prove Theorems 1 and 2as well as the convergence of the sequence (7) to the solution of (1). In precomputer times, the iteration was oflittle practical value because of the integrations.It can be shown that (3b) may be replaced by the weaker condition (4), which is knownas a Lipschitz condition.9However, continuity of is not enough to guarantee theuniqueness of the solution. This may be illustrated by the following example.E X A M P L E 2 NonuniquenessThe initial value problemhas the two solutionsandalthough is continuous for all y. The Lipschitz condition (4) is violated in any region that includesthe line , because for and positive we have(5)and this can be made as large as we please by choosing sufficiently small, whereas (4) requires that thequotient on the left side of (5) should not exceed a fixed constant M. ᭿y2(2y2 Ͼ 0)ƒ f(x, y2) Ϫ f(x, y1)ƒƒ y2 Ϫ y1 ƒϭ2y2y2ϭ12y2,y2y1 ϭ 0y ϭ 0f(x, y) ϭ 2 ƒyƒy* ϭ ex2>4 if x м 0Ϫx2>4 if x Ͻ 0y ϭ 0y(0) ϭ 0yr ϭ 2 ƒ yƒ.f(x, y)1. Linear ODE. If p and r in arecontinuous for all x in an interval showthat in this ODE satisfies the conditions of ourpresent theorems, so that a corresponding initial valueproblem has a unique solution. Do you actually needthese theorems for this ODE?2. Existence? Does the initial value problemhave a solution? Does yourresult contradict our present theorems?3. Vertical strip. If the assumptions of Theorems 1 and2 are satisfied not merely in a rectangle but in a verticalinfinite strip in what interval will thesolution of (1) exist?4. Change of initial condition. What happens in Prob.2 if you replace with5. Length of x-interval. In most cases the solution of aninitial value problem (1) exists in an x-interval larger thanthat guaranteed by the present theorems. Show this factfor by finding the best possible ayr ϭ 2y2, y(1) ϭ 1y(2) ϭ k?y(2) ϭ 1ƒx Ϫ x0 ƒ Ͻ a,(x Ϫ 2)yr ϭ y, y(2) ϭ 1f(x, y)ƒx Ϫ x0 ƒ Յ a,yr ϩ p(x)y ϭ r(x) (choosing b optimally) and comparing the result with theactual solution.6. CAS PROJECT. Picard Iteration. (a) Show that byintegrating the ODE in (1) and observing the initialcondition you obtain(6)This form (6) of (1) suggests Picard’s Iteration Method10which is defined by(7)It gives approximations of the unknownsolution y of (1). Indeed, you obtain by substitutingon the right and integrating—this is the firststep—then by substituting on the right andintegrating—this is the second step—and so on. Writey ϭ y1y2y ϭ y0y1y1, y2, y3, . . .yn(x) ϭ y0 ϩ Ύxx0f(t, yn؊1(t) dt, n ϭ 1, 2, Á .y(x) ϭ y0 ϩ Ύxx0f(t, y(t))dt.P R O B L E M S E T 1 . 7c01.qxd 7/30/10 8:15 PM Page 42
• Chapter 1 Review Questions and Problems 43a program of the iteration that gives a printout of thefirst approximations as well as theirgraphs on common axes. Try your program on twoinitial value problems of your own choice.(b) Apply the iteration to Alsosolve the problem exactly.(c) Apply the iteration to Alsosolve the problem exactly.(d) Find all solutions of Whichof them does Picard’s iteration approximate?(e) Experiment with the conjecture that Picard’siteration converges to the solution of the problem forany initial choice of y in the integrand in (7) (leavingoutside the integral as it is). Begin with a simple ODEand see what happens. When you are reasonably sure,take a slightly more complicated ODE and give it a try.y0yr ϭ 2 1y, y(1) ϭ 0.yr ϭ 2y2, y(0) ϭ 1.yr ϭ x ϩ y, y(0) ϭ 0.y0, y1, . . . , yN7. Maximum . What is the largest possible inExample 1 in the text?8. Lipschitz condition. Show that for a linear ODEwith continuous p and r ina Lipschitz condition holds. This isremarkable because it means that for a linear ODE thecontinuity of guarantees not only the existencebut also the uniqueness of the solution of an initialvalue problem. (Of course, this also follows directlyfrom (4) in Sec. 1.5.)9. Common points. Can two solution curves of the sameODE have a common point in a rectangle in which theassumptions of the present theorems are satisfied?10. Three possible cases. Find all initial conditions suchthat has no solution, preciselyone solution, and more than one solution.(x2Ϫ x)yr ϭ (2x Ϫ 1)yf(x, y)ƒx Ϫ x0 ƒ Ϲ ayr ϩ p(x)y ϭ r(x)aA14.15.16. Solve by Euler’s method(10 steps, ). Solve exactly and compute the error.17–21 GENERAL SOLUTIONFind the general solution. Indicate which method in thischapter you are using. Show the details of your work.17.18.19.20.21.22–26 INITIAL VALUE PROBLEM (IVP)Solve the IVP. Indicate the method used. Show the detailsof your work.22.23.24.25.26.27–30 MODELING, APPLICATIONS27. Exponential growth. If the growth rate of a cultureof bacteria is proportional to the number of bacteriapresent and after 1 day is 1.25 times the originalnumber, within what interval of time will the numberof bacteria (a) double, (b) triple?x sinh y dy ϭ cosh y dx, y(3) ϭ 03 sec y dx ϩ 13 sec x dy ϭ 0, y(0) ϭ 0yr ϩ 12 y ϭ y3, y(0) ϭ 13yr ϭ 21 Ϫ y2, y(0) ϭ 1>12yr ϩ 4xy ϭ eϪ2x2, y(0) ϭ Ϫ4.3(3xeyϩ 2y) dx ϩ (x2eyϩ x) dy ϭ 0yr ϭ ay ϩ by2(a 0)25yyr Ϫ 4x ϭ 0yr Ϫ 0.4y ϭ 29 sin xyr ϩ 2.5y ϭ 1.6xh ϭ 0.1yr ϭ y Ϫ y2, y(0) ϭ 0.2yr ϩ y ϭ 1.01 cos 10xxyr ϭ y ϩ x21. Explain the basic concepts ordinary and partialdifferential equations (ODEs, PDEs), order, generaland particular solutions, initial value problems (IVPs).Give examples.2. What is a linear ODE? Why is it easier to solve thana nonlinear ODE?3. Does every first-order ODE have a solution? A solutionformula? Give examples.4. What is a direction field? A numeric method for first-order ODEs?5. What is an exact ODE? Isalways exact?6. Explain the idea of an integrating factor. Give twoexamples.7. What other solution methods did we consider in thischapter?8. Can an ODE sometimes be solved by several methods?Give three examples.9. What does modeling mean? Can a CAS solve a modelgiven by a first-order ODE? Can a CAS set up a model?10. Give problems from mechanics, heat conduction, andpopulation dynamics that can be modeled by first-orderODEs.11–16 DIRECTION FIELD: NUMERIC SOLUTIONGraph a direction field (by a CAS or by hand) and sketchsome solution curves. Solve the ODE exactly and compare.In Prob. 16 use Euler’s method.11.12.13. yr ϭ y Ϫ 4y2yr ϭ 1 Ϫ y2yr ϩ 2y ϭ 0f(x) dx ϩ g(y) dy ϭ 0C H A P T E R 1 R E V I E W Q U E S T I O N S A N D P R O B L E M Sc01.qxd 7/30/10 8:15 PM Page 43
• 44 CHAP. 1 First-Order ODEs28. Mixing problem. The tank in Fig. 28 contains 80 lbof salt dissolved in 500 gal of water. The inflow perminute is 20 lb of salt dissolved in 20 gal of water. Theoutflow is 20 gal min of the uniform mixture. Find thetime when the salt content in the tank reaches 95%of its limiting value (as ).t : ϱy(t)>Fig. 28. Tank in Problem 2829. Half-life. If in a reactor, uranium loses 10% ofits weight within one day, what is its half-life? Howlong would it take for 99% of the original amount todisappear?30. Newton’s law of cooling. A metal bar whosetemperature is is placed in boiling water. Howlong does it take to heat the bar to practicallysay, to , if the temperature of the bar after 1 minof heating is First guess, then calculate.51.5°C?99.9°C100°C,20°C23797UThis chapter concerns ordinary differential equations (ODEs) of first order andtheir applications. These are equations of the form(1) or in explicit forminvolving the derivative of an unknown function y, given functions ofx, and, perhaps, y itself. If the independent variable x is time, we denote it by t.In Sec. 1.1 we explained the basic concepts and the process of modeling, that is,of expressing a physical or other problem in some mathematical form and solvingit. Then we discussed the method of direction fields (Sec. 1.2), solution methodsand models (Secs. 1.3–1.6), and, finally, ideas on existence and uniqueness ofsolutions (Sec. 1.7).A first-order ODE usually has a general solution, that is, a solution involving anarbitrary constant, which we denote by c. In applications we usually have to find aunique solution by determining a value of c from an initial condition .Together with the ODE this is called an initial value problem(2)and its solution is a particular solution of the ODE. Geometrically, a generalsolution represents a family of curves, which can be graphed by using directionfields (Sec. 1.2). And each particular solution corresponds to one of these curves.A separable ODE is one that we can put into the form(3) (Sec. 1.3)by algebraic manipulations (possibly combined with transformations, such as) and solve by integrating on both sides.y>x ϭ ug(y) dy ϭ f(x) dx(x0, y0 given numbers)y(x0) ϭ y0yr ϭ f(x, y),y(x0) ϭ y0yr ϭ dy>dxyr ϭ f(x, y)F(x, y, yr) ϭ 0SUMMARY OF CHAPTER 1First-Order ODEsc01.qxd 7/30/10 8:15 PM Page 44
• An exact ODE is of the form(4) (Sec. 1.4)where is the differentialof a function so that from we immediately get the implicit generalsolution This method extends to nonexact ODEs that can be made exactby multiplying them by some function called an integrating factor (Sec. 1.4).Linear ODEs(5)are very important. Their solutions are given by the integral formula (4), Sec. 1.5.Certain nonlinear ODEs can be transformed to linear form in terms of new variables.This holds for the Bernoulli equation(Sec. 1.5).Applications and modeling are discussed throughout the chapter, in particular inSecs. 1.1, 1.3, 1.5 (population dynamics, etc.), and 1.6 (trajectories).Picard’s existence and uniqueness theorems are explained in Sec. 1.7 (andPicard’s iteration in Problem Set 1.7).Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2immediately after this chapter, as indicated in the chapter opening.yr ϩ p(x)y ϭ g(x)yayr ϩ p(x)y ϭ r(x)F(x, y,),u(x, y) ϭ c.du ϭ 0u(x, y),du ϭ ux dx ϩ uy dyM dx ϩ N dyM(x, y) dx ϩ N(x, y) dy ϭ 0Summary of Chapter 1 45c01.qxd 7/30/10 8:15 PM Page 45
• 46C H A P T E R 2Second-Order Linear ODEsMany important applications in mechanical and electrical engineering, as shown in Secs.2.4, 2.8, and 2.9, are modeled by linear ordinary differential equations (linear ODEs) of thesecond order. Their theory is representative of all linear ODEs as is seen when comparedto linear ODEs of third and higher order, respectively. However, the solution formulas forsecond-order linear ODEs are simpler than those of higher order, so it is a natural progressionto study ODEs of second order first in this chapter and then of higher order in Chap. 3.Although ordinary differential equations (ODEs) can be grouped into linear and nonlinearODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which manybeautiful standard methods exist.Chapter 2 includes the derivation of general and particular solutions, the latter inconnection with initial value problems.For those interested in solution methods for Legendre’s, Bessel’s, and the hypergeometricequations consult Chap. 5 and for Sturm–Liouville problems Chap. 11.COMMENT. Numerics for second-order ODEs can be studied immediately after thischapter. See Sec. 21.3, which is independent of other sections in Chaps. 19–21.Prerequisite: Chap. 1, in particular, Sec. 1.5.Sections that may be omitted in a shorter course: 2.3, 2.9, 2.10.References and Answers to Problems: App. 1 Part A, and App. 2.2.1 Homogeneous Linear ODEs of Second OrderWe have already considered first-order linear ODEs (Sec. 1.5) and shall now define anddiscuss linear ODEs of second order. These equations have important engineeringapplications, especially in connection with mechanical and electrical vibrations (Secs. 2.4,2.8, 2.9) as well as in wave motion, heat conduction, and other parts of physics, as weshall see in Chap. 12.A second-order ODE is called linear if it can be written(1)and nonlinear if it cannot be written in this form.The distinctive feature of this equation is that it is linear in y and its derivatives, whereasthe functions p, q, and r on the right may be any given functions of x. If the equationbegins with, say, then divide by to have the standard form (1) with as thefirst term.ysf(x)f(x)ys,ys ϩ p(x)yr ϩ q(x)y ϭ r(x)c02.qxd 10/27/10 6:06 PM Page 46
• The definitions of homogeneous and nonhomogenous second-order linear ODEs arevery similar to those of first-order ODEs discussed in Sec. 1.5. Indeed, if (thatis, for all x considered; read “ is identically zero”), then (1) reduces to(2)and is called homogeneous. If then (1) is called nonhomogeneous. This issimilar to Sec. 1.5.An example of a nonhomogeneous linear ODE isand a homogeneous linear ODE iswritten in standard form .Finally, an example of a nonlinear ODE is.The functions p and q in (1) and (2) are called the coefficients of the ODEs.Solutions are defined similarly as for first-order ODEs in Chap. 1. A functionis called a solution of a (linear or nonlinear) second-order ODE on some open interval Iif h is defined and twice differentiable throughout that interval and is such that the ODEbecomes an identity if we replace the unknown y by h, the derivative by , and thesecond derivative by . Examples are given below.Homogeneous Linear ODEs: Superposition PrincipleSections 2.1–2.6 will be devoted to homogeneous linear ODEs (2) and the remainingsections of the chapter to nonhomogeneous linear ODEs.Linear ODEs have a rich solution structure. For the homogeneous equation the backboneof this structure is the superposition principle or linearity principle, which says that wecan obtain further solutions from given ones by adding them or by multiplying them withany constants. Of course, this is a great advantage of homogeneous linear ODEs. Let usfirst discuss an example.E X A M P L E 1 Homogeneous Linear ODEs: Superposition of SolutionsThe functions and are solutions of the homogeneous linear ODEfor all x. We verify this by differentiation and substitution. We obtain ; henceys ϩ y ϭ (cos x)s ϩ cos x ϭ Ϫcos x ϩ cos x ϭ 0.(cos x)s ϭ Ϫcos xys ϩ y ϭ 0y ϭ sin xy ϭ cos xhsyshryry ϭ h(x)ysy ϩ yr2ϭ 0ys ϩ1x yr ϩ y ϭ 0xys ϩ yr ϩ xy ϭ 0,ys ϩ 25y ϭ e؊xcos x,r(x) [ 0,ys ϩ p(x)yr ϩ q(x)y ϭ 0r(x)r(x) ϭ 0r(x) ϵ 0SEC. 2.1 Homogeneous Linear ODEs of Second Order 47c02.qxd 10/27/10 6:06 PM Page 47
• Similarly for (verify!). We can go an important step further. We multiply by any constant, forinstance, 4.7, and by, say, , and take the sum of the results, claiming that it is a solution. Indeed,differentiation and substitution givesIn this example we have obtained from and a function of the form(3) ( arbitrary constants).This is called a linear combination of and . In terms of this concept we can nowformulate the result suggested by our example, often called the superposition principleor linearity principle.T H E O R E M 1 Fundamental Theorem for the Homogeneous Linear ODE (2)For a homogeneous linear ODE (2), any linear combination of two solutions on anopen interval I is again a solution of (2) on I. In particular, for such an equation,sums and constant multiples of solutions are again solutions.P R O O F Let and be solutions of (2) on I. Then by substituting andits derivatives into (2), and using the familiar rule , etc.,we getsince in the last line, because and are solutions, by assumption. This showsthat y is a solution of (2) on I.CAUTION! Don’t forget that this highly important theorem holds for homogeneouslinear ODEs only but does not hold for nonhomogeneous linear or nonlinear ODEs, asthe following two examples illustrate.E X A M P L E 2 A Nonhomogeneous Linear ODEVerify by substitution that the functions and are solutions of the nonhomogeneouslinear ODEbut their sum is not a solution. Neither is, for instance, or .E X A M P L E 3 A Nonlinear ODEVerify by substitution that the functions and are solutions of the nonlinear ODEbut their sum is not a solution. Neither is , so you cannot even multiply by ! ᭿Ϫ1Ϫx2ysy Ϫ xyr ϭ 0,y ϭ 1y ϭ x2᭿5(1 ϩ sin x)2(1 ϩ cos x)ys ϩ y ϭ 1,y ϭ 1 ϩ sin xy ϭ 1 ϩ cos x᭿y2y1(Á) ϭ 0ϭ c1(ys1 ϩ pyr1 ϩ qy1) ϩ c2(ys2 ϩ pyr2 ϩ qy2) ϭ 0,ϭ c1 ys1 ϩ c2 ys2 ϩ p(c1 yr1 ϩ c2 yr2) ϩ q(c1 y1 ϩ c2 y2)ys ϩ pyr ϩ qy ϭ (c1 y1 ϩ c2 y2)s ϩ p(c1 y1 ϩ c2 y2)r ϩ q(c1 y1 ϩ c2 y2)(c1 y1 ϩ c2 y2)r ϭ c1 yr1 ϩ c2 yr2y ϭ c1 y1 ϩ c2 y2y2y1y2y1c1, c2y ϭ c1y1 ϩ c2y2y2 (ϭ sin x)y1 (ϭ cos x)᭿(4.7 cos x Ϫ 2 sin x)s ϩ (4.7 cos x Ϫ 2 sin x) ϭ Ϫ4.7 cos x ϩ 2 sin x ϩ 4.7 cos x Ϫ 2 sin x ϭ 0.Ϫ2sin xcos xy ϭ sin x48 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 48
• Initial Value Problem. Basis. General SolutionRecall from Chap. 1 that for a first-order ODE, an initial value problem consists of theODE and one initial condition . The initial condition is used to determine thearbitrary constant c in the general solution of the ODE. This results in a unique solution,as we need it in most applications. That solution is called a particular solution of theODE. These ideas extend to second-order ODEs as follows.For a second-order homogeneous linear ODE (2) an initial value problem consists of(2) and two initial conditions(4)These conditions prescribe given values and of the solution and its first derivative(the slope of its curve) at the same given in the open interval considered.The conditions (4) are used to determine the two arbitrary constants and in ageneral solution(5)of the ODE; here, and are suitable solutions of the ODE, with “suitable” to beexplained after the next example. This results in a unique solution, passing through thepoint with as the tangent direction (the slope) at that point. That solution iscalled a particular solution of the ODE (2).E X A M P L E 4 Initial Value ProblemSolve the initial value problemSolution. Step 1. General solution. The functions and are solutions of the ODE (by Example 1),and we takeThis will turn out to be a general solution as defined below.Step 2. Particular solution. We need the derivative . From this and theinitial values we obtain, since and ,This gives as the solution of our initial value problem the particular solutionFigure 29 shows that at it has the value 3.0 and the slope , so that its tangent intersectsthe x-axis at . (The scales on the axes differ!)Observation. Our choice of and was general enough to satisfy both initialconditions. Now let us take instead two proportional solutions andso that . Then we can write in the form.y ϭ c1 cos x ϩ c2(k cos x) ϭ C cos x where C ϭ c1 ϩ c2ky ϭ c1 y1 ϩ c2 y2y1/y2 ϭ 1/k ϭ consty2 ϭ k cos x,y1 ϭ cos xy2y1᭿x ϭ 3.0>0.5 ϭ 6.0Ϫ0.5x ϭ 0y ϭ 3.0 cos x Ϫ 0.5 sin x.y(0) ϭ c1 ϭ 3.0 and yr(0) ϭ c2 ϭ Ϫ0.5.sin 0 ϭ 0cos 0 ϭ 1yr ϭ Ϫc1 sin x ϩ c2 cos xy ϭ c1 cos x ϩ c2 sin x.sin xcos xys ϩ y ϭ 0, y(0) ϭ 3.0, yr(0) ϭ Ϫ0.5.K1(x0, K0)y2y1y ϭ c1 y1 ϩ c2 y2c2c1x ϭ x0K1K0y(x0) ϭ K0, yr(x0) ϭ K1.y(x0) ϭ y0SEC. 2.1 Homogeneous Linear ODEs of Second Order 492 4 6 108 x–3–2–10123yFig. 29. Particular solutionand initial tangent inExample 4c02.qxd 10/27/10 6:06 PM Page 49
• Hence we are no longer able to satisfy two initial conditions with only one arbitraryconstant C. Consequently, in defining the concept of a general solution, we must excludeproportionality. And we see at the same time why the concept of a general solution is ofimportance in connection with initial value problems.D E F I N I T I O N General Solution, Basis, Particular SolutionA general solution of an ODE (2) on an open interval I is a solution (5) in whichand are solutions of (2) on I that are not proportional, and and are arbitraryconstants. These , are called a basis (or a fundamental system) of solutionsof (2) on I.A particular solution of (2) on I is obtained if we assign specific values toand in (5).For the definition of an interval see Sec. 1.1. Furthermore, as usual, and are calledproportional on I if for all x on I,(6) (a) or (b)where k and l are numbers, zero or not. (Note that (a) implies (b) if and only if ).Actually, we can reformulate our definition of a basis by using a concept of generalimportance. Namely, two functions and are called linearly independent on aninterval I where they are defined if(7) everywhere on I implies .And and are called linearly dependent on I if (7) also holds for some constants ,not both zero. Then, if , we can divide and see that and areproportional,orIn contrast, in the case of linear independence these functions are not proportional becausethen we cannot divide in (7). This gives the followingD E F I N I T I O N Basis (Reformulated)A basis of solutions of (2) on an open interval I is a pair of linearly independentsolutions of (2) on I.If the coefficients p and q of (2) are continuous on some open interval I, then (2) has ageneral solution. It yields the unique solution of any initial value problem (2), (4). Itincludes all solutions of (2) on I; hence (2) has no singular solutions (solutions notobtainable from of a general solution; see also Problem Set 1.1). All this will be shownin Sec. 2.6.y2 ϭ Ϫk1k2y1.y1 ϭ Ϫk2k1y2y2y1k1 0 or k2 0k2k1y2y1k1 ϭ 0 and k2 ϭ 0k1y1(x) ϩ k2y2(x) ϭ 0y2y1k 0y2 ϭ ly1y1 ϭ ky2y2y1c2c1y2y1c2c1y2y150 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 50
• E X A M P L E 5 Basis, General Solution, Particular Solutionand in Example 4 form a basis of solutions of the ODE for all x because theirquotient is (or ). Hence is a general solution. The solutionof the initial value problem is a particular solution.E X A M P L E 6 Basis, General Solution, Particular SolutionVerify by substitution that and are solutions of the ODE . Then solve the initialvalue problem.Solution. and show that and are solutions. They are notproportional, . Hence , form a basis for all x. We now write down the correspondinggeneral solution and its derivative and equate their values at 0 to the given initial conditions,.By addition and subtraction, , so that the answer is . This is the particular solutionsatisfying the two initial conditions.Find a Basis if One Solution Is Known.Reduction of OrderIt happens quite often that one solution can be found by inspection or in some other way.Then a second linearly independent solution can be obtained by solving a first-order ODE.This is called the method of reduction of order.1We first show how this method worksin an example and then in general.E X A M P L E 7 Reduction of Order if a Solution Is Known. BasisFind a basis of solutions of the ODE.Solution. Inspection shows that is a solution because and , so that the first termvanishes identically and the second and third terms cancel. The idea of the method is to substituteinto the ODE. This givesux and –xu cancel and we are left with the following ODE, which we divide by x, order, and simplify,,This ODE is of first order in , namely, . Separation of variables and integrationgives, .ln ƒv ƒ ϭ ln ƒ x Ϫ 1 ƒ Ϫ 2 ln ƒ xƒ ϭ lnƒ x Ϫ 1ƒx2dvvϭ Ϫx Ϫ 2x2Ϫ xdx ϭ a1x Ϫ 1Ϫ2xb dx(x2Ϫ x)vr ϩ (x Ϫ 2)v ϭ 0v ϭ ur(x2Ϫ x)us ϩ (x Ϫ 2)ur ϭ 0.(x2Ϫ x)(usx ϩ 2ur) Ϫ x2ur ϭ 0(x2Ϫ x)(usx ϩ 2ur) Ϫ x(urx ϩ u) ϩ ux ϭ 0.y ϭ uy1 ϭ ux, yr ϭ urx ϩ u, ys ϭ usx ϩ 2urys1 ϭ 0yr1 ϭ 1y1 ϭ x(x2Ϫ x)ys Ϫ xyr ϩ y ϭ 0᭿y ϭ 2exϩ 4e؊xc1 ϭ 2, c2 ϭ 4y ϭ c1exϩ c2e؊x, yr ϭ c1exϪ c2e؊x, y(0) ϭ c1 ϩ c2 ϭ 6, yr(0) ϭ c1 Ϫ c2 ϭ Ϫ2e؊xexex/e؊xϭ e2xconste؊xex(e؊x)s Ϫ e؊xϭ 0(ex)s Ϫ exϭ 0ys Ϫ y ϭ 0, y(0) ϭ 6, yr(0) ϭ Ϫ2ys Ϫ y ϭ 0y2 ϭ e؊xy1 ϭ ex᭿y ϭ 3.0 cos x Ϫ 0.5 sin xy ϭ c1 cos x ϩ c2 sin xtan x constcot x constys ϩ y ϭ 0sin xcos xSEC. 2.1 Homogeneous Linear ODEs of Second Order 511Credited to the great mathematician JOSEPH LOUIS LAGRANGE (1736–1813), who was born in Turin,of French extraction, got his first professorship when he was 19 (at the Military Academy of Turin), becamedirector of the mathematical section of the Berlin Academy in 1766, and moved to Paris in 1787. His importantmajor work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique analytique,Paris, 1788), differential equations, approximation theory, algebra, and number theory.c02.qxd 10/27/10 6:06 PM Page 51
• We need no constant of integration because we want to obtain a particular solution; similarly in the nextintegration. Taking exponents and integrating again, we obtain, , hence .Since are linearly independent (their quotient is not constant), we have obtaineda basis of solutions, valid for all positive x.In this example we applied reduction of order to a homogeneous linear ODE [see (2)].Note that we now take the ODE in standard form, with not —this is essentialin applying our subsequent formulas. We assume a solution of (2), on an open intervalI, to be known and want to find a basis. For this we need a second linearly independentsolution of (2) on I. To get , we substitute, ,into (2). This gives(8)Collecting terms in and u, we have.Now comes the main point. Since is a solution of (2), the expression in the lastparentheses is zero. Hence u is gone, and we are left with an ODE in and . We dividethis remaining ODE by and set, thus .This is the desired first-order ODE, the reduced ODE. Separation of variables andintegration givesand .By taking exponents we finally obtain(9) .Here so that . Hence the desired second solution is.The quotient cannot be constant , so that and forma basis of solutions.y2y1(since U Ͼ 0)y2 /y1 ϭ u ϭ ͐U dxy2 ϭ y1u ϭ y1 ΎU dxu ϭ ͐U dxU ϭ ur,U ϭ1y21e؊͐p dxln ƒUƒ ϭ Ϫ2 ln ƒy1 ƒ Ϫ Ύp dxdUUϭ Ϫa2yr1y1ϩ pb dxUr ϩ a2yr1y1ϩ pb U ϭ 0us ϩ ur2yr1 ϩ py1y1ϭ 0ur ϭ U, us ϭ Ur,y1usury1usy1 ϩ ur(2yr1 ϩ py1) ϩ u(y1s ϩ pyr1 ϩ qy1) ϭ 0us, ur,usy1 ϩ 2ury1r ϩ uys1 ϩ p(ury1 ϩ uyr1) ϩ quy1 ϭ 0.ys ϭ y2s ϭ usy1 ϩ 2uryr1 ϩ uys1yr ϭ y2r ϭ ury1 ϩ uyr1y ϭ y2 ϭ uy1y2y2y1f(x)ysys,ys ϩ p(x)yr ϩ q(x)y ϭ 0᭿y1 ϭ x and y2 ϭ x ln ƒ xƒ ϩ 1y2 ϭ ux ϭ x ln ƒ xƒ ϩ 1u ϭ Ύv dx ϭ ln ƒ xƒ ϩ1xv ϭx Ϫ 1x2 ϭ1xϪ1x252 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 52
• SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 53REDUCTION OF ORDER is important because itgives a simpler ODE. A general second-order ODE, linear or not, can be reduced to firstorder if y does not occur explicitly (Prob. 1) or if x does notoccur explicitly (Prob. 2) or if the ODE is homogeneouslinear and we know a solution (see the text).1. Reduction. Show that can bereduced to first order in (from which y followsby integration). Give two examples of your own.2. Reduction. Show that can bereduced to a first-order ODE with y as the independentvariable and , where derive thisby the chain rule. Give two examples.3–10 REDUCTION OF ORDERReduce to first order and solve, showing each step in detail.3.4.5.6. ,7.8.9.10.11–14 APPLICATIONS OF REDUCIBLE ODEs11. Curve. Find the curve through the origin in thexy-plane which satisfies and whose tangentat the origin has slope 1.12. Hanging cable. It can be shown that the curveof an inextensible flexible homogeneous cable hangingbetween two fixed points is obtained by solvingy(x)ys ϭ 2yrys ϩ (1 ϩ 1/y)yr2ϭ 0x2ys Ϫ 5xyr ϩ 9y ϭ 0, y1 ϭ x3ys ϭ 1 ϩ yr2ys ϩ yr3sin y ϭ 0y1 ϭ (cos x)/xxys ϩ 2yr ϩ xy ϭ 0yys ϭ 3yr22xys ϭ 3yrys ϩ yr ϭ 0z ϭ yr;ys ϭ (dz/dy)zF(y, yr, ys) ϭ 0z ϭ yrF(x, yr, ys) ϭ 0F(x, y, yr, ys) ϭ 0, where the constant k depends on theweight. This curve is called catenary (from Latincatena = the chain). Find and graph , assuming thatand those fixed points are and ina vertical xy-plane.13. Motion. If, in the motion of a small body on astraight line, the sum of velocity and acceleration equalsa positive constant, how will the distance dependon the initial velocity and position?14. Motion. In a straight-line motion, let the velocity bethe reciprocal of the acceleration. Find the distancefor arbitrary initial position and velocity.15–19 GENERAL SOLUTION. INITIAL VALUEPROBLEM (IVP)(More in the next set.) (a) Verify that the given functionsare linearly independent and form a basis of solutions ofthe given ODE. (b) Solve the IVP. Graph or sketch thesolution.15.16.17.18.19.20. CAS PROJECT. Linear Independence. Write aprogram for testing linear independence and depen-dence. Try it out on some of the problems in this andthe next problem set and on examples of your own.e؊xsin xe؊xcos x,ys ϩ 2yr ϩ 2y ϭ 0, y(0) ϭ 0, yr(0) ϭ 15,x, x ln xx2ys Ϫ xyr ϩ y ϭ 0, y(1) ϭ 4.3, yr(1) ϭ 0.5,x3>2, x؊1>24x2ys Ϫ 3y ϭ 0, y(1) ϭ Ϫ3, yr(1) ϭ 0,e؊0.3x, xe؊0.3xyr(0) ϭ 0.14,ys ϩ 0.6yr ϩ 0.09y ϭ 0, y(0) ϭ 2.2,cos 2.5x, sin 2.5x4ys ϩ 25y ϭ 0, y(0) ϭ 3.0, yr(0) ϭ Ϫ2.5,y(t)y(t)(1, 0)(Ϫ1, 0)k ϭ 1y(x)ys ϭ k21 ϩ yr2P R O B L E M S E T 2 . 12.2 Homogeneous Linear ODEswith Constant CoefficientsWe shall now consider second-order homogeneous linear ODEs whose coefficients a andb are constant,(1) .These equations have important applications in mechanical and electrical vibrations, aswe shall see in Secs. 2.4, 2.8, and 2.9.To solve (1), we recall from Sec. 1.5 that the solution of the first-order linear ODE witha constant coefficient kyr ϩ ky ϭ 0ys ϩ ayr ϩ by ϭ 0c02.qxd 11/9/10 7:21 PM Page 53
• is an exponential function . This gives us the idea to try as a solution of (1) thefunction(2) .Substituting (2) and its derivativesandinto our equation (1), we obtain.Hence if is a solution of the important characteristic equation (or auxiliary equation)(3)then the exponential function (2) is a solution of the ODE (1). Now from algebra we recallthat the roots of this quadratic equation (3) are(4) ,(3) and (4) will be basic because our derivation shows that the functions(5) andare solutions of (1). Verify this by substituting (5) into (1).From algebra we further know that the quadratic equation (3) may have three kinds ofroots, depending on the sign of the discriminant , namely,a2Ϫ 4by2 ϭ el2xy1 ϭ el1xl2 ϭ 12 AϪa Ϫ 2a2Ϫ 4bB.l1 ϭ 12 AϪa ϩ 2a2Ϫ 4bBl2ϩ al ϩ b ϭ 0l(l2ϩ al ϩ b)elxϭ 0ys ϭ l2elxyr ϭ lelxy ϭ elxy ϭ ce؊kx54 CHAP. 2 Second-Order Linear ODEs(Case I) Two real roots if ,(Case II) A real double root if ,(Case III) Complex conjugate roots if .a2Ϫ 4b Ͻ 0a2Ϫ 4b ϭ 0a2Ϫ 4b Ͼ 0Case I. Two Distinct Real-Roots andIn this case, a basis of solutions of (1) on any interval isandbecause and are defined (and real) for all x and their quotient is not constant. Thecorresponding general solution is(6) .y ϭ c1el1xϩ c2el2xy2y1y2 ϭ el2xy1 ϭ el1xl2l1c02.qxd 10/27/10 6:06 PM Page 54
• E X A M P L E 1 General Solution in the Case of Distinct Real RootsWe can now solve in Example 6 of Sec. 2.1 systematically. The characteristic equation isIts roots are and . Hence a basis of solutions is and and gives the samegeneral solution as before,.E X A M P L E 2 Initial Value Problem in the Case of Distinct Real RootsSolve the initial value problem, , .Solution. Step 1. General solution. The characteristic equation is.Its roots areandso that we obtain the general solution.Step 2. Particular solution. Since , we obtain from the general solution and the initialconditionsHence and . This gives the answer . Figure 30 shows that the curve begins atwith a negative slope but note that the axes have different scales!), in agreement with the initialconditions. ᭿(Ϫ5,y ϭ 4y ϭ exϩ 3e؊2xc2 ϭ 3c1 ϭ 1yr(0) ϭ c1 Ϫ 2c2 ϭ Ϫ5.y(0) ϭ c1 ϩ c2 ϭ 4,yr(x) ϭ c1exϪ 2c2e؊2xy ϭ c1exϩ c2e؊2xl2 ϭ 12 (Ϫ1 Ϫ 19) ϭ Ϫ2l1 ϭ 12 (Ϫ1 ϩ 19) ϭ 1l2ϩ l Ϫ 2 ϭ 0yr(0) ϭ Ϫ5y(0) ϭ 4ys ϩ yr Ϫ 2y ϭ 0᭿y ϭ c1exϩ c2e؊xe؊xexl2 ϭ Ϫ1l1 ϭ 1l2Ϫ 1 ϭ 0.ys Ϫ y ϭ 0SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 552041 1.50.50 x682yCase II. Real Double RootIf the discriminant is zero, we see directly from (4) that we get only one root,, hence only one solution,.To obtain a second independent solution (needed for a basis), we use the method ofreduction of order discussed in the last section, setting . Substituting this and itsderivatives and into (1), we first have.(usy1 ϩ 2uryr1 ϩ uys1) ϩ a(ury1 ϩ uyr1) ϩ buy1 ϭ 0ys2yr2 ϭ ury1 ϩ uyr1y2 ϭ uy1y2y1 ϭ e؊(a/2)xl ϭ l1 ϭ l2 ϭ Ϫa/2a2Ϫ 4bl ϭ Ϫa/2Fig. 30. Solution in Example 2c02.qxd 10/27/10 6:06 PM Page 55
• Collecting terms in and u, as in the last section, we obtain.The expression in the last parentheses is zero, since is a solution of (1). The expressionin the first parentheses is zero, too, since.We are thus left with . Hence . By two integrations, . Toget a second independent solution , we can simply choose andtake . Then . Since these solutions are not proportional, they form a basis.Hence in the case of a double root of (3) a basis of solutions of (1) on any interval is.The corresponding general solution is(7)WARNING! If is a simple root of (4), then with is not a solutionof (1).E X A M P L E 3 General Solution in the Case of a Double RootThe characteristic equation of the ODE is . It has the doubleroot . Hence a basis is and . The corresponding general solution is .E X A M P L E 4 Initial Value Problem in the Case of a Double RootSolve the initial value problem, , .Solution. The characteristic equation is . It has the double rootThis gives the general solution.We need its derivative.From this and the initial conditions we obtain, ; hence .The particular solution of the initial value problem is . See Fig. 31. ᭿y ϭ (3 Ϫ 2x)e؊0.5xc2 ϭ Ϫ2yr(0) ϭ c2 Ϫ 0.5c1 ϭ 3.5y(0) ϭ c1 ϭ 3.0yr ϭ c2e؊0.5xϪ 0.5(c1 ϩ c2x)e؊0.5xy ϭ (c1 ϩ c2x)e؊0.5xl ϭ Ϫ0.5.l2ϩ l ϩ 0.25 ϭ (l ϩ 0.5) 2ϭ 0yr(0) ϭ Ϫ3.5y(0) ϭ 3.0ys ϩ yr ϩ 0.25y ϭ 0᭿y ϭ (c1 ϩ c2x)e؊3xxe؊3xe؊3xl ϭ Ϫ3l2ϩ 6l ϩ 9 ϭ (l ϩ 3)2ϭ 0ys ϩ 6yr ϩ 9y ϭ 0c2 0(c1 ϩ c2x)elxly ϭ (c1 ϩ c2x)e؊ax/2.e؊ax/2, xe؊ax/2y2 ϭ xy1u ϭ xc1 ϭ 1, c2 ϭ 0y2 ϭ uy1u ϭ c1x ϩ c2us ϭ 0usy1 ϭ 02yr1 ϭ Ϫae؊ax/2ϭ Ϫay1y1usy1 ϩ ur(2yr1 ϩ ay1) ϩ u(ys1 ϩ ayr1 ϩ by1) ϭ 0us, ur,56 CHAP. 2 Second-Order Linear ODEs1412108642 x–10123yFig. 31. Solution in Example 4c02.qxd 10/27/10 6:06 PM Page 56
• Case III. Complex RootsThis case occurs if the discriminant of the characteristic equation (3) is negative.In this case, the roots of (3) are the complex that give the complex solutionsof the ODE (1). However, we will show that we can obtain a basis of real solutions(8)where . It can be verified by substitution that these are solutions in thepresent case. We shall derive them systematically after the two examples by using thecomplex exponential function. They form a basis on any interval since their quotientis not constant. Hence a real general solution in Case III is(9) (A, B arbitrary).E X A M P L E 5 Complex Roots. Initial Value ProblemSolve the initial value problem.Solution. Step 1. General solution. The characteristic equation is . It has the rootsHence , and a general solution (9) is.Step 2. Particular solution. The first initial condition gives . The remaining expression is. We need the derivative (chain rule!).From this and the second initial condition we obtain . Hence . Our solution is.Figure 32 shows y and the curves of and (dashed), between which the curve of y oscillates.Such “damped vibrations” (with being time) have important mechanical and electrical applications, as weshall soon see (in Sec. 2.4). ᭿x ϭ tϪe؊0.2xe؊0.2xy ϭ e؊0.2xsin 3xB ϭ 1yr(0) ϭ 3B ϭ 3yr ϭ B(Ϫ0.2e؊0.2xsin 3x ϩ 3e؊0.2xcos 3x)y ϭ Be؊0.2xsin 3xy(0) ϭ A ϭ 0y ϭ e؊0.2x(A cos 3x ϩ B sin 3x)v ϭ 3Ϫ0.2 Ϯ 3i.l2ϩ 0.4l ϩ 9.04 ϭ 0ys ϩ 0.4yr ϩ 9.04y ϭ 0, y(0) ϭ 0, yr(0) ϭ 3y ϭ e؊ax/2(A cos vx ϩ B sin vx)cot vxv2ϭ b Ϫ 14 a2(v Ͼ 0)y1 ϭ e؊ax/2cos vx, y2 ϭ e؊ax/2sin vxl ϭ Ϫ12 a Ϯ iva2Ϫ 4bϪ12 a ϩ iv and Ϫ12 a Ϫ ivSEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 57Fig. 32. Solution in Example 5yx0 10 15 20 25 3050.51.0–0.5–1.0E X A M P L E 6 Complex RootsA general solution of the ODE( constant, not zero)isWith this confirms Example 4 in Sec. 2.1. ᭿v ϭ 1y ϭ A cos vx ϩ B sin vx.vys ϩ v2y ϭ 0c02.qxd 10/27/10 6:06 PM Page 57
• Summary of Cases I–III58 CHAP. 2 Second-Order Linear ODEsCase Roots of (2) Basis of (1) General Solution of (1)IDistinct realIIReal double rootComplex conjugateIII ,e؊ax>2sin vxl2 ϭ Ϫ12 a Ϫ ivy ϭ e؊ax>2(A cos vx ϩ B sin vx)e؊ax>2cos vxl1 ϭ Ϫ12 a ϩ ivy ϭ (c1 ϩ c2x)e؊ax>2e؊ax>2, xe؊ax>2l ϭ Ϫ12 ay ϭ c1el1xϩ c2el2xel1x, el2xl1, l2It is very interesting that in applications to mechanical systems or electrical circuits,these three cases correspond to three different forms of motion or flows of current,respectively. We shall discuss this basic relation between theory and practice in detail inSec. 2.4 (and again in Sec. 2.8).Derivation in Case III. Complex Exponential FunctionIf verification of the solutions in (8) satisfies you, skip the systematic derivation of thesereal solutions from the complex solutions by means of the complex exponential functionof a complex variable . We write , not because x and y occurin the ODE. The definition of in terms of the real functions , , and is(10) .This is motivated as follows. For real , hence , , , we getthe real exponential function . It can be shown that , just as in real. (Proofin Sec. 13.5.) Finally, if we use the Maclaurin series of with as well as, etc., and reorder the terms as shown (this is permissible, ascan be proved), we obtain the series(Look up these real series in your calculus book if necessary.) We see that we have obtainedthe formula(11)called the Euler formula. Multiplication by gives (10).ereitϭ cos t ϩ i sin t,ϭ cos t ϩ i sin t.ϭ 1 Ϫt22!ϩt44!Ϫ ϩ Á ϩ i at Ϫt33!ϩt55!Ϫ ϩ Ábeitϭ 1 ϩ it ϩ(it)22!ϩ(it)33!ϩ(it)44!ϩ(it) 55!ϩ Ái2ϭ Ϫ1, i3ϭ Ϫi, i4ϭ 1z ϭ itezez1ϩz2ϭ ez1ez2ersin 0 ϭ 0cos 0 ϭ 1t ϭ 0z ϭ rezϭ erϩitϭ ereitϭ er(cos t ϩ i sin t)sin tcos terezx ϩ iyr ϩ itz ϭ r ϩ itezc02.qxd 10/27/10 6:06 PM Page 58
• For later use we note that so that byaddition and subtraction of this and (11),(12) .After these comments on the definition (10), let us now turn to Case III.In Case III the radicand in (4) is negative. Hence is positive and,using , we obtain in (4)with defined as in (8). Hence in (4),and, similarly, .Using (10) with and , we thus obtainWe now add these two lines and multiply the result by . This gives as in (8). Thenwe subtract the second line from the first and multiply the result by . This givesas in (8). These results obtained by addition and multiplication by constants are againsolutions, as follows from the superposition principle in Sec. 2.1. This concludes thederivation of these real solutions in Case III.y21/(2i)y112el2xϭ e؊(a/2)xϪivxϭ e؊(a/2)x(cos vx Ϫ i sin vx).el1xϭ e؊(a/2)xϩivxϭ e؊(a/2)x(cos vx ϩ i sin vx)t ϭ vxr ϭ Ϫ12 axl2 ϭ 12 a Ϫ ivl1 ϭ 12 a ϩ ivv12 2a2Ϫ 4b ϭ 12 2Ϫ(4b Ϫ a2) ϭ 2Ϫ(b Ϫ 14 a2) ϭ i2b Ϫ 14 a2ϭ iv1Ϫ1 ϭ i4b Ϫ a2a2Ϫ 4bcos t ϭ 12 (eitϩ e؊it), sin t ϭ12i(eitϪ e؊it)e؊itϭ cos (Ϫt) ϩ i sin (Ϫt) ϭ cos t Ϫ i sin t,SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 591–15 GENERAL SOLUTIONFind a general solution. Check your answer by substitution.ODEs of this kind have important applications to bediscussed in Secs. 2.4, 2.7, and 2.9.1.2.3.4.5.6.7.8.9.10.11.12.13. 9ys Ϫ 30yr ϩ 25y ϭ 0ys ϩ 9yr ϩ 20y ϭ 04ys Ϫ 4yr Ϫ 3y ϭ 0100ys ϩ 240yr ϩ (196p2ϩ 144)y ϭ 0ys ϩ 1.8yr Ϫ 2.08y ϭ 0ys ϩ yr ϩ 3.25y ϭ 0ys ϩ 4.5yr ϭ 010ys Ϫ 32yr ϩ 25.6y ϭ 0ys ϩ 2pyr ϩ p2y ϭ 0ys ϩ 4yr ϩ (p2ϩ 4)y ϭ 0ys ϩ 6yr ϩ 8.96y ϭ 0ys ϩ 36y ϭ 04ys Ϫ 25y ϭ 0P R O B L E M S E T 2 . 214.15.16–20 FIND AN ODEfor the given basis.16. , 17. ,18. , 19. ,20. ,21–30 INITIAL VALUES PROBLEMSSolve the IVP. Check that your answer satisfies the ODE aswell as the initial conditions. Show the details of your work.21. ,22. The ODE in Prob. ,23. ,24. ,25. ,26. , yr(0) ϭ 1ys Ϫ k2y ϭ 0 (k 0), y(0) ϭ 1yr(0) ϭ Ϫ2ys Ϫ y ϭ 0, y(0) ϭ 2yr(Ϫ2) ϭ Ϫe>24ysϪ 4yr Ϫ 3y ϭ 0, y(Ϫ2) ϭ eyr(0) ϭ 0ys ϩ yr Ϫ 6y ϭ 0, y(0) ϭ 10yr(12) ϭ Ϫ24, y(12) ϭ 1yr(0) ϭ Ϫ1.2ys ϩ 25y ϭ 0, y(0) ϭ 4.6e؊3.1xsin 2.1xe؊3.1xcos 2.1xe(؊2؊i)xe(؊2ϩi)xsin 2pxcos 2pxxe؊25xe؊25xe؊4.3xe2.6xys ϩ ayr ϩ by ϭ 0ys ϩ 0.54yr ϩ (0.0729 ϩ p)y ϭ 0ys ϩ 2k2yr ϩ k4y ϭ 0c02.qxd 10/27/10 6:06 PM Page 59
• 27. The ODE in Prob. 5,,28. ,29. The ODE in Prob. ,30. ,31–36 LINEAR INDEPENDENCE is of basic impor-tance, in this chapter, in connection with general solutions,as explained in the text. Are the following functions linearlyindependent on the given interval? Show the details of yourwork.31. any interval32.33.34.35.36. , 0,37. Instability. Solve for the initial conditions, . Then change the initial conditionsto , and explain why thissmall change of 0.001 at causes a large change later,t ϭ 0yr(0) ϭ Ϫ0.999y(0) ϭ 1.001yr(0) ϭ Ϫ1y(0) ϭ 1ys Ϫ y ϭ 0Ϫ1 Ϲ x Ϲ 1e؊xcos 12 xsin 2x, cos x sin x, x Ͻ 0ln x, ln (x3), x Ͼ 1x2, x2ln x, x Ͼ 1eax, e؊ax, x Ͼ 0ekx, xekx,yr(0) ϭ 10.09ys Ϫ 30yr ϩ 25y ϭ 0, y(0) ϭ 3.3yr(0) ϭ 115, y(0) ϭ 0yr(0) ϭ Ϫ0.3258ys Ϫ 2yr Ϫ y ϭ 0, y(0) ϭ Ϫ0.2Ϫ4.5p Ϫ 1 ϭ 13.137yr(0) ϭy(0) ϭ 4.560 CHAP. 2 Second-Order Linear ODEse.g., 22 at . This is instability: a small initialdifference in setting a quantity (a current, for in-stance) becomes larger and larger with time t. This isundesirable.38. TEAM PROJECT. General Properties of Solutions(a) Coefficient formulas. Show how a and b in (1)can be expressed in terms of and . Explain howthese formulas can be used in constructing equationsfor given bases.(b) Root zero. Solve (i) by the presentmethod, and (ii) by reduction to first order. Can youexplain why the result must be the same in bothcases? Can you do the same for a general ODE(c) Double root. Verify directly that withis a solution of (1) in the case of a double root.Verify and explain why is a solution ofbut is not.(d) Limits. Double roots should be limiting cases ofdistinct roots , as, say, . Experiment withthis idea. (Remember l’Hôpital’s rule from calculus.)Can you arrive at ? Give it a try.xel1xl2 : l1l2l1xeϪ2xys Ϫ yr Ϫ 6y ϭ 0y ϭ e؊2xϪa>2l ϭxelxys ϩ ayr ϭ 0?ys ϩ 4yr ϭ 0l2l1t ϭ 102.3 Differential Operators. OptionalThis short section can be omitted without interrupting the flow of ideas. It will not beused subsequently, except for the notations , etc. to stand for , etc.Operational calculus means the technique and application of operators. Here, anoperator is a transformation that transforms a function into another function. Hencedifferential calculus involves an operator, the differential operator D, whichtransforms a (differentiable) function into its derivative. In operator notation we writeand(1) .Similarly, for the higher derivatives we write , and so on. For example,etc.For a homogeneous linear ODE with constant coefficients we cannow introduce the second-order differential operator,where I is the identity operator defined by . Then we can write that ODE as(2) .Ly ϭ P(D)y ϭ (D2ϩ aD ϩ bI)y ϭ 0Iy ϭ yL ϭ P(D) ϭ D2ϩ aD ϩ bIys ϩ ayr ϩ by ϭ 0D sin ϭ cos, D2sin ϭ Ϫsin,D2y ϭ D(Dy) ϭ ysDy ϭ yr ϭdydxD ϭ ddxyr, ysDy, D2yc02.qxd 10/27/10 6:06 PM Page 60
• P suggests “polynomial.” L is a linear operator. By definition this means that if Ly andexist (this is the case if y and w are twice differentiable), then exists forany constants c and k, and.Let us show that from (2) we reach agreement with the results in Sec. 2.2. Sinceand , we obtain(3)This confirms our result of Sec. 2.2 that is a solution of the ODE (2) if and only ifis a solution of the characteristic equation .is a polynomial in the usual sense of algebra. If we replace by the operator D,we obtain the “operator polynomial” . The point of this operational calculus is thatcan be treated just like an algebraic quantity. In particular, we can factor it.E X A M P L E 1 Factorization, Solution of an ODEFactor and solve .Solution. because . Now has thesolution . Similarly, the solution of is . This is a basis of on anyinterval. From the factorization we obtain the ODE, as expected,.Verify that this agrees with the result of our method in Sec. 2.2. This is not unexpected because we factoredin the same way as the characteristic polynomial .It was essential that L in (2) had constant coefficients. Extension of operator methods tovariable-coefficient ODEs is more difficult and will not be considered here.If operational methods were limited to the simple situations illustrated in this section,it would perhaps not be worth mentioning. Actually, the power of the operator approachappears in more complicated engineering problems, as we shall see in Chap. 6.᭿P(l) ϭ l2Ϫ 3l Ϫ 40P(D)ϭ ys ϩ 5yr Ϫ 8yr Ϫ 40y ϭ ys Ϫ 3r Ϫ 40y ϭ 0(D Ϫ 8I)(D ϩ 5I)y ϭ (D Ϫ 8I)(yr ϩ 5y) ϭ D(yr ϩ 5y) Ϫ 8(yr ϩ 5y)P(D)y ϭ 0y2 ϭ e؊5x(D ϩ 5I)y ϭ 0y1 ϭ e8x(D Ϫ 8I)y ϭ yr Ϫ 8y ϭ 0I2ϭ ID2Ϫ 3D Ϫ 40I ϭ (D Ϫ 8I)(D ϩ 5I)P(D)y ϭ 0P(D) ϭ D2Ϫ 3D Ϫ 40IP(D)P(D)lP(l)P(l) ϭ 0lelxϭ (l2ϩ al ϩ b)elxϭ P(l)elxϭ 0.Lel(x) ϭ P(D)el(x) ϭ (D2ϩ aD ϩ bI)el(x)(D2el)(x) ϭ l2elx(Del)(x) ϭ lelxL(cy ϩ kw) ϭ cLy ϩ kLwL(cy ϩ kw)LwSEC. 2.3 Differential Operators. Optional 611–5 APPLICATION OF DIFFERENTIALOPERATORSApply the given operator to the given functions. Show allsteps in detail.1.2.3.4.5. (D Ϫ 2I)(D ϩ 3I); e2x, xe2x, e؊3x(D ϩ 6I)2; 6x ϩ sin 6x, xe؊6x(D Ϫ 2I)2; e2x, xe2x, e؊2xD Ϫ 3I; 3x2ϩ 3x, 3e3x, cos 4x Ϫ sin 4xD2ϩ 2D; cosh 2x, e؊xϩ e2x, cos xP R O B L E M S E T 2 . 36–12 GENERAL SOLUTIONFactor as in the text and solve.6.7.8.9.10.11.12. (D2ϩ 3.0D ϩ 2.5I)y ϭ 0(D2Ϫ 4.00D ϩ 3.84I)y ϭ 0(D2ϩ 4.80D ϩ 5.76I)y ϭ 0(D2Ϫ 4.20D ϩ 4.41I)y ϭ 0(D2ϩ 3I)y ϭ 0(4D2Ϫ I)y ϭ 0(D2ϩ 4.00D ϩ 3.36I)y ϭ 0c02.qxd 10/27/10 6:06 PM Page 61
• 13. Linear operator. Illustrate the linearity of L in (2) bytaking , and .Prove that L is linear.14. Double root. If has distinct rootsand , show that a particular solution is. Obtain from this a solutionby letting and applying l’Hôpital’s rule.␮ : lxelxy ϭ (e␮xϪ elx)>(␮ Ϫ l)l␮D2ϩ aD ϩ bIw ϭ cos 2xc ϭ 4, k ϭ Ϫ6, y ϭ e2x62 CHAP. 2 Second-Order Linear ODEs15. Definition of linearity. Show that the definition oflinearity in the text is equivalent to the following. Ifand exist, then exists andand exist for all constants c and k, andas well asand .L[kw] ϭ kL[w]L[cy] ϭ cL[y]L[y ϩ w] ϭ L[y] ϩ L[w]L[kw]L[cy]L[y ϩ w]L[w]L[y]2.4 Modeling of Free Oscillationsof a Mass–Spring SystemLinear ODEs with constant coefficients have important applications in mechanics, as weshow in this section as well as in Sec. 2.8, and in electrical circuits as we show in Sec. 2.9.In this section we model and solve a basic mechanical system consisting of a mass on anelastic spring (a so-called “mass–spring system,” Fig. 33), which moves up and down.Setting Up the ModelWe take an ordinary coil spring that resists extension as well as compression. We suspendit vertically from a fixed support and attach a body at its lower end, for instance, an ironball, as shown in Fig. 33. We let denote the position of the ball when the systemis at rest (Fig. 33b). Furthermore, we choose the downward direction as positive, thusregarding downward forces as positive and upward forces as negative.y ϭ 02ROBERT HOOKE (1635–1703), English physicist, a forerunner of Newton with respect to the law ofgravitation.UnstretchedspringSystem atrestSystem inmotion(a) (b) (c)s0y(y = 0)Fig. 33. Mechanical mass–spring systemWe now let the ball move, as follows. We pull it down by an amount (Fig. 33c).This causes a spring force(1) (Hooke’s law2)proportional to the stretch y, with called the spring constant. The minus signindicates that points upward, against the displacement. It is a restoring force: It wantsto restore the system, that is, to pull it back to . Stiff springs have large k.y ϭ 0F1k (Ͼ0)F1 ϭ Ϫkyy Ͼ 0c02.qxd 10/27/10 6:06 PM Page 62
• Note that an additional force is present in the spring, caused by stretching it infastening the ball, but has no effect on the motion because it is in equilibrium withthe weight W of the ball, , whereis the constant of gravity at the Earth’s surface (not to be confused withthe universal gravitational constant , which weshall not need; here and are the Earth’s radius andmass, respectively).The motion of our mass–spring system is determined by Newton’s second law(2)where and “Force” is the resultant of all the forces acting on the ball. (Forsystems of units, see the inside of the front cover.)ODE of the Undamped SystemEvery system has damping. Otherwise it would keep moving forever. But if the dampingis small and the motion of the system is considered over a relatively short time, wemay disregard damping. Then Newton’s law with gives the modelthus(3) .This is a homogeneous linear ODE with constant coefficients. A general solution isobtained as in Sec. 2.2, namely (see Example 6 in Sec. 2.2)(4)This motion is called a harmonic oscillation (Fig. 34). Its frequency is Hertz3because and in (4) have the period . The frequency f is calledthe natural frequency of the system. (We write to reserve for Sec. 2.8.)vv02p>v0sincos(ϭ cycles>sec)f ϭ v0>2pv0 ϭBkm.y(t) ϭ A cos v0t ϩ B sin v0tmys ϩ ky ϭ 0mys ϭ ϪF1 ϭ Ϫky;F ϭ ϪF1ys ϭ d2y>dt2Mass ϫ Acceleration ϭ mys ϭ ForceM ϭ 5.98 # 1024kgR ϭ 6.37 # 106mG ϭ gR2>M ϭ 6.67 # 10؊11nt m2>kg232.17 ft>sec2g ϭ 980 cm>sec2ϭ 9.8 m>sec2ϭϪF0 ϭ W ϭ mgF0ϪF0SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 63yt123123PositiveZeroNegativeInitial velocityFig. 34. Typical harmonic oscillations (4) and with the same anddifferent initial velocities , positive 1 , zero 2 , negative 3yr(0) ϭ v0By(0) ϭ A(4*)3HEINRICH HERTZ (1857–1894), German physicist, who discovered electromagnetic waves, as the basisof wireless communication developed by GUGLIELMO MARCONI (1874–1937), Italian physicist (Nobel prizein 1909).c02.qxd 10/27/10 6:06 PM Page 63
• An alternative representation of (4), which shows the physical characteristics of amplitudeand phase shift of (4), is(4*)with and phase angle , where . This follows from theaddition formula (6) in App. 3.1.E X A M P L E 1 Harmonic Oscillation of an Undamped Mass–Spring SystemIf a mass–spring system with an iron ball of weight nt (about 22 lb) can be regarded as undamped, andthe spring is such that the ball stretches it 1.09 m (about 43 in.), how many cycles per minute will the systemexecute? What will its motion be if we pull the ball down from rest by 16 cm (about 6 in.) and let it start withzero initial velocity?Solution. Hooke’s law (1) with W as the force and 1.09 meter as the stretch gives ; thus. The mass is . Thisgives the frequency .From (4) and the initial conditions, . Hence the motion is(Fig. 35).If you have a chance of experimenting with a mass–spring system, don’t miss it. You will be surprised aboutthe good agreement between theory and experiment, usually within a fraction of one percent if you measurecarefully. ᭿y(t) ϭ 0.16 cos 3t [meter] or 0.52 cos 3t [ft]y(0) ϭ A ϭ 0.16 [meter] and yr(0) ϭ v0B ϭ 0v0>(2p) ϭ 2k>m>(2p) ϭ 3>(2p) ϭ 0.48 [Hz] ϭ 29 [cycles>min]m ϭ W>g ϭ 98>9.8 ϭ 10 [kg]98>1.09 ϭ 90 [kg>sec2] ϭ 90 [nt>meter]k ϭ W>1.09 ϭW ϭ 1.09kW ϭ 98tan d ϭ B>AdC ϭ 2A2ϩ B2y(t) ϭ C cos (v0t Ϫ d)64 CHAP. 2 Second-Order Linear ODEs102 4 6 8 t–0.1–0.200.10.2yFig. 35. Harmonic oscillation in Example 1ODE of the Damped SystemTo our model we now add a damping forceobtaining ; thus the ODE of the damped mass–spring system is(5) (Fig. 36)Physically this can be done by connecting the ball to a dashpot; see Fig. 36. We assumethis damping force to be proportional to the velocity . This is generally a goodapproximation for small velocities.yr ϭ dy>dtmys ϩ cyr ϩ ky ϭ 0.mys ϭ Ϫky Ϫ cyrF2 ϭ Ϫcyr,mys ϭ ϪkyFig. 36.Damped systemDashpotBallSpringkmcc02.qxd 10/27/10 6:06 PM Page 64
• SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 65Case I. . Distinct real roots . (Overdamping)Case II. . A real double root. (Critical damping)Case III. . Complex conjugate roots. (Underdamping)c2Ͻ 4mkc2ϭ 4mkl1, l2c2Ͼ 4mkThey correspond to the three Cases I, II, III in Sec. 2.2.Discussion of the Three CasesCase I. OverdampingIf the damping constant c is so large that , then are distinct real roots.In this case the corresponding general solution of (5) is(7) .We see that in this case, damping takes out energy so quickly that the body does notoscillate. For both exponents in (7) are negative because , and. Hence both terms in (7) approach zero as . Practicallyspeaking, after a sufficiently long time the mass will be at rest at the static equilibriumposition . Figure 37 shows (7) for some typical initial conditions.(y ϭ 0)t : ϱb2ϭ a2Ϫ k>m Ͻ a2a Ͼ 0, b Ͼ 0t Ͼ 0y(t) ϭ c1e؊(a؊b)tϩ c2e؊(a؉b)tl1 and l2c2Ͼ 4mkThe constant c is called the damping constant. Let us show that c is positive. Indeed,the damping force acts against the motion; hence for a downward motion wehave which for positive c makes F negative (an upward force), as it should be.Similarly, for an upward motion we have which, for makes positive (adownward force).The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solveit by the method in Sec. 2.2. The characteristic equation is (divide (5) by m).By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2,(6) , where and .It is now interesting that depending on the amount of damping present—whether a lot ofdamping, a medium amount of damping or little damping—three types of motions occur,respectively:b ϭ12m2c2Ϫ 4mka ϭc2ml1 ϭ Ϫa ϩ b, l2 ϭ Ϫa Ϫ bl2ϩcm l ϩkm ϭ 0F2c Ͼ 0yr Ͻ 0yr Ͼ 0F2 ϭ Ϫcyrc02.qxd 10/27/10 6:06 PM Page 65
• 66 CHAP. 2 Second-Order Linear ODEsty123(a)yt11232PositiveZeroNegativeInitial velocity3(b)Fig. 37. Typical motions (7) in the overdamped case(a) Positive initial displacement(b) Negative initial displacementCase II. Critical DampingCritical damping is the border case between nonoscillatory motions (Case I) and oscillations(Case III). It occurs if the characteristic equation has a double root, that is, if ,so that . Then the corresponding general solution of (5) is(8) .This solution can pass through the equilibrium position at most once becauseis never zero and can have at most one positive zero. If both are positive(or both negative), it has no positive zero, so that y does not pass through 0 at all. Figure 38shows typical forms of (8). Note that they look almost like those in the previous figure.c1 and c2c1 ϩ c2te؊aty ϭ 0y(t) ϭ (c1 ϩ c2t)e؊atb ϭ 0, l1 ϭ l2 ϭ Ϫac2ϭ 4mkyt123123PositiveZeroNegativeInitial velocityFig. 38. Critical damping [see (8)]c02.qxd 10/27/10 6:06 PM Page 66
• Case III. UnderdampingThis is the most interesting case. It occurs if the damping constant c is so small that. Then in (6) is no longer real but pure imaginary, say,(9) where .(We now write to reserve for driving and electromotive forces in Secs. 2.8 and 2.9.)The roots of the characteristic equation are now complex conjugates,with , as given in (6). Hence the corresponding general solution is(10)where , as in .This represents damped oscillations. Their curve lies between the dashed curvesin Fig. 39, touching them when is an integer multipleof because these are the points at which equals 1 or .The frequency is Hz (hertz, cycles/sec). From (9) we see that the smalleris, the larger is and the more rapid the oscillations become. If c approaches 0,then approaches , giving the harmonic oscillation (4), whose frequencyis the natural frequency of the system.v0>(2p)v0 ϭ 2k>mv*v*c (Ͼ0)v*>(2p)Ϫ1cos (v*t Ϫ d)pv*t Ϫ dy ϭ Ce؊atand y ϭ ϪCe؊at(4*)C2ϭ A2ϩ B2and tan d ϭ B>Ay(t) ϭ e؊at(A cos v*t ϩ B sin v*t) ϭ Ce؊atcos (v*t Ϫ d)a ϭ c>(2m)l1 ϭ Ϫa ϩ iv*, l2 ϭ Ϫa Ϫ iv*vv*(Ͼ0)v* ϭ12m24mk Ϫ c2ϭBkmϪc24m2b ϭ iv*bc2Ͻ 4mkSEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 67Fig. 39. Damped oscillation in Case III [see (10)]tyCe– tα–Ce– tαE X A M P L E 2 The Three Cases of Damped MotionHow does the motion in Example 1 change if we change the damping constant c from one to another of thefollowing three values, with as before?(I) , (II) , (III) .Solution. It is interesting to see how the behavior of the system changes due to the effect of the damping,which takes energy from the system, so that the oscillations decrease in amplitude (Case III) or even disappear(Cases II and I).(I) With , as in Example 1, the model is the initial value problem.10ys ϩ 100yr ϩ 90y ϭ 0, y(0) ϭ 0.16 [meter], yr(0) ϭ 0m ϭ 10 and k ϭ 90c ϭ 10 kg>secc ϭ 60 kg>secc ϭ 100 kg>secy(0) ϭ 0.16 and yr(0) ϭ 0c02.qxd 10/27/10 6:06 PM Page 67
• The characteristic equation is . It has the roots and . Thisgives the general solution. We also need .The initial conditions give . The solution is . Hence inthe overdamped case the solution is.It approaches 0 as . The approach is rapid; after a few seconds the solution is practically 0, that is, theiron ball is at rest.(II) The model is as before, with instead of 100. The characteristic equation now has the form. It has the double root . Hence the corresponding general solution is. We also need .The initial conditions give . Hence in the critical case thesolution is.It is always positive and decreases to 0 in a monotone fashion.(III) The model now is . Since is smaller than the critical c, we shall getoscillations. The characteristic equation is . It has the complexroots [see (4) in Sec. 2.2 with and ].This gives the general solution.Thus . We also need the derivative.Hence . This gives the solution.We see that these damped oscillations have a smaller frequency than the harmonic oscillations in Example 1 byabout (since 2.96 is smaller than 3.00 by about ). Their amplitude goes to zero. See Fig. 40. ᭿1%1%y ϭ e؊0.5t(0.16 cos 2.96t ϩ 0.027 sin 2.96t) ϭ 0.162e؊0.5tcos (2.96t Ϫ 0.17)yr(0) ϭ Ϫ0.5A ϩ 2.96B ϭ 0, B ϭ 0.5A>2.96 ϭ 0.027yr ϭ e؊0.5t(Ϫ0.5A cos 2.96t Ϫ 0.5B sin 2.96t Ϫ 2.96A sin 2.96t ϩ 2.96B cos 2.96t)y(0) ϭ A ϭ 0.16y ϭ e؊0.5t(A cos 2.96t ϩ B sin 2.96t)l ϭ Ϫ0.5 Ϯ 20.52Ϫ 9 ϭ Ϫ0.5 Ϯ 2.96ib ϭ 9a ϭ 110l2ϩ 10l ϩ 90 ϭ 10[(l ϩ 12) 2ϩ 9 Ϫ 14] ϭ 0c ϭ 1010ys ϩ 10yr ϩ 90y ϭ 0y ϭ (0.16 ϩ 0.48t)e؊3ty(0) ϭ c1 ϭ 0.16, yr(0) ϭ c2 Ϫ 3c1 ϭ 0, c2 ϭ 0.48yr ϭ (c2 Ϫ 3c1 Ϫ 3c2t)e؊3ty ϭ (c1 ϩ c2t)e؊3tϪ310l2ϩ 60l ϩ 90 ϭ 10(l ϩ 3) 2ϭ 0c ϭ 60t : ϱy ϭ Ϫ0.02e؊9tϩ 0.18e؊tc1 ϭ Ϫ0.02, c2 ϭ 0.18c1 ϩ c2 ϭ 0.16, Ϫ9c1 Ϫ c2 ϭ 0yr ϭ Ϫ9c1e؊9tϪ c2e؊ty ϭ c1e؊9tϩ c2e؊tϪ1Ϫ910l2ϩ 100l ϩ 90 ϭ 10(l ϩ 9)(l ϩ 1) ϭ 068 CHAP. 2 Second-Order Linear ODEs102 4 6 8 t–0.05–0.100.050.10.15yFig. 40. The three solutions in Example 2This section concerned free motions of mass–spring systems. Their models are homo-geneous linear ODEs. Nonhomogeneous linear ODEs will arise as models of forcedmotions, that is, motions under the influence of a “driving force.” We shall study themin Sec. 2.8, after we have learned how to solve those ODEs.c02.qxd 10/27/10 6:06 PM Page 68
• SEC. 2.4 Modeling of Free Oscillations of a Mass–Spring System 691–10 HARMONIC OSCILLATIONS(UNDAMPED MOTION)1. Initial value problem. Find the harmonic motion (4)that starts from with initial velocity . Graph orsketch the solutions for , and variousof your choice on common axes. At what t-valuesdo all these curves intersect? Why?2. Frequency. If a weight of 20 nt (about 4.5 lb) stretchesa certain spring by 2 cm, what will the frequency of thecorresponding harmonic oscillation be? The period?3. Frequency. How does the frequency of the harmonicoscillation change if we (i) double the mass, (ii) takea spring of twice the modulus? First find qualitativeanswers by physics, then look at formulas.4. Initial velocity. Could you make a harmonic oscillationmove faster by giving the body a greater initial push?5. Springs in parallel. What are the frequencies ofvibration of a body of mass kg (i) on a springof modulus , (ii) on a spring of modulus, (iii) on the two springs in parallel? SeeFig. 41.k2 ϭ 45 nt>mk1 ϭ 20 nt>mm ϭ 5v0v0 ϭ p, y0 ϭ 1v0y0P R O B L E M S E T 2 . 4The cylindrical buoy of diameter 60 cm in Fig. 43 isfloating in water with its axis vertical. When depresseddownward in the water and released, it vibrates withperiod 2 sec. What is its weight?Fig. 41. Parallel springs (Problem 5)Fig. 42. Pendulum (Problem 7)6. Spring in series. If a body hangs on a spring ofmodulus , which in turn hangs on a springof modulus , what is the modulus k of thiscombination of springs?7. Pendulum. Find the frequency of oscillation of apendulum of length L (Fig. 42), neglecting airresistance and the weight of the rod, and assumingto be so small that practically equals .usin uuk2 ϭ 12s2k1 ϭ 8s110. TEAM PROJECT. Harmonic Motions of SimilarModels. The unifying power of mathematical meth-ods results to a large extent from the fact that differentphysical (or other) systems may have the same or verysimilar models. Illustrate this for the following threesystems(a) Pendulum clock. A clock has a 1-meter pendulum.The clock ticks once for each time the pendulumcompletes a full swing, returning to its original position.How many times a minute does the clock tick?(b) Flat spring (Fig. 45). The harmonic oscillationsof a flat spring with a body attached at one end andhorizontally clamped at the other are also governed by(3). Find its motions, assuming that the body weighs8 nt (about 1.8 lb), the system has its static equilibrium1 cm below the horizontal line, and we let it start fromthis position with initial velocity 10 cm/sec.8. Archimedian principle. This principle states that thebuoyancy force equals the weight of the waterdisplaced by the body (partly or totally submerged).Fig. 44. Tube (Problem 9)9. Vibration of water in a tube. If 1 liter of water (about1.06 US quart) is vibrating up and down under theinfluence of gravitation in a U-shaped tube of diameter2 cm (Fig. 44), what is the frequency? Neglect friction.First guess.Fig. 43. Buoy (Problem 8)LθBody ofmass mWaterlevel( y = 0)yyFig. 45. Flat springc02.qxd 10/27/10 6:06 PM Page 69
• (c) Torsional vibrations (Fig. 46). Undampedtorsional vibrations (rotations back and forth) of awheel attached to an elastic thin rod or wire aregoverned by the equation , whereis the angle measured from the state of equilibrium.Solve this equation for , initialangle and initial angular velocity.20° sec؊1(ϭ 0.349 rad # sec؊1)30°(ϭ 0.5235 rad)K>I0 ϭ 13.69 sec؊2uI0us ϩ Ku ϭ 070 CHAP. 2 Second-Order Linear ODEs11–20 DAMPED MOTION11. Overdamping. Show that for (7) to satisfy initial condi-tions and we must haveand.12. Overdamping. Show that in the overdamped case, thebody can pass through at most once (Fig. 37).13. Initial value problem. Find the critical motion (8)that starts from with initial velocity . Graphsolution curves for and several suchthat (i) the curve does not intersect the t-axis, (ii) itintersects it at respectively.14. Shock absorber. What is the smallest value of thedamping constant of a shock absorber in the suspen-sion of a wheel of a car (consisting of a spring and anabsorber) that will provide (theoretically) an oscillation-free ride if the mass of the car is 2000 kg and the springconstant equals ?15. Frequency. Find an approximation formula for interms of by applying the binomial theorem in (9)and retaining only the first two terms. How good is theapproximation in Example 2, III?16. Maxima. Show that the maxima of an underdampedmotion occur at equidistant t-values and find thedistance.17. Underdamping. Determine the values of t corre-sponding to the maxima and minima of the oscillation. Check your result by graphing .18. Logarithmic decrement. Show that the ratio oftwo consecutive maximum amplitudes of a dampedoscillation (10) is constant, and the natural logarithmof this ratio called the logarithmic decrement,y(t)y(t) ϭ e؊tsin tv0v*4500 kg>sec2t ϭ 1, 2, . . . , 5,v0a ϭ 1, y0 ϭ 1v0y0y ϭ 0v0>b]>2c2 ϭ [(1 Ϫ a>b)y0 Ϫ[(1 ϩ a>b)y0 ϩ v0>b]>2c1 ϭv(0) ϭ v0y(0) ϭ y0equals . Find for the solutions of.19. Damping constant. Consider an underdamped motionof a body of mass . If the time between twoconsecutive maxima is 3 sec and the maximumamplitude decreases to its initial value after 10 cycles,what is the damping constant of the system?20. CAS PROJECT. Transition Between Cases I, II,III. Study this transition in terms of graphs of typicalsolutions. (Cf. Fig. 47.)(a) Avoiding unnecessary generality is part of goodmodeling. Show that the initial value problems (A)and (B),(A)(B) the same with different c and (insteadof 0), will give practically as much information as aproblem with other m, k, .(b) Consider (A). Choose suitable values of c,perhaps better ones than in Fig. 47, for the transitionfrom Case III to II and I. Guess c for the curves in thefigure.(c) Time to go to rest. Theoretically, this time isinfinite (why?). Practically, the system is at rest whenits motion has become very small, say, less than 0.1%of the initial displacement (this choice being up to us),that is in our case,(11) for all t greater than some .In engineering constructions, damping can often bevaried without too much trouble. Experimenting withyour graphs, find empirically a relation betweenand c.(d) Solve (A) analytically. Give a reason why thesolution c of , with the solution of, will give you the best possible c satisfying(11).(e) Consider (B) empirically as in (a) and (b). Whatis the main difference between (B) and (A)?yr(t) ϭ 0t2y(t2) ϭ Ϫ0.001t1t1ƒy(t)ƒ Ͻ 0.001y(0), yr(0)yr(0) ϭ Ϫ2ys ϩ cyr ϩ y ϭ 0, y(0) ϭ 1, yr(0) ϭ 012m ϭ 0.5 kgys ϩ 2yr ϩ 5y ϭ 0¢¢ ϭ 2pa>v*Fig. 47. CAS Project 20Fig. 46. Torsional vibrationsθ10.5–0.5–16 1084yt2c02.qxd 10/27/10 6:06 PM Page 70
• 2.5 Euler–Cauchy EquationsEuler–Cauchy equations4are ODEs of the form(1)with given constants a and b and unknown function . We substituteinto (1). This givesand we now see that was a rather natural choice because we have obtained a com-mon factor . Dropping it, we have the auxiliary equation or(2) . (Note: , not a.)Hence is a solution of (1) if and only if m is a root of (2). The roots of (2) are(3) .Case I. Real different roots give two real solutionsand .These are linearly independent since their quotient is not constant. Hence they constitutea basis of solutions of (1) for all x for which they are real. The corresponding generalsolution for all these x is(4) (c1, c2 arbitrary).E X A M P L E 1 General Solution in the Case of Different Real RootsThe Euler–Cauchy equation has the auxiliary equation . Theroots are 0.5 and . Hence a basis of solutions for all positive x is and and gives the generalsolution. ᭿(x Ͼ 0)y ϭ c1 1x ϩc2xy2 ϭ 1>xy1 ϭ x0.5Ϫ1m2ϩ 0.5m Ϫ 0.5 ϭ 0x2ys ϩ 1.5xyr Ϫ 0.5y ϭ 0y ϭ c1xm1ϩ c2xm2y2(x) ϭ xm2y1(x) ϭ xm1m1 and m2m1 ϭ 12 (1 Ϫ a) ϩ 214 (1 Ϫ a)2Ϫ b, m2 ϭ 12 (1 Ϫ a) Ϫ 214 (1 Ϫ a)2Ϫ by ϭ xma Ϫ 1m2ϩ (a Ϫ 1)m ϩ b ϭ 0m(m Ϫ 1) ϩ am ϩ b ϭ 0xmy ϭ xmx2m(m Ϫ 1)xmϪ2ϩ axmxmϪ1ϩ bxmϭ 0y ϭ xm, yr ϭ mxm؊1, ys ϭ m(m Ϫ 1)xm؊2y(x)x2ys ϩ axyr ϩ by ϭ 0SEC. 2.5 Euler–Cauchy Equations 714LEONHARD EULER (1707–1783) was an enormously creative Swiss mathematician. He madefundamental contributions to almost all branches of mathematics and its application to physics. His importantbooks on algebra and calculus contain numerous basic results of his own research. The great Frenchmathematician AUGUSTIN LOUIS CAUCHY (1789–1857) is the father of modern analysis. He is the creatorof complex analysis and had great influence on ODEs, PDEs, infinite series, elasticity theory, and optics.c02.qxd 10/27/10 6:06 PM Page 71
• Case II. A real double root occurs if and only if becausethen (2) becomes as can be readily verified. Then a solution is, and (1) is of the form(5) or .A second linearly independent solution can be obtained by the method of reduction oforder from Sec. 2.1, as follows. Starting from , we obtain for u the expression(9) Sec. 2.1, namely,whereFrom (5) in standard form (second ODE) we see that (not ax; this is essential!).Hence . Division bygives , so that by integration. Thus, , and andare linearly independent since their quotient is not constant. The general solutioncorresponding to this basis is(6) , .E X A M P L E 2 General Solution in the Case of a Double RootThe Euler–Cauchy equation has the auxiliary equation . It has thedouble root , so that a general solution for all positive x isCase III. Complex conjugate roots are of minor practical importance, and we discussthe derivation of real solutions from complex ones just in terms of a typical example.E X A M P L E 3 Real General Solution in the Case of Complex RootsThe Euler–Cauchy equation has the auxiliary equation .The roots are complex conjugate, and , where . We now use the trickof writing and obtainNext we apply Euler’s formula (11) in Sec. 2.2 with t ϭ 4 ln x to these two formulas. This givesWe now add these two formulas, so that the sine drops out, and divide the result by 2. Then we subtract thesecond formula from the first, so that the cosine drops out, and divide the result by 2i. This yieldsandrespectively. By the superposition principle in Sec. 2.2 these are solutions of the Euler–Cauchy equation (1).Since their quotient is not constant, they are linearly independent. Hence they form a basis of solutions,and the corresponding real general solution for all positive x is(8) .y ϭ x0.2[A cos (4 ln x) ϩ B sin (4 ln x)]cot (4 ln x)x0.2sin (4 ln x)x0.2cos (4 ln x)xm2ϭ x0.2[cos (4 ln x) Ϫ i sin (4 ln x)].xm1ϭ x0.2[cos (4 ln x) ϩ i sin (4 ln x)],xm2ϭ x0.2؊4iϭ x0.2(eln x)؊4iϭ x0.2e؊(4 ln x)i.xm1ϭ x0.2ϩ4iϭ x0.2(eln x)4iϭ x0.2e(4 ln x)i,x ϭ eln xi ϭ 1Ϫ1m2 ϭ 0.2 Ϫ 4im1 ϭ 0.2 ϩ 4im2Ϫ 0.4m ϩ 16.04 ϭ 0x2ys ϩ 0.6xyr ϩ 16.04y ϭ 0᭿y ϭ (c1 ϩ c2 ln x) x3.m ϭ 3m2Ϫ 6m ϩ 9 ϭ 0x2ys Ϫ 5xyr ϩ 9y ϭ 0m ϭ 12 (1 Ϫ a)y ϭ (c1 ϩ c2 ln x) xmy2y1y2 ϭ uy1 ϭ y1 ln xu ϭ ln xU ϭ 1>xy12ϭ x1Ϫaexp͐(Ϫp dx) ϭ exp (Ϫa ln x) ϭ exp (ln x؊a) ϭ 1>xap ϭ a>xU ϭ1y12exp aϪΎp dxb.u ϭ ΎU dxy2 ϭ uy1ys ϩaxyr ϩ(1 Ϫ a)24x2 y ϭ 0x2ys ϩ axyr ϩ 14 (1 Ϫ a)2y ϭ 0y1 ϭ x(1؊a)>2[m ϩ 12 (a Ϫ 1)]2,b ϭ 14 (a Ϫ 1)2m1 ϭ 12 (1 Ϫ a)72 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 72
• Figure 48 shows typical solution curves in the three cases discussed, in particular the real basis functions inExamples 1 and 3. ᭿SEC. 2.5 Euler–Cauchy Equations 73yx0Case I: Real rootsx1.5x ln xx–1.5x–1.5 ln xx0.5x0.5 ln x x0.2 sin (4 ln x)x0.2 cos (4 ln x)x–0.5x–0.5 ln xx1x–1Case II: Double root Case III: Complex roots1.02.03.0yx0 20.4 1.411.0–1.0–1.50.5–0.51.5yx01.0–1.0–1.50.5–0.51.5210.4 1.41 2Fig. 48. Euler–Cauchy equationsE X A M P L E 4 Boundary Value Problem. Electric Potential Field Between Two Concentric SpheresFind the electrostatic potential between two concentric spheres of radii cm and cmkept at potentials and , respectively.Physical Information. v(r) is a solution of the Euler–Cauchy equation , where .Solution. The auxiliary equation is . It has the roots 0 and . This gives the general solution. From the “boundary conditions” (the potentials on the spheres) we obtain.By subtraction, . From the second equation, . Answer:V. Figure 49 shows that the potential is not a straight line, as it would be for a potentialbetween two parallel plates. For example, on the sphere of radius 7.5 cm it is not V, but considerablyless. (What is it?) ᭿110>2 ϭ 55v(r) ϭ Ϫ110 ϩ 1100>rc1 ϭ Ϫc2>10 ϭ Ϫ110c2>10 ϭ 110, c2 ϭ 1100v(10) ϭ c1 ϩc210ϭ 0v(5) ϭ c1 ϩc25ϭ 110.v(r) ϭ c1 ϩ c2>rϪ1m2ϩ m ϭ 0vr ϭ dv>drrvs ϩ 2vr ϭ 0v2 ϭ 0v1 ϭ 110 Vr2 ϭ 10r1 ϭ 5v ϭ v(r)5 6 7 8 9 10 r100806040200vFig. 49. Potential in Example 4v(r)1. Double root. Verify directly by substitution thatis a solution of (1) if (2) has a double root,but and are not solutions of (1) if theroots m1 and m2 of (2) are different.2–11 GENERAL SOLUTIONFind a real general solution. Show the details of your work.2.3. 5x2ys ϩ 23xyr ϩ 16.2y ϭ 0x2ys Ϫ 20y ϭ 0xm2ln xxm1ln xx(1؊a)>2ln x4.5.6.7.8.9.10.11. (x2D2Ϫ 3xD ϩ 10I)y ϭ 0(x2D2Ϫ xD ϩ 5I)y ϭ 0(x2D2Ϫ 0.2xD ϩ 0.36I)y ϭ 0(x2D2Ϫ 3xD ϩ 4I)y ϭ 0(x2D2Ϫ 4xD ϩ 6I)y ϭ Cx2ys ϩ 0.7xyr Ϫ 0.1y ϭ 04x2ys ϩ 5y ϭ 0xys ϩ 2yr ϭ 0P R O B L E M S E T 2 . 5c02.qxd 10/27/10 6:06 PM Page 73
• 12–19 INITIAL VALUE PROBLEMSolve and graph the solution. Show the details of your work.12.13.14.15.16.17.18.19.yr(1) ϭ Ϫ4.5(x2D2Ϫ xD Ϫ 15I )y ϭ 0, y(1) ϭ 0.1,(9x2D2ϩ 3xD ϩ I)y ϭ 0, y(1) ϭ 1, yr(1) ϭ 0(x2D2ϩ xD ϩ I)y ϭ 0, y(1) ϭ 1, yr(1) ϭ 1(x2D2Ϫ 3xD ϩ 4I)y ϭ 0, y(1) ϭ Ϫp, yr(1) ϭ 2px2ys ϩ 3xyr ϩ y ϭ 0, y(1) ϭ 3.6, yr(1) ϭ 0.4x2ys ϩ xyr ϩ 9y ϭ 0, y(1) ϭ 0, yr(1) ϭ 2.5yr(1) ϭ Ϫ1.5x2ys ϩ 3xyr ϩ 0.75y ϭ 0, y(1) ϭ 1,x2ys Ϫ 4xyr ϩ 6y ϭ 0, y(1) ϭ 0.4, yr(1) ϭ 074 CHAP. 2 Second-Order Linear ODEs20. TEAM PROJECT. Double Root(a) Derive a second linearly independent solution of(1) by reduction of order; but instead of using (9), Sec.2.1, perform all steps directly for the present ODE (1).(b) Obtain by considering the solutionsand of a suitable Euler–Cauchy equation andletting .(c) Verify by substitution thatis a solution in the critical case.(d) Transform the Euler–Cauchy equation (1) intoan ODE with constant coefficients by setting.(e) Obtain a second linearly independent solution ofthe Euler–Cauchy equation in the “critical case” fromthat of a constant-coefficient ODE.x ϭ et(x Ͼ 0)m ϭ (1 Ϫ a)>2,xmln x,s : 0xmϩsxmxmln x2.6 Existence and Uniquenessof Solutions. WronskianIn this section we shall discuss the general theory of homogeneous linear ODEs(1)with continuous, but otherwise arbitrary, variable coefficients p and q. This will concernthe existence and form of a general solution of (1) as well as the uniqueness of the solutionof initial value problems consisting of such an ODE and two initial conditions(2)with given .The two main results will be Theorem 1, stating that such an initial value problemalways has a solution which is unique, and Theorem 4, stating that a general solution(3)includes all solutions. Hence linear ODEs with continuous coefficients have no “singularsolutions” (solutions not obtainable from a general solution).Clearly, no such theory was needed for constant-coefficient or Euler–Cauchy equationsbecause everything resulted explicitly from our calculations.Central to our present discussion is the following theorem.T H E O R E M 1 Existence and Uniqueness Theorem for Initial Value ProblemsIf and are continuous functions on some open interval I (see Sec. 1.1) andx0 is in I, then the initial value problem consisting of (1) and (2) has a uniquesolution on the interval I.y(x)q(x)p(x)(c1, c2 arbitrary)y ϭ c1y1 ϩ c2y2x0, K0, and K1y(x0) ϭ K0, yr(x0) ϭ K1ys ϩ p(x)yr ϩ q(x)y ϭ 0c02.qxd 10/27/10 6:06 PM Page 74
• The proof of existence uses the same prerequisites as the existence proof in Sec. 1.7and will not be presented here; it can be found in Ref. [A11] listed in App. 1. Uniquenessproofs are usually simpler than existence proofs. But for Theorem 1, even the uniquenessproof is long, and we give it as an additional proof in App. 4.Linear Independence of SolutionsRemember from Sec. 2.1 that a general solution on an open interval I is made up from abasis on I, that is, from a pair of linearly independent solutions on I. Here we calllinearly independent on I if the equation(4) .We call linearly dependent on I if this equation also holds for constantsnot both 0. In this case, and only in this case, are proportional on I, that is (seeSec. 2.1),(5) (a) or (b) for all on I.For our discussion the following criterion of linear independence and dependence ofsolutions will be helpful.T H E O R E M 2 Linear Dependence and Independence of SolutionsLet the ODE (1) have continuous coefficients and on an open interval I.Then two solutions of (1) on I are linearly dependent on I if and only iftheir “Wronskian”(6)is 0 at some in I. Furthermore, if at an in I, then on I;hence, if there is an in I at which W is not 0, then are linearly independenton I.P R O O F (a) Let be linearly dependent on I. Then (5a) or (5b) holds on I. If (5a) holds,thenSimilarly if (5b) holds.(b) Conversely, we let for some and show that this implies lineardependence of on I. We consider the linear system of equations in the unknowns(7)k1 y1r(x0) ϩ k2 y2r(x0) ϭ 0.k1 y1(x0) ϩ k2 y2(x0) ϭ 0k1, k2y1 and y2x ϭ x0W(y1, y2) ϭ 0W(y1, y2) ϭ y1y2r Ϫ y2y1r ϭ ky2y2r Ϫ y2ky2r ϭ 0.y1 and y2y1, y2x1W ϭ 0x ϭ x0W ϭ 0x0W(y1, y2) ϭ y1y2r Ϫ y2y1ry1 and y2q(x)p(x)y2 ϭ ly1y1 ϭ ky2y1 and y2k1, k2y1, y2k1y1(x) ϩ k2y2(x) ϭ 0 on I implies k1 ϭ 0, k2 ϭ 0y1, y2y1, y2SEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 75c02.qxd 10/27/10 6:06 PM Page 75
• To eliminate , multiply the first equation by and the second by and add theresulting equations. This gives.Similarly, to eliminate , multiply the first equation by and the second by andadd the resulting equations. This gives.If W were not 0 at , we could divide by W and conclude that . Since W is0, division is not possible, and the system has a solution for which are not both0. Using these numbers , we introduce the function.Since (1) is homogeneous linear, Fundamental Theorem 1 in Sec. 2.1 (the superpositionprinciple) implies that this function is a solution of (1) on I. From (7) we see that it satisfiesthe initial conditions . Now another solution of (1) satisfying thesame initial conditions is . Since the coefficients p and q of (1) are continuous,Theorem 1 applies and gives uniqueness, that is, , written outon I.Now since and are not both zero, this means linear dependence of , on I.(c) We prove the last statement of the theorem. If at an in I, we havelinear dependence of on I by part (b), hence by part (a) of this proof. Hencein the case of linear dependence it cannot happen that at an in I. If it doeshappen, it thus implies linear independence as claimed.For calculations, the following formulas are often simpler than (6).(6*) or (b)These formulas follow from the quotient rule of differentiation.Remark. Determinants. Students familiar with second-order determinants may havenoticed that.This determinant is called the Wronski determinant5or, briefly, the Wronskian, of twosolutions and of (1), as has already been mentioned in (6). Note that its four entriesoccupy the same positions as in the linear system (7).y2y1W(y1, y2) ϭ `y1 y2yr1 yr2` ϭ y1yr2 Ϫ y2yr1Ϫay1y2bry22 (y2 0).W(y1, y2) ϭ (a) ay2y1bry21 (y1 0)᭿x1W(x1) 0W ϵ 0y1, y2x0W(x0) ϭ 0y2y1k2k1k1y1 ϩ k2y2 ϵ 0y ϵ y*y* ϵ 0y(x0) ϭ 0, yr(x0) ϭ 0y(x) ϭ k1y1(x) ϩ k2y2(x)k1, k2k1 and k2k1 ϭ k2 ϭ 0x0k2W(y1(x0), y2(x0)) ϭ 0y1Ϫy1rk1k1y1(x0)y2r(x0) Ϫ k1y1r(x0)y2(x0) ϭ k1W(y1(x0), y2(x0)) ϭ 0Ϫy2yr2k276 CHAP. 2 Second-Order Linear ODEs5Introduced by WRONSKI (JOSEF MARIA HÖNE, 1776–1853), Polish mathematician.c02.qxd 10/27/10 6:06 PM Page 76
• E X A M P L E 1 Illustration of Theorem 2The functions and are solutions of . Their Wronskian is.Theorem 2 shows that these solutions are linearly independent if and only if . Of course, we can seethis directly from the quotient . For we have , which implies linear dependence(why?).E X A M P L E 2 Illustration of Theorem 2 for a Double RootA general solution of on any interval is . (Verify!). The correspondingWronskian is not 0, which shows linear independence of and on any interval. Namely,.A General Solution of (1) Includes All SolutionsThis will be our second main result, as announced at the beginning. Let us start withexistence.T H E O R E M 3 Existence of a General SolutionIf p(x) and q(x) are continuous on an open interval I, then (1) has a general solutionon I.P R O O F By Theorem 1, the ODE (1) has a solution on I satisfying the initial conditionsand a solution on I satisfying the initial conditionsThe Wronskian of these two solutions has at the valueHence, by Theorem 2, these solutions are linearly independent on I. They form a basis ofsolutions of (1) on I, and with arbitrary is a general solution of (1)on I, whose existence we wanted to prove. ᭿c1, c2y ϭ c1y1 ϩ c2˛y2W(y1(0), y2(0)) ϭ y1(x0)y2r(x0) Ϫ y2(x0)y1r(x0) ϭ 1.x ϭ x0y2r(x0) ϭ 1.y2(x0) ϭ 0,y2(x)y1r(x0) ϭ 0y1(x0) ϭ 1,y1(x)᭿W(x, xex) ϭ `exxexex(x ϩ 1)ex` ϭ (x ϩ 1)e2xϪ xe2xϭ e2x0xexexy ϭ (c1 ϩ c2x)exys Ϫ 2yr ϩ y ϭ 0᭿y2 ϭ 0v ϭ 0y2>y1 ϭ tan vxv 0W(cos vx, sin vx) ϭ `cos vx sin vxϪv sin vx v cos vx` ϭ y1y2r Ϫ y2y1r ϭ v cos2vx ϩ v sin2vx ϭ vys ϩ v2y ϭ 0y2 ϭ sin vxy1 ϭ cos vxSEC. 2.6 Existence and Uniqueness of Solutions. Wronskian 77c02.qxd 10/27/10 6:06 PM Page 77
• We finally show that a general solution is as general as it can possibly be.T H E O R E M 4 A General Solution Includes All SolutionsIf the ODE (1) has continuous coefficients p(x) and q(x) on some open interval I,then every solution of (1) on I is of the form(8)where is any basis of solutions of (1) on I and are suitable constants.Hence (1) does not have singular solutions (that is, solutions not obtainable froma general solution).P R O O F Let be any solution of (1) on I. Now, by Theorem 3 the ODE (1) has a generalsolution(9)on I. We have to find suitable values of such that on I. We choose anyin I and show first that we can find values of such that we reach agreement atthat is, and . Written out in terms of (9), this becomes(10)(a)(b)We determine the unknowns and . To eliminate we multiply (10a) by and(10b) by and add the resulting equations. This gives an equation for Then wemultiply (10a) by and (10b) by and add the resulting equations. This givesan equation for These new equations are as follows, where we take the values ofatSince is a basis, the Wronskian W in these equations is not 0, and we can solve forand We call the (unique) solution By substituting it into (9) weobtain from (9) the particular solutionNow since is a solution of (10), we see from (10) thatFrom the uniqueness stated in Theorem 1 this implies that y* and Y must be equaleverywhere on I, and the proof is complete. ᭿y*r(x0) ϭ Yr(x0).y*(x0) ϭ Y(x0),C1, C2y*(x) ϭ C1y1(x) ϩ C2y2(x).c1 ϭ C1, c2 ϭ C2.c2.c1y1, y2c2(y1y2r Ϫ y2y1r) ϭ c2W(y1, y2) ϭ y1Yr Ϫ Yy1r.c1(y1y2r Ϫ y2y1r) ϭ c1W(y1, y2) ϭ Yy2r Ϫ y2Yrx0.y1, y1r, y2, y2r, Y, Yrc2.y1(x0)Ϫy1r(x0)c1.Ϫy2(x0)y2r(x0)c2,c2c1c1y1r(x0) ϩ c2y2r(x0) ϭ Yr(x0).c1y1(x0) ϩ c2y2(x0) ϭ Y(x0)yr(x0) ϭ Yr(x0)y(x0) ϭ Y(x0)x0,c1, c2x0y(x) ϭ Y(x)c1, c2y(x) ϭ c1y1(x) ϩ c2y2(x)y ϭ Y(x)C1, C2y1, y2Y(x) ϭ C1y1(x) ϩ C2y2(x)y ϭ Y(x)78 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 78
• Reflecting on this section, we note that homogeneous linear ODEs with continuous variablecoefficients have a conceptually and structurally rather transparent existence and uniquenesstheory of solutions. Important in itself, this theory will also provide the foundation for ourstudy of nonhomogeneous linear ODEs, whose theory and engineering applications formthe content of the remaining four sections of this chapter.SEC. 2.7 Nonhomogeneous ODEs 791. Derive (6*) from (6).2–8 BASIS OF SOLUTIONS. WRONSKIANFind the Wronskian. Show linear independence by usingquotients and confirm it by Theorem 2.2.3.4.5.6.7.8.9–15 ODE FOR GIVEN BASIS. WRONSKIAN. IVP(a) Find a second-order homogeneous linear ODE forwhich the given functions are solutions. (b) Show linearindependence by the Wronskian. (c) Solve the initial valueproblem.9.10.11.12.13.14.15. cosh 1.8x, sinh 1.8x, y(0) ϭ 14.20, yr(0) ϭ 16.38yr(0) ϭ Ϫk Ϫ peϪkxcos px, eϪkxsin px, y(0) ϭ 1,1, eϪ2x, y(0) ϭ 1, yr(0) ϭ Ϫ1x2, x2ln x, y(1) ϭ 4, yr(1) ϭ 6yr(0) ϭ Ϫ7.5e؊2.5xcos 0.3x, e؊2.5xsin 0.3x, y(0) ϭ 3,xm1, xm2, y(1) ϭ Ϫ2, yr(1) ϭ 2m1 Ϫ 4m2cos 5x, sin 5x, y(0) ϭ 3, yr(0) ϭ Ϫ5xkcos (ln x), xksin (ln x)cosh ax, sinh axe؊xcos vx, e؊xsin vxx3, x2x, 1>xe؊0.4x, e؊2.6xe4.0x, e؊1.5x16. TEAM PROJECT. Consequences of the PresentTheory. This concerns some noteworthy generalproperties of solutions. Assume that the coefficients pand q of the ODE (1) are continuous on some openinterval I, to which the subsequent statements refer.(a) Solve (a) by exponential functions,(b) by hyperbolic functions. How are the constants inthe corresponding general solutions related?(b) Prove that the solutions of a basis cannot be 0 atthe same point.(c) Prove that the solutions of a basis cannot have amaximum or minimum at the same point.(d) Why is it likely that formulas of the form (6*)should exist?(e) Sketch if and 0 ifif and if Show linearindependence on What is theirWronskian? What Euler–Cauchy equation dosatisfy? Is there a contradiction to Theorem 2?(f) Prove Abel’s formula6where Apply it to Prob. 6. Hint:Write (1) for and for Eliminate q algebraicallyfrom these two ODEs, obtaining a first-order linearODE. Solve it.y2.y1c ϭ W(y1(x0), y2(x0)).W(y1(x), y2(x)) ϭ c exp cϪΎxx0p(t) dt dy1, y2Ϫ1 Ͻ x Ͻ 1.x Ͻ 0.x3x м 0y2(x) ϭ 0x Ͻ 0,x м 0y1(x) ϭ x3ys Ϫ y ϭ 0P R O B L E M S E T 2 . 66NIELS HENRIK ABEL (1802–1829), Norwegian mathematician.2.7 Nonhomogeneous ODEsWe now advance from homogeneous to nonhomogeneous linear ODEs.Consider the second-order nonhomogeneous linear ODE(1)where We shall see that a “general solution” of (1) is the sum of a generalsolution of the corresponding homogeneous ODEr(x) [ 0.ys ϩ p(x)yr ϩ q(x)y ϭ r(x)c02.qxd 10/27/10 6:06 PM Page 79
• (2)and a “particular solution” of (1). These two new terms “general solution of (1)” and“particular solution of (1)” are defined as follows.D E F I N I T I O N General Solution, Particular SolutionA general solution of the nonhomogeneous ODE (1) on an open interval I is asolution of the form(3)here, is a general solution of the homogeneous ODE (2) on I andis any solution of (1) on I containing no arbitrary constants.A particular solution of (1) on I is a solution obtained from (3) by assigningspecific values to the arbitrary constants and in .Our task is now twofold, first to justify these definitions and then to develop a methodfor finding a solution of (1).Accordingly, we first show that a general solution as just defined satisfies (1) and thatthe solutions of (1) and (2) are related in a very simple way.T H E O R E M 1 Relations of Solutions of (1) to Those of (2)(a) The sum of a solution y of (1) on some open interval I and a solution of(2) on I is a solution of (1) on I. In particular, (3) is a solution of (1) on I.(b) The difference of two solutions of (1) on I is a solution of (2) on I.P R O O F (a) Let denote the left side of (1). Then for any solutions y of (1) and of (2) on I,(b) For any solutions y and y* of (1) on I we haveNow for homogeneous ODEs (2) we know that general solutions include all solutions.We show that the same is true for nonhomogeneous ODEs (1).T H E O R E M 2 A General Solution of a Nonhomogeneous ODE Includes All SolutionsIf the coefficients p(x), q(x), and the function r(x) in (1) are continuous on someopen interval I, then every solution of (1) on I is obtained by assigning suitablevalues to the arbitrary constants and in a general solution (3) of (1) on I.P R O O F Let be any solution of (1) on I and any x in I. Let (3) be any general solution of(1) on I. This solution exists. Indeed, exists by Theorem 3 in Sec. 2.6yh ϭ c1y1 ϩ c2y2x0y*c2c1᭿r Ϫ r ϭ 0.L[y Ϫ y*] ϭ L[y] Ϫ L[y*] ϭL[y ϩ y~] ϭ L[y] ϩ L[y~] ϭ r ϩ 0 ϭ r.y~L[y]y~ypyhc2c1ypyh ϭ c1y1 ϩ c2y2y(x) ϭ yh(x) ϩ yp1x2;ys ϩ p(x)yr ϩ q(x)y ϭ 080 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 80
• because of the continuity assumption, and exists according to a construction to beshown in Sec. 2.10. Now, by Theorem 1(b) just proved, the difference is asolution of (2) on I. At we haveTheorem 1 in Sec. 2.6 implies that for these conditions, as for any other initial conditionsin I, there exists a unique particular solution of (2) obtained by assigning suitable valuesto in . From this and the statement follows.Method of Undetermined CoefficientsOur discussion suggests the following. To solve the nonhomogeneous ODE (1) or an initialvalue problem for (1), we have to solve the homogeneous ODE (2) and find any solutionof (1), so that we obtain a general solution (3) of (1).How can we find a solution of (1)? One method is the so-called method ofundetermined coefficients. It is much simpler than another, more general, method (givenin Sec. 2.10). Since it applies to models of vibrational systems and electric circuits to beshown in the next two sections, it is frequently used in engineering.More precisely, the method of undetermined coefficients is suitable for linear ODEswith constant coefficients a and b(4)when is an exponential function, a power of x, a cosine or sine, or sums or productsof such functions. These functions have derivatives similar to itself. This gives theidea. We choose a form for similar to , but with unknown coefficients to bedetermined by substituting that and its derivatives into the ODE. Table 2.1 on p. 82shows the choice of for practically important forms of . Corresponding rules areas follows.Choice Rules for the Method of Undetermined Coefficients(a) Basic Rule. If in (4) is one of the functions in the first column inTable 2.1, choose in the same line and determine its undeterminedcoefficients by substituting and its derivatives into (4).(b) Modification Rule. If a term in your choice for happens to be asolution of the homogeneous ODE corresponding to (4), multiply this termby x (or by if this solution corresponds to a double root of thecharacteristic equation of the homogeneous ODE).(c) Sum Rule. If is a sum of functions in the first column of Table 2.1,choose for the sum of the functions in the corresponding lines of thesecond column.The Basic Rule applies when is a single term. The Modification Rule helps in theindicated case, and to recognize such a case, we have to solve the homogeneous ODEfirst. The Sum Rule follows by noting that the sum of two solutions of (1) withand (and the same left side!) is a solution of (1) with . (Verify!)r ϭ r1 ϩ r2r ϭ r2r ϭ r1r(x)ypr(x)x2ypypypr(x)r(x)ypypr(x)ypr(x)r(x)ys ϩ ayr ϩ by ϭ r(x)ypyp᭿y* ϭ Y ϩ ypyhc1, c2Yr1x02 ϭ y*r1x02 Ϫ yrp1x02.Y1x02 ϭ y*1x02 Ϫ yp(x0).x0Y ϭ y* Ϫ ypypSEC. 2.7 Nonhomogeneous ODEs 81c02.qxd 10/27/10 6:06 PM Page 81
• The method is self-correcting. A false choice for or one with too few terms will leadto a contradiction. A choice with too many terms will give a correct result, with superfluouscoefficients coming out zero.Let us illustrate Rules (a)–(c) by the typical Examples 1–3.yp82 CHAP. 2 Second-Order Linear ODEsTerm in Choice forkeaxsin vxkeaxcos vxk sin vxk cos vxKnxnϩ KnϪ1xnϪ1ϩ Á ϩ K1x ϩ K0kxn(n ϭ 0, 1, Á )Cegxkegxyp(x)r(x)Table 2.1 Method of Undetermined Coefficientsfeax(K cos vx ϩ M sin vx)fK cos vx ϩ M sin vxE X A M P L E 1 Application of the Basic Rule (a)Solve the initial value problem(5)Solution. Step 1. General solution of the homogeneous ODE. The ODE has the general solutionStep 2. Solution of the nonhomogeneous ODE. We first try Then By substitution,For this to hold for all x, the coefficient of each power of must be the sameon both sides; thus and a contradiction.The second line in Table 2.1 suggests the choiceThenEquating the coefficients of on both sides, we have HenceThis gives andStep 3. Solution of the initial value problem. Setting and using the first initial condition giveshence By differentiation and from the second initial condition,andThis gives the answer (Fig. 50)Figure 50 shows y as well as the quadratic parabola about which y is oscillating, practically like a sine curvesince the cosine term is smaller by a factor of about ᭿1>1000.ypy ϭ 0.002 cos x ϩ 1.5 sin x ϩ 0.001x2Ϫ 0.002.yr(0) ϭ B ϭ 1.5.yr ϭ yrh ϩ yrp ϭ ϪA sin x ϩ B cos x ϩ 0.002xA ϭ 0.002.y(0) ϭ A Ϫ 0.002 ϭ 0,x ϭ 0y ϭ yh ϩ yp ϭ A cos x ϩ B sin x ϩ 0.001x2Ϫ 0.002.yp ϭ 0.001x2Ϫ 0.002,K0 ϭ Ϫ2K2 ϭ Ϫ0.002.K2 ϭ 0.001, K1 ϭ 0, 2K2 ϩ K0 ϭ 0.x2, x, x0ysp ϩ yp ϭ 2K2 ϩ K2x2ϩ K1x ϩ K0 ϭ 0.001x2.yp ϭ K2x2ϩ K1x ϩ K0.2K ϭ 0,K ϭ 0.001x (x2and x0)2K ϩ Kx2ϭ 0.001x2.ysp ϭ 2K.yp ϭ Kx2.ypyh ϭ A cos x ϩ B sin x.ys ϩ y ϭ 0yr(0) ϭ 1.5.y(0) ϭ 0,ys ϩ y ϭ 0.001x2,102–120 xy30 4010Fig. 50. Solution in Example 1c02.qxd 10/27/10 6:06 PM Page 82
• E X A M P L E 2 Application of the Modification Rule (b)Solve the initial value problem(6)Solution. Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneousODE is Hence the homogeneous ODE has the general solutionStep 2. Solution of the nonhomogeneous ODE. The function on the right would normally requirethe choice . But we see from that this function is a solution of the homogeneous ODE, whichcorresponds to a double root of the characteristic equation. Hence, according to the Modification Rule we haveto multiply our choice function by . That is, we choose. Then .We substitute these expressions into the given ODE and omit the factor . This yieldsComparing the coefficients of gives hence This gives the solution. Hence the given ODE has the general solutionStep 3. Solution of the initial value problem. Setting in y and using the first initial condition, we obtainDifferentiation of y givesFrom this and the second initial condition we have Hence This givesthe answer (Fig. 51)The curve begins with a horizontal tangent, crosses the x-axis at (where ) andapproaches the axis from below as x increases. ᭿1 ϩ 1.5x Ϫ 5x2ϭ 0x ϭ 0.6217y ϭ (1 ϩ 1.5x)e؊1.5xϪ 5x2e؊1.5xϭ (1 ϩ 1.5x Ϫ 5x2)e؊1.5x.c2 ϭ 1.5c1 ϭ 1.5.yr(0) ϭ c2 Ϫ 1.5c1 ϭ 0.yr ϭ (c2 Ϫ 1.5c1 Ϫ 1.5c2x)e؊1.5xϪ 10xe؊1.5xϩ 7.5x2e؊1.5x.y(0) ϭ c1 ϭ 1.x ϭ 0y ϭ yh ϩ yp ϭ (c1 ϩ c2x)e؊1.5xϪ 5x2e؊1.5x.yp ϭ Ϫ5x2e؊1.5xC ϭ Ϫ5.0 ϭ 0, 0 ϭ 0, 2C ϭ Ϫ10,x2, x, x0C(2 Ϫ 6x ϩ 2.25x2) ϩ 3C(2x Ϫ 1.5x2) ϩ 2.25Cx2ϭ Ϫ10.e؊1.5xysp ϭ C(2 Ϫ 3x Ϫ 3x ϩ 2.25x2)e؊1.5xyrp ϭ C(2x Ϫ 1.5x2)e؊1.5x,yp ϭ Cx2e؊1.5xx2yhCe؊1.5xe؊1.5xypyh ϭ (c1 ϩ c2˛x)e؊1.5x.l2ϩ 3l ϩ 2.25 ϭ (l ϩ 1.5)2ϭ 0.yr(0) ϭ 0.y(0) ϭ 1,ys ϩ 3yr ϩ 2.25y ϭ Ϫ10e؊1.5x,SEC. 2.7 Nonhomogeneous ODEs 83Fig. 51. Solution in Example 254321 x–0.5–1.000.51.0yE X A M P L E 3 Application of the Sum Rule (c)Solve the initial value problem(7)Solution. Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneousODE iswhich gives the general solution yh ϭ c1e؊x>2ϩ c2e؊3x>2.l2ϩ 2l ϩ 0.75 ϭ (l ϩ 12) (l ϩ 32) ϭ 0yr(0) ϭ Ϫ0.43.y(0) ϭ 2.78,ys ϩ 2yr ϩ 0.75y ϭ 2 cos x Ϫ 0.25 sin x ϩ 0.09x,c02.qxd 10/27/10 6:06 PM Page 83
• Step 2. Particular solution of the nonhomogeneous ODE. We write and, following Table 2.1,(C) and (B),andDifferentiation gives and Substitutionof into the ODE in (7) gives, by comparing the cosine and sine terms,hence and Substituting into the ODE in (7) and comparing the - and -terms givesthusHence a general solution of the ODE in (7) isStep 3. Solution of the initial value problem. From and the initial conditions we obtain.Hence This gives the solution of the IVP (Fig. 52)᭿y ϭ 3.1e؊x>2ϩ sin x ϩ 0.12x Ϫ 0.32.c1 ϭ 3.1, c2 ϭ 0.y(0) ϭ c1 ϩ c2 Ϫ 0.32 ϭ 2.78, yr(0) ϭ Ϫ12 c1 Ϫ 32 c2 ϩ 1 ϩ 0.12 ϭ Ϫ0.4y, yry ϭ c1e؊x>2ϩ c2e؊3x>2ϩ sin x ϩ 0.12x Ϫ 0.32.K1 ϭ 0.12, K0 ϭ Ϫ0.32.0.75K1 ϭ 0.09, 2K1 ϩ 0.75K0 ϭ 0,x0xyp2M ϭ 1.K ϭ 0ϪK ϩ 2M ϩ 0.75K ϭ 2, ϪM Ϫ 2K ϩ 0.75M ϭ Ϫ0.25,yp1yp2r ϭ 1, yp2s ϭ 0.yp1r ϭ ϪK sin x ϩ M cos x, yp1s ϭ ϪK cos x Ϫ M sin xyp2 ϭ K1x ϩ K0.yp1 ϭ K cos x ϩ M sin xyp ϭ yp1 ϩ yp284 CHAP. 2 Second-Order Linear ODEsFig. 52. Solution in Example 3x2 4 6 8 10 12 14 16 18 20y00.511.522.53–0.5Stability. The following is important. If (and only if) all the roots of the characteristicequation of the homogeneous ODE in (4) are negative, or have a negativereal part, then a general solution of this ODE goes to 0 as , so that the “transientsolution” of (4) approaches the “steady-state solution” . In this case thenonhomogeneous ODE and the physical or other system modeled by the ODE are calledstable; otherwise they are called unstable. For instance, the ODE in Example 1 is unstable.Applications follow in the next two sections.ypy ϭ yh ϩ ypx : ϱyhys ϩ ayr ϩ by ϭ 01–10 NONHOMOGENEOUS LINEAR ODEs:GENERAL SOLUTIONFind a (real) general solution. State which rule you areusing. Show each step of your work.1. ys ϩ 5yr ϩ 4y ϭ 10e؊3x2.3.4.5.6. ys ϩ yr ϩ (p2ϩ 14)y ϭ e؊x>2sin p xys ϩ 4yr ϩ 4y ϭ e؊xcos xys Ϫ 9y ϭ 18 cos pxys ϩ 3yr ϩ 2y ϭ 12x210ys ϩ 50yr ϩ 57.6y ϭ cos xP R O B L E M S E T 2 . 7c02.qxd 10/27/10 6:06 PM Page 84
• 7.8.9.10.11–18 NONHOMOGENEOUS LINEARODEs: IVPsSolve the initial value problem. State which rule you areusing. Show each step of your calculation in detail.11.12.13.14.15.16.17.yr(0) ϭ 0.35(D2ϩ 0.2D ϩ 0.26I)y ϭ 1.22e0.5x, y(0) ϭ 3.5,(D2Ϫ 2D)y ϭ 6e2xϪ 4e؊2x, y(0) ϭ Ϫ1, yr(0) ϭ 6yp ϭ ln xy(1) ϭ 0, yr(1) ϭ 1;(x2D2Ϫ 3xD ϩ 3I)y ϭ 3 ln x Ϫ 4,yr(0) ϭ Ϫ1.5ys ϩ 4yr ϩ 4y ϭ e؊2xsin 2x, y(0) ϭ 1,yr(0) ϭ 0.058ys Ϫ 6yr ϩ y ϭ 6 cosh x, y(0) ϭ 0.2,ys ϩ 4y ϭ Ϫ12 sin 2x, y(0) ϭ 1.8, yr(0) ϭ 5.0ys ϩ 3y ϭ 18x2, y(0) ϭ Ϫ3, yr(0) ϭ 0(D2ϩ 2D ϩ I)y ϭ 2x sin x(D2Ϫ 16I)y ϭ 9.6e4xϩ 30ex(3D2ϩ 27I)y ϭ 3 cos x ϩ cos 3x(D2ϩ 2D ϩ 34 I)y ϭ 3exϩ 92 xSEC. 2.8 Modeling: Forced Oscillations. Resonance 8518.19. CAS PROJECT. Structure of Solutions of InitialValue Problems. Using the present method, find,graph, and discuss the solutions y of initial valueproblems of your own choice. Explore effects onsolutions caused by changes of initial conditions.Graph separately, to see the separateeffects. Find a problem in which (a) the part of yresulting from decreases to zero, (b) increases,(c) is not present in the answer y. Study a problem withConsider a problem in whichyou need the Modification Rule (a) for a simple root,(b) for a double root. Make sure that your problemscover all three Cases I, II, III (see Sec. 2.2).20. TEAM PROJECT. Extensions of the Method ofUndetermined Coefficients. (a) Extend the methodto products of the function in Table 2.1, (b) Extendthe method to Euler–Cauchy equations. Comment onthe practical significance of such extensions.y(0) ϭ 0, yr(0) ϭ 0.yhyp, y, y Ϫ ypyr(0) ϭ Ϫ2.2y(0) ϭ 6.6,(D2ϩ 2D ϩ 10I)y ϭ 17 sin x Ϫ 37 sin 3x,2.8 Modeling: Forced Oscillations. ResonanceIn Sec. 2.4 we considered vertical motions of a mass–spring system (vibration of a massm on an elastic spring, as in Figs. 33 and 53) and modeled it by the homogeneous linearODE(1)Here as a function of time t is the displacement of the body of mass m from rest.The mass–spring system of Sec. 2.4 exhibited only free motion. This means no externalforces (outside forces) but only internal forces controlled the motion. The internal forcesare forces within the system. They are the force of inertia the damping force(if ), and the spring force ky, a restoring force.c Ͼ 0cyrmys,y(t)mys ϩ cyr ϩ ky ϭ 0.DashpotMassSpringkmcr(t)Fig. 53. Mass on a springc02.qxd 10/27/10 6:06 PM Page 85
• We now extend our model by including an additional force, that is, the external forceon the right. Then we have(2*)Mechanically this means that at each instant t the resultant of the internal forces is inequilibrium with The resulting motion is called a forced motion with forcing functionwhich is also known as input or driving force, and the solution to be obtainedis called the output or the response of the system to the driving force.Of special interest are periodic external forces, and we shall consider a driving forceof the formThen we have the nonhomogeneous ODE(2)Its solution will reveal facts that are fundamental in engineering mathematics and allowus to model resonance.Solving the Nonhomogeneous ODE (2)From Sec. 2.7 we know that a general solution of (2) is the sum of a general solutionof the homogeneous ODE (1) plus any solution of (2). To find we use the methodof undetermined coefficients (Sec. 2.7), starting from(3)By differentiating this function (chain rule!) we obtainSubstituting and into (2) and collecting the cosine and the sine terms, we getThe cosine terms on both sides must be equal, and the coefficient of the sine termon the left must be zero since there is no sine term on the right. This gives the twoequations(4)(k Ϫ mv2)b ϭ 0ϩϪvcaϭ F0vcb(k Ϫ mv2)a ϩ[(k Ϫ mv2)a ϩ vcb] cos vt ϩ [Ϫvca ϩ (k Ϫ mv2)b] sin vt ϭ F0 cos vt.yspyp, yrp,ysp ϭ Ϫv2a cos vt Ϫ v2b sin vt.yrp ϭ Ϫva sin vt ϩ vb cos vt,yp(t) ϭ a cos vt ϩ b sin vt.yp,ypyhmys ϩ cyr ϩ ky ϭ F0 cos vt.(F0 Ͼ 0, v Ͼ 0).r(t) ϭ F0 cos vty(t)r(t),r(t).mys ϩ cyr ϩ ky ϭ r(t).r(t),86 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 86
• for determining the unknown coefficients a and b. This is a linear system. We can solveit by elimination. To eliminate b, multiply the first equation by and the secondby and add the results, obtainingSimilarly, to eliminate a, multiply (the first equation by and the second byand add to getIf the factor is not zero, we can divide by this factor and solve for aand b,If we set as in Sec. 2.4, then and we obtain(5)We thus obtain the general solution of the nonhomogeneous ODE (2) in the form(6)Here is a general solution of the homogeneous ODE (1) and is given by (3) withcoefficients (5).We shall now discuss the behavior of the mechanical system, distinguishing betweenthe two cases (no damping) and (damping). These cases will correspond totwo basically different types of output.Case 1. Undamped Forced Oscillations. ResonanceIf the damping of the physical system is so small that its effect can be neglected over thetime interval considered, we can set Then (5) reduces toand Hence (3) becomes (use )(7)Here we must assume that ; physically, the frequency ofthe driving force is different from the natural frequency of the system, which isthe frequency of the free undamped motion [see (4) in Sec. 2.4]. From (7) and from (4*)in Sec. 2.4 we have the general solution of the “undamped system”(8)We see that this output is a superposition of two harmonic oscillations of the frequenciesjust mentioned.y(t) ϭ C cos (v0t Ϫ d) ϩF0m(v02Ϫ v2)cos vt.v0>(2p)v>(2p) [cycles>sec]v2v02yp(t) ϭF0m(v02Ϫ v2)cos vt ϭF0k[1 Ϫ (v>v0)2]cos vt.v02ϭ k>mb ϭ 0.a ϭ F0>[m(v02Ϫ v2)]c ϭ 0.c Ͼ 0c ϭ 0ypyhy(t) ϭ yh(t) ϩ yp(t).b ϭ F0vcm2(v02Ϫ v2)2ϩ v2c2.a ϭ F0m(v02Ϫ v2)m2(v02Ϫ v2)2ϩ v2c2,k ϭ mv022k>m ϭ v0 (Ͼ 0)b ϭ F0vc(k Ϫ mv2)2ϩ v2c2.a ϭ F0k Ϫ mv2(k Ϫ mv2)2ϩ v2c2,(k Ϫ mv2)2ϩ v2c2v2c2b ϩ (k Ϫ mv2)2b ϭ F0vc.k Ϫ mv2vc(k Ϫ mv2)2a ϩ v2c2a ϭ F0(k Ϫ mv2).Ϫvck Ϫ mv2SEC. 2.8 Modeling: Forced Oscillations. Resonance 87c02.qxd 10/27/10 6:06 PM Page 87
• Resonance. We discuss (7). We see that the maximum amplitude of is (put(9) wheredepends on and If , then and tend to infinity. This excitation of largeoscillations by matching input and natural frequencies is called resonance. iscalled the resonance factor (Fig. 54), and from (9) we see that is the ratioof the amplitudes of the particular solution and of the input We shall seelater in this section that resonance is of basic importance in the study of vibrating systems.In the case of resonance the nonhomogeneous ODE (2) becomes(10)Then (7) is no longer valid, and, from the Modification Rule in Sec. 2.7, we conclude thata particular solution of (10) is of the formyp(t) ϭ t(a cos v0t ϩ b sin v0t).ys ϩ v02y ϭF0m cos v0t.F0 cos vt.ypr>k ϭ a0>F0r(v ϭ v0)a0rv : v0v0.va0r ϭ11 Ϫ (v>v0)2.a0 ϭF0krcos vt ϭ 1)yp88 CHAP. 2 Second-Order Linear ODEsωρω0ω1Fig. 54. Resonance factor r(v)By substituting this into (10) we find and . Hence (Fig. 55)(11) yp(t) ϭF02mv0t sin v0t.b ϭ F0>(2mv0)a ϭ 0yptFig. 55. Particular solution in the case of resonanceWe see that, because of the factor t, the amplitude of the vibration becomes larger andlarger. Practically speaking, systems with very little damping may undergo large vibrationsc02.qxd 10/27/10 6:06 PM Page 88
• that can destroy the system. We shall return to this practical aspect of resonance later inthis section.Beats. Another interesting and highly important type of oscillation is obtained if isclose to . Take, for example, the particular solution [see (8)](12)Using (12) in App. 3.1, we may write this asSince is close to , the difference is small. Hence the period of the last sinefunction is large, and we obtain an oscillation of the type shown in Fig. 56, the dashedcurve resulting from the first sine factor. This is what musicians are listening to whenthey tune their instruments.v0 Ϫ vv0vy(t) ϭ2F0m(v02Ϫ v2)sin av0 ϩ v2tb sin av0 Ϫ v2tb.(v v0).y(t) ϭF0m(v02Ϫ v2)(cos vt Ϫ cos v0t)v0vSEC. 2.8 Modeling: Forced Oscillations. Resonance 89ytFig. 56. Forced undamped oscillation when the difference of the inputand natural frequencies is small (“beats”)Case 2. Damped Forced OscillationsIf the damping of the mass–spring system is not negligibly small, we have anda damping term in (1) and (2). Then the general solution of the homogeneousODE (1) approaches zero as t goes to infinity, as we know from Sec. 2.4. Practically,it is zero after a sufficiently long time. Hence the “transient solution” (6) of (2),given by approaches the “steady-state solution” . This proves thefollowing.T H E O R E M 1 Steady-State SolutionAfter a sufficiently long time the output of a damped vibrating system under a purelysinusoidal driving force [see (2)] will practically be a harmonic oscillation whosefrequency is that of the input.ypy ϭ yh ϩ yp,yhcyrc Ͼ 0c02.qxd 10/27/10 6:06 PM Page 89
• Amplitude of the Steady-State Solution. Practical ResonanceWhereas in the undamped case the amplitude of approaches infinity as approaches, this will not happen in the damped case. In this case the amplitude will always befinite. But it may have a maximum for some depending on the damping constant c.This may be called practical resonance. It is of great importance because if is not toolarge, then some input may excite oscillations large enough to damage or even destroythe system. Such cases happened, in particular in earlier times when less was known aboutresonance. Machines, cars, ships, airplanes, bridges, and high-rising buildings are vibratingmechanical systems, and it is sometimes rather difficult to find constructions that arecompletely free of undesired resonance effects, caused, for instance, by an engine or bystrong winds.To study the amplitude of as a function of , we write (3) in the form(13)C* is called the amplitude of and the phase angle or phase lag because it measuresthe lag of the output behind the input. According to (5), these quantities are(14)Let us see whether has a maximum and, if so, find its location and then its size.We denote the radicand in the second root in C* by R. Equating the derivative of C* tozero, we obtainThe expression in the brackets [. . .] is zero if(15)By reshuffling terms we haveThe right side of this equation becomes negative if so that then (15) has noreal solution and C* decreases monotone as increases, as the lowest curve in Fig. 57shows. If c is smaller, then (15) has a real solution where(15*)From (15*) we see that this solution increases as c decreases and approaches as capproaches zero. See also Fig. 57.v0vmax2ϭ v02Ϫc22m2.v ϭ vmax,c2Ͻ 2mk,vc2Ͼ 2mk,2m2v2ϭ 2m2v02Ϫ c2ϭ 2mk Ϫ c2.(v02ϭ k>m).c2ϭ 2m2(v02Ϫ v2)dC*dvϭ F0 aϪ12RϪ3>2b [2m2(v02Ϫ v2)(Ϫ2v) ϩ 2vc2].C*(v)tan h(v) ϭbaϭvcm(v02Ϫ v2).C*(v) ϭ 2a2ϩ b2ϭF02m2(v02Ϫ v2)2ϩ v2c2,hypyp(t) ϭ C* cos (vt Ϫ h).vypcvv0vyp90 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 90
• The size of is obtained from (14), with given by (15*). For thiswe obtain in the second radicand in (14) from (15*)andThe sum of the right sides of these two formulas isSubstitution into (14) gives(16)We see that is always finite when Furthermore, since the expressionin the denominator of (16) decreases monotone to zero as goes to zero, the maximumamplitude (16) increases monotone to infinity, in agreement with our result in Case 1. Figure 57shows the amplification (ratio of the amplitudes of output and input) as a function offor hence and various values of the damping constant c.Figure 58 shows the phase angle (the lag of the output behind the input), which is lessthan when and greater than for v Ͼ v0.p>2v Ͻ v0,p>2v0 ϭ 1,m ϭ 1, k ϭ 1,vC*>F0c2(Ͻ2mk)c24m2v02Ϫ c4ϭ c2(4mk Ϫ c2)c Ͼ 0.C*(vmax)C*(vmax) ϭ2mF0c24m2v02Ϫ c2.(c4ϩ 4m2v02c2Ϫ 2c4)>(4m2) ϭ c2(4m2v02Ϫ c2)>(4m2).vmax2c2ϭ av02Ϫc22m2b c2.m2(v02Ϫ vmax2)2ϭc44m2v2v2ϭ vmax2C*(vmax)SEC. 2.8 Modeling: Forced Oscillations. Resonance 9143200 1 2c = 1c = 2c = 1_4c = 1_2C*F01ωFig. 57. Amplification as a function offor and various values of thedamping constant cm ϭ 1, k ϭ 1,vC*>F0ηωc = 1/2__2c = 0c = 1c = 2ππ001 2Fig. 58. Phase lag as a function of forthus and various valuesof the damping constant cv0 ϭ 1,m ϭ 1, k ϭ 1,vh1. WRITING REPORT. Free and Forced Vibrations.Write a condensed report of 2–3 pages on the mostimportant similarities and differences of free and forcedvibrations, with examples of your own. No proofs.2. Which of Probs. 1–18 in Sec. 2.7 (with time t)can be models of mass–spring systems with a harmonicoscillation as steady-state solution?x ϭ3–7 STEADY-STATE SOLUTIONSFind the steady-state motion of the mass–spring systemmodeled by the ODE. Show the details of your work.3.4.5. (D2ϩ D ϩ 4.25I)y ϭ 22.1 cos 4.5tys ϩ 2.5yr ϩ 10y ϭ Ϫ13.6 sin 4tys ϩ 6yr ϩ 8y ϭ 42.5 cos 2tP R O B L E M S E T 2 . 8c02.qxd 10/27/10 6:06 PM Page 91
• 92 CHAP. 2 Second-Order Linear ODEsk = 1m = 1F = 0F = 1 – t2/π2F1π tFig. 59. Problem 24Fig. 60. Typical solution curves in CAS Experiment 256.7.8–15 TRANSIENT SOLUTIONSFind the transient motion of the mass–spring systemmodeled by the ODE. Show the details of your work.8.9.10.11.12.13.14.15.16–20 INITIAL VALUE PROBLEMSFind the motion of the mass–spring system modeled by theODE and the initial conditions. Sketch or graph the solutioncurve. In addition, sketch or graph the curve of tosee when the system practically reaches the steady state.16.17.18.19.20.21. Beats. Derive the formula after (12) from (12). Canwe have beats in a damped system?22. Beats. SolveHow does the graph of the solution changeif you change (a) (b) the frequency of the drivingforce?23. TEAM EXPERIMENT. Practical Resonance.(a) Derive, in detail, the crucial formula (16).(b) By considering show that in-creases as decreases.(c) Illustrate practical resonance with an ODE of yourown in which you vary c, and sketch or graphcorresponding curves as in Fig. 57.(d) Take your ODE with c fixed and an input of twoterms, one with frequency close to the practicalresonance frequency and the other not. Discuss andsketch or graph the output.(e) Give other applications (not in the book) in whichresonance is important.c (Ϲ 12mk)C*(vmax)dC*>dcy(0),(0) ϭ 0.yry(0) ϭ 2,ys ϩ 25y ϭ 99 cos 4.9t,yr(0) ϭ 0(D2ϩ5I)y ϭ cos pt Ϫ sin pt, y(0) ϭ 0,yr(0) ϭ 1(D2ϩ 2D ϩ 2I)y ϭ e؊t>2sin 12 t, y(0) ϭ 0,yr(0) ϭ 9.4(D2ϩ 8D ϩ 17I)y ϭ 474.5 sin 0.5t, y(0) ϭ Ϫ5.4,y(0) ϭ 0, yr(0) ϭ 335(D2ϩ 4I)y ϭ sin t ϩ 13 sin 3t ϩ 15 sin 5t,ys ϩ 25y ϭ 24 sin t, y(0) ϭ 1, yr(0) ϭ 1y Ϫ yp(D2ϩ 4D ϩ 8I)y ϭ 2 cos 2t ϩ sin 2t(D2ϩ I)y ϭ 5e؊tcos t(D2ϩ I)y ϭ cos vt, v21(D2ϩ 2D ϩ 5I)y ϭ 4 cos t ϩ 8 sin t(D2ϩ 2I)y ϭ cos 12t ϩ sin12tys ϩ 16y ϭ 56 cos 4tys ϩ 3yr ϩ 3.25y ϭ 3 cos t Ϫ 1.5 sin t2ys ϩ 4yr ϩ 6.5y ϭ 4 sin 1.5t(4D2ϩ 12D ϩ 9I)y ϭ 225 Ϫ 75 sin 3t(D2ϩ 4D ϩ 3I)y ϭ cos t ϩ 13 cos 3t 24. Gun barrel. Solve ifand 0 if here, Thismodels an undamped system on which a force F actsduring some interval of time (see Fig. 59), for instance,the force on a gun barrel when a shell is fired, the barrelbeing braked by heavy springs (and then damped by adashpot, which we disregard for simplicity). Hint: Atboth y and must be continuous.yrpy(0) ϭ 0, yr(0) ϭ 0.t : ϱ;t Ϲ p0 Ϲys ϩ y ϭ 1 Ϫ t2>p225. CAS EXPERIMENT. Undamped Vibrations.(a) Solve the initial value problemShow that the solutioncan be written(b) Experiment with the solution by changing tosee the change of the curves from those for smallto beats, to resonance, and to large values of(see Fig. 60).vv (Ͼ0)vy(t) ϭ21 Ϫ v2sin [12 (1 ϩ v)t] sin [12 (1 Ϫ v)t].v21, y(0) ϭ 0, yr(0) ϭ 0.ys ϩ y ϭ cos vt,10π 20π1–1ω = 0.220π10–10ω = 0.90.04–0.040.04ω = 610πc02.qxd 10/27/10 6:06 PM Page 92
• 2.9 Modeling: Electric CircuitsDesigning good models is a task the computer cannot do. Hence setting up models hasbecome an important task in modern applied mathematics. The best way to gain experiencein successful modeling is to carefully examine the modeling process in various fields andapplications. Accordingly, modeling electric circuits will be profitable for all students,not just for electrical engineers and computer scientists.Figure 61 shows an RLC-circuit, as it occurs as a basic building block of large electricnetworks in computers and elsewhere. An RLC-circuit is obtained from an RL-circuit byadding a capacitor. Recall Example 2 on the RL-circuit in Sec. 1.5: The model of theRL-circuit is It was obtained by KVL (Kirchhoff’s Voltage Law)7byequating the voltage drops across the resistor and the inductor to the EMF (electromotiveforce). Hence we obtain the model of the RLC-circuit simply by adding the voltage dropQ C across the capacitor. Here, C F (farads) is the capacitance of the capacitor. Q coulombsis the charge on the capacitor, related to the current bySee also Fig. 62. Assuming a sinusoidal EMF as in Fig. 61, we thus have the model ofthe RLC-circuitI(t) ϭdQdt, equivalently Q(t) ϭ ΎI(t) dt.>LIr ϩ RI ϭ E(t).SEC. 2.9 Modeling: Electric Circuits 937GUSTAV ROBERT KIRCHHOFF (1824–1887), German physicist. Later we shall also need Kirchhoff’sCurrent Law (KCL):At any point of a circuit, the sum of the inflowing currents is equal to the sum of the outflowing currents.The units of measurement of electrical quantities are named after ANDRÉ MARIE AMPÈRE (1775–1836),French physicist, CHARLES AUGUSTIN DE COULOMB (1736–1806), French physicist and engineer,MICHAEL FARADAY (1791–1867), English physicist, JOSEPH HENRY (1797–1878), American physicist,GEORG SIMON OHM (1789–1854), German physicist, and ALESSANDRO VOLTA (1745–1827), Italianphysicist.R LCE(t) = E0sin ωtωFig. 61. RLC-circuitFig. 62. Elements in an RLC-circuitNameOhm’s ResistorInductorCapacitorSymbol NotationR Ohm’s ResistanceL InductanceC CapacitanceUnitohms (⍀)henrys (H)farads (F)Voltage DropRILQ/CdIdtc02.qxd 10/27/10 6:06 PM Page 93
• This is an “integro-differential equation.” To get rid of the integral, we differentiatewith respect to t, obtaining(1)This shows that the current in an RLC-circuit is obtained as the solution of thisnonhomogeneous second-order ODE (1) with constant coefficients.In connection with initial value problems, we shall occasionally useobtained from andSolving the ODE (1) for the Current in an RLC-CircuitA general solution of (1) is the sum where is a general solution of thehomogeneous ODE corresponding to (1) and is a particular solution of (1). We firstdetermine by the method of undetermined coefficients, proceeding as in the previoussection. We substitute(2)into (1). Then we collect the cosine terms and equate them to on the right,and we equate the sine terms to zero because there is no sine term on the right,(Cosine terms)(Sine terms).Before solving this system for a and b, we first introduce a combination of L and C, calledthe reactance(3)Dividing the previous two equations by ordering them, and substituting S givesϪRa Ϫ Sb ϭ 0.ϪSa ϩ Rb ϭ E0v,S ϭ vL Ϫ1vC.Lv2(Ϫb) ϩ Rv(Ϫa) ϩ b>C ϭ 0Lv2(Ϫa) ϩ Rvb ϩ a>C ϭ E0vE0v cos vtIps ϭ v2(Ϫa cos vt Ϫ b sin vt)Ipr ϭ v(Ϫa sin vt ϩ b cos vt)Ip ϭ a cos vt ϩ b sin vtIpIpIhI ϭ Ih ϩ Ip,I ϭ Qr.(1r)LQs ϩ RQs ϩ1CQ ϭ E(t),(1s)LIs ϩ RIr ϩ1CI ϭ Er(t) ϭ E0v cos vt.(1r)LIr ϩ RI ϩ1C ΎI dt ϭ E(t) ϭ E0 sin vt.(1r)94 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 94
• We now eliminate b by multiplying the first equation by S and the second by R, andadding. Then we eliminate a by multiplying the first equation by R and the second byand adding. This givesWe can solve for a and b,(4)Equation (2) with coefficients a and b given by (4) is the desired particular solution ofthe nonhomogeneous ODE (1) governing the current I in an RLC-circuit with sinusoidalelectromotive force.Using (4), we can write in terms of “physically visible” quantities, namely, amplitudeand phase lag of the current behind the EMF, that is,(5)where [see (14) in App. A3.1]The quantity is called the impedance. Our formula shows that the impedanceequals the ratio This is somewhat analogous to (Ohm’s law) and, becauseof this analogy, the impedance is also known as the apparent resistance.A general solution of the homogeneous equation corresponding to (1) iswhere and are the roots of the characteristic equationWe can write these roots in the form and whereNow in an actual circuit, R is never zero (hence ). From this it follows thatapproaches zero, theoretically as but practically after a relatively short time. Hencethe transient current tends to the steady-state current and after some timethe output will practically be a harmonic oscillation, which is given by (5) and whosefrequency is that of the input (of the electromotive force).Ip,I ϭ Ih ϩ Ipt : ϱ,IhR Ͼ 0b ϭBR24L2Ϫ1LCϭ12L BR2Ϫ4LC.a ϭR2L,l2 ϭ Ϫa Ϫ b,l1 ϭ Ϫa ϩ bl2ϩRLl ϩ1LCϭ 0.l2l1Ih ϭ c1el1tϩ c2el2tE>I ϭ RE0>I0.2R2ϩ S2tan u ϭ ϪabϭSR.I0 ϭ 2a2ϩ b2ϭE02R2ϩ S2,Ip(t) ϭ I0 sin (vt Ϫ u)uI0IpIpb ϭE0RR2ϩ S2.a ϭϪE0SR2ϩ S2,(R2ϩ S2)b ϭ E0R.Ϫ(S2ϩ R2)a ϭ E0S,ϪS,SEC. 2.9 Modeling: Electric Circuits 95c02.qxd 10/27/10 6:06 PM Page 95
• E X A M P L E 1 RLC-CircuitFind the current in an RLC-circuit with (ohms), (henry), (farad), whichis connected to a source of EMF sin 377 t (hence 60 cycles sec, theusual in the U.S. and Canada; in Europe it would be 220 V and 50 Hz). Assume that current and capacitorcharge are 0 whenSolution. Step 1. General solution of the homogeneous ODE. Substituting R, L, C and the derivativeinto (1), we obtainHence the homogeneous ODE is Its characteristic equation isThe roots are and The corresponding general solution of the homogeneous ODE isStep 2. Particular solution of (1). We calculate the reactance and the steady-statecurrentwith coefficients obtained from (4) (and rounded)Hence in our present case, a general solution of the nonhomogeneous ODE (1) is(6)Step 3. Particular solution satisfying the initial conditions. How to use We finally determineand from the in initial conditions and From the first condition and (6) we have(7) henceWe turn to The integral in equals see near the beginning of this section. Hence forEq. becomesso thatDifferentiating (6) and setting we thus obtainThe solution of this and (7) is Hence the answer isYou may get slightly different values depending on the rounding. Figure 63 shows as well as whichpractically coincide, except for a very short time near because the exponential terms go to zero very rapidly.Thus after a very short time the current will practically execute harmonic oscillations of the input frequencycycles sec. Its maximum amplitude and phase lag can be seen from (5), which here takes the form᭿Ip(t) ϭ 2.824 sin (377t Ϫ 1.29).>60 Hz ϭ 60t ϭ 0Ip(t),I(t)I(t) ϭ Ϫ0.323e؊10tϩ 3.033e؊100tϪ 2.71 cos 377t ϩ 0.796 sin 377t.c1 ϭ Ϫ0.323, c2 ϭ 3.033.Ir(0) ϭ Ϫ10c1 Ϫ 100c2 ϩ 0 ϩ 0.796 # 377 ϭ 0, hence by (7), Ϫ10c1 ϭ 100(2.71 Ϫ c1) Ϫ 300.1.t ϭ 0,Ir(0) ϭ 0.LIr(0) ϩ R # 0 ϭ 0,(1r)t ϭ 0,͐I dt ϭ Q(t);(1r)Q(0) ϭ 0.c2 ϭ 2.71 Ϫ c1.I(0) ϭ c1 ϩ c2 Ϫ 2.71 ϭ 0,Q(0) ϭ 0.I(0) ϭ 0c2c1Q(0) ϭ 0?I(t) ϭ c1e؊10tϩ c2e؊100tϪ 2.71 cos 377t ϩ 0.796 sin 377t.a ϭϪ110 # 37.4112ϩ 37.42ϭ Ϫ2.71, b ϭ110 # 11112ϩ 37.42ϭ 0.796.Ip(t) ϭ a cos 377t ϩ b sin 377tS ϭ 37.7 Ϫ 0.3 ϭ 37.4IpIh(t) ϭ c1e؊10tϩ c2e؊100t.l2 ϭ Ϫ100.l1 ϭ Ϫ100.1l2ϩ 11l ϩ 100 ϭ 0.0.1Is ϩ 11Ir ϩ 100I ϭ 0.0.1Is ϩ 11Ir ϩ 100I ϭ 110 # 377 cos 377t.Er(t)t ϭ 0.>Hz ϭ 60E(t) ϭ 110 sin (60 # 2pt) ϭ 110C ϭ 10؊2FL ϭ 0.1 HR ϭ 11 ⍀I(t)96 CHAP. 2 Second-Order Linear ODEsc02.qxd 10/27/10 6:06 PM Page 96
• Analogy of Electrical and Mechanical QuantitiesEntirely different physical or other systems may have the same mathematical model.For instance, we have seen this from the various applications of the ODE inChap. 1. Another impressive demonstration of this unifying power of mathematics isgiven by the ODE (1) for an electric RLC-circuit and the ODE (2) in the last section fora mass–spring system. Both equationsandare of the same form. Table 2.2 shows the analogy between the various quantities involved.The inductance L corresponds to the mass m and, indeed, an inductor opposes a changein current, having an “inertia effect” similar to that of a mass. The resistance R correspondsto the damping constant c, and a resistor causes loss of energy, just as a damping dashpotdoes. And so on.This analogy is strictly quantitative in the sense that to a given mechanical system wecan construct an electric circuit whose current will give the exact values of the displacementin the mechanical system when suitable scale factors are introduced.The practical importance of this analogy is almost obvious. The analogy may be usedfor constructing an “electrical model” of a given mechanical model, resulting in substantialsavings of time and money because electric circuits are easy to assemble, and electricquantities can be measured much more quickly and accurately than mechanical ones.mys ϩ cyr ϩ ky ϭ F0 cos vtLIs ϩ RIr ϩ1CI ϭ E0v cos vtyr ϭ kySEC. 2.9 Modeling: Electric Circuits 97yt0 0.02 0.03 0.04 0.050.012–2–31–13I(t)Fig. 63. Transient (upper curve) and steady-state currents in Example 1Table 2.2 Analogy of Electrical and Mechanical QuantitiesElectrical System Mechanical SystemInductance L Mass mResistance R Damping constant cReciprocal 1 C of capacitance Spring modulus kDerivative of} Driving forceelectromotive forceCurrent Displacement y(t)I(t)F0 cos vtE0v cos vt>c02.qxd 10/27/10 6:06 PM Page 97
• Related to this analogy are transducers, devices that convert changes in a mechanicalquantity (for instance, in a displacement) into changes in an electrical quantity that canbe monitored; see Ref. [GenRef11] in App. 1.98 CHAP. 2 Second-Order Linear ODEs1–6 RLC-CIRCUITS: SPECIAL CASES1. RC-Circuit. Model the RC-circuit in Fig. 64. Find thecurrent due to a constant E.P R O B L E M S E T 2 . 9Fig. 64. RC-circuit2. RC-Circuit. Solve Prob. 1 when andR, C, , and are arbitrary.3. RL-Circuit. Model the RL-circuit in Fig. 66. Find ageneral solution when R, L, E are any constants. Graphor sketch solutions when H, , andE ϭ 48 V.R ϭ 10 ⍀L ϭ 0.25vE0E ϭ E0 sin vt4. RL-Circuit. Solve Prob. 3 when and R,L, and are arbitrary. Sketch a typical solution.E0,E ϭ E0 sin vt5. LC-Circuit. This is an RLC-circuit with negligiblysmall R (analog of an undamped mass–spring system).Find the current when , , and, assuming zero initial current and charge.E ϭ sin t VC ϭ 0.005 FL ϭ 0.5 H6. LC-Circuit. Find the current when ,F, , and initial current and chargezero.7–18 GENERAL RLC-CIRCUITS7. Tuning. In tuning a stereo system to a radio station,we adjust the tuning control (turn a knob) that changesC (or perhaps L) in an RLC-circuit so that the amplitudeof the steady-state current (5) becomes maximum. Forwhat C will this happen?8–14 Find the steady-state current in the RLC-circuitin Fig. 61 for the given data. Show the details of your work.8.9.10. R ϭ 2 ⍀, L ϭ 1 H, C ϭ 120 F, E ϭ 157 sin 3t VR ϭ 4 ⍀, L ϭ 0.1 H, C ϭ 0.05 F, E ϭ 110 VR ϭ 4 ⍀, L ϭ 0.5 H, C ϭ 0.1 F, E ϭ 500 sin 2t VE ϭ 2t2VC ϭ 0.005L ϭ 0.5 HE(t)CRFig. 65. Current 1 in Problem 1Current I(t)tcFig. 67. Currents in Problem 30.020 0.04 0.06 0.08 0.1Current I(t)t12345Fig. 68. Typical currentin Problem 4I ϭ e؊0.1tϩ sin (t Ϫ 14 p)0.5–0.5–111.52Current I(t)t12π4π 8πFig. 66. RL-circuitE(t)LRFig. 69. LC-circuitC LE(t)c02.qxd 10/27/10 6:06 PM Page 98
• 11.12.13.14. Prove the claim in the text that if (hencethen the transient current approaches as15. Cases of damping. What are the conditions for anRLC-circuit to be (I) overdamped, (II) critically damped,(III) underdamped? What is the critical resistance(the analog of the critical damping constant )?16–18 Solve the initial value problem for the RLC-circuit in Fig. 61 with the given data, assuming zero initialcurrent and charge. Graph or sketch the solution. Show thedetails of your work.21mkRcritt : ϱ.IpR Ͼ 0),R 0E ϭ 12,000 sin 25t VR ϭ 12, L ϭ 1.2 H, C ϭ 203# 10؊3F,R ϭ 0.2 ⍀, L ϭ 0.1 H, C ϭ 2 F, E ϭ 220 sin 314t VE ϭ 220 sin 10t VR ϭ 12 ⍀, L ϭ 0.4 H, C ϭ 180 F,SEC. 2.10 Solution by Variation of Parameters 9916.17.18.19. WRITING REPORT. Mechanic-Electric Analogy.Explain Table 2.2 in a 1–2 page report with examples,e.g., the analog (with ) of a mass–spring systemof mass 5 kg, damping constant 10 kg sec, spring constant, and driving force20. Complex Solution Method. Solveby substituting(K unknown) and its derivatives and taking the realpart of the solution . Show agreement with (2), (4).Hint: Use (11) cf. Sec. 2.2,and i2ϭ Ϫ1.eivtϭ cos vt ϩ i sin vt;I~pIpIp ϭ Keivti ϭ 1Ϫ1,I~>C ϭ E0eivt,LI~s ϩ RI~r ϩ220 cos 10t kg>sec.60 kg>sec2>L ϭ 1 HE ϭ 820 cos 10t VR ϭ 18 ⍀, L ϭ 1 H, C ϭ 12.5 # 10؊3F,E ϭ 600 (cos t ϩ 4 sin t) VR ϭ 6 ⍀, L ϭ 1 H, C ϭ 0.04 F,E ϭ 100 sin 10t VR ϭ 8 ⍀, L ϭ 0.2 H, C ϭ 12.5 # 10؊3F,2.10 Solution by Variation of ParametersWe continue our discussion of nonhomogeneous linear ODEs, that is(1)In Sec. 2.6 we have seen that a general solution of (1) is the sum of a general solutionof the corresponding homogeneous ODE and any particular solution of (1). To obtainwhen is not too complicated, we can often use the method of undetermined coefficients,as we have shown in Sec. 2.7 and applied to basic engineering models in Secs. 2.8 and 2.9.However, since this method is restricted to functions whose derivatives are of a formsimilar to itself (powers, exponential functions, etc.), it is desirable to have a method validfor more general ODEs (1), which we shall now develop. It is called the method of variationof parameters and is credited to Lagrange (Sec. 2.1). Here p, q, r in (1) may be variable(given functions of x), but we assume that they are continuous on some open interval I.Lagrange’s method gives a particular solution of (1) on I in the form(2)where form a basis of solutions of the corresponding homogeneous ODE(3)on I, and W is the Wronskian of(4) (see Sec. 2.6).CAUTION! The solution formula (2) is obtained under the assumption that the ODEis written in standard form, with as the first term as shown in (1). If it starts withdivide first by f(x).f(x)ys,ysW ϭ y1y2r Ϫ y2y1ry1, y2,ys ϩ p(x)yr ϩ q(x)y ϭ 0y1, y2yp(x) ϭ Ϫy1 Ύy2rWdx ϩ y2 Ύy1rWdxypr(x)r(x)r(x)ypypyhys ϩ p(x)yr ϩ q(x)y ϭ r(x).c02.qxd 10/27/10 6:06 PM Page 99
• The integration in (2) may often cause difficulties, and so may the determination ofif (1) has variable coefficients. If you have a choice, use the previous method. It issimpler. Before deriving (2) let us work an example for which you do need the newmethod. (Try otherwise.)E X A M P L E 1 Method of Variation of ParametersSolve the nonhomogeneous ODESolution. A basis of solutions of the homogeneous ODE on any interval is . This givesthe WronskianFrom (2), choosing zero constants of integration, we get the particular solution of the given ODE(Fig. 70)Figure 70 shows and its first term, which is small, so that essentially determines the shape of the curveof . (Recall from Sec. 2.8 that we have seen in connection with resonance, except for notation.) Fromand the general solution of the homogeneous ODE we obtain the answerHad we included integration constants in (2), then (2) would have given the additionalthat is, a general solution of the given ODE directly from (2). This willalways be the case. ᭿c1 cos x ϩ c2 sin x ϭ c1y1 ϩ c2y2,Ϫc1, c2y ϭ yh ϩ yp ϭ (c1 ϩ ln ƒ cos x ƒ) cos x ϩ (c2 ϩ x) sin x.yh ϭ c1y1 ϩ c2y2ypx sin xypx sin xypϭ cos x ln ƒ cos x ƒ ϩ x sin xyp ϭ Ϫcos xΎsin x sec x dx ϩ sin x Ύcos x sec x dxW(y1, y2) ϭ cos x cos x Ϫ sin x (Ϫsin x) ϭ 1.y1 ϭ cos x, y2 ϭ sin xys ϩ y ϭ sec x ϭ1cos x.y1, y2100 CHAP. 2 Second-Order Linear ODEsyx04 82510–5–106 10 12Fig. 70. Particular solution yp and its first term in Example 1Idea of the Method. Derivation of (2)What idea did Lagrange have? What gave the method the name? Where do we use thecontinuity assumptions?The idea is to start from a general solutionyh(x) ϭ c1y1(x) ϩ c2y2(x)c02.qxd 10/27/10 6:06 PM Page 100
• of the homogeneous ODE (3) on an open interval I and to replace the constants (“theparameters”) and by functions and this suggests the name of the method.We shall determine u and v so that the resulting function(5)is a particular solution of the nonhomogeneous ODE (1). Note that exists by Theorem3 in Sec. 2.6 because of the continuity of p and q on I. (The continuity of r will be usedlater.)We determine u and v by substituting (5) and its derivatives into (1). Differentiating (5),we obtainNow must satisfy (1). This is one condition for two functions u and v. It seems plausiblethat we may impose a second condition. Indeed, our calculation will show that we candetermine u and v such that satisfies (1) and u and v satisfy as a second condition theequation(6)This reduces the first derivative to the simpler form(7)Differentiating (7), we obtain(8)We now substitute and its derivatives according to (5), (7), (8) into (1). Collectingterms in u and terms in v, we obtainSince and are solutions of the homogeneous ODE (3), this reduces to(9a)Equation (6) is(9b)This is a linear system of two algebraic equations for the unknown functions andWe can solve it by elimination as follows (or by Cramer’s rule in Sec. 7.6). To eliminatewe multiply (9a) by and (9b) by and add, obtainingHere, W is the Wronskian (4) of To eliminate we multiply (9a) by and (9b)by and add, obtainingϪy1ry1,ury1, y2.ur(y1y2r Ϫ y2y1r) ϭ Ϫy2r, thus urW ϭ Ϫy2r.y2rϪy2vr,vr.urury1 ϩ vry2 ϭ 0.ury1r ϩ vry2r ϭ r.y2y1u(y1s ϩ py1r ϩ qy1) ϩ v(y2s ϩ py2r ϩ qy2) ϩ ury1r ϩ vry2r ϭ r.ypyps ϭ ury1r ϩ uy1s ϩ vry2r ϩ vy2s.ypr ϭ uy1r ϩ vy2r.yprury1 ϩ vry2 ϭ 0.ypypypr ϭ ury1 ϩ uy1r ϩ vry2 ϩ vy2r.yhyp(x) ϭ u(x)y1(x) ϩ v(x)y2(x)v(x);u(x)c2c1SEC. 2.10 Solution by Variation of Parameters 101c02.qxd 10/27/10 6:06 PM Page 101
• Since form a basis, we have (by Theorem 2 in Sec. 2.6) and can divide by W,(10)By integration,These integrals exist because is continuous. Inserting them into (5) gives (2) andcompletes the derivation. ᭿r(x)u ϭ ϪΎy2rWdx, v ϭ Ύy1rWdx.ur ϭ Ϫy2rW, vr ϭy1rW.W 0y1, y2vr(y1y2r Ϫ y2yr1) ϭ Ϫy1r, thus vrW ϭ y1r.102 CHAP. 2 Second-Order Linear ODEs1–13 GENERAL SOLUTIONSolve the given nonhomogeneous linear ODE by variationof parameters or undetermined coefficients. Show thedetails of your work.1.2.3.4.5.6.7.8.9.10. (D2ϩ 2D ϩ 2I)y ϭ 4e؊xsec3x(D2Ϫ 2D ϩ I)y ϭ 35x3>2ex(D2ϩ 4I)y ϭ cosh 2x(D2Ϫ 4D ϩ 4I)y ϭ 6e2x>x4(D2ϩ 6D ϩ 9I)y ϭ 16e؊3x>(x2ϩ 1)ys ϩ y ϭ cos x Ϫ sin xys Ϫ 4yr ϩ 5y ϭ e2xcsc xx2ys Ϫ 2xyr ϩ 2y ϭ x3sin xys ϩ 9y ϭ csc 3xys ϩ 9y ϭ sec 3x11.12.13.14. TEAM PROJECT. Comparison of Methods. Inven-tion. The undetermined-coefficient method should beused whenever possible because it is simpler. Compareit with the present method as follows.(a) Solve by both methods,showing all details, and compare.(b) Solveby applying each method to a suitable function onthe right.(c) Experiment to invent an undetermined-coefficientmethod for nonhomogeneous Euler–Cauchy equations.x2r2 ϭys Ϫ 2yr ϩ y ϭ r1 ϩ r2, r1 ϭ 35x3>2exys ϩ 4yr ϩ 3y ϭ 65 cos 2x(x2D2ϩ xD Ϫ 9I)y ϭ 48x5(D2Ϫ I)y ϭ 1>cosh x(x2D2Ϫ 4xD ϩ 6I)y ϭ 21x؊4P R O B L E M S E T 2 . 1 01. Why are linear ODEs preferable to nonlinear ones inmodeling?2. What does an initial value problem of a second-orderODE look like? Why must you have a general solutionto solve it?3. By what methods can you get a general solution of anonhomogeneous ODE from a general solution of ahomogeneous one?4. Describe applications of ODEs in mechanical systems.What are the electrical analogs of the latter?5. What is resonance? How can you remove undesirableresonance of a construction, such as a bridge, a ship,or a machine?6. What do you know about existence and uniqueness ofsolutions of linear second-order ODEs?7–18 GENERAL SOLUTIONFind a general solution. Show the details of your calculation.7.8.9.10.11.12.13.14.15.16.17.18. yys ϭ 2yr2(4D2Ϫ 12D ϩ 9I)y ϭ 2e1.5x(D2ϩ 2D ϩ 2I)y ϭ 3e؊xcos 2x(2D2Ϫ 3D Ϫ 2I)y ϭ 13 Ϫ 2x2(x2D2ϩ xD Ϫ 9I)y ϭ 0(x2D2ϩ 2xD Ϫ 12I)y ϭ 0(D2ϩ 4pD ϩ 4p2I)y ϭ 0(100D2Ϫ 160D ϩ 64I)y ϭ 0ys ϩ 0.20yr ϩ 0.17y ϭ 0ys ϩ 6yr ϩ 34y ϭ 0ys ϩ yr Ϫ 12y ϭ 04ys ϩ 32yr ϩ 63y ϭ 0C H A P T E R 2 R E V I E W Q U E S T I O N S A N D P R O B L E M Sc02.qxd 10/27/10 6:06 PM Page 102
• 19–22 INITIAL VALUE PROBLEMSSolve the problem, showing the details of your work.Sketch or graph the solution.19.20.21.22.23–30 APPLICATIONS23. Find the steady-state current in the RLC-circuit in Fig. 71when and(66 cycles sec).24. Find a general solution of the homogeneous linearODE corresponding to the ODE in Prob. 23.25. Find the steady-state current in the RLC-circuitin Fig. 71 when.E ϭ 200 sin 4t VR ϭ 50 ⍀, L ϭ 30 H, C ϭ 0.025 F,>E ϭ 110 sin 415t VRϭ2 k⍀ (2000 ⍀), Lϭ1 H,Cϭ4 # 10؊3F,yr(1) ϭ Ϫ11(x2D2ϩ 15xD ϩ 49I)y ϭ 0, y(1) ϭ 2,(x2D2ϩ xD Ϫ I)y ϭ 16x3, y(1) ϭ Ϫ1, yr(1) ϭ 1ys Ϫ 3yr ϩ 2y ϭ 10 sin x, y(0) ϭ 1, yr(0) ϭ Ϫ6ys ϩ 16y ϭ 17ex, y(0) ϭ 6, yr(0) ϭ Ϫ2Summary of Chapter 2 10327. Find an electrical analog of the mass–spring systemwith mass 4 kg, spring constant 10 dampingconstant 20 kg sec, and driving force28. Find the motion of the mass–spring system in Fig. 72with mass 0.125 kg, damping 0, spring constant1.125 and driving force ass-uming zero initial displacement and velocity. For whatfrequency of the driving force would you get resonance?cos t Ϫ 4 sin t nt,kg>sec2,100 sin 4t nt.>kg>sec2,29. Show that the system in Fig. 72 withand driving force exhibits beats.Hint: Choose zero initial conditions.30. In Fig. 72, let kg, kg sec,and nt. Determine w such that youget the steady-state vibration of maximum possibleamplitude. Determine this amplitude. Then find thegeneral solution with this and check whether the resultsare in agreement.vr(t) ϭ 10 cos vtkg>sec2,k ϭ 24>c ϭ 4m ϭ 161 cos 3.1tk ϭ 36,m ϭ 4, c ϭ 0,Fig. 71. RLC-circuitE(t)CR LFig. 72. Mass–spring systemDashpotMassSpringkmcSecond-order linear ODEs are particularly important in applications, for instance,in mechanics (Secs. 2.4, 2.8) and electrical engineering (Sec. 2.9). A second-orderODE is called linear if it can be written(1) (Sec. 2.1).(If the first term is, say, divide by to get the “standard form” (1) withas the first term.) Equation (1) is called homogeneous if is zero for all xconsidered, usually in some open interval; this is written Then(2)Equation (1) is called nonhomogeneous if (meaning is not zero forsome x considered).r(x)r(x) [ 0ys ϩ p(x)yr ϩ q(x)y ϭ 0.r(x) ϵ 0.r(x)ysf(x)f(x)ys,ys ϩ p(x)yr ϩ q(x)y ϭ r(x)SUMMARY OF CHAPTER 2Second-Order Linear ODEs26. Find the current in the RLC-circuit in Fig. 71when(50 cycles sec).>220 sin 314t VE ϭC ϭ 10؊4F,L ϭ 0.4 H,R ϭ 40 ⍀,c02.qxd 10/27/10 6:06 PM Page 103
• For the homogeneous ODE (2) we have the important superposition principle (Sec.2.1) that a linear combination of two solutions is again a solution.Two linearly independent solutions of (2) on an open interval I form a basis(or fundamental system) of solutions on I. and with arbitraryconstants a general solution of (2) on I. From it we obtain a particularsolution if we specify numeric values (numbers) for and usually by prescribingtwo initial conditions(3) given numbers; Sec. 2.1).(2) and (3) together form an initial value problem. Similarly for (1) and (3).For a nonhomogeneous ODE (1) a general solution is of the form(4) (Sec. 2.7).Here is a general solution of (2) and is a particular solution of (1). Such acan be determined by a general method (variation of parameters, Sec. 2.10) or inmany practical cases by the method of undetermined coefficients. The latter applieswhen (1) has constant coefficients p and q, and is a power of x, sine, cosine,etc. (Sec. 2.7). Then we write (1) as(5) (Sec. 2.7).The corresponding homogeneous ODE has solutionswhere is a root of(6)Hence there are three cases (Sec. 2.2):l2ϩ al ϩ b ϭ 0.ly ϭ elx,yr ϩ ayr ϩ by ϭ 0ys ϩ ayr ϩ by ϭ r(x)r(x)ypypyhy ϭ yh ϩ yp(x0, K0, K1yr(x0) ϭ K1y(x0) ϭ K0,c2,c1c1, c2y ϭ c1y1 ϩ c2y2y1, y2y1, y2y ϭ ky1 ϩ ly2104 CHAP. 2 Second-Order Linear ODEsCase Type of Roots General SolutionI Distinct realII DoubleIII Complex y ϭ e؊ax>2(A cos v*x ϩ B sin v*x)Ϫ12 a Ϯ iv*y ϭ (c1 ϩ c2x)eϪax>2Ϫ12 ay ϭ c1el1xϩ c2el2xl1, l2Here is used since is needed in driving forces.Important applications of (5) in mechanical and electrical engineering in connectionwith vibrations and resonance are discussed in Secs. 2.4, 2.7, and 2.8.Another large class of ODEs solvable “algebraically” consists of the Euler–Cauchyequations(7) (Sec. 2.5).These have solutions of the form where m is a solution of the auxiliary equation(8)Existence and uniqueness of solutions of (1) and (2) is discussed in Secs. 2.6and 2.7, and reduction of order in Sec. 2.1.m2ϩ (a Ϫ 1)m ϩ b ϭ 0.y ϭ xm,x2ys ϩ axyr ϩ by ϭ 0vv*c02.qxd 10/27/10 6:06 PM Page 104
• 105C H A P T E R 3Higher Order Linear ODEsThe concepts and methods of solving linear ODEs of order extend nicely to linearODEs of higher order n, that is, etc. This shows that the theory explained inChap. 2 for second-order linear ODEs is attractive, since it can be extended in astraightforward way to arbitrary n. We do so in this chapter and notice that the formulasbecome more involved, the variety of roots of the characteristic equation (in Sec. 3.2)becomes much larger with increasing n, and the Wronskian plays a more prominent role.The concepts and methods of solving second-order linear ODEs extend readily to linearODEs of higher order.This chapter follows Chap. 2 naturally, since the results of Chap. 2 can be readilyextended to that of Chap. 3.Prerequisite: Secs. 2.1, 2.2, 2.6, 2.7, 2.10.References and Answers to Problems: App. 1 Part A, and App. 2.3.1 Homogeneous Linear ODEsRecall from Sec. 1.1 that an ODE is of nth order if the nth derivative ofthe unknown function is the highest occurring derivative. Thus the ODE is of the formwhere lower order derivatives and y itself may or may not occur. Such an ODE is calledlinear if it can be written(1)(For this is (1) in Sec. 2.1 with and .) The coefficientsand the function r on the right are any given functions of x, and y is unknown. hascoefficient 1. We call this the standard form. (If you have divide byto get this form.) An nth-order ODE that cannot be written in the form (1) is callednonlinear.If is identically zero, (zero for all x considered, usually in some openinterval I), then (1) becomes(2) y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ 0r(x) ϵ 0r(x)pn(x)pn(x)y(n),y(n)p0, Á , pn؊1p0 ϭ qp1 ϭ pn ϭ 2y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ r(x).F(x, y, yr, Á , y(n)) ϭ 0y(x)y(n)ϭ dny>dxnn ϭ 3, 4,n ϭ 2c03.qxd 10/27/10 6:20 PM Page 105
• and is called homogeneous. If is not identically zero, then the ODE is callednonhomogeneous. This is as in Sec. 2.1.A solution of an nth-order (linear or nonlinear) ODE on some open interval I is afunction that is defined and n times differentiable on I and is such that the ODEbecomes an identity if we replace the unknown function y and its derivatives by h and itscorresponding derivatives.Sections 3.1–3.2 will be devoted to homogeneous linear ODEs and Section 3.3 tononhomogeneous linear ODEs.Homogeneous Linear ODE: Superposition Principle,General SolutionThe basic superposition or linearity principle of Sec. 2.1 extends to nth orderhomogeneous linear ODEs as follows.T H E O R E M 1 Fundamental Theorem for the Homogeneous Linear ODE (2)For a homogeneous linear ODE (2), sums and constant multiples of solutions onsome open interval I are again solutions on I. (This does not hold for anonhomogeneous or nonlinear ODE!)The proof is a simple generalization of that in Sec. 2.1 and we leave it to the student.Our further discussion parallels and extends that for second-order ODEs in Sec. 2.1.So we next define a general solution of (2), which will require an extension of linearindependence from 2 to n functions.D E F I N I T I O N General Solution, Basis, Particular SolutionA general solution of (2) on an open interval I is a solution of (2) on I of the form(3)where is a basis (or fundamental system) of solutions of (2) on I; thatis, these solutions are linearly independent on I, as defined below.A particular solution of (2) on I is obtained if we assign specific values to then constants in (3).D E F I N I T I O N Linear Independence and DependenceConsider n functions defined on some interval I.These functions are called linearly independent on I if the equation(4)implies that all are zero. These functions are called linearly dependenton I if this equation also holds on I for some not all zero.k1, Á , knk1, Á , knk1y1(x) ϩ Á ϩ knyn(x) ϭ 0 on Iy1(x), Á , yn(x)c1, Á , cny1, Á , yn(c1, Á , cn arbitrary)y(x) ϭ c1y1(x) ϩ Á ϩ cnyn(x)y ϭ h(x)r(x)106 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 106
• If and only if are linearly dependent on I, we can express (at least) one ofthese functions on I as a “linear combination” of the other functions, that is, as asum of those functions, each multiplied by a constant (zero or not). This motivates theterm “linearly dependent.” For instance, if (4) holds with we can divide byand express as the linear combinationNote that when these concepts reduce to those defined in Sec. 2.1.E X A M P L E 1 Linear DependenceShow that the functions are linearly dependent on any interval.Solution. . This proves linear dependence on any interval.E X A M P L E 2 Linear IndependenceShow that are linearly independent on any interval, for instance, onSolution. Equation (4) is Taking (a) (b) (c) we get(a) (b) (c)from Then from (c) (b). Then from (b). This proves linear independence.A better method for testing linear independence of solutions of ODEs will soon be explained.E X A M P L E 3 General Solution. BasisSolve the fourth-order ODE(where ).Solution. As in Sec. 2.2 we substitute . Omitting the common factor we obtain the characteristicequationThis is a quadratic equation in namely,The roots are and 4. Hence This gives four solutions. A general solution on anyinterval isprovided those four solutions are linearly independent. This is true but will be shown later.Initial Value Problem. Existence and UniquenessAn initial value problem for the ODE (2) consists of (2) and n initial conditions(5) ,with given in the open interval I considered, and given .K0, Á , Kn؊1x0y(n؊1)(x0) ϭ Kn؊1Áyr(x0) ϭ K1,y(x0) ϭ K0,᭿y ϭ c1e؊2xϩ c2e؊xϩ c3exϩ c4e2xl ϭ Ϫ2, Ϫ1, 1, 2.␮ ϭ 1␮2Ϫ 5␮ ϩ 4 ϭ (␮ Ϫ 1)(␮ Ϫ 4) ϭ 0.␮ ϭ l2,l4Ϫ 5l2ϩ 4 ϭ 0.elx,y ϭ elxyivϭ d4y>dx4yivϪ 5ys ϩ 4y ϭ 0᭿k1 ϭ 0Ϫ2k3 ϭ 0(a) ϩ (b).k2 ϭ 02k1 ϩ 4k2 ϩ 8k3 ϭ 0.k1 ϩ k2 ϩ k3 ϭ 0,Ϫk1 ϩ k2 Ϫ k3 ϭ 0,x ϭ 2,x ϭ 1,x ϭ Ϫ1,k1x ϩ k2x2ϩ k3x3ϭ 0.Ϫ1 Ϲ x Ϲ 2.y1 ϭ x, y2 ϭ x2, y3 ϭ x3᭿y2 ϭ 0y1 ϩ 2.5y3y1 ϭ x2, y2 ϭ 5x, y3 ϭ 2xn ϭ 2,y1 ϭ Ϫ1k1(k2y2 ϩ Á ϩ knyn).y1k1k1 0,n Ϫ 1y1, Á , ynSEC. 3.1 Homogeneous Linear ODEs 107c03.qxd 10/27/10 6:20 PM Page 107
• In extension of the existence and uniqueness theorem in Sec. 2.6 we now have thefollowing.T H E O R E M 2 Existence and Uniqueness Theorem for Initial Value ProblemsIf the coefficients of (2) are continuous on some open interval Iand is in I, then the initial value problem (2), (5) has a unique solution on I.Existence is proved in Ref. [A11] in App. 1. Uniqueness can be proved by a slightgeneralization of the uniqueness proof at the beginning of App. 4.E X A M P L E 4 Initial Value Problem for a Third-Order Euler–Cauchy EquationSolve the following initial value problem on any open interval I on the positive x-axis containingSolution. Step 1. General solution. As in Sec. 2.5 we try By differentiation and substitution,Dropping and ordering gives If we can guess the root We can divideby and find the other roots 2 and 3, thus obtaining the solutions which are linearly independenton I (see Example 2). [In general one shall need a root-finding method, such as Newton’s (Sec. 19.2), alsoavailable in a CAS (Computer Algebra System).] Hence a general solution isvalid on any interval I, even when it includes where the coefficients of the ODE divided by (to havethe standard form) are not continuous.Step 2. Particular solution. The derivatives are and From this, andy and the initial conditions, we get by setting(a)(b)(c)This is solved by Cramer’s rule (Sec. 7.6), or by elimination, which is simple, as follows. gives(d) Then (c) (d) gives Then (c) gives Finally from (a).Answer:Linear Independence of Solutions. WronskianLinear independence of solutions is crucial for obtaining general solutions. Although it canoften be seen by inspection, it would be good to have a criterion for it. Now Theorem 2in Sec. 2.6 extends from order to any n. This extended criterion uses the WronskianW of n solutions defined as the nth-order determinant(6) W(y1, Á , yn) ϭ 5y1 y2Á yny1r y2r Á ynr# # Á #y1(n؊1)y2(n؊1) Á yn(n؊1)5 .y1, Á , ynn ϭ 2᭿y ϭ 2x ϩ x2Ϫ x3.c1 ϭ 2c2 ϭ 1.c3 ϭ Ϫ1.Ϫ 2c2 ϩ 2c3 ϭ Ϫ1.(b) Ϫ (a)ys(1) ϭ 2c2 ϩ 6c3 ϭ Ϫ4.yr(1) ϭ c1 ϩ 2c2 ϩ 3c3 ϭ 1y(1) ϭ c1 ϩ c2 ϩ c3 ϭ 2x ϭ 1ys ϭ 2c2 ϩ 6c3x.yr ϭ c1 ϩ 2c2x ϩ 3c3x2x3x ϭ 0y ϭ c1x ϩ c2x2ϩ c3x3x, x2, x3,m Ϫ 1m ϭ 1.m3Ϫ 6m2ϩ 11m Ϫ 6 ϭ 0.xmm(m Ϫ 1)(m Ϫ 2)xmϪ 3m(m Ϫ 1)xmϩ 6mxmϪ 6xmϭ 0.y ϭ xm.ys(1) ϭ Ϫ4.yr(1) ϭ 1,y(1) ϭ 2,x3yt Ϫ 3x2ys ϩ 6xyr Ϫ 6y ϭ 0,x ϭ 1.y(x)x0p0(x), Á , pn؊1(x)108 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 108
• Note that W depends on x since do. The criterion states that these solutionsform a basis if and only if W is not zero; more precisely:T H E O R E M 3 Linear Dependence and Independence of SolutionsLet the ODE (2) have continuous coefficients on an open intervalI. Then n solutions of (2) on I are linearly dependent on I if and only if theirWronskian is zero for some in I. Furthermore, if W is zero for then Wis identically zero on I. Hence if there is an in I at which W is not zero, thenare linearly independent on I, so that they form a basis of solutions of (2) on I.P R O O F (a) Let be linearly dependent solutions of (2) on I. Then, by definition, thereare constants not all zero, such that for all x in I,(7)By differentiations of (7) we obtain for all x in I(8)(7), (8) is a homogeneous linear system of algebraic equations with a nontrivial solutionHence its coefficient determinant must be zero for every x on I, by Cramer’stheorem (Sec. 7.7). But that determinant is the Wronskian W, as we see from (6). HenceW is zero for every x on I.(b) Conversely, if W is zero at an in I, then the system (7), (8) with has asolution not all zero, by the same theorem. With these constants we definethe solution of (2) on I. By (7), (8) this solution satisfies theinitial conditions But another solution satisfying thesame conditions is Hence by Theorem 2, which applies since the coefficientsof (2) are continuous. Together, on I. This means lineardependence of on I.(c) If W is zero at an in I, we have linear dependence by (b) and then by (a).Hence if W is not zero at an in I, the solutions must be linearly independenton I.E X A M P L E 5 Basis, WronskianWe can now prove that in Example 3 we do have a basis. In evaluating W, pull out the exponential functionscolumnwise. In the result, subtract Column 1 from Columns 2, 3, 4 (without changing Column 1). Then expand byRow 1. In the resulting third-order determinant, subtract Column 1 from Column 2 and expand the result by Row 2:᭿W ϭ 6e؊2xe؊xexe2xϪ2e؊2xϪe؊xex2e2x4e؊2xe؊xex4e2xϪ8e؊2xϪe؊xex8e2x6 ϭ 61 1 1 1Ϫ2 Ϫ1 1 24 1 1 4Ϫ8 Ϫ1 1 86 ϭ 31 3 4Ϫ3 Ϫ3 07 9 163 ϭ 72.᭿y1, Á , ynx1W ϵ 0x0y1, Á , yny* ϭ k1*y1 ϩ Á ϩ kn*yn ϵ 0y* ϵ yy ϵ 0.y*(x0) ϭ 0, Á , y*(n؊1)(x0) ϭ 0.y* ϭ k1*y1 ϩ Á ϩ kn*ynk1*, Á , kn*,x ϭ x0x0k1, Á , kn.k1y1(n؊1)ϩ Á ϩ knyn(n؊1)ϭ 0....k1y1r ϩ Á ϩ knynr ϭ 0n Ϫ 1k1y1 ϩ Á ϩ knyn ϭ 0.k1, Á , kny1, Á , yny1, Á , ynx1x ϭ x0,x ϭ x0y1, Á , ynp0(x), Á , pn؊1(x)y1, Á , ynSEC. 3.1 Homogeneous Linear ODEs 109c03.qxd 10/27/10 6:20 PM Page 109
• A General Solution of (2) Includes All SolutionsLet us first show that general solutions always exist. Indeed, Theorem 3 in Sec. 2.6 extendsas follows.T H E O R E M 4 Existence of a General SolutionIf the coefficients of (2) are continuous on some open interval I,then (2) has a general solution on I.P R O O F We choose any fixed in I. By Theorem 2 the ODE (2) has n solutions wheresatisfies initial conditions (5) with and all other K’s equal to zero. TheirWronskian at equals 1. For instance, when thenand the other initial values are zero. Thus, as claimed,Hence for any n those solutions are linearly independent on I, by Theorem 3.They form a basis on I, and is a general solution of (2) on I.We can now prove the basic property that, from a general solution of (2), every solutionof (2) can be obtained by choosing suitable values of the arbitrary constants. Hence annth-order linear ODE has no singular solutions, that is, solutions that cannot be obtainedfrom a general solution.T H E O R E M 5 General Solution Includes All SolutionsIf the ODE (2) has continuous coefficients on some open intervalI, then every solution of (2) on I is of the form(9)where is a basis of solutions of (2) on I and are suitable constants.P R O O F Let Y be a given solution and a general solution of (2) on I. Wechoose any fixed in I and show that we can find constants for which y andits first derivatives agree with Y and its corresponding derivatives at That is,we should have at(10)But this is a linear system of equations in the unknowns Its coefficientdeterminant is the Wronskian W of at Since form a basis, theyy1, Á , ynx0.y1, Á , ync1, Á , cn.c1y1(n؊1)ϩ Á ϩ cnyn(n؊1)ϭ Y(n؊1)....c1y1r ϩ Á ϩ cnynr ϭ Y rc1y1 ϩ Á ϩ cnyn ϭ Yx ϭ x0x0.n Ϫ 1c1, Á , cnx0y ϭ c1y1 ϩ Á ϩ cnynC1, Á , Cny1, Á , ynY(x) ϭ C1y1(x) ϩ Á ϩ Cnyn(x)y ϭ Y(x)p0(x), Á , pn؊1(x)᭿y ϭ c1y1 ϩ Á ϩ cnyny1, Á , ynW(y1(x0), y2(x0), y3(x0)) ϭ 4y1(x0) y2(x0) y3(x0)y1r(x0) y2r(x0) y3r(x0)y1s(x0) y2s(x0) y3s(x0)4 ϭ 41 0 00 1 00 0 14 ϭ 1.y3s(x0) ϭ 1,y1(x0) ϭ 1, y2r(x0) ϭ 1,n ϭ 3,x0Kj؊1 ϭ 1yjy1, Á , yn,x0p0(x), Á , pn؊1(x)110 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 110
• are linearly independent, so that W is not zero by Theorem 3. Hence (10) has a uniquesolution (by Cramer’s theorem in Sec. 7.7). With these values weobtain the particular solutionon I. Equation (10) shows that and its first derivatives agree at with Y andits corresponding derivatives. That is, and Y satisfy, at , the same initial conditions.The uniqueness theorem (Theorem 2) now implies that on I. This proves thetheorem.This completes our theory of the homogeneous linear ODE (2). Note that for it isidentical with that in Sec. 2.6. This had to be expected.n ϭ 2᭿y* ϵ Yx0y*x0n Ϫ 1y*y*(x) ϭ C1y1(x) ϩ Á ϩ Cnyn(x)c1 ϭ C1, Á , cn ϭ CnSEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 1111–6 BASES: TYPICAL EXAMPLESTo get a feel for higher order ODEs, show that the givenfunctions are solutions and form a basis on any interval.Use Wronskians. In Prob. 6,1.2.3.4.5.6.7. TEAM PROJECT. General Properties of Solutionsof Linear ODEs. These properties are important inobtaining new solutions from given ones. Thereforeextend Team Project 38 in Sec. 2.2 to nth-order ODEs.Explore statements on sums and multiples of solutionsof (1) and (2) systematically and with proofs.Recognize clearly that no new ideas are needed in thisextension from to general n.8–15 LINEAR INDEPENDENCEAre the given functions linearly independent or dependenton the half-axis Give reason.8. 9. tan x, cot x, 1x2, 1>x2, 0x Ն 0?n ϭ 21, x2, x4, x2yt Ϫ 3xys ϩ 3yr ϭ 01, e؊xcos 2x, e؊xsin 2x, yt ϩ 2ys ϩ 5yr ϭ 0e؊4x, xe؊4x, x2e؊4x, ytϩ 12ysϩ 48yrϩ 64y ϭ 0cos x, sin x, x cos x, x sin x, yivϩ 2ys ϩ y ϭ 0ex, e؊x, e2x, yt Ϫ 2ys Ϫ yr ϩ 2y ϭ 01, x, x2, x3, yivϭ 0x Ͼ 0,P R O B L E M S E T 3 . 110. 11.12. 13.14. 15.16. TEAM PROJECT. Linear Independence andDependence. (a) Investigate the given question abouta set S of functions on an interval I. Give an example.Prove your answer.(1) If S contains the zero function, can S be linearlyindependent?(2) If S is linearly independent on a subinterval J of I,is it linearly independent on I?(3) If S is linearly dependent on a subinterval J of I,is it linearly dependent on I?(4) If S is linearly independent on I, is it linearlyindependent on a subinterval J?(5) If S is linearly dependent on I, is it linearlyindependent on a subinterval J?(6) If S is linearly dependent on I, and if T contains S,is T linearly dependent on I?(b) In what cases can you use the Wronskian fortesting linear independence? By what other means canyou perform such a test?cosh 2x, sinh 2x, e2xcos2x, sin2x, 2psin x, cos x, sin 2xsin2x, cos2x, cos 2xexcos x, exsin x, exe2x, xe2x, x2e2x3.2 Homogeneous Linear ODEswith Constant CoefficientsWe proceed along the lines of Sec. 2.2, and generalize the results from to arbitrary n.We want to solve an nth-order homogeneous linear ODE with constant coefficients,written as(1) y(n)ϩ an؊1y(n؊1)ϩ Á ϩ a1yr ϩ a0y ϭ 0n ϭ 2c03.qxd 10/27/10 6:20 PM Page 111
• where etc. As in Sec. 2.2, we substitute to obtain the characteristicequation(2)of (1). If is a root of (2), then is a solution of (1). To find these roots, you mayneed a numeric method, such as Newton’s in Sec. 19.2, also available on the usual CASs.For general n there are more cases than for We can have distinct real roots, simplecomplex roots, multiple roots, and multiple complex roots, respectively. This will be shownnext and illustrated by examples.Distinct Real RootsIf all the n roots of (2) are real and different, then the n solutions(3)constitute a basis for all x. The corresponding general solution of (1) is(4)Indeed, the solutions in (3) are linearly independent, as we shall see after the example.E X A M P L E 1 Distinct Real RootsSolve the ODESolution. The characteristic equation is It has the roots if you find oneof them by inspection, you can obtain the other two roots by solving a quadratic equation (explain!). Thecorresponding general solution (4) isLinear Independence of (3). Students familiar with nth-order determinants may verifythat, by pulling out all exponential functions from the columns and denoting their productby the Wronskian of the solutions in (3) becomes(5)ϭ E 71 1 Á 1l1 l2Á lnl12l22 Á ln2# # Á #l1n؊1l2n؊1 Á lnn؊17.W ϭ 7el1xel2x Á elnxl1el1xl2el2x Á lnelnxl12el1xl22el2x Á ln2elnx# # Á #l1n؊1el1xl2n؊1el2x Á lnn؊1elnx7E ϭ exp [l1 ϩ Á ϩ ln)x],᭿y ϭ c1e؊xϩ c2exϩ c3e2x.Ϫ1, 1, 2;l3Ϫ 2l2Ϫ l ϩ 2 ϭ 0.yt Ϫ 2ys Ϫ yr ϩ 2y ϭ 0.y ϭ c1el1xϩ Á ϩ cnelnx.y1 ϭ el1x, Á , yn ϭ elnx.l1, Á , lnn ϭ 2.y ϭ elxll(n)ϩ an؊1l(n؊1)ϩ Á ϩ a1l ϩ a0y ϭ 0y ϭ elxy(n)ϭ dny>dxn,112 CHAP. 3 Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 112
• The exponential function E is never zero. Hence if and only if the determinant onthe right is zero. This is a so-called Vandermonde or Cauchy determinant.1It can beshown that it equals(6)where V is the product of all factors with for instance, whenwe get This shows that the Wronskian is not zeroif and only if all the n roots of (2) are different and thus gives the following.T H E O R E M 1 BasisSolutions of (1) (with any real or complex ’s) form abasis of solutions of (1) on any open interval if and only if all n roots of (2) aredifferent.Actually, Theorem 1 is an important special case of our more general result obtainedfrom (5) and (6):T H E O R E M 2 Linear IndependenceAny number of solutions of (1) of the form are linearly independent on an openinterval I if and only if the corresponding are all different.Simple Complex RootsIf complex roots occur, they must occur in conjugate pairs since the coefficients of (1)are real. Thus, if is a simple root of (2), so is the conjugate andtwo corresponding linearly independent solutions are (as in Sec. 2.2, except for notation)E X A M P L E 2 Simple Complex Roots. Initial Value ProblemSolve the initial value problemSolution. The characteristic equation is It has the root 1, as can perhaps beseen by inspection. Then division by shows that the other roots are Hence a general solution andits derivatives (obtained by differentiation) areys ϭ c1exϪ 100A cos 10x Ϫ 100B sin 10x.yr ϭ c1exϪ 10A sin 10x ϩ 10B cos 10x,y ϭ c1exϩ A cos 10x ϩ B sin 10x,Ϯ10i.l Ϫ 1l3Ϫ l2ϩ 100l Ϫ 100 ϭ 0.ys(0) ϭ Ϫ299.yr(0) ϭ 11,y(0) ϭ 4,yt Ϫ ys ϩ 100yr Ϫ 100y ϭ 0,y2 ϭ egxsin vx.y1 ϭ egxcos vx,l ϭ g Ϫ iv,l ϭ g ϩ ivlelxljy1 ϭ el1x, Á , yn ϭ elnxϪV ϭ Ϫ(l1 Ϫ l2)(l1 Ϫ l3)(l2 Ϫ l3).n ϭ 3j Ͻ k (Ϲ n);lj Ϫ lk(Ϫ1)n(n؊1)>2VW ϭ 0SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 1131ALEXANDRE THÉOPHILE VANDERMONDE (1735–1796), French mathematician, who worked onsolution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5.c03.qxd 10/27/10 6:20 PM Page 113
• From this and the initial conditions we obtain, by setting ,(a) (b) (c)We solve this system for the unknowns A, B, Equation (a) minus Equation (c) givesThen from (a) and from (b). The solution is (Fig. 73)This gives the solution curve, which oscillates about (dashed in Fig. 73). ᭿exy ϭ exϩ 3 cos 10x ϩ sin 10x.B ϭ 1c1 ϭ 1101A ϭ 303, A ϭ 3.c1.c1 Ϫ 100A ϭ Ϫ299.c1 ϩ 10B ϭ 11,c1 ϩ A ϭ 4,x ϭ 0114 CHAP. 3 Higher Order Linear ODEs40010321 xy20Fig. 73. Solution in Example 2Multiple Real RootsIf a real double root occurs, say, then in (3), and we take and ascorresponding linearly independent solutions. This is as in Sec. 2.2.More generally, if is a real root of order m, then m corresponding linearly independentsolutions are(7)We derive these solutions after the next example and indicate how to prove their linearindependence.E X A M P L E 3 Real Double and Triple RootsSolve the ODESolution. The characteristic equation has the roots andand the answer is(8)Derivation of (7). We write the left side of (1) asLet Then by performing the differentiations we haveL[elx] ϭ (lnϩ an؊1ln؊1ϩ Á ϩ a0)elx.y ϭ elx.L[y] ϭ y(n)ϩ an؊1y(n؊1)ϩ Á ϩ a0y.᭿y ϭ c1 ϩ c2x ϩ (c3 ϩ c4x ϩ c5x2)ex.l5 ϭ 1,l3 ϭ l4 ϭl1 ϭ l2 ϭ 0,l5Ϫ 3l4ϩ 3l3Ϫ l2ϭ 0yvϪ 3yivϩ 3yt Ϫ ys ϭ 0.elx, xelx, x2elx, Á , xm؊1elx.lxy1y1y1 ϭ y2l1 ϭ l2,c03.qxd 10/27/10 6:20 PM Page 114
• Now let be a root of mth order of the polynomial on the right, where Forlet be the other roots, all different from Writing the polynomial inproduct form, we then havewith if and if Now comes thekey idea: We differentiate on both sides with respect to(9)The differentiations with respect to x and are independent and the resulting derivativesare continuous, so that we can interchange their order on the left:(10)The right side of (9) is zero for because of the factors (and sincewe have a multiple root!). Hence by (9) and (10). This proves that isa solution of (1).We can repeat this step and produce by another suchdifferentiations with respect to Going one step further would no longer give zero on theright because the lowest power of would then be multiplied byand because has no factors so we get precisely the solutions in (7).We finally show that the solutions (7) are linearly independent. For a specific n thiscan be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrarym we can pull out the exponential functions from the Wronskian. This givestimes a determinant which by “row operations” can be reduced to the Wronskian of 1,The latter is constant and different from zero (equal toThese functions are solutions of the ODE so that linear independence followsfrom Theroem 3 in Sec. 3.1.Multiple Complex RootsIn this case, real solutions are obtained as for complex simple roots above. Consequently,if is a complex double root, so is the conjugate Correspondinglinearly independent solutions are(11)The first two of these result from and as before, and the second two fromand in the same fashion. Obviously, the corresponding general solution is(12)For complex triple roots (which hardly ever occur in applications), one would obtaintwo more solutions and so on.x2egxcos vx, x2egxsin vx,y ϭ egx[(A1 ϩ A2x) cos vx ϩ (B1 ϩ B2x) sin vx].xelxxelxelxelxegxcos vx, egxsin vx, xegxcos vx, xegxsin vx.l ϭ g Ϫ iv.l ϭ g ϩ ivy(m)ϭ 0,1!2! Á (m Ϫ 1)!).x, Á , xm؊1.(elx)mϭ elmxl Ϫ l1;h(l)h(l1) 0m!h(l)(l Ϫ l1)0,l Ϫ l1l.m Ϫ 2x2el1x, Á , xm؊1el1xxel1xL[xel1x] ϭ 0m м 2l Ϫ l1l ϭ l100lL[elx] ϭ Lc00lelxd ϭ L[xelx].l00lL[elx] ϭ m(l Ϫ l1)m؊1h(l)elxϩ (l Ϫ l1)m 00l[h(l)elx].l,m Ͻ n.h(l) ϭ (l Ϫ lmϩ1) Á (l Ϫ ln)m ϭ n,h(l) ϭ 1L[elx] ϭ (l Ϫ l1)mh(l)elxl1.lm؉1, Á , lnm Ͻ nm Ϲ n.l1SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 115c03.qxd 10/27/10 6:20 PM Page 115
• 3.3 Nonhomogeneous Linear ODEsWe now turn from homogeneous to nonhomogeneous linear ODEs of nth order. We writethem in standard form(1)with as the first term, and As for second-order ODEs, a generalsolution of (1) on an open interval I of the x-axis is of the form(2)Here is a general solution of the correspondinghomogeneous ODE(3)on I. Also, is any solution of (1) on I containing no arbitrary constants. If (1) hascontinuous coefficients and a continuous on I, then a general solution of (1) existsand includes all solutions. Thus (1) has no singular solutions.r(x)ypy(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ 0yh(x) ϭ c1y1(x) ϩ Á ϩ cnyn(x)y(x) ϭ yh(x) ϩ yp(x).r(x) [ 0.y(n)ϭ dny>dxny(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ r(x)116 CHAP. 3 Higher Order Linear ODEs1–6 GENERAL SOLUTIONSolve the given ODE. Show the details of your work.1.2.3.4.5.6.7–13 INITIAL VALUE PROBLEMSolve the IVP by a CAS, giving a general solution and theparticular solution and its graph.7.8.9.10.11.12.ys(0) ϭ 11, yt(0) ϭ Ϫ23, yiv(0) ϭ 47yvϪ 5yt ϩ 4yr ϭ 0, y(0) ϭ 3, yr(0) ϭ Ϫ5,yt(0) ϭ 0ys(0) ϭ 41,yivϪ 9ys Ϫ 400y ϭ 0, y(0) ϭ 0, yr(0) ϭ 0,yt(0) ϭ Ϫ72yivϩ 4y ϭ 0, y(0) ϭ 12, yr(0) ϭ Ϫ32, ys(0) ϭ 52,ys(0) ϭ Ϫ39.75yr(0) ϭ Ϫ6.5,4yt ϩ 8ys ϩ 41yr ϩ 37y ϭ 0, y(0) ϭ 9,yr(0) ϭ Ϫ54.975, ys(0) ϭ 257.5125yt ϩ 7.5ys ϩ 14.25yr Ϫ 9.125y ϭ 0, y(0) ϭ 10.05,ys(0) ϭ 9.91yr(0) ϭ Ϫ4.6,yt ϩ 3.2ys ϩ 4.81yr ϭ 0, y(0) ϭ 3.4,(D5ϩ 8D3ϩ 16D)y ϭ 0(D4ϩ 10D2ϩ 9I)y ϭ 0(D3Ϫ D2Ϫ D ϩ I)y ϭ 0yivϩ 4ys ϭ 0yivϩ 2ys ϩ y ϭ 0yt ϩ 25yr ϭ 013.14. PROJECT. Reduction of Order. This is of practicalinterest since a single solution of an ODE can often beguessed. For second order, see Example 7 in Sec. 2.1.(a) How could you reduce the order of a linearconstant-coefficient ODE if a solution is known?(b) Extend the method to a variable-coefficient ODEAssuming a solution to be known, show that anothersolution is with andz obtained by solving(c) Reduceusing (perhaps obtainable by inspection).15. CAS EXPERIMENT. Reduction of Order. Startingwith a basis, find third-order linear ODEs with variablecoefficients for which the reduction to second orderturns out to be relatively simple.y1 ϭ xx3yt Ϫ 3x2ys ϩ (6 Ϫ x2)xyr Ϫ (6 Ϫ x2)y ϭ 0,y1zsϩ (3y1r ϩ p2y1)zr ϩ (3y1sϩ 2p2y1r ϩ p1y1)z ϭ 0.u(x) ϭ ͐z(x) dxy2(x) ϭ u(x)y1(x)y1yt ϩ p2(x)ys ϩ p1(x)yr ϩ p0(x)y ϭ 0.yt(0) ϭ Ϫ1.458675y(0) ϭ 17.4, yr(0) ϭ Ϫ2.82, ys(0) ϭ 2.0485,yivϩ 0.45yt Ϫ 0.165ys ϩ 0.0045yrϪ 0.00175y ϭ 0,P R O B L E M S E T 3 . 2c03.qxd 10/27/10 6:20 PM Page 116
• An initial value problem for (1) consists of (1) and n initial conditions(4)with in I. Under those continuity assumptions it has a unique solution. The ideas ofproof are the same as those for in Sec. 2.7.Method of Undetermined CoefficientsEquation (2) shows that for solving (1) we have to determine a particular solution of (1).For a constant-coefficient equation(5)( constant) and special as in Sec. 2.7, such a can be determined bythe method of undetermined coefficients, as in Sec. 2.7, using the following rules.(A) Basic Rule as in Sec. 2.7.(B) Modification Rule. If a term in your choice for is a solution of thehomogeneous equation (3), then multiply this term by where k is the smallestpositive integer such that this term times is not a solution of (3).(C) Sum Rule as in Sec. 2.7.The practical application of the method is the same as that in Sec. 2.7. It suffices toillustrate the typical steps of solving an initial value problem and, in particular, the newModification Rule, which includes the old Modification Rule as a particular case (withor 2). We shall see that the technicalities are the same as for except perhapsfor the more involved determination of the constants.E X A M P L E 1 Initial Value Problem. Modification RuleSolve the initial value problem(6)Solution. Step 1. The characteristic equation is It has the triple rootHence a general solution of the homogeneous ODE isStep 2. If we try we get which has no solution. Try andThe Modification Rule calls forThenypt ϭ C(6 Ϫ 18x ϩ 9x2Ϫ x3)e؊x.yps ϭ C(6x Ϫ 6x2ϩ x3)e؊x,ypr ϭ C(3x2Ϫ x3)e؊x,yp ϭ Cx3e؊x.Cx2e؊x.Cxe؊xϪC ϩ 3C Ϫ 3C ϩ C ϭ 30,yp ϭ Ce؊x,ϭ (c1 ϩ c2x ϩ c3x2)e؊x.yh ϭ c1e؊xϩ c2xe؊xϩ c3x2e؊xl ϭ Ϫ1.l3ϩ 3l2ϩ 3l ϩ 1 ϭ (l ϩ 1)3ϭ 0.yt ϩ 3ys ϩ 3yr ϩ y ϭ 30e؊x, y(0) ϭ 3, yr(0) ϭ Ϫ3, ys(0) ϭ Ϫ47.n ϭ 2,k ϭ 1xkxk,yp(x)yp(x)r(x)a0, Á , an؊1y(n)ϩ an؊1y(n؊1)ϩ Á ϩ a1yr ϩ a0y ϭ r(x)n ϭ 2x0y(x0) ϭ K0, yr(x0) ϭ K1, Á , y(n؊1)(x0) ϭ Kn؊1SEC. 3.3 Nonhomogeneous Linear ODEs 117c03.qxd 10/27/10 6:20 PM Page 117
• Substitution of these expressions into (6) and omission of the common factor givesThe linear, quadratic, and cubic terms drop out, and Hence This givesStep 3. We now write down the general solution of the given ODE. From it we find by thefirst initial condition. We insert the value, differentiate, and determine from the second initial condition, insertthe value, and finally determine from and the third initial condition:Hence the answer to our problem is (Fig. 73)The curve of y begins at (0, 3) with a negative slope, as expected from the initial values, and approaches zeroas The dashed curve in Fig. 74 is ᭿yp.x : ϱ.y ϭ (3 Ϫ 25x2)e؊xϩ 5x3e؊x.ys ϭ [3 ϩ 2c3 ϩ (30 Ϫ 4c3)x ϩ (Ϫ30 ϩ c3)x2ϩ 5x3]e؊x, ys(0) ϭ 3 ϩ 2c3 ϭ Ϫ47, c3 ϭ Ϫ25.yr ϭ [Ϫ3 ϩ c2 ϩ (Ϫc2 ϩ 2c3)x ϩ (15 Ϫ c3)x2Ϫ 5x3]e؊x, yr(0) ϭ Ϫ3 ϩ c2 ϭ Ϫ3, c2 ϭ 0y ϭ yh ϩ yp ϭ (c1 ϩ c2x ϩ c3x2)e؊xϩ 5x3e؊x, y(0) ϭ c1 ϭ 3ys(0)c3c2c1y ϭ yh ϩ yp,yp ϭ 5x3e؊x.C ϭ 5.6C ϭ 30.C(6 Ϫ 18x ϩ 9x2Ϫ x3) ϩ 3C(6x Ϫ 6x2ϩ x3) ϩ 3C(3x2Ϫ x3) ϩ Cx3ϭ 30.e؊x118 CHAP. 3 Higher Order Linear ODEs–5505 xy10Fig. 74. y and (dashed) in Example 1ypMethod of Variation of ParametersThe method of variation of parameters (see Sec. 2.10) also extends to arbitrary order n.It gives a particular solution for the nonhomogeneous equation (1) (in standard formwith as the first term!) by the formula(7)on an open interval I on which the coefficients of (1) and are continuous. In (7) thefunctions form a basis of the homogeneous ODE (3), with Wronskian W, andis obtained from W by replacing the jth column of W by the columnThus, when this becomes identical with (2) in Sec. 2.10,W ϭ `y1 y2y1r y2r` , W1 ϭ `0 y21 y2r` ϭ Ϫy2, W2 ϭ `y1 0y1r 1` ϭ y1.n ϭ 2,[0 0 Á 0 1]T.Wj ( j ϭ 1, Á , n)y1, Á , ynr(x)ϭ y1(x)ΎW1(x)W(x)r(x) dx ϩ Á ϩ yn(x)ΎWn(x)W(x)r(x) dxyp(x) ϭ ankϭ1yk(x)ΎWk(x)W(x)r(x) dxy(n)ypc03.qxd 10/27/10 6:20 PM Page 118
• The proof of (7) uses an extension of the idea of the proof of (2) in Sec. 2.10 and canbe found in Ref [A11] listed in App. 1.E X A M P L E 2 Variation of Parameters. Nonhomogeneous Euler–Cauchy EquationSolve the nonhomogeneous Euler–Cauchy equationSolution. Step 1. General solution of the homogeneous ODE. Substitution of and the derivativesinto the homogeneous ODE and deletion of the factor givesThe roots are 1, 2, 3 and give as a basisHence the corresponding general solution of the homogeneous ODE isStep 2. Determinants needed in (7). These areStep 3. Integration. In (7) we also need the right side of our ODE in standard form, obtained by divisionof the given equation by the coefficient of thus, In (7) we have the simplequotients Hence (7) becomesSimplification gives Hence the answer isFigure 75 shows Can you explain the shape of this curve? Its behavior near The occurrence of a minimum?Its rapid increase? Why would the method of undetermined coefficients not have given the solution? ᭿x ϭ 0?yp.y ϭ yh ϩ yp ϭ c1x ϩ c2x2ϩ c3x3ϩ 16 x4(ln x Ϫ 116 ).yp ϭ 16 x4(ln x Ϫ 116 ).ϭx2ax33ln x Ϫx39b Ϫ x2ax22ln x Ϫx24b ϩx32(x ln x Ϫ x).yp ϭ x Ύx2x ln x dx Ϫ x2Ύx ln x dx ϩ x3Ύ12xx ln x dxW1>W ϭ x>2, W2>W ϭ Ϫ1, W3>W ϭ 1>(2x).r(x) ϭ (x4ln x)>x3ϭ x ln x.yt;x3r(x)W3 ϭ 4x x201 2x 00 2 14 ϭ x2.W2 ϭ 4x 0 x31 0 3x20 1 6x4 ϭ Ϫ2x3W1 ϭ 40 x2x30 2x 3x21 2 6x4 ϭ x4W ϭ 3x x2x31 2x 3x20 2 6x3 ϭ 2x3yh ϭ c1x ϩ c2x2ϩ c3x3.y1 ϭ x, y2 ϭ x2, y3 ϭ x3.m(m Ϫ 1)(m Ϫ 2) Ϫ 3m(m Ϫ 1) ϩ 6m Ϫ 6 ϭ 0.xmy ϭ xm(x Ͼ 0).x3yt Ϫ 3x2ys ϩ 6xyr Ϫ 6y ϭ x4ln xSEC. 3.3 Nonhomogeneous Linear ODEs 119c03.qxd 10/27/10 6:20 PM Page 119
• Application: Elastic BeamsWhereas second-order ODEs have various applications, of which we have discussed someof the more important ones, higher order ODEs have much fewer engineering applications.An important fourth-order ODE governs the bending of elastic beams, such as wooden oriron girders in a building or a bridge.A related application of vibration of beams does not fit in here since it leads to PDEsand will therefore be discussed in Sec. 12.3.E X A M P L E 3 Bending of an Elastic Beam under a LoadWe consider a beam B of length L and constant (e.g., rectangular) cross section and homogeneous elasticmaterial (e.g., steel); see Fig. 76. We assume that under its own weight the beam is bent so little that it ispractically straight. If we apply a load to B in a vertical plane through the axis of symmetry (the x-axis inFig. 76), B is bent. Its axis is curved into the so-called elastic curve C (or deflection curve). It is shown inelasticity theory that the bending moment is proportional to the curvature of C. We assume the bendingto be small, so that the deflection and its derivative (determining the tangent direction of C) are small.Then, by calculus, HenceEI is the constant of proportionality. E is Young’s modulus of elasticity of the material of the beam. I is themoment of inertia of the cross section about the (horizontal) z-axis in Fig. 76.Elasticity theory shows further that where is the load per unit length. Together,(8) EIyivϭ f(x).f(x)Ms(x) ϭ f(x),M(x) ϭ EIys(x).k ϭ ys>(1 ϩ yr2)3>2Ϸ ys.yr(x)y(x)k(x)M(x)120 CHAP. 3 Higher Order Linear ODEs–20205 xy3010–10100Fig. 75. Particular solution of the nonhomogeneousEuler–Cauchy equation in Example 2ypLUndeformed beamDeformed beamunder uniform load(simply supported)xzyxzyFig. 76. Elastic beamc03.qxd 10/27/10 6:20 PM Page 120
• In applications the most important supports and corresponding boundary conditions are as follows and shownin Fig. 77.(A) Simply supported at and L(B) Clamped at both ends at and L(C) Clamped at , free atThe boundary condition means no displacement at that point, means a horizontal tangent,means no bending moment, and means no shear force.Let us apply this to the uniformly loaded simply supported beam in Fig. 76. The load isThen (8) is(9)This can be solved simply by calculus. Two integrations givegives Then (since ). HenceIntegrating this twice, we obtainwith from ThenInserting the expression for k, we obtain as our solutionSince the boundary conditions at both ends are the same, we expect the deflection to be “symmetric” withrespect to that is, Verify this directly or set and show that y becomes aneven function of u,From this we can see that the maximum deflection in the middle at is Recallthat the positive direction points downward. ᭿5f0L4>(16 # 24EI).u ϭ 0(x ϭ L>2)y ϭf024EIau2Ϫ14L2b au2Ϫ54L2b.x ϭ u ϩ L>2y(x) ϭ y(L Ϫ x).L>2,y(x)y ϭf024EI(x4Ϫ 2Lx3ϩ L3x).y(L) ϭkL2aL312ϪL36ϩ c3b ϭ 0, c3 ϭL312.y(0) ϭ 0.c4 ϭ 0y ϭk2a112x4ϪL6x3ϩ c3x ϩ c4bys ϭk2(x2Ϫ Lx).L 0ys(L) ϭ L(12 kL ϩ c1) ϭ 0, c1 ϭ ϪkL>2c2 ϭ 0.ys(0) ϭ 0ys ϭk2x2ϩ c1x ϩ c2.yivϭ k, k ϭf0EI.f(x) ϵ f0 ϭ const.yt ϭ 0ys ϭ 0yr ϭ 0y ϭ 0y(0) ϭ yr(0) ϭ 0, ys(L) ϭ yt(L) ϭ 0.x ϭ Lx ϭ 0x ϭ 0y ϭ yr ϭ 0x ϭ 0y ϭ ys ϭ 0SEC. 3.3 Nonhomogeneous Linear ODEs 121xx = 0 x = Lx = 0 x = Lx = 0 x = L(A) Simply supported(B) Clamped at bothends(C) Clamped at the leftend, free at theright endFig. 77. Supports of a beamc03.qxd 10/27/10 6:20 PM Page 121
• 122 CHAP. 3 Higher Order Linear ODEs1–7 GENERAL SOLUTIONSolve the following ODEs, showing the details of yourwork.1.2.3.4.5.6.7.8–13 INITIAL VALUE PROBLEMSolve the given IVP, showing the details of your work.8.9.10.11.12.ys(0) ϭ 17.2yr(0) ϭ 8.8,y(0) ϭ 4.5,(D3Ϫ 2D2Ϫ 9D ϩ 18I)y ϭ e2x,ys(0) ϭ Ϫ5.2yr(0) ϭ 3.2,y(0) ϭ Ϫ1.4,(D3Ϫ 2D2Ϫ 3D)y ϭ 74e؊3xsin x,ys(1) ϭ 14yr(1) ϭ 3,y(1) ϭ 1,x3yt ϩ xyr Ϫ y ϭ x2,yt(0) ϭ Ϫ32ys(0) ϭ Ϫ1,yr(0) ϭ 2,y(0) ϭ 1,yivϩ 5ys ϩ 4y ϭ 90 sin 4x,yt(0) ϭ 0ys(0) ϭ 0,yr(0) ϭ 0,y(0) ϭ 1,yivϪ 5ys ϩ 4y ϭ 10e؊3x,(D3Ϫ 9D2ϩ 27D Ϫ 27I)y ϭ 27 sin 3x(D3ϩ 4D)y ϭ sin x(x3D3ϩ x2D2Ϫ 2xD ϩ 2I)y ϭ x؊2(D3ϩ 3D2Ϫ 5D Ϫ 39I)y ϭ Ϫ300 cos x(D4ϩ 10D2ϩ 9I)y ϭ 6.5 sinh 2xyt ϩ 2ys Ϫ yr Ϫ 2y ϭ 1 Ϫ 4x3yt ϩ 3ys ϩ 3yr ϩ y ϭ exϪ x Ϫ 1P R O B L E M S E T 3 . 313.14. CAS EXPERIMENT. Undetermined Coefficients.Since variation of parameters is generally complicated,it seems worthwhile to try to extend the other method.Find out experimentally for what ODEs this is possibleand for what not. Hint: Work backward, solving ODEswith a CAS and then looking whether the solutioncould be obtained by undetermined coefficients. Forexample, considerand15. WRITING REPORT. Comparison of Methods. Writea report on the method of undetermined coefficients andthe method of variation of parameters, discussing andcomparing the advantages and disadvantages of eachmethod. Illustrate your findings with typical examples.Try to show that the method of undetermined coefficients,say, for a third-order ODE with constant coefficients andan exponential function on the right, can be derived fromthe method of variation of parameters.x3yt ϩ x2ys Ϫ 2xyr ϩ 2y ϭ x3ln x.yt Ϫ 3ys ϩ 3yr Ϫ y ϭ x1>2exys(0) ϭ Ϫ1yr(0) ϭ Ϫ2,y(0) ϭ 3,(D3Ϫ 4D)y ϭ 10 cos x ϩ 5 sin x,1. What is the superposition or linearity principle? Forwhat nth-order ODEs does it hold?2. List some other basic theorems that extend fromsecond-order to nth-order ODEs.3. If you know a general solution of a homogeneous linearODE, what do you need to obtain from it a generalsolution of a corresponding nonhomogeneous linearODE?4. What form does an initial value problem for an nth-order linear ODE have?5. What is the Wronskian? What is it used for?6–15 GENERAL SOLUTIONSolve the given ODE. Show the details of your work.6.7.8.9.10. x2yt ϩ 3xys Ϫ 2yr ϭ 0(D4Ϫ 16I)y ϭ Ϫ15 cosh xyt Ϫ 4ys Ϫ yr ϩ 4y ϭ 30e2xyt ϩ 4ys ϩ 13yr ϭ 0yivϪ 3ys Ϫ 4y ϭ 0C H A P T E R 3 R E V I E W Q U E S T I O N S A N D P R O B L E M S11.12.13.14.15.16–20 INITIAL VALUE PROBLEMSolve the IVP. Show the details of your work.16.17.18.19.20.ys(0) ϭ 5yr(0) ϭ Ϫ3,y(0) ϭ Ϫ1,(D3ϩ 3D2ϩ 3D ϩ I)y ϭ 8 sin x,D2y(0) ϭ 189Dy(0) ϭ Ϫ41,y(0) ϭ 9,(D3ϩ 9D2ϩ 23D ϩ 15I)y ϭ 12exp(Ϫ4x),D3y(0) ϭ Ϫ130D2y(0) ϭ 34,Dy(0) ϭ Ϫ6,y(0) ϭ 12.16,(D4Ϫ 26D2ϩ 25I)y ϭ 50(x ϩ 1)2,ys ϭ Ϫ24yr(0) ϭ Ϫ3.95,y(0) ϭ 1.94,yt ϩ 5ys ϩ 24yr ϩ 20y ϭ x,D2y(0) ϭ 0Dy(0) ϭ 1,y(0) ϭ 0,(D3Ϫ D2Ϫ D ϩ I)y ϭ 0,4x3yt ϩ 3xyr Ϫ 3y ϭ 10(D4Ϫ 13D2ϩ 36I)y ϭ 12ex(D3ϩ 6D2ϩ 12D ϩ 8I)y ϭ 8x2(D3Ϫ D)y ϭ sinh 0.8xyt ϩ 4.5ys ϩ 6.75yr ϩ 3.375y ϭ 0c03.qxd 10/27/10 6:20 PM Page 122
• Summary of Chapter 3 123Compare with the similar Summary of Chap. 2 (the case ).Chapter 3 extends Chap. 2 from order to arbitrary order n. An nth-orderlinear ODE is an ODE that can be written(1)with as the first term; we again call this the standard form. Equation(1) is called homogeneous if on a given open interval I considered,nonhomogeneous if on I. For the homogeneous ODE(2)the superposition principle (Sec. 3.1) holds, just as in the case A basis orfundamental system of solutions of (2) on I consists of n linearly independentsolutions of (2) on I. A general solution of (2) on I is a linear combinationof these,(3) ( arbitrary constants).A general solution of the nonhomogeneous ODE (1) on I is of the form(4) (Sec. 3.3).Here, is a particular solution of (1) and is obtained by two methods (undeterminedcoefficients or variation of parameters) explained in Sec. 3.3.An initial value problem for (1) or (2) consists of one of these ODEs and ninitial conditions (Secs. 3.1, 3.3)(5)with given in I and given If are continuous on I,then general solutions of (1) and (2) on I exist, and initial value problems (1), (5)or (2), (5) have a unique solution.p0, Á , pn؊1, rK0, Á , Kn؊1.x0y(x0) ϭ K0, yr(x0) ϭ K1, Á , y(n؊1)(x0) ϭ Kn؊1ypy ϭ yh ϩ ypc1, Á , cny ϭ c1y1 ϩ Á ϩ cnyny1, Á , ynn ϭ 2.y(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ 0r(x) [ 0r(x) ϵ 0y(n)ϭ dny>dxny(n)ϩ pn؊1(x)y(n؊1)ϩ Á ϩ p1(x)yr ϩ p0(x)y ϭ r(x)n ϭ 2n ‫؍‬ 2SUMMARY OF CHAPTER 3Higher Order Linear ODEsc03.qxd 10/27/10 6:20 PM Page 123
• 124C H A P T E R 4Systems of ODEs. Phase Plane.Qualitative MethodsTying in with Chap. 3, we present another method of solving higher order ODEs inSec. 4.1. This converts any nth-order ODE into a system of n first-order ODEs. We alsoshow some applications. Moreover, in the same section we solve systems of first-orderODEs that occur directly in applications, that is, not derived from an nth-order ODE butdictated by the application such as two tanks in mixing problems and two circuits inelectrical networks. (The elementary aspects of vectors and matrices needed in this chapterare reviewed in Sec. 4.0 and are probably familiar to most students.)In Sec. 4.3 we introduce a totally different way of looking at systems of ODEs. Themethod consists of examining the general behavior of whole families of solutions of ODEsin the phase plane, and aptly is called the phase plane method. It gives information on thestability of solutions. (Stability of a physical system is desirable and means roughly that asmall change at some instant causes only a small change in the behavior of the system atlater times.) This approach to systems of ODEs is a qualitative method because it dependsonly on the nature of the ODEs and does not require the actual solutions. This can be veryuseful because it is often difficult or even impossible to solve systems of ODEs. In contrast,the approach of actually solving a system is known as a quantitative method.The phase plane method has many applications in control theory, circuit theory,population dynamics and so on. Its use in linear systems is discussed in Secs. 4.3, 4.4,and 4.6 and its even more important use in nonlinear systems is discussed in Sec. 4.5 withapplications to the pendulum equation and the Lokta–Volterra population model. Thechapter closes with a discussion of nonhomogeneous linear systems of ODEs.NOTATION. We continue to denote unknown functions by y; thus, —analogous to Chaps. 1–3. (Note that some authors use x for functions, whendealing with systems of ODEs.)Prerequisite: Chap. 2.References and Answers to Problems: App. 1 Part A, and App. 2.4.0 For Reference:Basics of Matrices and VectorsFor clarity and simplicity of notation, we use matrices and vectors in our discussionof linear systems of ODEs. We need only a few elementary facts (and not the bulk ofthe material of Chaps. 7 and 8). Most students will very likely be already familiarx1(t), x2(t)y1(t), y2(t)c04.qxd 10/27/10 9:32 PM Page 124
• with these facts. Thus this section is for reference only. Begin with Sec. 4.1 and consult4.0 as needed.Most of our linear systems will consist of two linear ODEs in two unknown functions,(1)(perhaps with additional given functions on the right in the two ODEs).Similarly, a linear system of n first-order ODEs in n unknown functionsis of the form(2)(perhaps with an additional given function on the right in each ODE).Some Definitions and TermsMatrices. In (1) the (constant or variable) coefficients form a 2 2 matrix A, that is,an array(3) , for example, .Similarly, the coefficients in (2) form an n n matrix(4)The are called entries, the horizontal lines rows, and the vertical lines columns.Thus, in (3) the first row is , the second row is , and the first andsecond columns areand .In the “double subscript notation” for entries, the first subscript denotes the row and thesecond the column in which the entry stands. Similarly in (4). The main diagonal is thediagonal in (4), hence in (3).a22a11a11 a22Á annca12a22dca11a21d[a21 a22][a11 a12]a11, a12, ÁA ϭ [ajk] ϭ Ea11 a12Á a1na21 a22Á a2n# # Á #an1 an2Á annU .؋A ϭ cϪ5 213 12dA ϭ [ajk] ϭ ca11 a12a21 a22d؋yr1 ϭ a11y1 ϩ a12y2 ϩ Á ϩ a1nynyr2 ϭ a21y1 ϩ a22y2 ϩ Á ϩ a2nyn. . . . . . . . . . . . . . . . . . . . . . . . . . . . .yrn ϭ an1y1 ϩ an2y2 ϩ Á ϩ annyny1(t), Á , yn(t)g1(t), g2(t)yr1 ϭ a11y1 ϩ a12y2, yr1 ϭ Ϫ5y1 ϩ 2y2for example,yr2 ϭ a21y1 ϩ a22y2, yr2 ϭ 13y1 ϩ 12 y2y1(t), y2(t)SEC. 4.0 For Reference: Basics of Matrices and Vectors 125c04.qxd 10/27/10 9:32 PM Page 125
• We shall need only square matrices, that is, matrices with the same number of rowsand columns, as in (3) and (4).Vectors. A column vector x with n components is of the formthus if .Similarly, a row vector v is of the form, thus if , then .Calculations with Matrices and VectorsEquality. Two n n matrices are equal if and only if corresponding entries are equal.Thus for , letand .Then A B if and only if.Two column vectors (or two row vectors) are equal if and only if they both have ncomponents and corresponding components are equal. Thus, let. Then if and only ifAddition is performed by adding corresponding entries (or components); here, matricesmust both be n n, and vectors must both have the same number of components. Thusfor ,(5) .Scalar multiplication (multiplication by a number c) is performed by multiplying eachentry (or component) by c. For example, ifA ϭ c9 3Ϫ2 0d, then Ϫ7A ϭ cϪ63 Ϫ2114 0d.A ϩ B ϭ ca11 ϩ b11 a12 ϩ b12a21 ϩ b21 a22 ϩ b22d, v ϩ x ϭ cv1 ϩ x1v2 ϩ x2dn ϭ 2ϫv1 ϭ x1v2 ϭ x2.v ϭ xv ϭ cv1v2d and x ϭ cx1x2da21 ϭ b21, a22 ϭ b22a11 ϭ b11, a12 ϭ b12ϭB ϭ cb11 b12b21 b22dA ϭ ca11 a12a21 a22dn ϭ 2ϫv ϭ [v1 v2]n ϭ 2v ϭ [v1Á vn]x ϭ cx1x2dn ϭ 2,x ϭ Ex1x2oxnU ,x1, Á , xn126 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 126
• If.Matrix Multiplication. The product (in this order) of two n n matricesis the n n matrix with entries(6)that is, multiply each entry in the jth row of A by the corresponding entry in the kth columnof B and then add these n products. One says briefly that this is a “multiplication of rowsinto columns.” For example,CAUTION! Matrix multiplication is not commutative, in general. In ourexample,Multiplication of an n n matrix A by a vector x with n components is defined by thesame rule: is the vector with the n components.For example,Systems of ODEs as Vector EquationsDifferentiation. The derivative of a matrix (or vector) with variable entries (orcomponents) is obtained by differentiating each entry (or component). Thus, if.y(t) ϭ cy1(t)y2(t)d ϭ ce؊2tsin td, then yr(t) ϭ cyr1(t)yr2(t)d ϭ cϪ2e؊2tcos tdc12 7Ϫ8 3d cx1x2d ϭ c12x1 ϩ 7x2Ϫ8x1 ϩ 3x2d.j ϭ 1, Á , nvj ϭ anmϭ1ajmxmv ϭ Axϫϭ c17 38 6d.c1 Ϫ42 5d c9 3Ϫ2 0d ϭ c1 ؒ 9 ϩ (Ϫ4) ؒ (Ϫ2) 1 ؒ 3 ϩ (Ϫ4) ؒ 02 ؒ 9 ϩ 5 ؒ (Ϫ2) 2 ؒ 3 ϩ 5 ؒ 0dAB BAϭ c15 Ϫ21Ϫ2 8d.c9 3Ϫ2 0d c1 Ϫ42 5d ϭ c9 ؒ 1 ϩ 3 ؒ 2 9 ؒ (Ϫ4) ϩ 3 ؒ 5Ϫ2 ؒ 1 ϩ 0 ؒ 2 (Ϫ2) ؒ (Ϫ4) ϩ 0 ؒ 5d,j ϭ 1, Á , nk ϭ 1, Á , n,cjk ϭ anmϭ1ajmbmkC ϭ [cjk]ϫA ϭ [ajk] and B ϭ [bjk]ϫC ϭ ABv ϭ c0.4Ϫ13d, then 10v ϭ c4Ϫ130dSEC. 4.0 For Reference: Basics of Matrices and Vectors 127c04.qxd 10/27/10 9:32 PM Page 127
• Using matrix multiplication and differentiation, we can now write (1) as(7) .Similarly for (2) by means of an n n matrix A and a column vector y with n components,namely, . The vector equation (7) is equivalent to two equations for thecomponents, and these are precisely the two ODEs in (1).Some Further Operations and TermsTransposition is the operation of writing columns as rows and conversely and is indicatedby T. Thus the transpose of the 2 2 matrixis .The transpose of a column vector, say,, is a row vector, ,and conversely.Inverse of a Matrix. The n n unit matrix I is the n n matrix with main diagonaland all other entries zero. If, for a given n n matrix A, there is an n nmatrix B such that , then A is called nonsingular and B is called the inverseof A and is denoted by ; thus(8) .The inverse exists if the determinant det A of A is not zero.If A has no inverse, it is called singular. For ,(9)where the determinant of A is(10) .(For general n, see Sec. 7.7, but this will not be needed in this chapter.)Linear Independence. r given vectors with n components are called alinearly independent set or, more briefly, linearly independent, if(11) c1v(1)ϩ Á ϩ crv(r)ϭ 0v(1), Á , v(r)det A ϭ 2a11 a12a21 a222 ϭ a11a22 Ϫ a12a21A؊1ϭ1det Aca22 Ϫa12Ϫa21 a11d,n ϭ 2AA؊1ϭ A؊1A ϭ IA؊1AB ϭ BA ϭ Iϫϫ1, 1, Á , 1ϫϫvTϭ [v1 v2]v ϭ cv1v2dATϭ ca11 a21a12 a22d ϭ cϪ5 132 12dA ϭ ca11 a12a21 a22d ϭ cϪ5 213 12dϫATyr ϭ Ayϫyr ϭ cyr1yr2d ϭ Ay ϭ ca11 a12a21 a22d cy1y2d, e.g., yr ϭ cϪ5 213 12d cy1y2d128 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 128
• implies that all scalars must be zero; here, 0 denotes the zero vector, whose ncomponents are all zero. If (11) also holds for scalars not all zero (so that at least one ofthese scalars is not zero), then these vectors are called a linearly dependent set or, briefly,linearly dependent, because then at least one of them can be expressed as a linearcombination of the others; that is, if, for instance, in (11), then we can obtainEigenvalues, EigenvectorsEigenvalues and eigenvectors will be very important in this chapter (and, as a matter offact, throughout mathematics).Let be an n n matrix. Consider the equation(12)where is a scalar (a real or complex number) to be determined and x is a vector to bedetermined. Now, for every , a solution is . A scalar such that (12) holds forsome vector is called an eigenvalue of A, and this vector is called an eigenvectorof A corresponding to this eigenvalue .We can write (12) as or(13) .These are n linear algebraic equations in the n unknowns (the componentsof x). For these equations to have a solution , the determinant of the coefficientmatrix must be zero. This is proved as a basic fact in linear algebra (Theorem 4in Sec. 7.7). In this chapter we need this only for . Then (13) is(14) ;in components,Now is singular if and only if its determinant , called the characteristicdeterminant of A (also for general n), is zero. This gives(15)ϭ l2Ϫ (a11 ϩ a22)l ϩ a11a22 Ϫ a12a21 ϭ 0.ϭ (a11 Ϫ l)(a22 Ϫ l) Ϫ a12a21det (A Ϫ lI) ϭ 2a11 Ϫ l a12a21 a22 Ϫ l2det (A Ϫ lI)A Ϫ lIa21 x1 ϩ (a22 Ϫ l)x2 ϭ 0.(a11 Ϫ l)x1 ϩ a12 x2 ϭ 0(14*)ca11 Ϫ l a12a21 a22 Ϫ ld cx1x2d ϭ c00dn ϭ 2A Ϫ lIx 0x1, Á , xn(A Ϫ lI)x ϭ 0Ax Ϫ lx ϭ 0lx 0lx ϭ 0llAx ϭ lxϫA ϭ [ajk]v(1)ϭ Ϫ1c1(c2v(2)ϩ Á ϩ crv(r)).c1 0c1, Á , crSEC. 4.0 For Reference: Basics of Matrices and Vectors 129c04.qxd 10/27/10 9:32 PM Page 129
• This quadratic equation in is called the characteristic equation of A. Its solutions arethe eigenvalues of A. First determine these. Then use with todetermine an eigenvector of A corresponding to . Finally use withto find an eigenvector of A corresponding to . Note that if x is an eigenvector ofA, so is kx with any .E X A M P L E 1 Eigenvalue ProblemFind the eigenvalues and eigenvectors of the matrix(16)Solution. The characteristic equation is the quadratic equation.It has the solutions . These are the eigenvalues of A.Eigenvectors are obtained from . For we have fromA solution of the first equation is . This also satisfies the second equation. (Why?) Hence aneigenvector of A corresponding to is(17) . Similarly,is an eigenvector of A corresponding to , as obtained from with . Verify this.4.1 Systems of ODEs as Modelsin Engineering ApplicationsWe show how systems of ODEs are of practical importance as follows. We first illustratehow systems of ODEs can serve as models in various applications. Then we show how ahigher order ODE (with the highest derivative standing alone on one side) can be reducedto a first-order system.E X A M P L E 1 Mixing Problem Involving Two TanksA mixing problem involving a single tank is modeled by a single ODE, and you may first review thecorresponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two tanks. Themodel will be a system of two first-order ODEs.Tank and in Fig. 78 contain initially 100 gal of water each. In the water is pure, whereas 150 lb offertilizer are dissolved in . By circulating liquid at a rate of and stirring (to keep the mixture uniform)the amounts of fertilizer in and in change with time t. How long should we let the liquid circulateso that will contain at least half as much fertilizer as there will be left in ?T2T1T2y2(t)T1y1(t)2 gal>minT2T1T2T1᭿l ϭ l2(14*)l2 ϭ Ϫ0.8x(2)ϭ c10.8dx(1)ϭ c21dl1 ϭ Ϫ2.0x1 ϭ 2, x2 ϭ 1Ϫ1.6x1 ϩ (1.2 ϩ 2.0)x2 ϭ 0.(Ϫ4.0 ϩ 2.0)x1 ϩ 4.0x2 ϭ 0(14*)l ϭ l1 ϭ Ϫ2(14*)l1 ϭ Ϫ2 and l2 ϭ Ϫ0.8det ƒ A Ϫ lIƒ ϭ 2Ϫ4 Ϫ l 4Ϫ1.6 1.2 Ϫ l2 ϭ l2ϩ 2.8l ϩ 1.6 ϭ 0A ϭ cϪ4.0 4.0Ϫ1.6 1.2dk 0l2x(2)l ϭ l2(14*)l1x(1)l ϭ l1(14*)l1 and l2l130 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 130
• Solution. Step 1. Setting up the model. As for a single tank, the time rate of change of equalsinflow minus outflow. Similarly for tank . From Fig. 78 we see that(Tank )(Tank ).Hence the mathematical model of our mixture problem is the system of first-order ODEs(Tank )(Tank ).As a vector equation with column vector and matrix A this becomes.Step 2. General solution. As for a single equation, we try an exponential function of t,(1) .Dividing the last equation by and interchanging the left and right sides, we obtain.We need nontrivial solutions (solutions that are not identically zero). Hence we have to look for eigenvaluesand eigenvectors of A. The eigenvalues are the solutions of the characteristic equation(2) .We see that (which can very well happen—don’t get mixed up—it is eigenvectors that must not be zero)and . Eigenvectors are obtained from in Sec. 4.0 with and . For our presentA this gives [we need only the first equation in ]and ,(Ϫ0.02 ϩ 0.04)x1 ϩ 0.02x2 ϭ 0Ϫ0.02x1 ϩ 0.02x2 ϭ 0(14*)l ϭ Ϫ0.04l ϭ 0(14*)l2 ϭ Ϫ0.04l1 ϭ 0det (A Ϫ lI) ϭ 2Ϫ0.02 Ϫ l 0.020.02 Ϫ0.02 Ϫ l2 ϭ (Ϫ0.02 Ϫ l)2Ϫ 0.022ϭ l(l ϩ 0.04) ϭ 0Ax ϭ lxeltlxeltϭ Axelty ϭ xelt. Then yr ϭ lxeltϭ Axeltyr ϭ Ay, where A ϭ cϪ0.02 0.020.02 Ϫ0.02dy ϭ cy1y2dT2yr2 ϭ 0.02y1 Ϫ 0.02y2T1yr1 ϭ Ϫ0.02y1 ϩ 0.02y2T2yr2 ϭ Inflow>min Ϫ Outflow>min ϭ2100y1 Ϫ2100y2T1yr1 ϭ Inflow>min Ϫ Outflow>min ϭ2100y2 Ϫ2100y1T2y1(t)yr1(t)SEC. 4.1 Systems of ODEs as Models in Engineering Applications 131T15001005027.50System of tanks ty(t)y1(t)y2(t)10075150T22 gal/min2 gal/minFig. 78. Fertilizer content in Tanks (lower curve) and T2T1c04.qxd 10/27/10 9:32 PM Page 131
• respectively. Hence and , respectively, and we can take and .This gives two eigenvectors corresponding to and , respectively, namely,and .From (1) and the superposition principle (which continues to hold for systems of homogeneous linear ODEs)we thus obtain a solution(3)where are arbitrary constants. Later we shall call this a general solution.Step 3. Use of initial conditions. The initial conditions are (no fertilizer in tank ) and .From this and (3) with we obtainIn components this is . The solution is . This gives the answer.In components,(Tank , lower curve)(Tank , upper curve).Figure 78 shows the exponential increase of and the exponential decrease of to the common limit 75 lb.Did you expect this for physical reasons? Can you physically explain why the curves look “symmetric”? Wouldthe limit change if initially contained 100 lb of fertilizer and contained 50 lb?Step 4. Answer. contains half the fertilizer amount of if it contains of the total amount, that is,50 lb. Thus.Hence the fluid should circulate for at least about half an hour.E X A M P L E 2 Electrical NetworkFind the currents and in the network in Fig. 79. Assume all currents and charges to be zero at ,the instant when the switch is closed.t ϭ 0I2(t)I1(t)᭿y1 ϭ 75 Ϫ 75e؊0.04tϭ 50, e؊0.04tϭ 13 , t ϭ (ln 3)>0.04 ϭ 27.51>3T2T1T2T1y2y1T2y2 ϭ 75 ϩ 75e؊0.04tT1y1 ϭ 75 Ϫ 75e؊0.04ty ϭ 75x(1)Ϫ 75x(2)e؊0.04tϭ 75c11d Ϫ 75c1Ϫ1d e؊0.04tc1 ϭ 75, c2 ϭ Ϫ75c1 ϩ c2 ϭ 0, c1 Ϫ c2 ϭ 150y(0) ϭ c1 c11d ϩ c2 c1Ϫ1d ϭ cc1 ϩ c2c1 Ϫ c2d ϭ c0150d.t ϭ 0y2(0) ϭ 150T1y1(0) ϭ 0c1 and c2y ϭ c1x(1)el1tϩ c2x(2)el2tϭ c1 c11d ϩ c2 c1Ϫ1de؊0.04tx(2)ϭ c1Ϫ1dx(1)ϭ c11dl2 ϭ Ϫ0.04l1 ϭ 0x1 ϭ Ϫx2 ϭ 1x1 ϭ x2 ϭ 1x1 ϭ Ϫx2x1 ϭ x2132 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative MethodsSolution. Step 1. Setting up the mathematical model. The model of this network is obtained fromKirchhoff’s Voltage Law, as in Sec. 2.9 (where we considered single circuits). Let and be the currentsI2(t)I1(t)Switcht = 0E = 12 voltsL = 1 henry C = 0.25 faradR1= 4 ohmsR2= 6 ohmsI1I1I1I2I2I2Fig. 79. Electrical network in Example 2c04.qxd 10/27/10 9:32 PM Page 132
• in the left and right loops, respectively. In the left loop, the voltage drops are over the inductorand over the resistor, the difference because and flow through the resistor inopposite directions. By Kirchhoff’s Voltage Law the sum of these drops equals the voltage of the battery; thatis, , hence(4a) .In the right loop, the voltage drops are and over the resistors andover the capacitor, and their sum is zero,or .Division by 10 and differentiation gives .To simplify the solution process, we first get rid of , which by (4a) equals .Substitution into the present ODE givesand by simplification(4b) .In matrix form, (4) is (we write J since I is the unit matrix)(5) , where .Step 2. Solving (5). Because of the vector g this is a nonhomogeneous system, and we try to proceed as for asingle ODE, solving first the homogeneous system (thus ) by substituting . Thisgives, hence .Hence, to obtain a nontrivial solution, we again need the eigenvalues and eigenvectors. For the present matrixA they are derived in Example 1 in Sec. 4.0:, ; ,Hence a “general solution” of the homogeneous system is.For a particular solution of the nonhomogeneous system (5), since g is constant, we try a constant columnvector with components . Then , and substitution into (5) gives ; in components,The solution is ; thus . Hence(6) ;in components,I2 ϭ c1e؊2tϩ 0.8c2e؊0.8t.I1 ϭ 2c1e؊2tϩ c2e؊0.8tϩ 3J ϭ Jh ϩ Jp ϭ c1x(1)e؊2tϩ c2x(2)e؊0.8tϩ aa ϭ c30da1 ϭ 3, a2 ϭ 0Ϫ1.6a1 ϩ 1.2a2 ϩ 4.8 ϭ 0.Ϫ4.0a1 ϩ 4.0a2 ϩ 12.0 ϭ 0Aa ϩ g ϭ 0Jrp ϭ 0a1, a2Jp ϭ aJh ϭ c1x(1)e؊2tϩ c2x(2)e؊0.8tx(2)ϭ c10.8d.l2 ϭ Ϫ0.8x(1)ϭ c21dl1 ϭ Ϫ2Ax ϭ lxJr ϭ lxeltϭ AxeltJ ϭ xeltJr Ϫ AJ ϭ 0Jr ϭ AJJ ϭ cI1I2d, A ϭ cϪ4.0 4.0Ϫ1.6 1.2d, g ϭ c12.04.8dJr ϭ AJ ϩ gIr2 ϭ Ϫ1.6I1 ϩ 1.2I2 ϩ 4.8Ir2 ϭ 0.4Ir1 Ϫ 0.4I2 ϭ 0.4(Ϫ4I1 ϩ 4I2 ϩ 12) Ϫ 0.4I20.4(Ϫ4I1 ϩ 4I2 ϩ 12)0.4Ir1Ir2 Ϫ 0.4Ir1 ϩ 0.4I2 ϭ 010I2 Ϫ 4I1 ϩ 4Ύ I2 dt ϭ 06I2 ϩ 4(I2 Ϫ I1) ϩ 4Ύ I2 dt ϭ 0(I>C)͐ I2 dt ϭ 4͐ I2 dt [V]R1(I2 Ϫ I1) ϭ 4(I2 Ϫ I1) [V]R2I2 ϭ 6I2 [V]Ir1 ϭ Ϫ4I1 ϩ 4I2 ϩ 12Ir1 ϩ 4(I1 Ϫ I2) ϭ 12I2I1R1(I1 Ϫ I2) ϭ 4(I1 Ϫ I2) [V]LIr1 ϭ Ir1 [V]SEC. 4.1 Systems of ODEs as Models in Engineering Applications 133c04.qxd 10/27/10 9:32 PM Page 133
• The initial conditions giveHence and . As the solution of our problem we thus obtain(7)In components (Fig. 80b),Now comes an important idea, on which we shall elaborate further, beginning in Sec. 4.3. Figure 80a showsand as two separate curves. Figure 80b shows these two currents as a single curve in the-plane. This is a parametric representation with time t as the parameter. It is often important to know inwhich sense such a curve is traced. This can be indicated by an arrow in the sense of increasing t, as is shown.The -plane is called the phase plane of our system (5), and the curve in Fig. 80b is called a trajectory. Weshall see that such “phase plane representations” are far more important than graphs as in Fig. 80a becausethey will give a much better qualitative overall impression of the general behavior of whole families of solutions,not merely of one solution as in the present case. ᭿I1I2I1I2[I1(t), I2(t)]I2(t)I1(t)I2 ϭ Ϫ4e؊2tϩ 4e؊0.8t.I1 ϭ Ϫ8e؊2tϩ 5e؊0.8tϩ 3J ϭ Ϫ4x(1)e؊2tϩ 5x(2)e؊0.8tϩ a.c2 ϭ 5c1 ϭ Ϫ4I2(0) ϭ c1 ϩ 0.8c2 ϭ 0.I1(0) ϭ 2c1 ϩ c2 ϩ 3 ϭ 0134 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods0.5015432101.5I1I210254321034tI(t)(a) Currents I1(upper curve)and I2(b) Trajectory [I1(t), I2(t)]in the I1I2-plane(the “phase plane”)I1(t)I2(t)Fig. 80. Currents in Example 2Remark. In both examples, by growing the dimension of the problem (from one tank totwo tanks or one circuit to two circuits) we also increased the number of ODEs (from oneODE to two ODEs). This “growth” in the problem being reflected by an “increase” in themathematical model is attractive and affirms the quality of our mathematical modeling andtheory.Conversion of an nth-Order ODE to a SystemWe show that an nth-order ODE of the general form (8) (see Theorem 1) can be convertedto a system of n first-order ODEs. This is practically and theoretically important—practically because it permits the study and solution of single ODEs by methods forsystems, and theoretically because it opens a way of including the theory of higher orderODEs into that of first-order systems. This conversion is another reason for the importanceof systems, in addition to their use as models in various basic applications. The idea ofthe conversion is simple and straightforward, as follows.c04.qxd 10/27/10 9:32 PM Page 134
• T H E O R E M 1 Conversion of an ODEAn nth-order ODE(8)can be converted to a system of n first-order ODEs by setting(9) .This system is of the form(10) .P R O O F The first of these n ODEs follows immediately from (9) by differentiation. Also,by (9), so that the last equation in (10) results from the given ODE (8).E X A M P L E 3 Mass on a SpringTo gain confidence in the conversion method, let us apply it to an old friend of ours, modeling the free motionsof a mass on a spring (see Sec. 2.4)For this ODE (8) the system (10) is linear and homogeneous,Setting , we get in matrix formThe characteristic equation isdet (A Ϫ lI) ϭ 4Ϫl 1ϪkmϪcmϪ l4 ϭ l2ϩcml ϩkmϭ 0.yr ϭ Ay ϭ D0 1ϪkmϪcmT cy1y2d.y ϭ cy1y2dyr2 ϭ Ϫkmy1 Ϫcmy2.yr1 ϭ y2mys ϩ cyr ϩ ky ϭ 0 or ys ϭ Ϫcmyr Ϫkmy.᭿yrn ϭ y(n)n Ϫ 1yr1 ϭ y2yr2 ϭ y3oyrn؊1 ϭ ynyrn ϭ F(t, y1, y2, Á , yn).y1 ϭ y, y2 ϭ yr, y3 ϭ ys, Á , yn ϭ y(n؊1)y(n)ϭ F(t, y, yr, Á , y(n؊1))SEC. 4.1 Systems of ODEs as Models in Engineering Applications 135c04.qxd 10/27/10 9:32 PM Page 135
• 136 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods15. CAS EXPERIMENT. Electrical Network. (a) InExample 2 choose a sequence of values of C thatincreases beyond bound, and compare the correspondingsequences of eigenvalues of A. What limits of thesesequences do your numeric values (approximately)suggest?(b) Find these limits analytically.(c) Explain your result physically.(d) Below what value (approximately) must you decreaseC to get vibrations?k1= 3k2= 2 (Net change inspring length= y2– y1)System inmotionSystem instaticequilibriumm1= 1(y1= 0)(y2= 0) m2= 1y1y2y2y1Fig. 81. Mechanical system in Team Project1–6 MIXING PROBLEMS1. Find out, without calculation, whether doubling theflow rate in Example 1 has the same effect as halfingthe tank sizes. (Give a reason.)2. What happens in Example 1 if we replace by a tankcontaining 200 gal of water and 150 lb of fertilizerdissolved in it?3. Derive the eigenvectors in Example 1 without consultingthis book.4. In Example 1 find a “general solution” for any ratio, tank sizes being equal.Comment on the result.5. If you extend Example 1 by a tank of the same sizeas the others and connected to by two tubes withflow rates as between and , what system of ODEswill you get?6. Find a “general solution” of the system in Prob. 5.7–9 ELECTRICAL NETWORKIn Example 2 find the currents:7. If the initial currents are 0 A and A (minus meaningthat flows against the direction of the arrow).8. If the capacitance is changed to . (Generalsolution only.)9. If the initial currents in Example 2 are 28 A and 14 A.10–13 CONVERSION TO SYSTEMSFind a general solution of the given ODE (a) by first convertingit to a system, (b), as given. Show the details of your work.10. 11.12.13. ys ϩ 2yr Ϫ 24y ϭ 0yt ϩ 2ys Ϫ yr Ϫ 2y ϭ 04ys Ϫ 15yr Ϫ 4y ϭ 0ys ϩ 3yr ϩ 2y ϭ 0C ϭ 5>27 FI2(0)Ϫ3T2T1T2T3a ϭ (flow rate)>(tank size)T114. TEAM PROJECT. Two Masses on Springs. (a) Setup the model for the (undamped) system in Fig. 81.(b) Solve the system of ODEs obtained. Hint. Tryand set . Proceed as in Example 1 or2. (c) Describe the influence of initial conditions on thepossible kind of motions.v2ϭ ly ϭ xevtP R O B L E M S E T 4 . 1It agrees with that in Sec. 2.4. For an illustrative computation, let , and . ThenThis gives the eigenvalues and . Eigenvectors follow from the first equation inwhich is . For this gives , say, , . For it gives, say, , . These eigenvectorsgiveThis vector solution has the first componentwhich is the expected solution. The second component is its derivative᭿y2 ϭ yr1 ϭ yr ϭ Ϫc1e؊0.5tϪ 1.5c2e؊1.5t.y ϭ y1 ϭ 2c1e؊0.5tϩ c2e؊1.5ty ϭ c1 c2Ϫ1d e؊0.5tϩ c2 c1Ϫ1.5d e؊1.5t.x(1)ϭ c2Ϫ1d, x(2)ϭ c1Ϫ1.5dx2 ϭ Ϫ1.5x1 ϭ 11.5x1 ϩ x2 ϭ 0l2 ϭ Ϫ1.5x2 ϭ Ϫ1x1 ϭ 20.5x1 ϩ x2 ϭ 0l1Ϫlx1 ϩ x2 ϭ 0A Ϫ lI ϭ 0,l2 ϭ Ϫ1.5l1 ϭ Ϫ0.5l2ϩ 2l ϩ 0.75 ϭ (l ϩ 0.5)(l ϩ 1.5) ϭ 0.k ϭ 0.75m ϭ 1,c ϭ 2c04.qxd 10/27/10 9:32 PM Page 136
• 4.2 Basic Theory of Systems of ODEs.WronskianIn this section we discuss some basic concepts and facts about system of ODEs that arequite similar to those for single ODEs.The first-order systems in the last section were special cases of the more general system(1)We can write the system (1) as a vector equation by introducing the column vectorsand (where means transposition and saves usthe space that would be needed for writing y and f as columns). This gives(1)This system (1) includes almost all cases of practical interest. For it becomesor, simply, , well known to us from Chap. 1.A solution of (1) on some interval is a set of n differentiable functionson that satisfy (1) throughout this interval. In vector from, introducing the“solution vector” (a column vector!) we can writeAn initial value problem for (1) consists of (1) and n given initial conditions(2)in vector form, , where is a specified value of t in the interval considered andthe components of are given numbers. Sufficient conditions for theexistence and uniqueness of a solution of an initial value problem (1), (2) are stated inthe following theorem, which extends the theorems in Sec. 1.7 for a single equation. (Fora proof, see Ref. [A7].)T H E O R E M 1 Existence and Uniqueness TheoremLet in (1) be continuous functions having continuous partial derivativesin some domain R of -spacecontaining the point . Then (1) has a solution on some intervalsatisfying (2), and this solution is unique.t0 Ϫ a Ͻ t Ͻ t0 ϩ a(t0, K1, Á , Kn)ty1 y2Á yn0f1>0y1, Á , 0f1>0yn, Á , 0fn>0ynf1, Á , fnK ϭ [K1Á Kn]Tt0y(t0) ϭ Ky1(t0) ϭ K1, y2(t0) ϭ K2, Á , yn(t0) ϭ Kn,y ϭ h(t).h ϭ [h1Á hn]Ta Ͻ t Ͻ by1 ϭ h1(t), Á , yn ϭ hn(t)a Ͻ t Ͻ byr ϭ f(t, y)yr1 ϭ f1(t, y1)n ϭ 1yr ϭ f(t, y).Tf ϭ [ f1Á fn]Ty ϭ [y1Á yn]Tyr1 ϭ f1(t, y1, Á , yn)yr2 ϭ f2(t, y1, Á , yn)Áyrn ϭ fn(t, y1, Á , yn).SEC. 4.2 Basic Theory of Systems of ODEs. Wronskian 137c04.qxd 10/27/10 9:32 PM Page 137
• Linear SystemsExtending the notion of a linear ODE, we call (1) a linear system if it is linear inthat is, if it can be written(3)As a vector equation this becomes(3)whereThis system is called homogeneous if so that it is(4)If then (3) is called nonhomogeneous. For example, the systems in Examples 1 and3 of Sec. 4.1 are homogeneous. The system in Example 2 of that section is nonhomogeneous.For a linear system (3) we have in Theorem 1.Hence for a linear system we simply obtain the following.T H E O R E M 2 Existence and Uniqueness in the Linear CaseLet the ’s and ’s in (3) be continuous functions of t on an open intervalcontaining the point Then (3) has a solution y(t) on this intervalsatisfying (2), and this solution is unique.As for a single homogeneous linear ODE we haveT H E O R E M 3 Superposition Principle or Linearity PrincipleIf and are solutions of the homogeneous linear system (4) on some interval,so is any linear combination .P R O O F Differentiating and using (4), we obtain᭿ϭ A(c1 y(1)ϩ c2 y(2)) ϭ Ay.ϭ c1Ay(1)ϩ c2Ay(2)ϭ c1y(1)r ϩ c2y(2)ryr ϭ [c1 y(1)ϩ c1 y(2)]ry ϭ c1 y(1)ϩ c1 y(2)y(2)y(1)t ϭ t0.a Ͻ t Ͻ bgjajk0f1 >0y1 ϭ a11(t), Á , 0fn >0yn ϭ ann(t)g 0,yr ϭ Ay.g ϭ 0,A ϭ Da11Á a1n. Á .an1Á annT, y ϭ Dy1oynT, g ϭ Dg1ognT.yr ϭ Ay ϩ gyr1 ϭ a11(t)y1 ϩ Á ϩ a1n(t)yn ϩ g1(t)oyrn ϭ an1(t)y1 ϩ Á ϩ ann(t)yn ϩ gn(t).y1, Á , yn;138 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 138
• The general theory of linear systems of ODEs is quite similar to that of a single linearODE in Secs. 2.6 and 2.7. To see this, we explain the most basic concepts and facts. Forproofs we refer to more advanced texts, such as [A7].Basis. General Solution. WronskianBy a basis or a fundamental system of solutions of the homogeneous system (4) on someinterval J we mean a linearly independent set of n solutions of (4) on thatinterval. (We write J because we need I to denote the unit matrix.) We call a correspondinglinear combination(5)a general solution of (4) on J. It can be shown that if the (t) in (4) are continuous onJ, then (4) has a basis of solutions on J, hence a general solution, which includes everysolution of (4) on J.We can write n solutions of (4) on some interval J as columns of anmatrix(6)The determinant of Y is called the Wronskian of , written(7)The columns are these solutions, each in terms of components. These solutions form abasis on J if and only if W is not zero at any in this interval. W is either identicallyzero or nowhere zero in J. (This is similar to Secs. 2.6 and 3.1.)If the solutions in (5) form a basis (a fundamental system), then (6) isoften called a fundamental matrix. Introducing a column vectorwe can now write (5) simply as(8)Furthermore, we can relate (7) to Sec. 2.6, as follows. If y and z are solutions of asecond-order homogeneous linear ODE, their Wronskian isTo write this ODE as a system, we have to set and similarly for z(see Sec. 4.1). But then becomes (7), except for notation.W(y, z)y ϭ y1, yr ϭ y1r ϭ y2W(y, z) ϭ 2y zyr zr2.y ϭ Yc.c ϭ [c1 c2Á cn]T,y(1), Á , y(n)t1W(y(1), Á , y(n)) ϭ 5y1(1)y1(2) Á y1(n)y2(1)y2(2) Á y2(n)# # Á #yn(1)yn(2) Á yn(n)5.y(1), Á , y(n)Y ϭ [y(1) Á y(n)].n ϫ ny(1), Á , y(n)ajk(c1, Á , cn arbitrary)y ϭ c1y(1) Á ϩ cny(n)y(1), Á , y(n)SEC. 4.2 Basic Theory of Systems of ODEs. Wronskian 139c04.qxd 10/27/10 9:32 PM Page 139
• 4.3 Constant-Coefficient Systems.Phase Plane MethodContinuing, we now assume that our homogeneous linear system(1)under discussion has constant coefficients, so that the matrix has entriesnot depending on t. We want to solve (1). Now a single ODE has the solution. So let us try(2)Substitution into (1) gives . Dividing by , we obtain theeigenvalue problem(3)Thus the nontrivial solutions of (1) (solutions that are not zero vectors) are of the form(2), where is an eigenvalue of A and x is a corresponding eigenvector.We assume that A has a linearly independent set of n eigenvectors. This holds in mostapplications, in particular if A is symmetric or skew-symmetricor has n different eigenvalues.Let those eigenvectors be and let them correspond to eigenvalues(which may be all different, or some––or even all––may be equal). Then thecorresponding solutions (2) are(4)Their Wronskian [(7) in Sec. 4.2] is given byOn the right, the exponential function is never zero, and the determinant is not zero eitherbecause its columns are the n linearly independent eigenvectors. This proves the followingtheorem, whose assumption is true if the matrix A is symmetric or skew-symmetric, or ifthe n eigenvalues of A are all different.W ϭ (y(1), Á , y(n)) ϭ 5x1(1)el1t Á x1(n)elntx2(1)el1t Á x2(n)elnt# Á #xn(1)el1t Á xn(n)elnt5 ϭ el1tϩ Á ϩlnt5x1(1) Á x1(n)x2(1) Á x2(n)# Á #xn(1) Á xn(n)5.W ϭ W(y(1), Á , y(n))y(4)ϭ x(1)el1t, Á , y(n)ϭ x(n)elnt.l1, Á , lnx(1), Á , x(n)(akj ϭ Ϫajk)(akj ϭ ajk)lAx ϭ lx.eltyr ϭ lxeltϭ Ay ϭ Axelty ϭ xelt.y ϭ Cektyr ϭ kyA ϭ [ajk]n ϫ ny؅ ϭ Ay140 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 140
• T H E O R E M 1 General SolutionIf the constant matrix A in the system (1) has a linearly independent set of neigenvectors, then the corresponding solutions in (4) form a basis ofsolutions of (1), and the corresponding general solution is(5)How to Graph Solutions in the Phase PlaneWe shall now concentrate on systems (1) with constant coefficients consisting of twoODEs(6) in components,Of course, we can graph solutions of (6),(7)as two curves over the t-axis, one for each component of y(t). (Figure 80a in Sec. 4.1 showsan example.) But we can also graph (7) as a single curve in the -plane. This is a parametricrepresentation (parametric equation) with parameter t. (See Fig. 80b for an example. Manymore follow. Parametric equations also occur in calculus.) Such a curve is called a trajectory(or sometimes an orbit or path) of (6). The -plane is called the phase plane.1If we fillthe phase plane with trajectories of (6), we obtain the so-called phase portrait of (6).Studies of solutions in the phase plane have become quite important, along withadvances in computer graphics, because a phase portrait gives a good general qualitativeimpression of the entire family of solutions. Consider the following example, in whichwe develop such a phase portrait.E X A M P L E 1 Trajectories in the Phase Plane (Phase Portrait)Find and graph solutions of the system.In order to see what is going on, let us find and graph solutions of the system(8) thusy1r ϭ Ϫ3y1 ϩ y2y2r ϭ y1 Ϫ 3y2.yr ϭ Ay ϭ cϪ3 11 Ϫ3d y,y1 y2y1 y2y(t) ϭ cy1(t)y2(t)d,y1r ϭ a11 y1 ϩ a12 y2y2r ϭ a21 y1 ϩ a22 y2.y؅ ϭ Ay;y ϭ c1x(1)el1tϩ Á ϩ cnx(n)elnt.y(1), Á , y(n)SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 1411A name that comes from physics, where it is the y-(mv)-plane, used to plot a motion in terms of position yand velocity yЈ ϭ v (m ϭ mass); but the name is now used quite generally for the y1 y2-plane.The use of the phase plane is a qualitative method, a method of obtaining general qualitative informationon solutions without actually solving an ODE or a system. This method was created by HENRI POINCARÉ(1854–1912), a great French mathematician, whose work was also fundamental in complex analysis, divergentseries, topology, and astronomy.c04.qxd 10/27/10 9:32 PM Page 141
• Solution. By substituting and and dropping the exponential function we getThe characteristic equation isThis gives the eigenvalues and . Eigenvectors are then obtained fromFor this is . Hence we can take . For this becomesand an eigenvector is . This gives the general solutionFigure 82 shows a phase portrait of some of the trajectories (to which more trajectories could be added if sodesired). The two straight trajectories correspond to and and the others to other choices ofThe method of the phase plane is particularly valuable in the frequent cases when solvingan ODE or a system is inconvenient of impossible.Critical Points of the System (6)The point in Fig. 82 seems to be a common point of all trajectories, and we wantto explore the reason for this remarkable observation. The answer will follow by calculus.Indeed, from (6) we obtain(9)This associates with every point a unique tangent direction of thetrajectory passing through P, except for the point , where the right side of (9)becomes . This point , at which becomes undetermined, is called a criticalpoint of (6).Five Types of Critical PointsThere are five types of critical points depending on the geometric shape of the trajectoriesnear them. They are called improper nodes, proper nodes, saddle points, centers, andspiral points. We define and illustrate them in Examples 1–5.E X A M P L E 1 (Continued ) Improper Node (Fig. 82)An improper node is a critical point at which all the trajectories, except for two of them, have the samelimiting direction of the tangent. The two exceptional trajectories also have a limiting direction of the tangentat which, however, is different.The system (8) has an improper node at 0, as its phase portrait Fig. 82 shows. The common limiting directionat 0 is that of the eigenvector because goes to zero faster than as t increases. The twoexceptional limiting tangent directions are those of and . ᭿Ϫx(2)ϭ [Ϫ1 1]Tx(2)ϭ [1 Ϫ1]Te؊2te؊4tx(1)ϭ [1 1]TP0P0dy2>dy1P00>0P ϭ P0 :(0, 0)dy2>dy1P: (y1, y2)dy2dy1ϭy2r dty1r dtϭy2ry1rϭa21 y1 ϩ a22 y2a11 y1 ϩ a12 y2.y ϭ 0᭿c1, c2.c2 ϭ 0c1 ϭ 0y ϭ cy1y2d ϭ c1 y(1)ϩ c2 y(2)ϭ c1 c11d e؊2tϩ c2 c1Ϫ1d e؊4t.x(2)ϭ [1 Ϫ1]Tx1 ϩ x2 ϭ 0,l2 ϭ Ϫ4x(1)ϭ [1 1]TϪx1 ϩ x2 ϭ 0l1 ϭ Ϫ2(Ϫ3 Ϫ l)x1 ϩ x2 ϭ 0.l2 ϭ Ϫ4l1 ϭ Ϫ2ϭ l2ϩ 6l ϩ 8 ϭ 0.det (A Ϫ lI) ϭ 2Ϫ3 Ϫ l 11 Ϫ3 Ϫ l2Ax ϭ lx.yr ϭ lxelty ϭ xelt142 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsc04.qxd 10/27/10 9:32 PM Page 142
• E X A M P L E 2 Proper Node (Fig. 83)A proper node is a critical point at which every trajectory has a definite limiting direction and for any givendirection d at there is a trajectory having d as its limiting direction.The system(10)has a proper node at the origin (see Fig. 83). Indeed, the matrix is the unit matrix. Its characteristic equationhas the root . Any is an eigenvector, and we can take and . Hencea general solution is᭿y ϭ c1 c10d etϩ c2 c01d etory1 ϭ c1ety2 ϭ c2etor c1 y2 ϭ c2 y1.[0 1]T[1 0]Tx 0l ϭ 1(1 Ϫ l)2ϭ 0yr ϭ c1 00 1d y, thusy1r ϭ y1y2r ϭ y2P0P0SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 143y2y1y(1)(t)y(2)(t)Fig. 82. Trajectories of the system (8)(Improper node)y2y1Fig. 83. Trajectories of the system (10)(Proper node)E X A M P L E 3 Saddle Point (Fig. 84)A saddle point is a critical point at which there are two incoming trajectories, two outgoing trajectories, andall the other trajectories in a neighborhood of bypass .The system(11)has a saddle point at the origin. Its characteristic equation has the roots and. For an eigenvector is obtained from the second row of that is,. For the first row gives . Hence a general solution isThis is a family of hyperbolas (and the coordinate axes); see Fig. 84. ᭿y ϭ c1 c10d etϩ c2 c01d e؊tory1 ϭ c1ety2 ϭ c2e؊tor y1 y2 ϭ const.[0 1]Tl2 ϭ Ϫ10x1 ϩ (Ϫ1 Ϫ 1)x2 ϭ 0(A Ϫ lI)x ϭ 0,[1 0]Tl ϭ 1l2 ϭ Ϫ1l1 ϭ 1(1 Ϫ l)(Ϫ1 Ϫ l) ϭ 0yr ϭ c1 00 Ϫ1d y, thusy1r ϭ y1y1r ϭ Ϫy2P0P0P0c04.qxd 10/27/10 9:32 PM Page 143
• E X A M P L E 4 Center (Fig. 85)A center is a critical point that is enclosed by infinitely many closed trajectories.The system(12)has a center at the origin. The characteristic equation gives the eigenvalues 2i and . For 2i aneigenvector follows from the first equation of , say, . For thatequation is and gives, say, . Hence a complex general solution is(12 )A real solution is obtained from (12 ) by the Euler formula or directly from (12) by a trick. (Remember thetrick and call it a method when you apply it again.) Namely, the left side of (a) times the right side of (b) is. This must equal the left side of (b) times the right side of (a). Thus,. By integration, .This is a family of ellipses (see Fig. 85) enclosing the center at the origin. ᭿2y12ϩ 12 y22ϭ constϪ4y1y1r ϭ y2y2rϪ4y1y1r*y ϭ c1 c12id e2itϩ c2 c1Ϫ2id e؊2it, thusy1 ϭ c1e2itϩ c2e؊2ity2 ϭ 2ic1e2itϪ 2ic2e؊2it.*[1 Ϫ2i]TϪ(Ϫ2i)x1 ϩ x2 ϭ 0l ϭ Ϫ2i[1 2i]T(A Ϫ lI)x ϭ 0Ϫ2ix1 ϩ x2 ϭ 0Ϫ2il2ϩ 4 ϭ 0yr ϭ c0 1Ϫ4 0d y, thus(a)(b)y1r ϭ y2y2r ϭ Ϫ4y1144 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsy2y1Fig. 84. Trajectories of the system (11)(Saddle point)y2y1Fig. 85. Trajectories of the system (12)(Center)E X A M P L E 5 Spiral Point (Fig. 86)A spiral point is a critical point about which the trajectories spiral, approaching as (or tracing thesespirals in the opposite sense, away from ).The system(13)has a spiral point at the origin, as we shall see. The characteristic equation is . It gives theeigenvalues and . Corresponding eigenvectors are obtained from . For(Ϫ1 Ϫ l)x1 ϩ x2 ϭ 0Ϫ1 Ϫ iϪ1 ϩ il2ϩ 2l ϩ 2 ϭ 0yr ϭ cϪ1 1Ϫ1 Ϫ1d y, thusy1r ϭ Ϫy1 ϩ y2y2r ϭ Ϫy1 Ϫ y2P0t : ϱP0P0c04.qxd 10/27/10 9:32 PM Page 144
• this becomes and we can take as an eigenvector. Similarly, an eigenvectorcorresponding to is . This gives the complex general solutionThe next step would be the transformation of this complex solution to a real general solution by the Eulerformula. But, as in the last example, we just wanted to see what eigenvalues to expect in the case of a spiralpoint. Accordingly, we start again from the beginning and instead of that rather lengthy systematic calculationwe use a shortcut. We multiply the first equation in (13) by , the second by , and add, obtaining.We now introduce polar coordinates r, t, where . Differentiating this with respect to t gives. Hence the previous equation can be written, Thus, , , .For each real c this is a spiral, as claimed (see Fig. 86). ᭿r ϭ ce؊tln ƒ rƒ ϭ Ϫt ϩ c*,dr>r ϭ Ϫdtrr ϭ Ϫrrrr ϭ Ϫr22rrr ϭ 2y1 yr1 ϩ 2y2 yr2r 2ϭ y12ϩ y22y1 yr1 ϩ y2 yr2 ϭ Ϫ(y12ϩ y22)y2y1y ϭ c1 c1id e(؊1؉i)tϩ c2 c1Ϫid e(؊1؊i)t.[1 Ϫi]TϪ1 Ϫ i[1 i]TϪix1 ϩ x2 ϭ 0l ϭ Ϫ1 ϩ iSEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 145y2y1Fig. 86. Trajectories of the system (13) (Spiral point)E X A M P L E 6 No Basis of Eigenvectors Available. Degenerate Node (Fig. 87)This cannot happen if A in (1) is symmetric , as in Examples 1–3) or skew-symmetricthus . And it does not happen in many other cases (see Examples 4 and 5). Hence it suffices to explainthe method to be used by an example.Find and graph a general solution of(14)Solution. A is not skew-symmetric! Its characteristic equation is.det (A Ϫ lI) ϭ 24 Ϫ l 1Ϫ1 2 Ϫ l2 ϭ l2Ϫ 6l ϩ 9 ϭ (l Ϫ 3)2ϭ 0yr ϭ Ay ϭ c4 1Ϫ1 2d y.ajj ϭ 0)(akj ϭ Ϫajk,(akj ϭ ajkc04.qxd 10/27/10 9:32 PM Page 145
• It has a double root . Hence eigenvectors are obtained from , thus fromsay, and nonzero multiples of it (which do not help). The method now is to substitutewith constant into (14). (The xt-term alone, the analog of what we did in Sec. 2.2 in the caseof a double root, would not be enough. Try it.) This gives.On the right, . Hence the terms cancel, and then division by gives, thus .Here and , so that, thusA solution, linearly independent of , is . This yields the answer (Fig. 87)The critical point at the origin is often called a degenerate node. gives the heavy straight line, withthe lower part and the upper part of it. gives the right part of the heavy curve from 0 throughthe second, first, and—finally—fourth quadrants. gives the other part of that curve. ᭿Ϫy(2)y(2)c1 Ͻ 0c1 Ͼ 0c1y(1)y ϭ c1y(1)ϩ c2y(2)ϭ c1 c1Ϫ1d e3tϩ c2 £c1Ϫ1d t ϩ c01d≥ e3t.u ϭ [0 1]Tx ϭ [1 Ϫ1]Tu1 ϩ u2 ϭ 1Ϫu1 Ϫ u2 ϭ Ϫ1.(A Ϫ 3I)u ϭ c4 Ϫ 3 1Ϫ1 2 Ϫ 3d u ϭ c1Ϫ1dx ϭ [1 Ϫ1]Tl ϭ 3(A Ϫ lI)u ϭ xx ϩ lu ϭ AueltlxteltAx ϭ lxy(2)r ϭ xeltϩ lxteltϩ lueltϭ Ay(2)ϭ Axteltϩ Aueltu ϭ [u1 u2]Ty(2)ϭ xteltϩ ueltx(1)ϭ [1 Ϫ1]Tx1 ϩ x2 ϭ 0,(4 Ϫ l)x1 ϩ x2 ϭ 0l ϭ 3146 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methodsy2y1y(1)y(2)Fig. 87. Degenerate node in Example 6We mention that for a system (1) with three or more equations and a triple eigenvaluewith only one linearly independent eigenvector, one will get two solutions, as justdiscussed, and a third linearly independent one fromwith v from u ϩ lv ϭ Av.y(3)ϭ 12 xt2eltϩ uteltϩ veltc04.qxd 10/27/10 9:32 PM Page 146
• SEC. 4.3 Constant-Coefficient Systems. Phase Plane Method 1471–9 GENERAL SOLUTIONFind a real general solution of the following systems. Showthe details.1.2.3.4.5.6.7.8.9.10–15 IVPsSolve the following initial value problems.10.11.12.13.y1(0) ϭ 0, y2(0) ϭ 2y2r ϭ y1y1r ϭ y2y1(0) ϭ 12, y2(0) ϭ 2y2r ϭ 13 y1 ϩ y2y1r ϭ y1 ϩ 3y2y1(0) ϭ Ϫ12, y2(0) ϭ 0y2r ϭ Ϫ12 y1 Ϫ 32 y2y1r ϭ 2y1 ϩ 5y2y1(0) ϭ 0, y2(0) ϭ 7y2r ϭ 5y1 Ϫ y2y1r ϭ 2y1 ϩ 2y2y3r ϭ Ϫ4y1 Ϫ 14y2 Ϫ 2y3y2r ϭ Ϫ10y1 ϩ y2 Ϫ 14y3y1r ϭ 10y1 Ϫ 10y2 Ϫ 4y3y2r ϭ y1 ϩ 10y2y1r ϭ 8y1 Ϫ y2y3r ϭ Ϫy2y2r ϭ Ϫy1 ϩ y3y1r ϭ y2y2r ϭ 2y1 ϩ 2y2y1r ϭ 2y1 Ϫ 2y2y2r ϭ 5y1 ϩ 12.5y2y1r ϭ 2y1 ϩ 5y2y2r ϭ 2y1 Ϫ 4y2y1r ϭ Ϫ8y1 Ϫ 2y2y2r ϭ 12 y1 ϩ y2y1r ϭ y1 ϩ 2y2y2r ϭ y1 ϩ 6y2y1r ϭ 6y1 ϩ 9y2y2r ϭ 3y1 Ϫ y2y1r ϭ y1 ϩ y214.15.16–17 CONVERSIONFind a general solution by conversion to a single ODE.16. The system in Prob. 8.17. The system in Example 5 of the text.18. Mixing problem, Fig. 88. Each of the two tankscontains 200 gal of water, in which initially 100 lb(Tank ) and 200 lb (Tank ) of fertilizer are dissolved.The inflow, circulation, and outflow are shown inFig. 88. The mixture is kept uniform by stirring. Findthe fertilizer contents in and in .T2y2(t)T1y1(t)T2T1y1(0) ϭ 0.5, y2(0) ϭ Ϫ0.5y2r ϭ 2y1 ϩ 3y2y1r ϭ 3y1 ϩ 2y2y1(0) ϭ 1, y2(0) ϭ 0y2r ϭ y1 Ϫ y2y1r ϭ Ϫy1 Ϫ y2P R O B L E M S E T 4 . 3Fig. 88. Tanks in Problem 184 gal/min16 gal/min 12 gal/min12 gal/min(Pure water)T1T219. Network. Show that a model for the currents andin Fig. 89 is, .Find a general solution, assuming that ,.C ϭ 1>12 FL ϭ 4 H,R ϭ 3 ⍀LIr2 ϩ R(I2 Ϫ I1) ϭ 01C ΎI1 dt ϩ R(I1 Ϫ I2) ϭ 0I2(t)I1(t)Fig. 89. Network in Problem 19I1CRLI220. CAS PROJECT. Phase Portraits. Graph some ofthe figures in this section, in particular Fig. 87 on thedegenerate node, in which the vector depends on t.In each figure highlight a trajectory that satisfies aninitial condition of your choice.y(2)c04.qxd 10/27/10 9:32 PM Page 147
• 148 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative Methods4.4 Criteria for Critical Points. StabilityWe continue our discussion of homogeneous linear systems with constant coefficients (1).Let us review where we are. From Sec. 4.3 we have(1) in components,From the examples in the last section, we have seen that we can obtain an overview offamilies of solution curves if we represent them parametrically asand graph them as curves in the -plane, called the phase plane. Such a curve is calleda trajectory of (1), and their totality is known as the phase portrait of (1).Now we have seen that solutions are of the form. Substitution into (1) gives .Dropping the common factor , we have(2)Hence is a (nonzero) solution of (1) if is an eigenvalue of A and x a correspondingeigenvector.Our examples in the last section show that the general form of the phase portrait isdetermined to a large extent by the type of critical point of the system (1) defined as apoint at which becomes undetermined, ; here [see (9) in Sec. 4.3](3)We also recall from Sec. 4.3 that there are various types of critical points.What is now new, is that we shall see how these types of critical points are relatedto the eigenvalues. The latter are solutions and of the characteristic equation(4) .This is a quadratic equation with coefficients p, q and discriminantgiven by(5) , , .From algebra we know that the solutions of this equation are(6) , .l2 ϭ 12 (p Ϫ 1¢)l1 ϭ 12 (p ϩ 1¢)¢ ϭ p2Ϫ 4qq ϭ det A ϭ a11a22 Ϫ a12a21p ϭ a11 ϩ a22¢l2Ϫ pl ϩ q ϭ 0det (A Ϫ lI) ϭ 2a11 Ϫ l a12a21 a22 Ϫ l2 ϭ l 2Ϫ (a11 ϩ a22)l ϩ det A ϭ 0l2l ϭ l1dy2dy1ϭyr2 dtyr1 dtϭa21 y1 ϩ a22 y2a11 y1 ϩ a12 y2.0>0dy2>dy1ly(t)Ax ϭ lx.eltyr(t) ϭ lxeltϭ Ay ϭ Axelty(t) ϭ xelty1 y2y(t) ϭ [y1(t) y2(t)]Tyr1 ϭ a11 y1 ϩ a12 y2yr2 ϭ a21 y1 ϩ a22 y2.yr ϭ Ay ϭ ca11 a12a21 a22d y,c04.qxd 10/27/10 9:32 PM Page 148
• Furthermore, the product representation of the equation gives.Hence p is the sum and q the product of the eigenvalues. Also from (6).Together,(7) , , .This gives the criteria in Table 4.1 for classifying critical points. A derivation will beindicated later in this section.¢ ϭ (l1 Ϫ l2)2q ϭ l1l2p ϭ l1 ϩ l2l1 Ϫ l2 ϭ 1¢l2Ϫ pl ϩ q ϭ (l Ϫ l1)(l Ϫ l2) ϭ l2Ϫ (l1 ϩ l2)l ϩ l1l2SEC. 4.4 Criteria for Critical Points. Stability 149Table 4.1 Eigenvalue Criteria for Critical Points(Derivation after Table 4.2)Name Comments on(a) Node Real, same sign(b) Saddle point Real, opposite signs(c) Center Pure imaginary(d) Spiral point Complex, not pureimaginary¢ Ͻ 0p 0q Ͼ 0p ϭ 0q Ͻ 0¢ м 0q Ͼ 0l1, l2¢ ϭ (l1 Ϫ l2)2q ϭ l1l2p ϭ l1 ϩ l2StabilityCritical points may also be classified in terms of their stability. Stability concepts are basicin engineering and other applications. They are suggested by physics, where stabilitymeans, roughly speaking, that a small change (a small disturbance) of a physical systemat some instant changes the behavior of the system only slightly at all future times t. Forcritical points, the following concepts are appropriate.D E F I N I T I O N S Stable, Unstable, Stable and AttractiveA critical point of (1) is called stable2if, roughly, all trajectories of (1) that atsome instant are close to remain close to at all future times; precisely: if forevery disk of radius with center there is a disk of radius withcenter such that every trajectory of (1) that has a point (corresponding tosay) in has all its points corresponding to in . See Fig. 90.is called unstable if is not stable.is called stable and attractive (or asymptotically stable) if is stable andevery trajectory that has a point in approaches as . See Fig. 91.Classification criteria for critical points in terms of stability are given in Table 4.2. Bothtables are summarized in the stability chart in Fig. 92. In this chart region of instabilityis dark blue.t : ϱP0DdP0P0P0P0DPt м t1Ddt ϭ t1,P1P0d Ͼ 0DdP0P Ͼ 0DPP0P0P02In the sense of the Russian mathematician ALEXANDER MICHAILOVICH LJAPUNOV (1857–1918),whose work was fundamental in stability theory for ODEs. This is perhaps the most appropriate definition ofstability (and the only we shall use), but there are others, too.c04.qxd 10/27/10 9:32 PM Page 149
• We indicate how the criteria in Tables 4.1 and 4.2 are obtained. If , bothof the eigenvalues are positive or both are negative or complex conjugates. If also, both are negative or have a negative real part. Hence is stable andattractive. The reasoning for the other two lines in Table 4.2 is similar.If , the eigenvalues are complex conjugates, say, andIf also , this gives a spiral point that is stable and attractive. If, this gives an unstable spiral point.If , then and . If also , then , sothat , and thus , must be pure imaginary. This gives periodic solutions, their trajectoriesbeing closed curves around , which is a center.E X A M P L E 1 Application of the Criteria in Tables 4.1 and 4.2In Example 1, Sec 4.3, we have a node by Table 4.1(a), which isstable and attractive by Table 4.2(a). ᭿yr ϭ cϪ3 11 Ϫ3d y, p ϭ Ϫ6, q ϭ 8, ¢ ϭ 4,P0l2l1l12ϭ Ϫq Ͻ 0q Ͼ 0q ϭ l1l2 ϭ Ϫl12l2 ϭ Ϫl1p ϭ 0p ϭ 2a Ͼ 0p ϭ l1 ϩ l2 ϭ 2a Ͻ 0l2 ϭ a Ϫ ib.l1 ϭ a ϩ ib¢ Ͻ 0P0p ϭ l1 ϩ l2 Ͻ 0q ϭ l1l2 Ͼ 0150 CHAP. 4 Systems of ODEs. Phase Plane. Qualitative MethodsP1P0∈ δFig. 90. Stable critical point P0 of (1)(The trajectory initiating at P1 staysin the disk of radius ⑀.)P0∈δFig. 91. Stable and attractive criticalpoint P0 of (1)Table 4.2 Stability Criteria for Critical PointsType of Stability(a) Stable and attractive(b) Stable(c) Unstable OR q Ͻ 0p Ͼ 0q Ͼ 0p Ϲ 0q Ͼ 0p Ͻ 0q ϭ l1l2p ϭ l1 ϩ l2qpΔ=0Δ > 0 Δ < 0 Δ < 0 Δ > 0Δ=0SpiralpointSpiralpointNode NodeSaddle pointFig. 92. Stability chart of the system (1) with p, q, ⌬ defined in (5).Stable and attractive: The second quadrant without the q-axis.Stability also on the positive q-axis (which corresponds to centers).Unstable: Dark blue regionc04.qxd 10/27/10 9:32 PM Page 150
• E X A M P L E 2 Free Motions of a Mass on a SpringWhat kind of critical point does in Sec. 2.4 have?Solution. Division by m gives . To get a system, set (see Sec. 4.1).Then . Hence, .We see that . From this and Tables 4.1 and 4.2 we obtain the followingresults. Note that in the last three cases the discriminant plays an essential role.No damping. , a center.Underdamping. , a stable and attractive spiral point.Critical damping. , a stable and attractive node.Overdamping. , a stable and attractive node. ᭿c2Ͼ 4mk, p Ͻ 0, q Ͼ 0, ¢ Ͼ 0c2ϭ 4mk, p Ͻ 0, q Ͼ 0, ¢ ϭ 0c2Ͻ 4mk, p Ͻ 0, q Ͼ 0, ¢ Ͻ 0c ϭ 0, p ϭ 0, q Ͼ 0¢p ϭ Ϫc>m, q ϭ k>m, ¢ ϭ (c>m)2Ϫ 4k>mdet (A Ϫ lI) ϭ 2Ϫl 1Ϫk>m Ϫc>m Ϫ l2 ϭ l2ϩcml ϩkmϭ 0yr ϭ c0 1Ϫk>m Ϫc>md yyr2 ϭ ys ϭ Ϫ(k>m)y1 Ϫ (c>m)y2y1 ϭ y, y2 ϭ yrys ϭ Ϫ(k>m)y Ϫ (c>m)yrmys ϩ cyr ϩ ky ϭ 0SEC. 4.4 Criteria for Critical Points. Stability 1511–10 TYPE AND STABILITY OFCRITICAL POINTDetermine the type and stability of the critical point. Thenfind a real general solution and sketch or graph some of thetrajectories in the phase plane. Show the details of your work.1. 2.3. 4.5. 6.7. 8.9. 10.11–18 TRAJECTORIES OF SYSTEMS ANDSECOND-ORDER ODEs. CRITICALPOINTS11. Damped oscillations. Solve . Whatkind of curves are the trajectories?12. Harmonic oscillations. Solve Find thetrajectories. Sketch or graph some of them.13. Types of critical points. Discuss the critical points in(10)–(13) of Sec. 4.3 by using Tables 4.1 and 4.2.14. Transformation of parameter. What happens to thecritical point in Example 1 if you introduce asa new independent variable?t ϭ Ϫtys ϩ 19 y ϭ 0.ys ϩ 2yr ϩ 2y