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1. 1. Licensed to: iChapters User ADVANCED ENGINEERING MATHEMATICS International Student Edition PETER V. O’NEIL University of Alabama at Birmingham Australia Canada Mexico Singapore Spain United Kingdom United States Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
3. 3. Licensed to: iChapters User PRELIMINARY CONCEPTS SEPARABLE EQUATIONS HOMOGENEOUS, BERNOULLI, AND RICCATI EQUA- CHAPTER 1 TIONS APPLICATIONS TO MECHANICS, ELECTRICAL CIRCUITS, AND ORTHOGONAL TRAJECTORIES EXI First-Order Differential Equations 1.1 Preliminary Concepts Before developing techniques for solving various kinds of differential equations, we will develop some terminology and geometric insight. 1.1.1 General and Particular Solutions A first-order differential equation is any equation involving a first derivative, but no higher derivative. In its most general form, it has the appearance F x y y =0 (1.1) in which y x is the function of interest and x is the independent variable. Examples are y − y 2 − ey = 0 y −2 = 0 and y − cos x = 0 Note that y must be present for an equation to qualify as a first-order differential equation, but x and/or y need not occur explicitly. A solution of equation (1.1) on an interval I is a function that satisfies the equation for all x in I. That is, Fx x x =0 for all x in I. For example, x = 2 + ke−x 3 Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
4. 4. Licensed to: iChapters User 4 CHAPTER 1 First-Order Differential Equations is a solution of y +y = 2 for all real x, and for any number k. Here I can be chosen as the entire real line. And x = x ln x + cx is a solution of y y = +1 x for all x > 0, and for any number c. In both of these examples, the solution contained an arbitrary constant. This is a symbol independent of x and y that can be assigned any numerical value. Such a solution is called the general solution of the differential equation. Thus x = 2 + ke−x is the general solution of y + y = 2. Each choice of the constant in the general solution yields a particular solution. For example, f x = 2 + e−x g x = 2 − e−x and √ h x = 2 − 53e−x are all particular solutions of y + y = 2, obtained by choosing, respectively, k = 1, −1 and √ − 53 in the general solution. 1.1.2 Implicitly Defined Solutions Sometimes we can write a solution explicitly giving y as a function of x. For example, y = ke−x is the general solution of y = −y as can be verified by substitution. This general solution is explicit, with y isolated on one side of an equation, and a function of x on the other. By contrast, consider 2xy3 + 2 y =− 3x2 y2 + 8e4y We claim that the general solution is the function y x implicitly defined by the equation x2 y3 + 2x + 2e4y = k (1.2) in which k can be any number. To verify this, implicitly differentiate equation (1.2) with respect to x, remembering that y is a function of x. We obtain 2xy3 + 3x2 y2 y + 2 + 8e4y y = 0 and solving for y yields the differential equation. In this example we are unable to solve equation (1.2) explicitly for y as a function of x, isolating y on one side. Equation (1.2), implicitly defining the general solution, was obtained by a technique we will develop shortly, but this technique cannot guarantee an explicit solution. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
5. 5. Licensed to: iChapters User 1.1 Preliminary Concepts 5 1.1.3 Integral Curves A graph of a solution of a first-order differential equation is called an integral curve of the equation. If we know the general solution, we obtain an infinite family of integral curves, one for each choice of the arbitrary constant. EXAMPLE 1.1 We have seen that the general solution of y +y = 2 is y = 2 + ke−x for all x. The integral curves of y + y = 2 are graphs of y = 2 + ke−x for different choices of k. Some of these are shown in Figure 1.1. y 30 kϭ6 20 kϭ3 10 k ϭ 0 ( y ϭ 2) x Ϫ2 Ϫ1 0 1 2 3 4 5 6 k ϭ Ϫ3 Ϫ10 k ϭ Ϫ6 Ϫ20 FIGURE 1.1 Integral curves of y + y = 2 for k = 0 3 −3 6, and −6. EXAMPLE 1.2 It is routine to verify that the general solution of y y + = ex x is 1 y= xex − ex + c x Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
6. 6. Licensed to: iChapters User 6 CHAPTER 1 First-Order Differential Equations for x = 0. Graphs of some of these integral curves, obtained by making choices for c, are shown in Figure 1.2. y 40 c ϭ 20 30 20 10 cϭ5 cϭ0 x 0.5 1.0 1.5 2.0 2.5 3.0 Ϫ10 c ϭ Ϫ6 Ϫ20 c ϭ Ϫ10 FIGURE 1.2 Integral curves of y + x y = ex for c = 0 5 20 −6, and 1 −10. We will see shortly how these general solutions are obtained. For the moment, we simply want to illustrate integral curves. Although in simple cases integral curves can be sketched by hand, generally we need computer assistance. Computer packages such as MAPLE, MATHEMATICA and MATLAB are widely available. Here is an example in which the need for computing assistance is clear. EXAMPLE 1.3 The differential equation y + xy = 2 has general solution x y x = e−x d + ke−x 2 /2 2 /2 2 /2 2e 0 Figure 1.3 shows computer-generated integral curves corresponding to k = 0, 4, 13, −7, −15 and −11. 1.1.4 The Initial Value Problem The general solution of a first-order differential equation F x y y = 0 contains an arbitrary constant, hence there is an infinite family of integral curves, one for each choice of the constant. If we specify that a solution is to pass through a particular point x0 y0 , then we must find that particular integral curve (or curves) passing through this point. This is called an initial value problem. Thus, a first order initial value problem has the form F x y y =0 y x0 = y0 in which x0 and y0 are given numbers. The condition y x0 = y0 is called an initial condition. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
7. 7. Licensed to: iChapters User 1.1 Preliminary Concepts 7 y k ϭ 13 10 5 kϭ4 Ϫ2 kϭ0 0 x Ϫ4 k ϭ Ϫ7 2 4 Ϫ5 Ϫ10 k ϭ Ϫ11 k ϭ Ϫ15 Ϫ15 FIGURE 1.3 Integral curves of y + xy = 2 for k = 0 4 13 −7 −15, and −11. EXAMPLE 1.4 Consider the initial value problem y +y = 2 y 1 = −5 From Example 1.1, the general solution of y + y = 2 is y = 2 + ke−x Graphs of this equation are the integral curves. We want the one passing through 1 −5 . Solve for k so that y 1 = 2 + ke−1 = −5 obtaining k = −7e The solution of this initial value problem is y = 2 − 7ee−x = 2 − 7e− x−1 As a check, y 1 = 2 − 7 = −5 The effect of the initial condition in this example was to pick out one special integral curve as the solution sought. This suggests that an initial value problem may be expected to have a unique solution. We will see later that this is the case, under mild conditions on the coefficients in the differential equation. 1.1.5 Direction Fields Imagine a curve, as in Figure 1.4. If we choose some points on the curve and, at each point, draw a segment of the tangent to the curve there, then these segments give a rough outline of the shape of the curve. This simple observation is the key to a powerful device for envisioning integral curves of a differential equation. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
8. 8. Licensed to: iChapters User 8 CHAPTER 1 First-Order Differential Equations y x FIGURE 1.4 Short tangent segments suggest the shape of the curve. The general first-order differential equation has the form F x y y =0 Suppose we can solve for y and write the differential equation as y =f x y Here f is a known function. Suppose f x y is defined for all points x y in some region R of the plane. The slope of the integral curve through a given point x0 y0 of R is y x0 , which equals f x0 y0 . If we compute f x y at selected points in R, and draw a small line segment having slope f x y at each x y , we obtain a collection of segments which trace out the shapes of the integral curves. This enables us to obtain important insight into the behavior of the solutions (such as where solutions are increasing or decreasing, limits they might have at various points, or behavior as x increases). A drawing of the plane, with short line segments of slope f x y drawn at selected points x y , is called a direction field of the differential equation y = f x y . The name derives from the fact that at each point the line segment gives the direction of the integral curve through that point. The line segments are called lineal elements. EXAMPLE 1.5 Consider the equation y = y2 Here f x y = y2 , so the slope of the integral curve through x y is y2 . Select some points and, through each, draw a short line segment having slope y2 . A computer generated direction field is shown in Figure 1.5(a). The lineal elements form a profile of some integral curves and give us some insight into the behavior of solutions, at least in this part of the plane. Figure 1.5(b) reproduces this direction field, with graphs of the integral curves through 0 1 , 0 2 , 0 3 , 0 −1 , 0 −2 and 0 −3 . By a method we will develop, the general solution of y = y2 is 1 y=− x+k so the integral curves form a family of hyperbolas, as suggested by the curves sketched in Figure 1.5(b). Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
9. 9. Licensed to: iChapters User 1.1 Preliminary Concepts 9 y 4 2 x Ϫ4 Ϫ2 0 2 4 Ϫ2 Ϫ4 FIGURE 1.5(a) A direction field for y = y2 . y 4 2 2 4 x Ϫ4 Ϫ2 0 Ϫ2 Ϫ4 FIGURE 1.5(b) Direction field for y = y2 and integral curves through 0 1 , 0 2 , 0 3 0 −1 , 0 −2 , and 0 −3 . Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
10. 10. Licensed to: iChapters User 10 CHAPTER 1 First-Order Differential Equations EXAMPLE 1.6 Figure 1.6 shows a direction field for y = sin xy together with the integral curves through 0 1 , 0 2 , 0 3 , 0 −1 , 0 −2 and 0 −3 . In this case, we cannot write a simple expression for the general solution, and the direction field provides information about the behavior of solutions that is not otherwise readily apparent. y 4 2 x Ϫ4 Ϫ2 0 2 4 Ϫ2 Ϫ4 FIGURE 1.6 Direction field for y = sin xy and integral curves through 0 1 , 0 2 , 0 3 , 0 −1 , 0 −2 , and 0 −3 . With this as background, we will begin a program of identifying special classes of first- order differential equations for which there are techniques for writing the general solution. This will occupy the next five sections. SECTION 1.1 PROBLEMS In each of Problems 1 through 6, determine whether the x2 − 3 5. xy = x − y x = for x = 0 given function is a solution of the differential equation. 2x √ 6. y + y = 1 x = 1 + Ce−x 1. 2yy = 1 x = x − 1 for x > 1 2. y + y = 0 x = Ce−x In each of Problems 7 through 11, verify by implicit differentiation that the given equation implicitly defines a 2y + e x C −e x 3. y = − for x > 0 x = solution of the differential equation. 2x 2x 2xy √ C 7. y2 + xy − 2x2 − 3x − 2y = C 4. y = for x = ± 2 x = 2−x 2 x 2 −2 y − 4x − 2 + x + 2y − 2 y = 0 Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
11. 11. Licensed to: iChapters User 1.2 Separable Equations 11 8. xy3 − y = C y3 + 3xy2 − 1 y = 0 17. y = x + y y 2 = 2 9. y − 4x + e = C 8x − ye − 2y + xe 2 2 xy xy xy y =0 18. y = x − xy y 0 = −1 10. 8 ln x − 2y + 4 − 2x + 6y = C; 19. y = xy y 0 = 2 x − 2y 20. y = x − y + 1 y 0 = 1 y = 3x − 6y + 4 In each of Problems 21 through 26, generate a direction −1 2x3 + 2xy2 − y x field and some integral curves for the differential equation. 11. tan y/x + x = C2 + 2 y =0 x2 + y 2 x + y2 Also draw the integral curve representing the solution of the initial value problem. These problems should be done In each of Problems 12 through 16, solve the initial value by a software package. problem and graph the solution. Hint: Each of these dif- ferential equations can be solved by direct integration. Use 21. y = sin y y 1 = /2 the initial condition to solve for the constant of integration. 22. y = x cos 2x − y y 1 = 0 12. y = 2x y 2 = 1 23. y = y sin x − 3x2 y 0 = 1 −x 13. y = e y 0 =2 24. y = ex − y y −2 = 1 14. y = 2x + 2 y −1 = 1 25. y − y cos x = 1 − x2 y 2 = 2 15. y = 4 cos x sin x y /2 = 0 26. y = 2y + 3 y 0 = 1 16. y = 8x + cos 2x y 0 = −3 27. Show that, for the differential equation y + p x y = In each of Problems 17 through 20 draw some lineal q x , the lineal elements on any vertical line x = x0 , elements of the differential equation for −4 ≤ x ≤ 4, with p x0 = 0, all pass through the single point −4 ≤ y ≤ 4. Use the resulting direction field to sketch a , where graph of the solution of the initial value problem. (These 1 q x0 problems can be done by hand.) = x0 + and = p x0 p x0 1.2 Separable Equations DEFINITION 1.1 Separable Differential Equation A differential equation is called separable if it can be written y =A x B y In this event, we can separate the variables and write, in differential form, 1 dy = A x dx By wherever B y = 0. We attempt to integrate this equation, writing 1 dy = A x dx By This yields an equation in x, y, and a constant of integration. This equation implicitly defines the general solution y x . It may or may not be possible to solve explicitly for y x . Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
12. 12. Licensed to: iChapters User 12 CHAPTER 1 First-Order Differential Equations EXAMPLE 1.7 y = y2 e−x is separable. Write dy = y2 e−x dx as 1 dx = e−x dx y2 for y = 0. Integrate this equation to obtain 1 − = −e−x + k y an equation that implicitly defines the general solution. In this example we can explicitly solve for y, obtaining the general solution 1 y= e−x − k Now recall that we required that y = 0 in order to separate the variables by dividing by y2 . In fact, the zero function y x = 0 is a solution of y = y2 ex , although it cannot be obtained from the general solution by any choice of k. For this reason, y x = 0 is called a singular solution of this equation. Figure 1.7 shows graphs of particular solutions obtained by choosing k as 0, 3, −3, 6 and −6. y 3 2 kϭ0 1 k ϭ Ϫ3 k ϭ Ϫ6 x Ϫ2 Ϫ1 1 2 kϭ6 kϭ3 Ϫ1 FIGURE 1.7 Integral curves of y = y2 e−x for k = 0 3 −3 6, and −6. Whenever we use separation of variables, we must be alert to solutions potentially lost through conditions imposed by the algebra used to make the separation. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
13. 13. Licensed to: iChapters User 1.2 Separable Equations 13 EXAMPLE 1.8 x2 y = 1 + y is separable, and we can write 1 1 dy = 2 dx 1+y x The algebra of separation has required that x = 0 and y = −1, even though we can put x = 0 and y = −1 into the differential equation to obtain the correct equation 0 = 0. Now integrate the separated equation to obtain 1 ln 1 + y = − + k x This implicitly defines the general solution. In this case, we can solve for y x explicitly. Begin by taking the exponential of both sides to obtain 1 + y = ek e−1/x = Ae−1/x in which we have written A = ek . Since k could be any number, A can be any positive number. Then 1 + y = ±Ae−1/x = Be−1/x in which B = ±A can be any nonzero number. The general solution is y = −1 + Be−1/x in which B is any nonzero number. Now revisit the assumption that x = 0 and y = −1. In the general solution, we actually obtain y = −1 if we allow B = 0. Further, the constant function y x = −1 does satisfy x2 y = 1 + y. Thus, by allowing B to be any number, including 0, the general solution y x = −1 + Be−1/x contains all the solutions we have found. In this example, y = −1 is a solution, but not a singular solution, since it occurs as a special case of the general solution. Figure 1.8 shows graphs of solutions corresponding to B = −8 −5 0 4 and 7. y Bϭ7 4 Bϭ4 2 x 0 1 2 3 4 5 Ϫ2 Bϭ0 Ϫ4 B ϭ Ϫ5 Ϫ6 B ϭ Ϫ8 Ϫ8 FIGURE 1.8 Integral curves of x2 y = 1 + y for B = 0 4 7 −5, and −8. We often solve an initial value problem by finding the general solution of the differential equation, then solving for the appropriate choice of the constant. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
14. 14. Licensed to: iChapters User 14 CHAPTER 1 First-Order Differential Equations EXAMPLE 1.9 Solve the initial value problem y = y2 e−x y 1 = 4 We know from Example 1.7 that the general solution of y = y2 e−x is 1 yx = e−x − k Now we need to choose k so that 1 y1 = =4 e−1 − k from which we get 1 k = e−1 − 4 The solution of the initial value problem is 1 yx = e−x + 4 − e−1 1 EXAMPLE 1.10 The general solution of x−1 2 y =y y+3 is implicitly defined by 1 y + 3 ln y = x−1 3 +k (1.3) 3 To obtain the solution satisfying y 3 = −1, put x = 3 and y = −1 into equation (1.3) to obtain 1 3 −1 = 2 +k 3 hence 11 k=− 3 The solution of this initial value problem is implicitly defined by 1 11 y + 3 ln y = x−1 3 − 3 3 1.2.1 Some Applications of Separable Differential Equations Separable equations arise in many contexts, of which we will discuss three. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
15. 15. Licensed to: iChapters User 1.2 Separable Equations 15 EXAMPLE 1.11 (The Mathematical Policewoman) A murder victim is discovered, and a lieutenant from the forensic science laboratory is summoned to estimate the time of death. The body is located in a room that is kept at a constant 68 degrees Fahrenheit. For some time after the death, the body will radiate heat into the cooler room, causing the body’s temperature to decrease. Assuming (for want of better information) that the victim’s temperature was a “normal” 98 6 at the time of death, the lieutenant will try to estimate this time by observing the body’s current temperature and calculating how long it would have had to lose heat to reach this point. According to Newton’s law of cooling, the body will radiate heat energy into the room at a rate proportional to the difference in temperature between the body and the room. If T t is the body temperature at time t, then for some constant of proportionality k, T t = k T t − 68 The lieutenant recognizes this as a separable differential equation and writes 1 dT = k dt T − 68 Upon integrating, she gets ln T − 68 = kt + C Taking exponentials, she gets T − 68 = ekt+C = Aekt in which A = eC . Then T − 68 = ±Aekt = Bekt Then T t = 68 + Bekt Now the constants k and B must be determined, and this requires information. The lieutenant arrived at 9:40 p.m. and immediately measured the body temperature, obtaining 94 4 degrees. Letting 9:40 be time zero for convenience, this means that T 0 = 94 4 = 68 + B and so B = 26 4. Thus far, T t = 68 + 26 4ekt To determine k, the lieutenant makes another measurement. At 11:00 she finds that the body temperature is 89 2 degrees. Since 11:00 is 80 minutes past 9:40, this means that T 80 = 89 2 = 68 + 26 4e80k Then 21 2 e80k = 26 4 so 21 2 80k = ln 26 4 Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
16. 16. Licensed to: iChapters User 16 CHAPTER 1 First-Order Differential Equations and 1 21 2 k= ln 80 26 4 The lieutenant now has the temperature function: T t = 68 + 26 4eln 21 2/26 4 t/80 In order to find when last time when the body was 98 6 (presumably the time of death), solve for the time in T t = 98 6 = 68 + 26 4eln 21 2/26 4 t/80 To do this, the lieutenant writes 30 6 = eln 21 2/26 4 t/80 26 4 and takes the logarithm of both sides to obtain 30 6 t 21 2 ln = ln 26 4 80 26 4 Therefore the time of death, according to this mathematical model, was 80 ln 30 6/26 4 t= ln 21 2/26 4 which is approximately −53 8 minutes. Death occurred approximately 53 8 minutes before (because of the negative sign) the first measurement at 9:40, which was chosen as time zero. This puts the murder at about 8:46 p.m. EXAMPLE 1.12 (Radioactive Decay and Carbon Dating) In radioactive decay, mass is converted to energy by radiation. It has been observed that the rate of change of the mass of a radioactive substance is proportional to the mass itself. This means that, if m t is the mass at time t, then for some constant of proportionality k that depends on the substance, dm = km dt This is a separable differential equation. Write it as 1 dm = k dt m and integrate to obtain ln m = kt + c Since mass is positive, m = m and ln m = kt + c Then m t = ekt+c = Aekt in which A can be any positive number. Copyright 2007 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.