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Higher order differential equation

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Higher order differential equation

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Higher order differential equation

  1. 1. AdvancedEngineeringMathematics(2131904)
  2. 2. Enrollmentnumber. Name Electronic and communication Batch B
  3. 3. Content β€’ Introduction β€’ Stepsto solve Higher Order Differential Equation β€’ Auxiliary Equation (A.E) β€’ Complementary function (C.F.) β€’ Particular Integral (P.I.) β€’ Linear Differential eqn with Constantcoefficient β€’ General Method β€’ Shortcut Method β€’ Method of UndeterminedCoefficient β€’ Method of Variation Parameter (WronkianMethod) β€’ Linear Differential eqn with Variablecoefficient β€’ Cauchy-EulerMethod β€’ Legendre’sMethod (Variablecoefficient)
  4. 4. Linear Differential Equation:- It isintheformof, 𝒅 𝒏 π’š 𝒅 π’βˆ’πŸ π’š 𝒅 π’šπ’…π’™ 𝒏 + 𝒂 π’βˆ’πŸ 𝒅𝒙 π’βˆ’πŸ +β‹―+ 𝒂 𝟏 𝒅𝒙 + 𝒂 𝟎 π’š =𝑹(𝒙) 𝒅 𝒏 π’š 𝒅 π’βˆ’πŸ π’š 𝒅 π’šπ’…π’™ 𝒏 +(𝑿 +𝒂 π’βˆ’πŸ) 𝒅𝒙 π’βˆ’πŸ +β‹―+(𝑿 +𝒂 𝟏) 𝒅𝒙 +(𝑿 +𝒂 𝟎 π’š) = 𝑹(𝒙) constantcoefficient Vairablecoefficient
  5. 5. Non-homogenous Linear D.E. β€’ In this R.H.Sof D.E.is not zero/is having 𝑓(π‘₯) i.e 𝒅𝒙 𝒏 𝒅 𝒏 π’š +𝒂 π’βˆ’πŸ 𝒅𝒙 π’βˆ’πŸ 𝟏 𝒅𝒙 𝟎 𝒅 π’βˆ’πŸ π’š +β‹―+𝒂 π’…π’š +𝒂 π’š =𝒇(𝒙) Example :- 𝑑2 𝑦 𝑑𝑦 (1) 𝑑π‘₯2 +9 𝑑π‘₯ + 𝑦 =cos π‘₯ (2) 𝑦′′ +39𝑦′ + 𝑦 = 𝑒 π‘₯ (3) 𝑦4+ 𝑦3+3𝑦2βˆ’9𝑦1=log π‘₯+sin π‘₯cos π‘₯+π‘₯βˆ’2
  6. 6. Non - Linear DifferentialEquation β€’ The term homogenous and non homogenous have no meaningfor nonlinearequation. Examples:- (1) 𝑑2 𝑦 = π‘₯ 1+ 𝑑𝑦 𝑑π‘₯2 𝑑π‘₯ 2 3 2 (2) 𝑑2 πœƒ + 𝑔 sin πœƒ=0 𝑑𝑑2 𝑙
  7. 7. StepstosolveLinear D.E. -IdentifyAuxiliaryEquation(A.E.),Byputting 𝑑 𝑛 𝑑 π‘₯ 𝑛 𝑑 π‘₯2 = 𝐷 𝑛 i.e. 𝑑2 𝑦 = 𝐷2 𝑦 - FindtherootsofA.E.byputtingD=minit andequatingwithitzero. i.e.A.E.=0 - Accordingorootsobtainedfind, ComplimentaryFunction (C.F.)= 𝑦𝑐 - Find Particular Integral(P.I.)= 𝑦𝑝,fromtheR.H.S.oflinear NonHomogenous Equation. - Findcompletesolution/ GeneralSolution(𝑦) =𝑦𝑐+𝑦𝑝
  8. 8. Auxiliary Equation(A.E.) 𝑑2 𝑦 𝑑𝑦 (1) +2 + 𝑦 =sin(𝑒 π‘₯) 𝑑π‘₯2 𝑑π‘₯  𝐷2 𝑦 +2𝐷𝑦 +𝑦 =sin(𝑒 π‘₯)  𝑫 𝟐 + πŸπ‘« + 𝟏 𝑦 =sin(𝑒 π‘₯) A. E.
  9. 9. Formulae for FindingRoots ο‚§ π‘Ž2 Β±2π‘Žπ‘ + 𝑏2 = π‘Ž ±𝑏 2 ο‚§ π‘Ž3 + 𝑏3 +3π‘Žπ‘ π‘Ž +𝑏 ο‚§ π‘Ž3 βˆ’ 𝑏3 βˆ’3π‘Žπ‘ π‘Ž βˆ’π‘ = π‘Ž3 + 𝑏3 +3π‘Ž2 𝑏 + 3π‘Žπ‘2 = 𝒂 + 𝒃 3 = π‘Ž3 βˆ’ 𝑏3 βˆ’3π‘Ž2 𝑏 + 3π‘Žπ‘2 = π‘Ž βˆ’ 𝑏 3 ο‚§ π‘Ž2 βˆ’π‘2 = π‘Ž +𝑏 π‘Ž βˆ’π‘ ο‚§ 𝒂 𝟐 + 𝒃 𝟐 β‡’ 𝒂 𝟐 =βˆ’π’ƒ 𝟐 β‡’ 𝒂 =Β±π’ƒπ’Š ο‚§ π‘Ž3 +𝑏3 = π‘Ž +𝑏 π‘Ž2 βˆ’ π‘Žπ‘ +𝑏2 ο‚§ π‘Ž3 βˆ’ 𝑏3 =(π‘Ž βˆ’ 𝑏)(π‘Ž2 + π‘Žπ‘ +𝑏2)
  10. 10. 𝒂 πŸ’ βˆ’π’ƒ πŸ’ = π‘Ž2 2 βˆ’ 𝑏2 2 β–  = π‘Ž2 βˆ’π‘2 π‘Ž2+𝑏2 β–  = π‘Žβˆ’π‘ π‘Ž+ 𝑏(π‘Ž2 +𝑏2) β€’ 𝒂 πŸ’ + 𝒃 πŸ’ = π‘Ž4+ 𝑏4+2π‘Ž2 𝑏2βˆ’2π‘Ž2 𝑏2 (FindMiddleTerm) = π‘Ž2 2 +2π‘Ž2 𝑏2 + 𝑏2 2 βˆ’ 2π‘Ž2 𝑏2 = π‘Ž2 +𝑏2 2 βˆ’ 2 π‘Žπ‘ 2 =(π‘Ž2 + 𝑏2 βˆ’ 2 π‘Žπ‘)(π‘Ž2 + 𝑏2 + 2π‘Žπ‘) If equationisinformof,𝑨𝒙 𝟐 +𝑩𝒙 +π‘ͺ then,𝒙 =βˆ’π‘© Β± 𝑩 πŸβˆ’ πŸ’π‘¨π‘ͺ 𝟐 𝑨 ORSeparatethemiddleterm(Bπ‘₯) in suchwaythattheir addition or substractionbethemultiple ofA&C.
  11. 11. Solved Example (1)Find therootsof:- πŸ‘π’šβ€²β€² βˆ’ π’šβ€² βˆ’ πŸπ’š= 𝒆 𝒙 𝑑 π‘₯2 𝑑π‘₯  3 𝑑2 𝑦 βˆ’ 𝑑𝑦 βˆ’2𝑦 =𝑒 π‘₯   3π‘š +2 π‘š βˆ’1 =0 3π‘š +2=0 and 3𝐷2 𝑦 βˆ’ 𝐷𝑦 βˆ’2𝑦 =𝑒 π‘₯ 3𝐷2 βˆ’ 𝐷 βˆ’2 𝑦 = 𝑒 π‘₯ Let,A.E.=0 and put D= m  3π‘š2 βˆ’ π‘š βˆ’2=0  3π‘š2 βˆ’3π‘š +2π‘š βˆ’2=0  3π‘š π‘š βˆ’1 +2 π‘š βˆ’1 =0    π’Ž 𝟏 =βˆ’ 𝟐 πŸ‘ π‘š βˆ’1=0 and π’Ž 𝟐 =𝟏 2Γ—3=6 2 3 -1 =-3 +2
  12. 12. (2) Find the rootsof : 𝑫 πŸ’ + π’Œ πŸ’ π’š =𝟎LetA.E.=0adputD=m  π‘š4 + π‘˜4 =0  π‘š2 2 +2π‘š2 π‘˜2 + π‘˜2 2 βˆ’ 2π‘š2 π‘˜2 =0  π‘š2 +π‘˜2 2 βˆ’ 2π‘šπ‘˜ 2 =0 (π‘š2 + π‘˜2 βˆ’ οœπ‘š2 + π‘˜2 βˆ’ 2 π‘šπ‘˜)(π‘š2 + π‘˜2 + 2 π‘šπ‘˜ =0 π‘š2+π‘˜2 + 2 π‘šπ‘˜ =0  π‘š = 2π‘˜Β± 2π‘˜2βˆ’4π‘˜2 2 π‘š2 =βˆ’ 2π‘˜Β± 2π‘˜2βˆ’4π‘˜2 21  π’Ž 𝟏 = 𝟐 𝟐 π’Œ Β± π’Œ π’Š 2 π‘šπ‘˜) =0 and and and π’Ž 𝟐 =βˆ’π’Œ Β± π’Œ π’Š 𝟐 𝟐
  13. 13. ComplimentaryFunction β€’ FromtherootsofA.E.,C.F.( 𝑦𝑐) ofD.E.is decided.C.F.is alwaysin termsof 𝑦𝑐= 𝐢1 𝑦1+𝐢2 𝑦2 - If therootsarereal&district(unequal),then π’šπ’„ = 𝒄 𝟏 𝒆 π’Ž 𝟏 𝒙 + 𝒄 𝟐 𝒆 π’Ž 𝟐 𝒙 +β‹―+ 𝒄 𝒏 𝒆 π’Ž 𝒏 𝒙 Example:- If roots are π‘š1 =2& π‘š2 =βˆ’3then, 𝑦𝑐= 𝑐1 𝑒2π‘₯+ 𝑐2 π‘’βˆ’3π‘₯ - Iftherootsarereal&equalthen, π’šπ’„ = 𝒄 𝟏 + 𝒄 𝟐 𝒙 + 𝒄 πŸ‘ 𝒙 𝟐 +β‹― 𝒆 π’Ž 𝟏 𝒙 Example:- If roots are π‘š1 = π‘š2 =βˆ’3then, 𝑦𝑐=(𝑐1+ 𝑐2 π‘₯) π‘’βˆ’3π‘₯
  14. 14. - If therootsarecomplexthen,i.e.rootsin theformof(𝛼±𝛽𝑖) π’šπ’„ = 𝒆 πœΆπ’™(𝒄 𝟏 𝐜𝐨𝐬 𝒙 + 𝒄 𝟐 𝐬𝐒 𝐧 𝒙) Example:- 1 (1) If rootsis π‘š =2 Β± 3𝑖then, 𝑦 = 𝑒 1 π‘₯ 𝑐 cos 3π‘₯+𝑐 sin 3π‘₯ 𝑐 2 1 2 (2) If root is π‘š =Β±3𝑖then, 𝑦𝑐 = 𝑒0π‘₯ 𝑐1cos3π‘₯+ 𝑐2sin3π‘₯ = 𝑐1cos3π‘₯+ 𝑐2sin3π‘₯ - If therootsarecomplex&repeatedthen, π’šπ’„ = 𝒆 πœΆπ’™ 𝒄 𝟏 + 𝒄 𝟐 𝒙 𝒄𝒐𝒔 𝒙 + 𝒄 πŸ‘ + 𝒄 πŸ’ 𝒙 π’”π’Š 𝒏 𝒙 - If therootsarecomplex&realboththen, π’šπ’„ = 𝒄 𝟏 𝒆 π’Ž 𝟏 𝒙 + 𝒄 𝟐 𝒆 π’Ž 𝟐 𝒙 + 𝒆 πœΆπ’™ (𝒄 πŸ‘ 𝐜𝐨𝐬 𝒙 + 𝒄 πŸ’ 𝐬𝐒 𝐧 𝒙)
  15. 15. NOTE :- β€’ IftheR.H.S.=0ofgivenD.E.i.e.forHomogenousLinear D.E. π’š 𝒑 = 𝟎 andhencethegeneralsolution/finalsolutionisgivenby, π’š = π’šπ’„
  16. 16. MethodsforFindingParticular Integral β€’ Linear Differential eqn with Constant coefficient β€’ General Method β€’ Shortcut Method β€’ Method of UndeterminedCoefficient β€’ Method of Variation Parameter (WronkianMethod) β€’ Linear Differential eqn with Variablecoefficient β€’ Cauchy-Euler Method β€’ Legendre’s Method (Variable coefficient)
  17. 17. General Method
  18. 18. Solve by using general method:- (1) 𝐷2 +3𝐷 +2 𝑦 =𝑒 𝑒 π‘₯ (2) 𝐷2 +1 𝑦 =sec2 π‘₯
  19. 19. ShortcutMethod
  20. 20. SolvedExample 𝒅 𝟐 π’š πŸ”π’…π’š (1) Solve :- + + πŸ—π’š = πŸ“ 𝒆 πŸ‘π’™ 𝒅𝒙 𝟐 𝒅𝒙
  21. 21. MethodofVariation Parameter β€’ Stepsto solve linearD.E. - Find out 𝑦𝑐 - Compared with it 𝑦𝑐= 𝑐1 𝑦1+ 𝑐2 𝑦2and find 𝑦1&𝑦2 - Solve π‘Š = 𝑦1β€² 𝑦1 𝑦2 𝑦2β€² 0 𝑦2 , W1= 1 𝑦2β€² , π‘Š 2= 𝑦1β€² 𝑦1 0 1 - Find 𝑦𝑝 = 𝑦1∫ 𝑀1 𝑅 π‘₯ 𝑑π‘₯ + 𝑦2 ∫ 𝑀2 𝑅 π‘₯ 𝑑π‘₯ 𝑀 𝑀
  22. 22. SolvedExample:- (1) Solve by Variation parameter method:- 𝒅 𝟐 π’šπ’… 𝒙 𝟐 + π’š = π¬πžπœπ’™
  23. 23. Exercise:-
  24. 24. Exercise :-

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