Higher order differential equation

1. AdvancedEngineeringMathematics(2131904)

2. Enrollmentnumber. Name
Electronic and communication
Batch B

3. Content
• Introduction
• Stepsto solve Higher Order Differential Equation
• Auxiliary Equation (A.E)
• Complementary function (C.F.)
• Particular Integral (P.I.)
• Linear Differential eqn with Constantcoefficient
• General Method
• Shortcut Method
• Method of UndeterminedCoefficient
• Method of Variation Parameter (WronkianMethod)
• Linear Differential eqn with Variablecoefficient
• Cauchy-EulerMethod
• Legendre’sMethod (Variablecoefficient)

4. Linear Differential Equation:-
It isintheformof,
𝒅 𝒏
𝒚
𝒅 𝒏−𝟏 𝒚 𝒅
𝒚𝒅𝒙 𝒏 + 𝒂 𝒏−𝟏
𝒅𝒙 𝒏−𝟏 +⋯+ 𝒂 𝟏
𝒅𝒙
+ 𝒂 𝟎 𝒚 =𝑹(𝒙)
𝒅 𝒏
𝒚
𝒅 𝒏−𝟏 𝒚 𝒅
𝒚𝒅𝒙 𝒏 +(𝑿 +𝒂 𝒏−𝟏)
𝒅𝒙 𝒏−𝟏 +⋯+(𝑿 +𝒂 𝟏)
𝒅𝒙
+(𝑿 +𝒂 𝟎 𝒚) = 𝑹(𝒙)
constantcoefficient
Vairablecoefficient

5. Non-homogenous Linear D.E.
• In this R.H.Sof D.E.is not zero/is having 𝑓(𝑥) i.e
𝒅𝒙
𝒏
𝒅 𝒏 𝒚
+𝒂 𝒏−𝟏 𝒅𝒙 𝒏−𝟏 𝟏 𝒅𝒙 𝟎
𝒅 𝒏−𝟏 𝒚
+⋯+𝒂 𝒅𝒚
+𝒂 𝒚 =𝒇(𝒙)
Example :-
𝑑2 𝑦 𝑑𝑦
(1) 𝑑𝑥2 +9 𝑑𝑥
+ 𝑦 =cos 𝑥
(2) 𝑦′′ +39𝑦′ + 𝑦 = 𝑒 𝑥
(3) 𝑦4+ 𝑦3+3𝑦2−9𝑦1=log 𝑥+sin 𝑥cos 𝑥+𝑥−2

6. Non - Linear DifferentialEquation
• The term homogenous and non homogenous have no
meaningfor nonlinearequation.
Examples:-
(1) 𝑑2 𝑦
= 𝑥 1+
𝑑𝑦
𝑑𝑥2 𝑑𝑥
2
3
2
(2) 𝑑2 𝜃
+ 𝑔
sin 𝜃=0
𝑑𝑡2 𝑙

7. StepstosolveLinear D.E.
-IdentifyAuxiliaryEquation(A.E.),Byputting
𝑑 𝑛
𝑑 𝑥 𝑛 𝑑 𝑥2
= 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. Auxiliary Equation(A.E.)
𝑑2 𝑦 𝑑𝑦
(1) +2 + 𝑦 =sin(𝑒 𝑥)
𝑑𝑥2 𝑑𝑥
 𝐷2 𝑦 +2𝐷𝑦 +𝑦 =sin(𝑒 𝑥)
 𝑫 𝟐 + 𝟐𝑫 + 𝟏 𝑦 =sin(𝑒 𝑥)
A. E.

9. Formulae for FindingRoots
 𝑎2 ±2𝑎𝑏 + 𝑏2 = 𝑎 ±𝑏 2
 𝑎3 + 𝑏3 +3𝑎𝑏 𝑎 +𝑏
 𝑎3 − 𝑏3 −3𝑎𝑏 𝑎 −𝑏
= 𝑎3 + 𝑏3 +3𝑎2 𝑏 + 3𝑎𝑏2 = 𝒂 + 𝒃 3
= 𝑎3 − 𝑏3 −3𝑎2 𝑏 + 3𝑎𝑏2 = 𝑎 − 𝑏 3
 𝑎2 −𝑏2 = 𝑎 +𝑏 𝑎 −𝑏
 𝒂 𝟐 + 𝒃 𝟐 ⇒ 𝒂 𝟐 =−𝒃 𝟐
⇒ 𝒂 =±𝒃𝒊
 𝑎3 +𝑏3 = 𝑎 +𝑏 𝑎2 − 𝑎𝑏 +𝑏2
 𝑎3 − 𝑏3 =(𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 +𝑏2)

10. 𝒂 𝟒 −𝒃 𝟒 = 𝑎2 2 − 𝑏2 2
■ = 𝑎2 −𝑏2 𝑎2+𝑏2
■ = 𝑎−𝑏 𝑎+ 𝑏(𝑎2 +𝑏2)
• 𝒂 𝟒 + 𝒃 𝟒 = 𝑎4+ 𝑏4+2𝑎2 𝑏2−2𝑎2 𝑏2 (FindMiddleTerm)
= 𝑎2 2 +2𝑎2 𝑏2 + 𝑏2 2 − 2𝑎2 𝑏2
= 𝑎2 +𝑏2 2 − 2 𝑎𝑏
2
=(𝑎2 + 𝑏2 − 2 𝑎𝑏)(𝑎2 + 𝑏2 + 2𝑎𝑏)
If equationisinformof,𝑨𝒙 𝟐 +𝑩𝒙 +𝑪 then,𝒙 =−𝑩 ± 𝑩 𝟐− 𝟒𝑨𝑪
𝟐
𝑨
ORSeparatethemiddleterm(B𝑥) in suchwaythattheir addition or
substractionbethemultiple ofA&C.

11. Solved Example
(1)Find therootsof:- 𝟑𝒚′′ − 𝒚′ − 𝟐𝒚= 𝒆 𝒙
𝑑 𝑥2 𝑑𝑥
 3 𝑑2 𝑦
− 𝑑𝑦
−2𝑦 =𝑒 𝑥


3𝑚 +2 𝑚 −1 =0
3𝑚 +2=0 and
3𝐷2 𝑦 − 𝐷𝑦 −2𝑦 =𝑒 𝑥
3𝐷2 − 𝐷 −2 𝑦 = 𝑒 𝑥
Let,A.E.=0 and put D= m
 3𝑚2 − 𝑚 −2=0
 3𝑚2 −3𝑚 +2𝑚 −2=0
 3𝑚 𝑚 −1 +2 𝑚 −1 =0


 𝒎 𝟏 =− 𝟐
𝟑
𝑚 −1=0
and 𝒎 𝟐 =𝟏
2×3=6
2 3
-1 =-3 +2

12. (2) Find the rootsof : 𝑫 𝟒 +
𝒌 𝟒 𝒚 =𝟎LetA.E.=0adputD=m
 𝑚4 + 𝑘4 =0
 𝑚2 2 +2𝑚2 𝑘2 + 𝑘2 2 − 2𝑚2 𝑘2 =0
 𝑚2 +𝑘2 2 − 2𝑚𝑘
2
=0
(𝑚2 + 𝑘2 −
𝑚2 + 𝑘2 −
2 𝑚𝑘)(𝑚2 + 𝑘2 +
2 𝑚𝑘 =0 𝑚2+𝑘2 + 2 𝑚𝑘 =0
 𝑚 =
2𝑘± 2𝑘2−4𝑘2
2
𝑚2 =− 2𝑘± 2𝑘2−4𝑘2
21
 𝒎 𝟏 =
𝟐 𝟐
𝒌
± 𝒌
𝒊
2 𝑚𝑘) =0
and
and
and 𝒎 𝟐 =−𝒌
± 𝒌
𝒊
𝟐 𝟐

13. ComplimentaryFunction
• FromtherootsofA.E.,C.F.( 𝑦𝑐) ofD.E.is decided.C.F.is alwaysin termsof 𝑦𝑐=
𝐶1 𝑦1+𝐶2 𝑦2
- If therootsarereal&district(unequal),then
𝒚𝒄 = 𝒄 𝟏 𝒆 𝒎 𝟏 𝒙 + 𝒄 𝟐 𝒆 𝒎 𝟐 𝒙 +⋯+ 𝒄 𝒏 𝒆 𝒎 𝒏 𝒙
Example:- If roots are 𝑚1 =2& 𝑚2 =−3then, 𝑦𝑐= 𝑐1 𝑒2𝑥+ 𝑐2 𝑒−3𝑥
- Iftherootsarereal&equalthen,
𝒚𝒄 = 𝒄 𝟏 + 𝒄 𝟐 𝒙 + 𝒄 𝟑 𝒙 𝟐 +⋯ 𝒆 𝒎 𝟏 𝒙
Example:- If roots are 𝑚1 = 𝑚2 =−3then, 𝑦𝑐=(𝑐1+ 𝑐2 𝑥) 𝑒−3𝑥

14. - If therootsarecomplexthen,i.e.rootsin theformof(𝛼±𝛽𝑖)
𝒚𝒄 = 𝒆 𝜶𝒙(𝒄 𝟏 𝐜𝐨𝐬 𝒙 + 𝒄 𝟐 𝐬𝐢 𝐧 𝒙)
Example:-
1
(1) If rootsis 𝑚 =2
± 3𝑖then, 𝑦 = 𝑒
1
𝑥
𝑐 cos 3𝑥+𝑐 sin 3𝑥
𝑐 2 1 2
(2) If root is 𝑚 =±3𝑖then, 𝑦𝑐 = 𝑒0𝑥 𝑐1cos3𝑥+ 𝑐2sin3𝑥
= 𝑐1cos3𝑥+ 𝑐2sin3𝑥
- If therootsarecomplex&repeatedthen,
𝒚𝒄 = 𝒆 𝜶𝒙 𝒄 𝟏 + 𝒄 𝟐 𝒙 𝒄𝒐𝒔 𝒙 + 𝒄 𝟑 + 𝒄 𝟒 𝒙 𝒔𝒊 𝒏 𝒙
- If therootsarecomplex&realboththen,
𝒚𝒄 = 𝒄 𝟏 𝒆 𝒎 𝟏 𝒙 + 𝒄 𝟐 𝒆 𝒎 𝟐 𝒙 + 𝒆 𝜶𝒙 (𝒄 𝟑 𝐜𝐨𝐬 𝒙 + 𝒄 𝟒 𝐬𝐢 𝐧 𝒙)

15. NOTE :-
• IftheR.H.S.=0ofgivenD.E.i.e.forHomogenousLinear
D.E. 𝒚 𝒑 = 𝟎 andhencethegeneralsolution/finalsolutionisgivenby, 𝒚 =
𝒚𝒄

16. MethodsforFindingParticular Integral
• Linear Differential eqn with Constant coefficient
• General Method
• Shortcut Method
• Method of UndeterminedCoefficient
• Method of Variation Parameter (WronkianMethod)
• Linear Differential eqn with Variablecoefficient
• Cauchy-Euler Method
• Legendre’s Method (Variable coefficient)

17. General Method

18. Solve by using general method:-
(1) 𝐷2 +3𝐷 +2 𝑦 =𝑒 𝑒 𝑥
(2) 𝐷2 +1 𝑦 =sec2 𝑥

19. ShortcutMethod

21. SolvedExample
𝒅 𝟐 𝒚 𝟔𝒅𝒚
(1) Solve :- + + 𝟗𝒚 = 𝟓 𝒆 𝟑𝒙
𝒅𝒙 𝟐 𝒅𝒙

22. MethodofVariation Parameter
• Stepsto solve linearD.E.
- Find out 𝑦𝑐
- Compared with it 𝑦𝑐= 𝑐1 𝑦1+ 𝑐2 𝑦2and find 𝑦1&𝑦2
- Solve 𝑊 = 𝑦1′
𝑦1 𝑦2
𝑦2′
0 𝑦2
, W1= 1 𝑦2′ , 𝑊 2= 𝑦1′
𝑦1 0
1
- Find 𝑦𝑝 = 𝑦1∫ 𝑤1
𝑅 𝑥 𝑑𝑥 + 𝑦2 ∫ 𝑤2
𝑅 𝑥 𝑑𝑥
𝑤 𝑤

23. SolvedExample:-
(1) Solve by Variation parameter method:-
𝒅 𝟐
𝒚𝒅
𝒙
𝟐 + 𝒚 = 𝐬𝐞𝐜𝒙

25. Exercise:-

33. Exercise :-