Higher Order Differential Equation
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The document provides an overview of advanced engineering mathematics concepts for differential equations. It covers topics such as homogeneous and non-homogeneous linear differential equations with constant and variable coefficients. Methods for solving differential equations are discussed, including finding the auxiliary equation, complementary function, and particular integral. Specific solving techniques like the Cauchy-Euler and Legendre methods for variable coefficient equations are also mentioned. Examples of different types of differential equations are provided throughout.
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Higher Order Differential Equation
- 1. Advanced Engineering Mathematics (2131904) B.E. MECH – Sem III rd Prepared by, Patel Shrey B (170953119031) DD 42 – D2D – B2 Guided by, Miss. Chaitali Shah (Applied Mathematics Deptt.)
- 2. Content • Introduction • Steps to solve Higher Order Differential Equation • Auxiliary Equation (A.E) • Complementary function (C.F.) • Particular Integral (P.I.) • Linear Differential eqn with Constant coefficient • General Method • Shortcut Method • Method of Undetermined Coefficient • Method of Variation Parameter (Wronkian Method) • Linear Differential eqn with Variable coefficient • Cauchy-Euler Method • Legendre’s Method (Variable coefficient)
- 3. Linear Differential Equation :- It is in the form of, 𝒅 𝒏 𝒚 𝒅𝒙 𝒏 + 𝒂 𝒏−𝟏 𝒅 𝒏−𝟏 𝒚 𝒅𝒙 𝒏−𝟏 + ⋯ + 𝒂 𝟏 𝒅𝒚 𝒅𝒙 + 𝒂 𝟎 𝒚 = 𝑹(𝒙) 𝒅 𝒏 𝒚 𝒅𝒙 𝒏 + (𝑿 + 𝒂 𝒏−𝟏) 𝒅 𝒏−𝟏 𝒚 𝒅𝒙 𝒏−𝟏 + ⋯ + (𝑿 + 𝒂 𝟏) 𝒅𝒚 𝒅𝒙 + (𝑿 + 𝒂 𝟎 𝒚) = 𝑹(𝒙) constant coefficient Vairable coefficient
- 4. Homogenous Linear D.E. • In this R.H.S of D.E. is zero i.e 𝒅 𝒏 𝒚 𝒅𝒙 𝒏 + 𝒂 𝒏−𝟏 𝒅 𝒏−𝟏 𝒚 𝒅𝒙 𝒏−𝟏 + ⋯ + 𝒂 𝟏 𝒅𝒚 𝒅𝒙 + 𝒂 𝟎 𝒚 = 𝟎 Example :- (1) 𝑑2 𝑦 𝑑𝑥2 + 9 𝑑𝑦 𝑑𝑥 + 𝑦 = 0 (2) 𝑦′′ + 39𝑦′ + 𝑦 = 0 (3) 𝑦4 + 𝑦3 + 3𝑦2 − 9𝑦1 = 0
- 5. Non-homogenous Linear D.E. • In this R.H.S of D.E. is not zero/is having 𝑓(𝑥) i.e 𝒅 𝒏 𝒚 𝒅𝒙 𝒏 + 𝒂 𝒏−𝟏 𝒅 𝒏−𝟏 𝒚 𝒅𝒙 𝒏−𝟏 + ⋯ + 𝒂 𝟏 𝒅𝒚 𝒅𝒙 + 𝒂 𝟎 𝒚 = 𝒇(𝒙) Example :- (1) 𝑑2 𝑦 𝑑𝑥2 + 9 𝑑𝑦 𝑑𝑥 + 𝑦 = cos 𝑥 (2) 𝑦′′ + 39𝑦′ + 𝑦 = 𝑒 𝑥 (3) 𝑦4 + 𝑦3 + 3𝑦2 − 9𝑦1 = log 𝑥 + sin 𝑥 cos 𝑥 + 𝑥−2
- 6. Non - Linear Differential Equation • The term homogenous and non homogenous have no meaning for non linear equation. Examples :- (1) 𝑑2 𝑦 𝑑𝑥2 = 𝑥 1 + 𝑑𝑦 𝑑𝑥 2 3 2 (2) 𝑑2 𝜃 𝑑𝑡2 + 𝑔 𝑙 sin 𝜃 = 0
- 7. Steps to solve Linear D.E. - Identify Auxiliary Equation (A.E.) , By putting 𝑑 𝑛 𝑑𝑥 𝑛 = 𝐷 𝑛 i.e. 𝑑2 𝑦 𝑑𝑥2 = 𝐷2 𝑦 - Find the roots of A.E. by putting D = m in it and equating with it zero. i.e. A.E. = 0 - According o roots obtained find, Complimentary Function (C.F.) = 𝑦𝑐 - Find Particular Integral (P.I.) = 𝑦𝑝 , from the R.H.S. of linear Non Homogenous Equation. - Find complete solution / General Solution (𝑦) = 𝑦𝑐 + 𝑦𝑝
- 8. Auxiliary Equation (A.E.) (1) 𝑑2 𝑦 𝑑𝑥2 + 2 𝑑𝑦 𝑑𝑥 + 𝑦 = sin(𝑒 𝑥) 𝐷2 𝑦 + 2𝐷𝑦 + 𝑦 = sin(𝑒 𝑥) 𝑫 𝟐 + 𝟐𝑫 + 𝟏 𝑦 = sin(𝑒 𝑥) A. E.
- 9. Formulae for Finding Roots ▪ 𝑎2 ± 2𝑎𝑏 + 𝑏2 = 𝑎 ± 𝑏 2 ▪ 𝑎3 + 𝑏3 + 3𝑎𝑏 𝑎 + 𝑏 = 𝑎3 + 𝑏3 + 3𝑎2 𝑏 + 3𝑎𝑏2 = 𝒂 + 𝒃 3 ▪ 𝑎3 − 𝑏3 − 3𝑎𝑏 𝑎 − 𝑏 = 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 (Find Middle Term) = 𝑎2 2 + 2𝑎2 𝑏2 + 𝑏2 2 − 2𝑎2 𝑏2 = 𝑎2 + 𝑏2 2 − 2 𝑎𝑏 2 = (𝑎2 + 𝑏2 − 2 𝑎𝑏)(𝑎2 + 𝑏2 + 2 𝑎𝑏) If equation is in form of, 𝑨𝒙 𝟐 + 𝑩𝒙 + 𝑪 then, 𝒙 = −𝑩 ± 𝑩 𝟐− 𝟒𝑨𝑪 𝟐𝑨 OR Separate the middle term (B𝑥) in such way that their addition or substraction be the multiple of A & C .
- 11. Solved Example (1)Find the roots of :- 𝟑𝒚′′ − 𝒚′ − 𝟐𝒚 = 𝒆 𝒙 3 𝑑2 𝑦 𝑑𝑥2 − 𝑑𝑦 𝑑𝑥 − 2𝑦 = 𝑒 𝑥 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 3𝑚 + 2 𝑚 − 1 = 0 3𝑚 + 2 = 0 and 𝑚 − 1 = 0 𝒎 𝟏 = − 𝟐 𝟑 and 𝒎 𝟐 = 𝟏 2×3 = 6 2 3 -1 = -3 +2
- 12. (2) Find the roots of : 𝑫 𝟒 + 𝒌 𝟒 𝒚 = 𝟎 Let A.E. = 0 ad put D = m 𝑚4 + 𝑘4 = 0 𝑚2 2 + 2𝑚2 𝑘2 + 𝑘2 2 − 2𝑚2 𝑘2 = 0 𝑚2 + 𝑘2 2 − 2 𝑚𝑘 2 = 0 (𝑚2 + 𝑘2 − 2 𝑚𝑘)(𝑚2 + 𝑘2 + 2 𝑚𝑘) = 0 𝑚2 + 𝑘2 − 2 𝑚𝑘 = 0 and 𝑚2+𝑘2 + 2 𝑚𝑘 = 0 𝑚1 = 2𝑘± 2𝑘2−4𝑘2 2 and 𝑚2 = − 2𝑘± 2𝑘2−4𝑘2 2 𝒎 𝟏 = 𝒌 𝟐 ± 𝒌 𝟐 𝒊 and 𝒎 𝟐 = −𝒌 𝟐 ± 𝒌 𝟐 𝒊
- 13. Exercise • Find the roots of given Differential Equation :- (1) 𝐷2 + 1 𝑦 = 0 (2) y’’’– y’’ + 100y’ – 100y = 0 (3) 𝑑4 𝑦 𝑑𝑥4 − 𝑑3 𝑦 𝑑𝑥3 − 9 𝑑2 𝑦 𝑑𝑥2 − 11 𝑑𝑦 𝑑𝑥 − 4𝑦 = 0 (4) 𝐷4 + 𝑘4 𝑦 = 0 (5) (𝐷4 − 𝑘4)𝑦 = 0 (6) (𝐷2 + 6𝐷 + 4)𝑦 = 0 (7) 𝐷2 + 1 3 𝐷2 + 𝐷 + 1 2 𝑦 = 0 (8) 𝑦2 − 𝑦1 − 2𝑦 = sinh 2𝑥
- 14. Complimentary Function • From the roots of A.E. , C.F. ( 𝑦𝑐) of D.E. is decided. C.F. is always in terms of 𝑦𝑐= 𝐶1 𝑦1 + 𝐶2 𝑦2 - If the roots are real & district (unequal), then 𝒚 𝒄 = 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 = −3 then, 𝑦𝑐 = 𝑐1 𝑒2𝑥 + 𝑐2 𝑒−3𝑥 - If the roots are real & equal then, 𝒚 𝒄 = 𝒄 𝟏 + 𝒄 𝟐 𝒙 + 𝒄 𝟑 𝒙 𝟐 + ⋯ 𝒆 𝒎 𝟏 𝒙 Example :- If roots are 𝑚1 = 𝑚2 = −3 then, 𝑦𝑐 = (𝑐1 + 𝑐2 𝑥) 𝑒−3𝑥
- 15. - If the roots are complex then, i.e. roots in the form of (𝛼 ± 𝛽𝑖) 𝒚 𝒄 = 𝒆 𝜶𝒙 (𝒄 𝟏 𝐜𝐨𝐬 𝒙 + 𝒄 𝟐 𝐬𝐢𝐧 𝒙) Example :- (1) If roots is 𝑚 = 1 2 ± 3𝑖 then, 𝑦𝑐 = 𝑒 1 2 𝑥 𝑐1 cos 3𝑥 + 𝑐2 sin 3𝑥 (2) If root is 𝑚 = ±3𝑖 then, 𝑦𝑐 = 𝑒0𝑥 𝑐1 cos 3𝑥 + 𝑐2 sin 3𝑥 = 𝑐1 cos 3𝑥 + 𝑐2 sin 3𝑥 - If the roots are complex & repeated then, 𝒚 𝒄 = 𝒆 𝜶𝒙 𝒄 𝟏 + 𝒄 𝟐 𝒙 𝒄𝒐𝒔 𝒙 + 𝒄 𝟑 + 𝒄 𝟒 𝒙 𝒔𝒊𝒏 𝒙 - If the roots are complex & real both then, 𝒚 𝒄 = 𝒄 𝟏 𝒆 𝒎 𝟏 𝒙 + 𝒄 𝟐 𝒆 𝒎 𝟐 𝒙 + 𝒆 𝜶𝒙 (𝒄 𝟑 𝐜𝐨𝐬 𝒙 + 𝒄 𝟒 𝐬𝐢𝐧 𝒙)
- 16. NOTE :- • If the R.H.S. = 0 of given D.E. i.e. for Homogenous Linear D.E. 𝒚 𝒑 = 𝟎 and hence the general solution/final solution is given by, 𝒚 = 𝒚 𝒄
- 17. Solved Example (1) Solve :- 𝒅 𝟐 𝒙 𝒅𝒕 𝟐 + 𝟔 𝒅𝒙 𝒅𝒕 + 𝟗𝒙 = 𝟎 𝐷2 𝑥 + 6𝐷𝑥 + 9𝑥 = 0 𝐷2 + 6𝐷 + 9 𝑥 = 0 Let A.E. = 0 & put D = m 𝑚2 + 6𝑚 + 9 = 0 𝑚 + 3 2 = 0 𝑚1 = 𝑚2 = −3 - Roots are real and equal then, C.F. is given by, 𝑦𝑐 = 𝑐1 + 𝑐2 𝑡 𝑒−3𝑡 - Here R.H.S. = 0 then, 𝑦𝑝 = 0 & complete soln is given by, 𝒚 = 𝒚 𝒄 = 𝒄 𝟏 + 𝒄 𝟐 𝒕 𝒆−𝟑𝒕
- 18. (2) Solve :- 𝑫 𝟐 𝒚 + 𝟒𝑫𝒚 + 𝟓𝒚 = 𝟎 & Find the value of 𝒄 𝟏 & 𝒄 𝟐 if 𝒚 = 𝟐 & 𝒚 𝟐 = 𝒚 when 𝒙 = 𝟎
- 20. Methods for Finding Particular Integral • Linear Differential eqn with Constant coefficient • General Method • Shortcut Method • Method of Undetermined Coefficient • Method of Variation Parameter (Wronkian Method) • Linear Differential eqn with Variable coefficient • Cauchy-Euler Method • Legendre’s Method (Variable coefficient)
- 21. General Method
- 22. Solve by using general method :- (1) 𝐷2 + 3𝐷 + 2 𝑦 = 𝑒 𝑒 𝑥 (2) 𝐷2 + 1 𝑦 = sec2 𝑥
- 23. Shortcut Method
- 25. Solved Example (1) Solve :- 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 + 𝟔𝒅𝒚 𝒅𝒙 + 𝟗𝒚 = 𝟓 𝒆 𝟑𝒙
- 26. (2) Solve :- 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 − 𝟔 𝒅𝒚 𝒅𝒙 + 𝟗𝒚 = 𝟔𝒆 𝟑𝒙 + 𝟕𝒆−𝟐𝒙 − 𝐥𝐨𝐠 𝟐
- 27. (3) Solve :- 𝒅 𝟐 𝒙 𝒅𝒕 𝟐 + 𝒈 𝒕 𝒙 = 𝒈 𝒍 𝑳, where 𝒈, 𝒍, 𝑳 are constants subjected to condition, 𝒙 = 𝒂, 𝒅𝒙 𝒅𝒕 = 𝟎 𝒂𝒕 𝒕 = 𝟎.
- 29. (3) Solve :- 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 − 𝟐 𝒅𝒚 𝒅𝒙 + 𝒚 = 𝒙 𝒆 𝒙 𝒔𝒊𝒏 𝒙
- 30. (4) Solve :- 𝒅 𝟑 𝒚 𝒅𝒙 𝟑 − 𝟑 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 + 𝟒 𝒅𝒚 𝒅𝒙 − 𝟐𝒚 = 𝒆 𝒙 + 𝐜𝐨𝐬 𝒙
- 31. (5) Solve :- (𝑫 𝟐 − 𝟒𝑫 + 𝟑)𝒚 = 𝐬𝐢𝐧 𝟑𝒙 𝐜𝐨𝐬 𝟐𝒙
- 32. (6)Solve :- 𝒅 𝟑 𝒚 𝒅𝒙 𝟑 − 𝟕 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 + 𝟏𝟎 𝒅𝒚 𝒅𝒙 = 𝒆 𝟐𝒙 𝐬𝐢𝐧 𝒙
- 34. Method of Variation Parameter • Steps to solve linear D.E. - Find out 𝑦𝑐 - Compared with it 𝑦𝑐 = 𝑐1 𝑦1 + 𝑐2 𝑦2 and find 𝑦1 & 𝑦2 - Solve 𝑊 = 𝑦1 𝑦2 𝑦1′ 𝑦2′ , W1 = 0 𝑦2 1 𝑦2′ , 𝑊2= 𝑦1 0 𝑦1′ 1 - Find 𝑦𝑝 = 𝑦1 ∫ 𝑤1 𝑤 𝑅 𝑥 𝑑𝑥 + 𝑦2 ∫ 𝑤2 𝑤 𝑅 𝑥 𝑑𝑥
- 35. Solved Example :- (1) Solve by Variation parameter method :- 𝒅 𝟐 𝒚 𝒅𝒙 𝟐 + 𝒚 = 𝐬𝐞𝐜 𝒙
- 37. Exercise :-
- 45. Exercise :-