Exercise 5.2 (method of variation of parameters)

EXERCISE 5.2

Solve the following by using method of variation of parameters :

1. (D2+1) y = x. 

[Ans. y = x + C₁ cos x + C₂ sinx.]

2. y'' + 3y' + 2y = x². 

[Ans. y = (1/2 x² – 3/2 x+ 7/4) + C₁e^-2x+ C₂e^-x]

3. x2y'' + 4xy' + 2y = e^x.

[Ans. y = C₁x^-2 + C₂x^-1 + x^-2 e^x]

4. (D² + 1) y = sinx + cos x

5. (D² + 4) y = tan 2x [AU, Nov. 2001]

6. y'' + 4y = 4 sec² 2x [A.U A/M 2017 R-13]

[Ans. y = C₁ cos 2x + C₂ sin 2x - 1 + (sin 2x) log (sec 2x + tan 2x)]

7. y' + y =1 / 1 + sin x

[Ans. y = C₁ cos x + C₂ sin x − (x cos x – sin x + 1) + sin x log (1 + sin x)]

8. d₂y/dx₂ + y =  sec x tan x

[Ans. y = C₁ cos x + C₂ sin x + x cos x - sin x + (sin x) log sec x]

9. y"- 2y' + 2y = e^x tan x

[Ans. y = e^x [C₁ cosx + C₂ sin x – cos x log (sec x + tanx)]]

10. y" - 2y' = e^x. sin x 

[ Ans. y = C₁ + C₂ e^2x – ½ e^x . sinx ]

11. y'' - 2y' + y = e^x/x 

[ Ans. y = ( C₁ + C₂x) e^x – xe^x + ( x log x) e^x]

12. y"+2y' + y = logx 

13. d²y/dx² – y = 2/1 + e^x  

14. y'' + y' – 2y = 1/1 - e^x

15. d²y/dx² + 3 dy/dx + 2y = 

16. y'- 3y' + 2y = cos (e^-x) 

[ Ans. y = C₁e^x + C₂e^2x – e^2x – e^2x cos (e^-x)]

17. (D² - 1) y = e ̄^-2x sin (e^{- ̄x}) 

[Ans. C₁e^x + C₂e^2x –sin( e^-x )– e^x cos (e^-x)]

18. (D² + 2D + 2) y = e^-x sec³ x 

19. x²d²y/dx² + x dy/dx – y = x² log x 

[ Ans. y = C₁x + C₂/x + x²/9 (3 log x – 4) ]

20. x²d²y/dx² - 4x dy/dx + 6y = sin (log x) 

[ Ans. y = C₁x² + C₂x³ + 1/10 [sin( log x) + cos (log x)]

21. x²d²y/dx² + 2 dy/dx + y = e^-x /x²  [A.U M/J 2013]  

[ Ans. C₁e^-x + C₂e^-x – e^-x log x ]

22. x²d²y/dx² + x dy/dx – y = x² log x 

23. y'' + 7y' - 8y = e^2x  [ Ans. y = Ae^x + Be^-8x + 1/10 e^2x ]