Need help with 37 please U1₂ =Ssin x dx = -cos x.solution is(Do you see why we choose the constants of integration to be zero?) Thus our particularyp(x) = u₁(x)yı (x) + u₂(x)y2(x)= (sin x - In [sec x + tan x) cos x + (- cos x) (sin x);that is,yp(x) = −(cos x) In [sec x + tan.x].3.5 ProblemsIn Problems 1 through 20, find a particular solution yp of thegiven equation. In all these problems, primes denote deriva-tives with respect to x.33. y" +9y = sin 2x; y (0) = 1, y' (0) = 034. y" + y = cos x; y (0) = 1, y' (0) = -135. y" - 2y' + 2y = x + 1; y(0) = 3, y'(0) = 01. y" + 16y=e³x3. y" - y' - 6y = 2 sin 3x5. y" + y + y = sin² x2. y" - y' - 2y = 3x + 44. 4y" + 4y' + y = 3xex6. 2y" + 4y' + 7y = x²8. y" - 4y = cosh 2x36. y(4) - 4y" = x2; y(0) = y'(0) = 1, y" (0) = y(3) (0) = -137. y(3) - 2y"+y' = 1 + xe*; y(0) = y'(0) = 0, y" (0) = 138. y" + 2y + 2y = sin 3x; y (0) = 2, y'(0) = 039. y(3) + y = x + ex; y(0) = 1, y'(0) = 0, y" (0) = 140. y(4) - y = 5; y(0) = y'(0) = y" (0) = y(3) (0) = 041. Find a particular solution of the equation-7. y” – 4y = sinh x9. y" + 2y' - 3y = 1 + xe*10. y" +9y = 2 cos 3x + 3 sin 3.x11. y(3) + 4y= 3x - 1y (4) - y(3) - y" - y' - 2y = 8x5.13. y" + 2y+ 5y = e* sin x15. y(5) + 5y (4) - y = 1712. y(3) + y' = 2 — sin x14. y(4) - 2y" + y = xex16. y" +9y = 2x²e³x +542. Find the solution of the initial value problem consistingof the differential equation of Problem 41 and the initialconditions17. y" + y = sin x + x cos xy (0) = y'(0) = y" (0) = y(3) (0) = 0.18. y(4) – 5y” + 4y = ex -xe2x19. y(5) + 2y (3) + 2y" = 3x² - 120. y(3) - y = e* + 743. (a) Writecos 3x + i sin 3x = e3ix = (cosx + i sin.x)³In Problems 21 through 30, set up the appropriate form of aparticular solution yp, but do not determine the values of thecoefficients.by Euler's formula, expand, and equate real and imag-inary parts to derive the identitiescos³x = cos x+cos3x.21. y" - 2y' + 2y = e* sin x22. y(5) - y(3) = ex + 2x² - 5sin³ x =sinx-sin 3x.23. y" + 4y = 3x cos 2x(b) Use the result of part (a) to find a general solution of24. y(3) - y" - 12y' = x - 2xe-3xy" + 4y = cos³x.25. y" + 3y' + 2y = x(e-* -e-2x)26. y" - 6y' + 13y = xe³x sin 2xUse trigonometric identities to find general solutions of theequations in Problems 44 through 46.27. y(4) + 5y" + 4y = sin x + cos2x28. y(4) +9y" = (x² + 1) sin 3x44. y"+y' + y = sin x sin 3x29. (D-1)³ (D² - 4)y = xe* + e2x + e-2x45. y" +9y = sin* x30. y(4) - 2y" + y = x² cos x46. y" + y = x cos³xSolve the initial value problems in Problems 31 through 40.In Problems 47 through 56, use the method of variation of pa-rameters to find a particular solution of the given differentialequation.31. y" + 4y = 2x; y (0) = 1, y'(0) = 232. y" + 3y' + 2y = e*; y(0) = 0, y'(0) = 319647. y" + 3y' + 2y = 4e*49. y" - 4y + 4y = 2e²x51. y" + 4y = cos 3.x53. y" +9y = 2 sec 3xIn Problems 58 through 62, a nonhomogeneous second-orderlinear equation and a complementary function yc are given.Apply the method of Problem 57 to find a particular solutionof the equation.Chapter 3 Linear Equations of Higher Order=48. y"-2y'-8y = 3e-2x50. y” – 4y = sinh 2x52. y" +9y = sin 3x54. y" + y = csc² x

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Need help with 37 please U1₂ =Ssin x dx = -cos x.solution is(Do you see why we choose the constants of integration to be zero?) Thus our particularyp(x) = u₁(x)yı (x) + u₂(x)y2(x)= (sin x - In [sec x + tan x) cos x + (- cos x) (sin x);that is,yp(x) = −(cos x) In [sec x + tan.x].3.5 ProblemsIn Problems 1 through 20, find a particular solution yp of thegiven equation. In all these problems, primes denote deriva-tives with respect to x.33. y&#34; +9y = sin 2x; y (0) = 1, y' (0) = 034. y&#34; + y = cos x; y (0) = 1, y' (0) = -135. y&#34; - 2y' + 2y = x + 1; y(0) = 3, y'(0) = 01. y&#34; + 16y=e³x3. y&#34; - y' - 6y = 2 sin 3x5. y&#34; + y + y = sin² x2. y&#34; - y' - 2y = 3x + 44. 4y&#34; + 4y' + y = 3xex6. 2y&#34; + 4y' + 7y = x²8. y&#34; - 4y = cosh 2x36. y(4) - 4y&#34; = x2; y(0) = y'(0) = 1, y&#34; (0) = y(3) (0) = -137. y(3) - 2y&#34;+y' = 1 + xe*; y(0) = y'(0) = 0, y&#34; (0) = 138. y&#34; + 2y + 2y = sin 3x; y (0) = 2, y'(0) = 039. y(3) + y = x + ex; y(0) = 1, y'(0) = 0, y&#34; (0) = 140. y(4) - y = 5; y(0) = y'(0) = y&#34; (0) = y(3) (0) = 041. Find a particular solution of the equation-7. y” – 4y = sinh x9. y&#34; + 2y' - 3y = 1 + xe*10. y&#34; +9y = 2 cos 3x + 3 sin 3.x11. y(3) + 4y= 3x - 1y (4) - y(3) - y&#34; - y' - 2y = 8x5.13. y&#34; + 2y+ 5y = e* sin x15. y(5) + 5y (4) - y = 1712. y(3) + y' = 2 — sin x14. y(4) - 2y&#34; + y = xex16. y&#34; +9y = 2x²e³x +542. Find the solution of the initial value problem consistingof the differential equation of Problem 41 and the initialconditions17. y&#34; + y = sin x + x cos xy (0) = y'(0) = y&#34; (0) = y(3) (0) = 0.18. y(4) – 5y” + 4y = ex -xe2x19. y(5) + 2y (3) + 2y&#34; = 3x² - 120. y(3) - y = e* + 743. (a) Writecos 3x + i sin 3x = e3ix = (cosx + i sin.x)³In Problems 21 through 30, set up the appropriate form of aparticular solution yp, but do not determine the values of thecoefficients.by Euler's formula, expand, and equate real and imag-inary parts to derive the identitiescos³x = cos x+cos3x.21. y&#34; - 2y' + 2y = e* sin x22. y(5) - y(3) = ex + 2x² - 5sin³ x =sinx-sin 3x.23. y&#34; + 4y = 3x cos 2x(b) Use the result of part (a) to find a general solution of24. y(3) - y&#34; - 12y' = x - 2xe-3xy&#34; + 4y = cos³x.25. y&#34; + 3y' + 2y = x(e-* -e-2x)26. y&#34; - 6y' + 13y = xe³x sin 2xUse trigonometric identities to find general solutions of theequations in Problems 44 through 46.27. y(4) + 5y&#34; + 4y = sin x + cos2x28. y(4) +9y&#34; = (x² + 1) sin 3x44. y&#34;+y' + y = sin x sin 3x29. (D-1)³ (D² - 4)y = xe* + e2x + e-2x45. y&#34; +9y = sin* x30. y(4) - 2y&#34; + y = x² cos x46. y&#34; + y = x cos³xSolve the initial value problems in Problems 31 through 40.In Problems 47 through 56, use the method of variation of pa-rameters to find a particular solution of the given differentialequation.31. y&#34; + 4y = 2x; y (0) = 1, y'(0) = 232. y&#34; + 3y' + 2y = e*; y(0) = 0, y'(0) = 319647. y&#34; + 3y' + 2y = 4e*49. y&#34; - 4y + 4y = 2e²x51. y&#34; + 4y = cos 3.x53. y&#34; +9y = 2 sec 3xIn Problems 58 through 62, a nonhomogeneous second-orderlinear equation and a complementary function yc are given.Apply the method of Problem 57 to find a particular solutionof the equation.Chapter 3 Linear Equations of Higher Order=48. y&#34;-2y'-8y = 3e-2x50. y” – 4y = sinh 2x52. y&#34; +9y = sin 3x54. y&#34; + y = csc² x

Accepted Answer

Given: y(3)-2y''+y'=1+xexAuxilliary equation: m3-2m2+m=0Therefore, m=0,1,1Complimentary…