Practice Exercises for Chapter 4 Pauline Chow Use the reduction of order method to find the second linearly independent solution of the differential equation. b( x) dx e a( x) Formula: a( x) y "( x) b( x) y '( x) c( x) 0; y2 y1 ( x) 2 dx y1 ( x) 1. y” + 9y = 0; y1 = sin 3x 3. y” + 2y’ + y = 0; y1 = e –x Solve the higher order ODE: 5. y”’ – 6y” + 12y’ – 8y = 0 8. y(5) – 2y(4) + 17y”’ = 0 6. 9. 4x2 y” + y = 0: y1 = x1/2 x2y” – 3xy’ + 5y = 0; y1 = x2 cos (ln x) 2. 4. y”’ + 3y” + 3y’ + y = 0 3y”’ – 14y” + 20y’ – 8y = 0 Solve the higher order Cauchy-Euler ODE: 11. x2 y” – 7xy’ + 41y = 0 13 x3 y’’’– 2x2y” +4xy’ – 4y = 0 15. x3 y’’’ + xy’ – y = 0 17. x4 y(4) + 6x3 y”’ + 9x2 y” + 3x y’ + y = 0 7. 10. y”’ + 5y” = 0 y(4) – 9y = 0 x3 y’’’– 2x2y” – 2xy’ + 8y = 0 x3 y’’’ + 6y = 0 x4 y(4) + 6x3 y”’ = 0 x3 y’’’+ 5x2y” + 7xy’ + 8y = 0 12. 14. 16. 18. Use undetermined coefficients method and the annihilator method, if applicable, to solve. 19. y”’ – 6y” = 3 – cos x 20. y”’ – 2y” – 4y’ + 8y = 6e2x x 21. y”’ – 3y” + 3y’ – y = x – 4e 22. y”’ – y” – 4y’ + 4y = 8 – 3ex + 4e2x –2x 23. y”’ + 8y = 24x – 16 + 24e 24. 2y” + 5y’ – 3y = –7e–3x – 42 sin 3x + 30 cos 3x 2x 2 2 25. y” – 2y’ + 5y = 15e + 25x 26. x y” + 5xy’ – 21y = –54x2 – 48x – 84, y(1) = 13, y’(1) = 7 2 2 2 27. x y” – xy’ + y = 2x ln x + 10x Use parameter of variations method to solve. 28. x3 y’’’ – 3x2y” + 6xy’ – 6y = 4 + ln x 30. y”’ + 4y’ = sec 2x 32. 2y”’ – 6y” = x2 34. x2 y” – 4xy’ + 6y = 1/x 36. y” – 6y’ + 9y = 27x 38. x2 y” + 4xy’ – 10y = 28 lnx 29. 31. 33. 35. 37. y”’ + y’ = tan x y”’ – 2y” – y’ + 2y = e3x y” – 4y’ + 4y = (12x2 – 6x) e2x y” + y = cot x y” – 4y’ + 13y = 36e2x Use any method to solve: 39. y(4) + 2y” + y = (x – 1)2 40. 41. y”’ – 2y” + y’ = 2 – 24ex + 40e5x, y(0) = ½, y’(0) = 5/2, y”(0) = –9/2 y(4) – y” = 4x + 2xe–x Solutions: 1. y2 = –(1/3) cos 3x 2. y2 = x 1/2 ln x 3. y2 = xe–x 4. y2 = x2 sin (ln x) 5. (r – 2)3 = 0; y = C1 e2x + C2 xe2x + C3 x2 e2x 6. (r + 1)3 = 0; y = C1 e–x + C2 xe–x + C3 x2 e–x 2 –5x 8. r3(r2 – 2r + 17) = 0; y = C1 + C2 x + C3 x2 + C4 ex cos 4x + C5 ex sin 4x 7. r ( r + 5) = 0; y = C1 + C2 x + C3 e 2 2x 2x 2/3x 9. (r – 2) (3r – 2) = 0; y = C1 e + C2 xe + C3 e 10. (r2 + 3)(r2 – 3) = 0; y C1 cos 3 x C2 sin 3 x C3 e 3 x C4 e 3 x 11. y = C1 x4 cos (5 ln x) + C2 x4 sin (5 ln x) 12. r3 – 5r2 + 2r + 8 = (r + 1)(r – 2)(r – 4) = 0; y = C1 x-1 + C2 x2 + C3 x4 13. r3 – 5r2 + 8r – 4 = (r – 1)(r – 2)2 = 0; y = C1 x + C2 x2 + C3 x2 ln x 14. r3 – 3r2 + 2r + 6 = (r + 1)(r2 – 4r + 6) = 0; y C1 x 2 cos 3 2 ln x C2 x 2 sin 2 ln x C3 x 1 2 16. r(r – 1)(r – 2)(r + 3) = 0; y = C1 + C2 x + C3 x2 + C4 x–3 15. (r – 1) = 0; y = C1 x + C2 x ln x + C3 x (ln x) 4 2 2 2 17. r + 2r + 1 = (r + 1) = 0; y = C1 cos (ln x) + C2 sin (ln x) + C3 (ln x) cos (ln x) + C4 (ln x) sin (ln x) 18. r3 + 2r2 + 4r + 8 = (r + 2)(r2 + 4) = 0; y = C1x–2 + C2 cos (2 lnx) + C3 sin (2 lnx) 1 6 1 3 19. y C1 C2 x C3 e6 x x 2 cos x sin x 20. y C1e 2 x C2 e 2 x C3 xe 2 x x 2 e 2 x 4 37 37 4 2 3 x x x 2 x x 2 x 2x 21. y C1e C2 xe C3 x e 3 x x e 22, y C1e C2 e C3 e 2 xe x xe 2 x 3 23. y C1e x cos 3 x C2 e x sin 3 x C3 e 2 x 3 x 2 2e 2 x 2 x x 2x 2 25. y C1e cos 2 x C2 e sin 2 x 3e 5 x 4 x 5 24. y C1e x/2 C2 e 3 x xe 3 x 26. y = –2x3 + 2x–7x + 6x2 + 3x + 4 2 sin 3 x 35 1 ln x 36 6 29. y = C1 + C2 cos x + C3 sin x – ln | cos x| | + 1 – sin x ln | sec x + tan x | 1 1 1 30. y C1 C2 cos x C3 sin x ln | sec x tan x | x cos 2 x sin 2 x ln | cos 2 x | 8 4 4 1 1 4 1 3 1 2 1 1 32. y C1 x C2 x C3 e3 x x x x x 31. W 6e 2 x ; y C1e x C2 e x C3 e 2 x e3 x 3 72 54 54 81 243 1 1 2x 2x 4 3 2x 2 3 34. y C1 x C2 x x 35. y = C1 cos x + C2 sin x – sin x ln | csc x + cot x | 33. y C1e C2 xe ( x x )e 12 14 5 2 14 36. y = C1 e3x + C2 xe3x + 3x + 2 37. y = C1 e2x cos 3x + C2 e2x sin 3x + 4e2x 38. y C1 x C2 x ln x 25 5 39. y = C1 cos x + C2 sin x + C3 x cos x + C4 x sin x + x2 – 2x – 3 2 5 1 1 40. y C1 C2 x C3 e x C4 e x x3 xe x x 2 e x 41. y 11 11e x 9 xe x 2 x 12 x 2 e x e5 x 2 3 2 2 More questions: 28. y C1 x C2 x 2 C3 x 3 27. y = C1x + C2 x lnx + 2x2 lnx + 6x2 Find a differential operator that annihilates the given function. 2. 6e2x cos 3x + 5e2x sin 3x 1. e4x + 4xe4x + 3x – 9 –x 4. x3 + 4x – 10 3. xe cos 5x Solutions: 1. D2 (D – 4)2 2. D2 – 4D + 13 3. (D2 + 2D + 26)2 4. D4 Given the homogenous solutions, find the particular solution using an appropriate method. 1. x4 y(4) + 2x3 y’” – 3x2 y” + 19xy’ + 13y = x4 /16; yh = C1 x3 cos (2ln x) + C2 x3 sin (2 ln x) + C3 x –1 + C4 x –1 ln x 2. x2 y” – 9xy’ + 25y = 0; yh = C1 x5 + C2 x5 ln x 3. x2 y” – xy’ – 3y = 0; yh = C1 x3 + C2 x –1 4. y(4) – 4y’” + 2y” + 20y’ + 13y = 90e2x; yh = C1 e3x cos 2x + C2 e3x sin 2x + C3 e –x + C4 xe –x 5. ; y(4) – 4y’” + 28y” + 4y’ – 29y = 21ex cos x + 52ex sin x; yh= C1 e2x cos 5x + C2 e2x sin 5x + C3 e –x + C4 ex 6. y” – 10y’ + 25y = 82 sin 2x; yh = C1 e5x + C2 xe5x; 7. y” – 2y’ – 3y = 6xe2x; yh = C1 e3x + C2 e –x 8. y” – 8y’ + 20y = 60x3 + 8x2 – 6x – 8; yh = C1 e4x cos 2x + C2 e4x sin 2x x 9. y’” – 2y” + 4y’ = 36x2 + 4x ; yh C1e cos Solutions: 5. yp = –ex cos x 4. yp = 2e2x 3 2 9. yp = 3x + 5x + x x 3 x C2 e sin 6. yp = 2 sin 2x 3 x C3 7. yp = –2xe2x + (4/3)e2x Use the Elimination Method to find a general solution for the given linear system. 1. x’ = 2x – y y’ = x 2. x’ = –y + t y’ = x – t 3. x’ = 4x + 7y y’ = x – 2y 4. x’ – 4y = 1 x + y’ = 2 5. x’ + y’ = et –x” + x’ + x + y = 0 6. x’ = –5x – y y’ = 4x – y x(1) = 0, y(1) = 1 8. yp = 3x3 + 4x2 + 2x 7. x’ = y – 1 y’ = –3x + 2y x(0) = 0, y(0) = 0 8. D2 x – Dy = 1 (D + 3)x + (D + 3)y = 2 9. (D2 – 1)x – y = 0 (D – 1)x + Dy = 0 10. Dx + D2 y = e3t (D + 1)x + (D – 1)y = 4e3t 11. x’ – 6y = 0 x – y’ + z = 0 x + y – z’ = 0 12. x’ = –x + z y’ = –y + z z’ = –x + y 13. 2Dx + (D – 1)y = t Dx + Dy = t2 14. Dx – 2Dy = t2 (D – 1)x – 2(D + 1)y = 1 Solutions: 1. x = C1 et + C2 tet, y = (C1 – C2)et + C2 tet 2. x = C1 cos t + C2 sin t + t + 1, y = C1 sin t – C2 cos t – 1 + t 1 3. x = C1 e5t + C2 e–3t, y C1e5t C2 e3t 7 1 1 1 4. x = C1 cos 2t + C2 sin 2t + 2, y C1 sin 2t C2 cos 2t 2 2 4 5. x = C1 + C2 t + C3 et + tet, y = (–C1 + C2) – C2 t + (1 – C3) et – tet 6. x = (1 – t)e3 – 3t, y = (2t – 1)e3 – 3t 1 2 t 2 e sin 2t , y 1 et cos 2t 7. x et cos 2t 3 3 3 2 8. x = C1 + C2 e–3t + C3 e–t + t, y C1 3C2 e 3t C3 e t t 3 3 3 1 3 1 3 t C3 e 1/ 2t sin t , y (3C2 3C3 )e 1/ 2t cos t (3 C3 3C2 )e 1/ 2t sin t 2 2 2 2 2 2 17 4 10. x C1 + C2 sin t C3 cos t e3t , y C1 + C2 cos t C3 sin t e3t 15 15 1 1 1 5 1 1 11. x = C1 e–t + C2 e3t + C3 e–2t, y C1e t C2 e3t C3 e 2t , z C1e t C2 e3t C3 e 2t 6 2 3 6 2 3 12. x = C3 e–t + C1 + C2 te–t, y = (C3 – C2)e–t + C1 + C2 te–t, z = C1 + C2 e–t 9. x C1et C2 e 1/ 2t cos 13. x = (1/3)t3 – C1 e–t + C2 – 2t2 + 5t, y = C1 e–t + 2t2 – 5t + 5 14. x 1 2 1 3 1 1 1 t t C1 , y t 2 t 3 (1 C1 ) 2 6 4 12 2