9.3E: Undetermined Coefficients for Higher Order Equations (Exercises)
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Q9.3.1
In Exercises 9.3.1-9.3.59 find a particular solution.
1. y‴−6y″+11y′−6y=−e−x(4+76x−24x2)
2. y‴−2y″−5y′+6y=e−3x(32−23x+6x2)
3. 4y‴+8y″−y′−2y=−ex(4+45x+9x2)
4. y‴+3y″−y′−3y=e−2x(2−17x+3x2)
5. y‴+3y″−y′−3y=ex(−1+2x+24x2+16x3)
6. y‴+y″−2y=ex(14+34x+15x2)
7. 4y‴+8y″−y′−2y=−e−2x(1−15x)
8. y‴−y″−y′+y=ex(7+6x)
9. 2y‴−7y″+4y′+4y=e2x(17+30x)
10. y‴−5y″+3y′+9y=2e3x(11−24x2)
11. y‴−7y″+8y′+16y=2e4x(13+15x)
12. 8y‴−12y″+6y′−y=ex/2(1+4x)
13. y(4)+3y‴−3y″−7y′+6y=−e−x(12+8x−8x2)
14. y(4)+3y‴+y″−3y′−2y=−3e2x(11+12x)
15. y(4)+8y‴+24y″+32y′=−16e−2x(1+x+x2−x3)
16. 4y(4)−11y″−9y′−2y=−ex(1−6x)
17. y(4)−2y‴+3y′−y=ex(3+4x+x2)
18. y(4)−4y‴+6y″−4y′+2y=e2x(24+x+x4)
19. 2y(4)+5y‴−5y′−2y=18ex(5+2x)
20. y(4)+y‴−2y″−6y′−4y=−e2x(4+28x+15x2)
21. 2y(4)+y‴−2y′−y=3e−x/2(1−6x)
22. y(4)−5y″+4y=ex(3+x−3x2)
23. y(4)−2y‴−3y″+4y′+4y=e2x(13+33x+18x2)
24. y(4)−3y‴+4y′=e2x(15+26x+12x2)
25. y(4)−2y‴+2y′−y=ex(1+x)
26. 2y(4)−5y‴+3y″+y′−y=ex(11+12x)
27. y(4)+3y‴+3y″+y′=e−x(5−24x+10x2)
28. y(4)−7y‴+18y″−20y′+8y=e2x(3−8x−5x2)
29. y‴−y″−4y′+4y=e−x[(16+10x)cosx+(30−10x)sinx]
30. y‴+y″−4y′−4y=e−x[(1−22x)cos2x−(1+6x)sin2x]
31. y‴−y″+2y′−2y=e2x[(27+5x−x2)cosx+(2+13x+9x2)sinx]
32. y‴−2y″+y′−2y=−ex[(9−5x+4x2)cos2x−(6−5x−3x2)sin2x]
33. y‴+3y″+4y′+12y=8cos2x−16sin2x
34. y‴−y″+2y=ex[(20+4x)cosx−(12+12x)sinx]
35. y‴−7y″+20y′−24y=−e2x[(13−8x)cos2x−(8−4x)sin2x]
36. y‴−6y″+18y′=−e3x[(2−3x)cos3x−(3+3x)sin3x]
37. y(4)+2y‴−2y″−8y′−8y=ex(8cosx+16sinx)
38. y(4)−3y‴+2y″+2y′−4y=ex(2cos2x−sin2x)
39. y(4)−8y‴+24y″−32y′+15y=e2x(15xcos2x+32sin2x)
40. y(4)+6y‴+13y″+12y′+4y=e−x[(4−x)cosx−(5+x)sinx]
41. y(4)+3y‴+2y″−2y′−4y=−e−x(cosx−sinx)
42. y(4)−5y‴+13y″−19y′+10y=ex(cos2x+sin2x)
43. y(4)+8y‴+32y″+64y′+39y=e−2x[(4−15x)cos3x−(4+15x)sin3x]
44. y(4)−5y‴+13y″−19y′+10y=ex[(7+8x)cos2x+(8−4x)sin2x]
45. y(4)+4y‴+8y″+8y′+4y=−2e−x(cosx−2sinx)
46. y(4)−8y‴+32y″−64y′+64y=e2x(cos2x−sin2x)
47. y(4)−8y‴+26y″−40y′+25y=e2x[3cosx−(1+3x)sinx]
48. y‴−4y″+5y′−2y=e2x−4ex−2cosx+4sinx
49. y‴−y″+y′−y=5e2x+2ex−4cosx+4sinx
50. y‴−y′=−2(1+x)+4ex−6e−x+96e3x
51. y‴−4y″+9y′−10y=10e2x+20exsin2x−10
52. y‴+3y″+3y′+y=12e−x+9cos2x−13sin2x
53. y‴+y″−y′−y=4e−x(1−6x)−2xcosx+2(1+x)sinx
54. y(4)−5y″+4y=−12ex+6e−x+10cosx
55. y(4)−4y‴+11y″−14y′+10y=−ex(sinx+2cos2x)
56. y(4)+2y‴−3y″−4y′+4y=2ex(1+x)+e−2x
57. y(4)+4y=sinhxcosx−coshxsinx
58. y(4)+5y‴+9y″+7y′+2y=e−x(30+24x)−e−2x
59. y(4)−4y‴+7y″−6y′+2y=ex(12x−2cosx+2sinx)
Q9.3.2
In Exercises 9.3.60-9.3.68 find the general solution.
60. y‴−y″−y′+y=e2x(10+3x)
61. y‴+y″−2y=−e3x(9+67x+17x2)
62. y‴−6y″+11y′−6y=e2x(5−4x−3x2)
63. y‴+2y″+y′=−2e−x(7−18x+6x2)
64. y‴−3y″+3y′−y=ex(1+x)
65. y(4)−2y″+y=−e−x(4−9x+3x2)
66. y‴+2y″−y′−2y=e−2x[(23−2x)cosx+(8−9x)sinx]
67. y(4)−3y‴+4y″−2y′=ex[(28+6x)cos2x+(11−12x)sin2x]
68. y(4)−4y‴+14y″−20y′+25y=ex[(2+6x)cos2x+3sin2x]
Q9.3.3
In Exercises 9.3.69-9.3.74 solve the initial value problem and graph the solution.
69. y‴−2y″−5y′+6y=2ex(1−6x),y(0)=2,y′(0)=7,y″(0)=9
70. y‴−y″−y′+y=−e−x(4−8x),y(0)=2,y′(0)=0,y″(0)=0
71. 4y‴−3y′−y=e−x/2(2−3x),y(0)=−1,y′(0)=15,y″(0)=−17
72. y(4)+2y‴+2y″+2y′+y=e−x(20−12x),y(0)=3,y′(0)=−4,y″(0)=7,y‴(0)=−22
73. y‴+2y″+y′+2y=30cosx−10sinx,y(0)=3,y′(0)=−4,y″(0)=16
74. y(4)−3y‴+5y″−2y′=−2ex(cosx−sinx),y(0)=2,y′(0)=0,y″(0) = −1,y‴(0)=−5
Q9.3.4
75. Prove: A function y is a solution of the constant coefficient nonhomogeneous equation
a0y(n)+a1y(n−1)+⋯+any=eαxG(x)
if and only if y=ueαx, where u satisfies the differential equation
a0u(n)+p(n−1)(α)(n−1)!u(n−1)+p(n−2)(α)(n−2)!u(n−2)+⋯+p(α)u=G(x)
and
p(r)=a0rn+a1rn−1+⋯+an
is the characteristic polynomial of the complementary equation
a0y(n)+a1y(n−1)+⋯+any=0.
76. Prove:
- The equation a0u(n)+p(n−1)(α)(n−1)!u(n−1)+p(n−2)(α)(n−2)!u(n−2)+⋯+P(α)u=(p0+p1x+⋯+pkxk)cosωx+(q0+q1x+⋯+qkxk)sinωx has a particular solution of the form up=xm(u0+u1x+⋯+ukxk)cosωx+(v0+v1x+⋯+vkxk)sinωx.
- If λ+iω is a zero of p with multiplicity m≥1, then (A) can be written as a(u″+ω2u)=(p0+p1x+⋯+pkxk)cosωx+(q0+q1x+⋯+qkxk)sinωx, which has a particular solution of the form up=U(x)cosωx+V(x)sinωx, where U(x)=u0x+u1x2+⋯+ukxk+1,V(x)=v0x+v1x2+⋯+vkxk+1 and a(U″(x)+2ωV′(x))=p0+p1x+⋯+pkxka(V″(x)−2ωU′(x))=q0+q1x+⋯+qkxk.