Not For Sale
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
4.5 Undetermined Coefficients—Annihilator Approachq
4.5 Undetermined Coefficients—Annihilator Approachq
1. (9D2 − 4)y = (3D − 2)(3D + 2)y = sin x
2. (D2 − 5)y = (D −
√
5 )(D +
√
5 )y = x2 − 2x
3. (D2 − 4D − 12)y = (D − 6)(D + 2)y = x − 6
4. (2D2 − 3D − 2)y = (2D + 1)(D − 2)y = 1
5. (D3 + 10D2 + 25D)y = D(D + 5)2 y = ex
6. (D3 + 4D)y = D(D2 + 4)y = ex cos 2x
7. (D3 + 2D2 − 13D + 10)y = (D − 1)(D − 2)(D + 5)y = xe−x
8. (D3 + 4D2 + 3D)y = D(D + 1)(D + 3)y = x2 cos x − 3x
9. (D4 + 8D)y = D(D + 2)(D2 − 2D + 4)y = 4
10. (D4 − 8D2 + 16)y = (D − 2)2 (D + 2)2 y = (x3 − 2x)e4x
11. D4 y = D4 (10x3 − 2x) = D3 (30x2 − 2) = D2 (60x) = D(60) = 0
12. (2D − 1)y = (2D − 1)4ex/2 = 8Dex/2 − 4ex/2 = 4ex/2 − 4ex/2 = 0
13. (D − 2)(D + 5)(e2x + 3e−5x ) = (D − 2)(2e2x − 15e−5x + 5e2x + 15e−5x )
(D − 2)(D + 5)(e2x + 3e−5x ) = (D − 2)7e2x = 14e2x − 14e2x = 0
14. (D2 + 64)(2 cos 8x − 5 sin 8x) = D(−16 sin 8x − 40 cos 8x) + 64(2 cos 8x − 5 sin 8x)
(D2 + 64)(2 cos 8x − 5 sin 8x) = −128 cos 8x + 320 sin 8x + 128 cos 8x − 320 sin 8x = 0
15. D4 because of x3
16. D5 because of x4
17. D(D − 2) because of 1 and e2x
18. D2 (D − 6)2 because of x and xe6x
19. D2 + 4 because of cos 2x
20. D(D2 + 1) because of 1 and sin x
21. D3 (D2 + 16) because of x2 and sin 4x
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
170
4.5
Undetermined Coefficients—Annihilator Approachq
22. D2 (D2 + 1)(D2 + 25) because of x, sin x, and cos 5x
23. (D + 1)(D − 1)3 because of e−x and x2 ex
24. D(D − 1)(D − 2) because of 1, ex , and e2x
25. D(D2 − 2D + 5) because of 1 and ex cos 2x
26. (D2 + 2D + 2)(D2 − 4D + 5) because of e−x sin x and e2x cos x
© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
27. 1, x, x2 , x3 , x4
28. D2 + 4D = D(D + 4);
1, e−4x
29. e6x , e−3x/2
30. D2 − 9D − 36 = (D − 12)(D + 3);
31. cos
e12x , e−3x
√
√
5 x, sin 5 x
32. D2 − 6D + 10 = D2 − 2(3)D + (32 + 12 ); e3x cos x, e3x sin x
33. D3 − 10D2 + 25D = D(D − 5)2 ;
1, e5x , xe5x
34. 1, x, e5x , e7x
35. Applying D to the differential equation we obtain
D(D2 − 9)y = 0.
Then
y = c1 e3x + c2 e−3x + c3
|
{z
}
yc
and yp = A. Substituting yp into the differential equation yields −9A = 54 or A = −6. The
general solution is
y = c1 e3x + c2 e−3x − 6.
36. Applying D to the differential equation we obtain
D(2D2 − 7D + 5)y = 0.
Then
y = c1 e5x/2 + c2 ex + c3
|
{z
}
yc
and yp = A. Substituting yp into the differential equation yields 5A = −29 or A = −29/5. The
general solution is
y = c1 e5x/2 + c2 ex −
29
.
5
Not For Sale
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
171