望高手指教..solve differential equation

Solve y'' + y = x cos x.

The auxiliary equation is λ² + 1 = 0
　　　　　　　　　　　　　λ² = -1
　　　　　　　　　　　　　λ = ± i

∴ y_c = C_1 sin x + C_2 cos x

（Originally, we let y_p = Ax sin x + Bx cos x + C sin x + D cos x. However, C sin x and D cos x is doubled in y_c (∵y_c = C_1 sin x + C_2 cos x). So we should let y_p = Ax² sin x + Bx² cos x + Cx sin x + Dx cos x.）

Let y_p = Ax² sin x + Bx² cos x + Cx sin x + Dx cos x
(y_p)' = Ax² cos x + 2Ax sin x - Bx² sin x + 2Bx cos x + Cx cos x + C sin x - Dx sin x + D cos x
= - Bx² sin x + Ax² cos x + (2A - D)x sin x + (2B + C)x cos x + C sin x + D cos x
(y_p)'' = - Bx² cos x - 2Bx sin x - Ax² sin x + 2Ax cos x + (2A - D)x cos x + (2A - D)sin x - (2B + C)x sin x + (2B + C)cos x + C cos x - D sin x
= - Ax² sin x - Bx² cos x - (4B + C)x sin x + (4A - D)x cos x + (2A - 2D)sin x + (2B + 2C)cos x

Sub. into the original equation:
- Ax² sin x - Bx² cos x - (4B + C)x sin x + (4A - D)x cos x + (2A - 2D)sin x + (2B + 2C)cos x + Ax² sin x + Bx² cos x + Cx sin x + Dx cos x ≡ x cos x
- 4Bx sin x + 4Ax cos x + (2A - 2D)sin x + (2B + 2C)cos x ≡ x cos x

Comparing the coefficients of x² sin x, x² cos x, x sin x and x cos x:
　　　4A = 1 , 　　∴ A = 1/4
　　- 4B = 0 , 　　∴ B = 0
2B + 2C = 0 , 　　∴ C = 0
2A - 2D = 0 , 　　∴ D = 1/4

∴ y = y_c + y_p
　　y = (x² sin x)/4 + (x cos x)/4 + C_1 sin x + C_2 cos x

2007-12-10 04:48:32 補充：
Sorry,「(y_p)'' = - Bx² cos x - 2Bx sin x - Ax² sin x 2Ax cos x (2A - D)x cos x (2A - D)sin x - (2B C)x sin x (2B C)cos x C cos x - D sin x」應該改成

2007-12-10 04:49:03 補充：
「(y_p)'' = - Bx² cos x - 2Bx sin x - Ax² sin x 2Ax cos x (2A - D)x cos x (2A - D)sin x - (2B C)x sin x (2B C)cos x C cos x - D sin x」

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