Help please in solving with this differential equation?

Note that the homogeneous part of this DE has characteristic equation

r^2 - 2r + 1 = 0 ==> r = 1, 1 (double root).

So, y_h = (C₁ + C₂ x) e^x.

Now, for y_p.

Assume that y_p = (Ax + B) sin x + (Cx + D) cos x.

y'_p = A sin x + (Ax + B) cos x + C cos x - (Cx + D) sin x

y''_p = 2A cos x - (Ax + B) sin x - 2C sin x - (Cx + D) cos x.

Substituting this into the DE yields

[2A cos x - (Ax + B) sin x - 2C sin x - (Cx + D) cos x] - 2[A sin x + (Ax + B) cos x + C cos x - (Cx + D) sin x] + [(Ax + B) sin x + (Cx + D) cos x] = x sin x

Simplifying:

(2A - 2B - 2C) cos x + (-2C - 2A + 2D) sin x - 2Ax cos x + 2Cx sin x = x sin x

Equating like coefficients,

x sin x coefficients: 2C = 1 ==> C = 1/2

x cos x coefficients: -2A = 0 ==> A = 0

sin x coefficients: -2 * 1/2 - 2 * 0 + 2D = 0 ==> D = 1/2

cos x coefficients: 2 * 0 - 2B - 2 * 1/2 = 0 ==> B = -1/2.

Hence, the general solution is

y = (C₁ + C₂ x) e^x - (1/2) sin x + (1/2)(x + 1) cos x.

Double check:

http://www.wolframalpha.com/input/?i=y%27%27+-+2y%...

I hope this helps!