How do you use the Undetermined coefficients method for solving this non-homogeneous DE? 
I do not know where to go from here. Do I even have this equation right? Or is it supposed to be $y_p = (Ax+B)(C\cos(x) + D\sin(x))$
I've tried both methods and I still get stuck. 
 A: Normally, the particular solution to guess for a right hand side of $2x \sin x$ would be $y_p = C_1 x \cos x + C_2 x \sin x + C_3 \cos x + C_4 \sin x$.  However, in this case, the coefficient $i$ corresponding to the $\cos x, \sin x$ "space" is a single root of the characteristic equation $\lambda^2 + 1$, so we need to multiply by $x$, and the particular solution will be of the form $y_p = C_1 x^2 \cos x + C_2 x^2 \sin x + C_3 x \cos x + C_4 x \sin x$.

A useful mnemonic in figuring out the form of a particular solution, in equations where the method of undetermined coefficients will apply, is sometimes called the "annihilator method".  To apply this method to your example, first rewrite it as:
$$(D^2+1)y = 2x \sin x.$$
Now, we want to apply a constant-coefficient differential operator which will make the equation become homogenous.  In this case, $2x \sin x$ is annihilated by $(D^2+1)^2$; so applying $(D^2+1)^2$ to both sides, we get
$$(D^2+1)^3 y = 0.$$
The general solution to this equation is:
$$y = C_1 x^2 \cos x + C_2 x^2 \sin x + C_3 x \cos x + C_4 x \sin x + C_5 \cos x + C_6 \cos x.$$
Now, the last two terms are the homogeneous part of the general solution to the original equation; so the remaining terms, in this method, will become the form of the particular solution.
(Note that applying $(D^2+1)^2$ to both sides introduced numerous extraneous solutions, which is why we now need to go back to the original equation and substitute to find out what values of $C_1, \ldots, C_4$ will result in a solution.)
A: I used the variation of parameters method to solve this ODE. So, the particular solution is
$$y_p = \frac{1}{2} (x \sin x - x^2 \cos x)$$
So, if you had guessed $y_p = A x \sin x + B x^2 \cos x$ or $y_p = (Ax + Bx^2) (C\sin x + D \cos x)$, you would have got the desired solution. However, it is not straightforward at all to guess this solution type.
Your $y_p = (Ax + B) (C\sin x + D \cos x)$ guess was a good try; but, since $BC\sin x + BD \cos x$ is already the homogeneous solution, it did not work.
