How to obtain the form of a particular solution of an ODE if we have a trigonometric function as f(x) We have $y'' -3y' + 2y = sin(4x)$
There is no problem to obtain $y_h$. However, I do not understand properly how to obtain the form of the particular solution of this ODE. As I know, if we had, for instance, $... = 3x^2$, we would have $y_p = Ax^2$. But for this particular case, $... = sin(4x)$, I do not know how to proceed, since my hand book says the form should be $y_p = Asin(4x) + Bcos(4x)$.
I think the problem here is that I am not proceeding correctly to obtain the form of particular solutions in general, so I would really appreciate if you can clarify how to obtain that form for any case, so then we will be able to solve this and any other kind of particular solution.
Thank you very much.

If $y_p = A\sin(4x) + B\cos(4x)$ then
$$y_p'=4A\cos(4x)-4B\sin(4x)$$ and
$$y_p''=-16A\sin(4x)-16B\cos(4x).$$ So
$$y_p''-3y_p'+2y_p=\sin(4x)$$ if and only if $$(-16A+12B+2A)\sin (4x)+(-16B-12A+2B)\cos (4x)=\sin (4x).$$
Solve the system 
\begin{cases}-16A+12B+2A=1\\ -16B-12A+2B=0\end{cases} and you are done.
