How to obtain the form of a particular solution of an ODE if we have a trigonometric function as f(x)

Question

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.

Answer 1

You must ask yourself "what function can give a sine when you form a linear combination of its derivatives up to second order" ?

If you have in mind the successive derivatives of the sine, you know that they are sines and cosines, and similarly for the cosine function.

As $A\sin(4x)$ alone doesn't work, we are lead to try a mixture like $A\sin(4x)+B\cos(4x)$, which does work.

Answer 2

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.