For $\omega > 0$, find the general solution to $y'' + \omega^2 y = \sin \omega x$
My attempt: $$\lambda^2 + \omega^2 = 0 \implies \lambda = \pm i\omega \\ \therefore y(x) = C_1 \cos (\omega x) + C_2 \sin (\omega x)$$
Sorry but I don't have any idea... what should be the next step?
There are two answers: private and homogeneous. Since the input frequency is the same as roots of the characteristic equation $\lambda^2+\omega^2=0$ so the answer must be in the form $y=Ax\sin\omega x+Bx\cos\omega x$. Putting this in equation and knowing that $y$ is an odd function in this case we obtain:$$y''+\omega^2 y=\omega^2Bx\cos\omega x-B\omega\sin\omega x-B\omega\sin\omega x-B\omega^2x\cos \omega x=-2B\omega\sin\omega x=\sin\omega x$$therefore $B=-\dfrac{1}{2}$. The homogeneous answer is obviously $y=A\sin\omega x+B\cos\omega x$. Where $A$ and $B$ are constants. So the final answer is:$$y=A\sin\omega x+B\cos\omega x-\dfrac{1}{2}x\cos\omega x$$