Forced wave equation If a system satisfies the equation $$\nu^2 {\partial^2 \psi\over \partial x^2}={\partial^2 \psi\over \partial t^2}+a{\partial \psi\over \partial t}-b\sin\left({\pi x \over L}\right)\cos\left({\pi \nu t\over L}\right)$$
subjected to conditions: $\psi(0,t)=\psi(L,t)={\partial \psi(x,0)\over \partial t}=0$ and $\psi(x,0)=c\sin\left({\pi x\over L}\right)$,
how might I solve this? I can solve the equation $$\nu^2 {\partial^2 \psi\over \partial x^2}={\partial^2 \psi\over \partial t^2}+a{\partial \psi\over \partial t}$$ by separation of variables. But I don't know how to deal with the $$b\sin\left({\pi x \over L}\right)\cos\left({\pi \nu t\over L}\right)$$ term. Also, what is the "forced component" of $\psi(x,t)$?
Thanks.

It is convenient to assume that $\nu = 1 ,L=\pi$.And $\psi (x,t)= c \phi (t) sin(x)$. So, the equation becomes $\ddot{\phi} + a \dot{\phi} + \phi = b cos(t)$, with initial value $\dot{\phi}(t=0)=0$; $\phi (t=0) = 1$, and this equation maybe easy to solve now.
