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.

$$\nu^2\dfrac{\partial^2\psi(x,t)}{\partial x^2}=\dfrac{\partial^2\psi(x,t)}{\partial t^2}+a\dfrac{\partial\psi(x,t)}{\partial t}-b\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
$$\nu^2\dfrac{\partial^2\psi(x,t)}{\partial x^2}-\dfrac{\partial^2\psi(x,t)}{\partial t^2}-a\dfrac{\partial\psi(x,t)}{\partial t}=-b\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
It is possible that the subsititution $\psi(x,t)=\psi_c(x,t)+\psi_p(x,t)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $\psi_p(x,t)$ can be found.
For this question, the form of $\psi_p(x,t)$ is not difficult to guess, just $$A_1\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+A_2\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}+A_3\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+A_4\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
Let $$\psi(x,t)=\psi_c(x,t)+A_1\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+A_2\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}+A_3\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+A_4\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$ 
Then $$\dfrac{\partial\psi(x,t)}{\partial x}=\dfrac{\partial\psi_c(x,t)}{\partial x}+\dfrac{A_1\pi}{L}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_2\pi}{L}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi}{L}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi}{L}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
$$\dfrac{\partial^2\psi(x,t)}{\partial x^2}=\dfrac{\partial^2\psi_c(x,t)}{\partial x^2}-\dfrac{A_1\pi^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
$$\dfrac{\partial\psi(x,t)}{\partial t}=\dfrac{\partial\psi_c(x,t)}{\partial t}+\dfrac{A_1\pi\nu}{L}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi\nu}{L}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_3\pi\nu}{L}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi\nu}{L}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}$$
$$\dfrac{\partial^2\psi(x,t)}{\partial t^2}=\dfrac{\partial^2\psi_c(x,t)}{\partial t^2}-\dfrac{A_1\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
$$\therefore \nu^2\Biggl(\dfrac{\partial^2\psi_c(x,t)}{\partial x^2}-\dfrac{A_1\pi^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}\Biggr)-\Biggl(\dfrac{\partial^2\psi_c(x,t)}{\partial t^2}-\dfrac{A_1\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}\Biggr)-a\Biggl(\dfrac{\partial\psi_c(x,t)}{\partial t}+\dfrac{A_1\pi\nu}{L}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi\nu}{L}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_3\pi\nu}{L}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi\nu}{L}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}\Biggr)=-b\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$$
$\nu^2\dfrac{\partial^2\psi_c(x,t)}{\partial x^2}-\dfrac{A_1\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{\partial^2\psi_c(x,t)}{\partial t^2}+\dfrac{A_1\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_2\pi^2\nu^2}{L^2}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}+\dfrac{A_3\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_4\pi^2\nu^2}{L^2}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-a\dfrac{\partial\psi_c(x,t)}{\partial t}-\dfrac{A_1\pi a\nu}{L}\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}+\dfrac{A_2\pi a\nu}{L}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}-\dfrac{A_3\pi a\nu}{L}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}+\dfrac{A_4\pi a\nu}{L}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}=-b\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}$
$\nu^2\dfrac{\partial^2\psi_c(x,t)}{\partial x^2}-\dfrac{\partial^2\psi_c(x,t)}{\partial t^2}-a\dfrac{\partial\psi_c(x,t)}{\partial t}=\left(\dfrac{A_1\pi a\nu}{L}-b\right)\sin\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_2\pi a\nu}{L}\sin\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}+\dfrac{A_3\pi a\nu}{L}\cos\dfrac{\pi x}{L}\cos\dfrac{\pi\nu t}{L}-\dfrac{A_4\pi a\nu}{L}\cos\dfrac{\pi x}{L}\sin\dfrac{\pi\nu t}{L}$
$\therefore A_1=\dfrac{bL}{\pi a\nu}$ , $A_2=0$ , $A_3=0$ , $A_4=0$
The above calculation is only suitable for $a\neq0$ . For $a=0$ , the situation is more complicated and I still have no idea until now.
