$$u_{tt}-4u_{xx}=sin(3x\pi)-7sin(5x\pi),0\le x\le1,t\ge0$$ $u(0,t)=0$, $u(1,t)=0$
$u(x,0)=0$, $u_t(x,0)=0$
I'm having a hard time solving this question. My first attempt was writing $u(x,t)$ as $\sum_{n=1}^\infty\alpha_n(t)sin\frac{nx\pi}{b}$ and to determine $\alpha_n$ (Since the space variable part gives a orthonormal basis for $u(x,t)$.). I'm stuck here and cannot find anything useful to proceed.
Any help would be greatly appreciated. Thanks!
This is a hyperbolic, wave-like equation which is not in canonical form. To reduce it to such, introduce characteristic variables: $\xi = x-2t$, $\eta = x+2t$. Writing $u(x,t) = v(\xi,\eta)$, your PDE becomes: \begin{equation} -16v_{\xi \eta} =\sin \left( \frac{3 \pi}{2}(\xi+\eta) \right)- 7\sin \left( \frac{5 \pi}{2}(\xi+\eta) \right). \end{equation} At first it may not be obvious how to integrate this, but recall you may use the sine addition formulae $\sin(A+B) = \sin A \cos B + \sin B \cos A$, which make the right hand side a sum of separable functions of $\xi$ and $\eta$. As such, we may integrate with respect to either of these variables keeping the other constant, thus solve the PDE.
Finally, observe what the boundary conditions map to within the new coordinate system. For instance, $u(0,t) = 0$ becomes $v(s,-s) =0$. Via the chain rule, $u_t(x,0) = 0$ becomes $v_\xi(s,s) = v_ \eta(s,s)$ etc.