$u_{tt}-c^2u_{xx}=0$, $0<x<1, t>0$, $u(x,0)=0, u_t(x,0)=1$, $0\leq x\leq 1$, $u(0,t)=u(1,t)=0$, $t\geq 0$.
My attempt: Separation of variables. $c^2 \frac{X''}{X} = \frac{T''}{T}=\lambda$
case i: $\lambda=0$. So $X=c_1x+c_2$, $T=c_3t+c_4$. $u(x,0)=u(0,t)=u(1,t)=0 \implies X=0$. Not possible
case ii: $\lambda>0$. $X=c_1 \exp(\frac{\sqrt{\lambda}}{c}x) + c_2 \exp(\frac{-\sqrt{\lambda}}{c}x)$. $T=c_3 \exp(\sqrt{\lambda}t) + c_4 \exp(-\sqrt{\lambda}t)$. $u(x,0)=u(0,t)=0 \implies X=2c_1\sinh(\frac{\sqrt{\lambda}}{c}x), T= 2c_3\sinh(-\sqrt{\lambda}t)$ $u(1,t)=0 \implies \sinh(\frac{\sqrt{\lambda}}{c}x) = 0$. not possible.
case iii: $\lambda<0$
$X=c_1\sin(\frac{\sqrt{-\lambda}}{c}x) + c_2\cos(\frac{\sqrt{-\lambda}}{c}x)$, $T=c_3\sin(\sqrt{-\lambda}t) +c_4\cos(\sqrt{-\lambda}t)$. $u(x,0)=u(0,t)=0 \implies X=c_1\sin(\frac{\sqrt{-\lambda}}{c}x), T=c_3\sin(\sqrt{-\lambda}t).$ $u(1,t)=0 \implies \sin(\frac{\sqrt{-\lambda}}{c}) = 0 \implies \sqrt{-\lambda} = cn\pi$. $u(x,t) = \sum_{n=1}^{\infty} d_n \sin(n\pi x)\sin(cn \pi t)$. $u_t(x,0)=1 \implies \sum_{n=1}^{\infty} cn\pi d_n \sin(n\pi x) =1$.
I don't know how to solve for $d_n$ in $\sum_{n=1}^{\infty} cn\pi d_n \sin(n\pi x) =1$. Fourier series probably isn't applicable.
Please let me know. Thanks