Chapter 8.2, Problem 1E

Prove that the functions (a) $u(x,t)=\sin \pi x\cos 4\pi t$, (b) $u(x,t)=e^{-x-2t}$ (c) $u(x,t)=\ln (1+x+t)$ are solutions of the wave equation with me specified initial-boundary conditions:

$\begin{array}{l}\text{a}\text{. }\{\begin{array}{c}{u}_{tt}=16u_{xx}\\ u(x,0)=\sin \pi x\text{ for 0}\le x\le 1\\ u_t(x,0)=0\text{ for 0}\le x\le 1\\ \begin{array}{l}u(0,t)=0\text{ for 0}\le t\le 1\\ u(1,t)=0\text{ for 0}\le t\le 1\end{array}\end{array}\\ \text{b}\text{. }\{\begin{array}{c}{u}_{tt}=4u_{xx}\\ u(x,0)=e^{-x}\text{ for 0}\le x\le 1\\ u_t(x,0)=-2e^{-x}\text{ for 0}\le x\le 1\\ \begin{array}{l}u(0,t)=e^{-2t}\text{ for 0}\le t\le 1\\ u(1,t)=e^{-1-2t}\text{ for 0}\le t\le 1\end{array}\end{array}\\ \text{c}\text{. }\{\begin{array}{c}{u}_{tt}=u_{xx}\\ u(x,0)=\ln (1+x)\text{ for 0}\le x\le 1\\ u_t(x,0)=1/(1+x)\text{ for 0}\le x\le 1\\ \begin{array}{l}u(0,t)=\ln (1+t)\text{for 0}\le t\le 1\\ u(1,t)=\ln (2+t)\text{for 0}\le t\le 1\end{array}\end{array}\end{array}$