- Thread starter
- #1

Solve the following:

$\begin{align*}

& {{u}_{tt}}={{u}_{xx}}+1+x,\text{ }0<x<1,\text{ }t>0. \\

& u(x,0)=\frac{1}{6}{{x}^{3}}-\frac{1}{2}{{x}^{2}}+\frac{1}{3},\text{ }{{u}_{t}}(x,0)=0,\text{ }0<x<1. \\

& {{u}_{x}}(0,t)=u(1,t)=0,\text{ }t>0.

\end{align*}$

I think it can be solved by using $u(x,t)=v(x,t)+a(x)$ and then applying D'Alembert later, does this work?