- #1
Hurry
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Hi, I have a test tomorrow and I'd like you to guys help me please.
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?
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?