MHB PDE and more boundary conditions

Markov2
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Solve

$\begin{aligned} & {{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)=0=u(1,t),\text{ }t>0.
\end{aligned}
$

Here's something new for me, the boundary condition $u_x.$ I've always seen the $u_t$ condition, but what to do in this case?
 
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Try a "Fourier series" solution of the form
$\sum_{n=0}^\infty A_n(t)cos(n\frac{\pi}{2}t)$
Do you see why that will work?
 
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Not actually. I thought this can be solved by using another function, etc, don't know how to make it yet. :(
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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