PDE27

 $u_{tt} - c^2 u_{xx}=2t=f(x,t)\,$ $u(x,0)=x^2=g(x)\,$ $u_t(x,0)=1=h(x)\,$

In general, combining Duhammel's principle and D'Alembert's formula, the solution is:

$u(x,t)=\frac{1}{2}\left[g(x+ct)+g(x-ct)\right]+\frac{1}{2c}\int_{x-ct}^{x+ct}h(\xi)d\xi + \frac{1}{2c}\int_0^t\int_{x-c(t-s)}^{x+c(t-s)} f(\xi,s)\,d\xi\,ds\,$.

In this particular case,

$u(x,t) = \frac{1}{2}\left[(x-ct)^2+(x+ct)^2\right]+\frac{1}{2c}\int_{x-ct}^{x+ct} d\xi + \frac{1}{2c}\int_0^t\int_{x-c(t-s)}^{x+c(t-s)} 2s \,d\xi\,ds\,$

$=\frac{1}{2}\left[2x^2+2c^2t^2\right] + \frac{1}{2c}\left[(x+ct)-(x-ct)\right] + \frac{1}{c}\int_0^t s\left[x+c(t-s)-x+c(t-s)\right] ds\,$

$=x^2 + c^2t^2 + \frac{1}{2c}2ct + \frac{1}{c}\int_0^t 2sc(t-s)ds\,$

And after simplifying, the soution is

$u(x,t)=x^2+c^2t^2+t+\frac{1}{3}t^3\,$

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