Solve: $$u_{tt}=c^2u_{xx}+x t,\quad u(x,0)=0, \quad u_t(x,0)=0$$
The final answer should be $u=xt^3/6$. I keep getting $xt^3/2$.
How I did the problem: 1/2[phi(x)+phi(-x)]+1/2c int(x+ct,x-ct, 0) dy +1/2c int(0,t)int(x+c(t-s),x-c(t-s), xt) dxds =(xt)int(t,0, t-s) ds =xt[t^2-1/2t^2-0] =1/2xt^3
Sorry if this is hard to read. First time using this forum for help and not sure how to type everything out on here yet. Any help is appreciated. I have been stuck on it for weeks.
The solution to the general non-homogeneous equation $$u_{tt} - c^2 u_{xx} = f(x,t),\quad u(x,0) = \phi(x), \quad u_t(x,0) = \psi(x)$$ Is given by: $$u(x,t) = \frac{\phi(x-ct)+\phi(x+ct)}{2} + \frac{1}{2} \int_{x-ct}^{x+ct} \psi(s)ds +\frac{1}{2c} \int_{0}^{t}\int_{x-c(t-s)}^{x+c(t-s)}f(r,s)dsdr$$ In your case, $\psi = \phi = 0$, so that: $$u(x,t) = \frac{1}{2c} \int_{0}^{t}\int_{x-c(t-s)}^{x+c(t-s)}rs\ dsdr = \int_{0}^{t} sx(t-s) ds = \frac{xt^3}{6}$$