Solving a PDE with the Poisson equation New to PDEs. This one is apparently solved with the Poisson equation but I have no idea how. I would appreciate any help.
$$u_t=u_{xx}+e^{-t}\sin t$$
and
$$u|_{t=0}=e^{-x^2}$$
 A: Hint: The steady state form of this equation is the 1D Poisson equation.
A: Here is one approach:
We can write $u(x,t)$ in terms of a Fourier transform,
$$ u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \hat{u}(k,t) dk, $$
substituting this into the PDE we get,$u_t-u_{xx}=e^{-t}\sin(t)$ we get,
$$ \frac{d}{dt} \hat{u}_k(t) - k^2 \hat{u}_k = \color{blue}{0} + \color{green}{e^{-t}\sin(t)}$$
$$ \Rightarrow \hat{u}(k,t) = \color{blue}{c(k) e^{k^2 t}}+\color{green}{\frac{e^{-t}e^{it}}{2i(-1+i-k^2)} -\frac{e^{-t}e^{-it}}{2i(-1-i-k^2)}}. $$
We have that $u(x,0) = e^{-x^2}$, this can be written as,
$$ u(x,0)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \color{red}{\frac{1}{\sqrt{2}} e^{-k^2/4} } dk, $$
which tells us that,
$$\color{blue}{c(k)} = \color{red}{\frac{1}{\sqrt{2}}e^{-k^2/4}}- \left(\color{green}{\frac{1}{2i(-1+i-k^2)} -\frac{1}{2i(-1-i-k^2)}}\right).$$
All that is left is to evaluate the integral,
$$ u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ikx} \hat{u}(k,t) dk. $$
