Find the solution u to the initial value problem $$u_t − u_{xx} =sin(x), u(x,0)=0$$
I know that the particular solution is $u_p = -sin(x)$ So
$u(x,t) = h(x,t) + u_p$
$h(x,t) = u(x,t) - u_p$
$h(x,0) = u(x,0) - u_p(x,0)$
$h(x,0) = 0 + sin(x) = sin(x)$
Now I need to plug $h(x,0)$ into the solution $\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}e^{\frac{-y^2}{4t}}h(x-y,0)dy$
First of all, am I on the right track?
Second of all, how would I simplify this integral? Using $sin(x-y)= sin(x)cos(y) - sin(y)cos(x)$?
$u(x,t) = -\sin(x)$ is a solution to your inhomogeneous PDE, so you want $u(x,t) = h(x,t) - \sin(x)$ where $h(x,t)$ is a solution to the homogeneous PDE $u_t - u_{xx} = 0$ with the initial condition $h(x,0) = u(x,0) + \sin(x) = \sin(x)$. Now from separation of variables $\sin(a x) e^{-a^2 t}$ is a solution to the homogeneous PDE for any constant $a$. Taking $a=1$, we get $h(x,t) = \sin(x) e^{-t}$ which does satisfy the initial condition. Thus a solution to your initial value problem is
$$ u(x,t) = -\sin(x) + \sin(x) e^{-t} $$
I say "a solution" rather than "the solution" because solutions are nonunique unless you restrict their growth at $\infty$: see e.g. here.