Solve $u_t=2u_{xx}+t\sin(x)$ $t>0,0<x<\pi$
boundary conditions are:
$u(0,t)=u(\pi,t)=0$ and $u(x,0)=\sin(x)\cos(x)$
my attempt:
solving for the non homog. first and I got that $u(x,t)=\frac{1}{2}e^{-8t}\sin(2x)$ now i tried to take this and derivate and put it back into the first equation to get the non-homog. part and I got
$$\frac{-1}{2}8e^{-8t}\sin(2x)=2(\frac{1}{2}4e^{-8t}\sin(2x))+t\sin(x)$$
But from here I didn't know how to proceed any hints?
The homogeneous equation is $u_{t}=2u_{xx}$. We are seeking a solution of
the form $u(x,t)=X(x)T(t)$.
Then $-\lambda \dot{T}X(x)=-2\lambda \ddot{X}(x)T(t)$ and the equation is
satisfied when $\ddot{X}(x)+\lambda X(x)=0$ ,$ X(0)=X(\pi )=0$ which is a Sturm-Liouville problem.
The general solution is $X(x)=c_{1}cos\sqrt{\lambda }\,x+c_{2}sin\sqrt{\lambda }\,x$
and from the boundary conditions we get
$c_{1}=0$ and $sin\sqrt{\lambda }\,\pi =0$ which gives a sequence of solutions with
$\lambda _{n}=n^{2}$, $X_{n}(x)=sinnx$.
By the second equation $\dot{T}(t)=-2\lambda T(t)$ we get
$T(t)=ce^{-2\lambda t}$ leading to a sequence
of solutions $T_{n}(t)=c_{n}e^{-2n^{2} t}$. The function
$u(x,t)=\sum_{1}^{\infty }X_{n}(x)T_{n}(t)=\sum_{1}^{\infty }c_{n}e^{-2n^{2}t}sinnx$ is a solution of the problem.
We now expand $f(t,x)=tsinx$ to the form
$\sum_{1}^{\infty }f_{n}(t)X_{n}(x)$ which is simply
$tsinx$ (the rest of the terms are zero). Putting this expansion back to the original equation we get
$u(t,x)=\sum_{1}^{\infty }((e^{-2n^{2}t})\int_{0}^{t}f_{n}(s)e^{2n^{2}s}ds+T_{n}(0)e^{-2n^{2}t})sinnx$.
Where $T_{1}(0)=0$ and $T_{2}(0)=\frac{1}{2}$ (the rest are zero) and by the formula we get
$u(t,x)=(\frac{t}{2}-\frac{1}{4}+\frac{1}{4}e^{-2t})sinx+\frac{1}{2}e^{-8t}sin2x$
which is the solution of the BVP.!!! For more details
https://huanqbui.com/LaTeX%20projects/HuanBui_PDE/HuanBui_PDE.pdf