I am trying to solve this problem: $$u_t(x,t)=u_{xx}(x,t)+e^{-t}u(x,t),x\in (0,\pi),t>0\\u(0,t)=u(\pi,t)=0,t>0\\u(x,0)=\sin(2x),x\in(0,\pi)$$
but I don't know how to deal with the $e^{-t}$ term. If it was a constant coefficient of $u$, I would substitute: $U = Ve^{ax +bt} $ where $V_t = V_{xx}$ with homogenous boundary conditions.
P.S. (it's my first post here so I am sorry if the question is unclear. Also, this is not a homework question.)
Substituting the ansatz $u(x,t)=f(t)\sin(2x)$ in the PDE $u_t=u_{xx}+e^{-t}u$, we obtain \begin{align} f'(t)\sin(2x)&=-4f(t)\sin(2x)+e^{-t}f(t)\sin(2x) \\ &\implies f'(t)=(e^{-t}-4)f(t) \\ &\implies f(t)=C\exp(-e^{-t}-4t), \tag{1} \end{align} hence $$ u(x,t)=C\exp(-e^{-t}-4t)\sin(2x). \tag{2} $$ This function satisfies the boundary conditions $u(0,t)=u(\pi,t)=0$, and the initial condition $u(0,t)=\sin(2x)$ if $C=e$. Therefore, the solution to the problem is given by $$ u(x,t)=\exp(1-e^{-t}-4t)\sin(2x). \tag{3} $$