I'm trying to work around my way of this problem
$$a\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial u}{\partial r})+be^{-y(r-a)}=\frac{\partial u}{\partial t}$$
$$\left.\frac{\partial u}{\partial r}\right\rvert_{r=1}=-[bu+c]$$
$$u(a,t)=0$$ $$u(r,0)=1$$
I tried to use $u(r,t)=w(r,t)+v(r)$ but the thing is the only method that I know of to solve non homogeneous PDE involves doing separation of variables first, but I run into the problem of the non homogeneous BC. So I'm getting confused on how exactly to proceed, because I can try to solve the homogeneous PDE with $u(r,t)=w(r,t)+v(r)$, and get a solution, but how then should I proceed with the non homogeneous PDE? With other problems I'd try to express $f(r,t)$ with the functions that result from the separation of variables, in this case I can't proceed in the same way
Is there another method or am I doing it wrong?
You are doing nothing wrong. Write $u(r,t)=w(r,t)+v(r)$ and choose $v(r)$ as the solution to the ODE $$ a\frac{1}{r}\frac{d}{dr}\left(r\frac{dv}{dr}\right)+be^{-y(r-a)}=0 \tag{1} $$ with boundary conditions \begin{align} v'(1)&=-bv(1)-c, \tag{2.1}\\ v(a)&=0. \tag{2.2} \end{align} With this choice, $w(r,t)$ will satisfy the homogeneous PDE $$ a\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial w}{\partial r}\right)=\frac{\partial w}{\partial t} \tag{3} $$ with the homogeneous boundary conditions \begin{align} w_r(1,t)&=-bw(1,t), \tag{4.1} \\ w(a,t)&=0, \tag{4.2} \\ \end{align} and the modified initial condition $$ w(r,0)=1-v(r). \tag{5} $$ The problem for $w$ can be solved with the usual method of separation of variables.