The problem: Solve $$ \frac{dT}{dt} + k \frac{d^{2}T}{dx^2} = \exp \Bigl[-\alpha t\Bigr] \sin\left( \frac{2\pi x}{L}\right)\text{ with }T(0,t) = T(L,t) = 0,\ T(x,0) = f(x) $$ with $k$, $\alpha > 0$ constants, and $f(x)$ piecewise smooth. Hint: look for the solution as a sum of a forced (particular) and homogeneous solution. There is a solution for $\alpha \neq \alpha_0$ and one when $\alpha = \alpha_0$; you need to find $\alpha_0$.
Here is my work.I have used separation of Variables From the hint, the homogeneous equation would be just an ordinary heat equation with Dirchlet BC's which the only eigenvalues and eigenfunctions produced were $\lambda_n = \frac{n \pi x}{L} $, and $\phi_n (x) = \sin(\frac{n \pi x}{L}) $ respectively. By principle of Superposition, the function, $$ u(x,t)= \sum_ {n=1}^{\infty} \exp \left[ \left(\frac{k n \pi x }{L}\right)^{2} \right] \sin\left(\frac{n \pi x}{L}\right). $$ I am just not sure how to get the particular solution of this equation.
Thank you for all of your help.
The only reasonable guess for a particular solution is that it should be something that looks like the forcing term. (If it does not, we are out of luck). So, the first thing to do is to take the forcing term $$g(x,t)=\exp \Bigl[-\alpha t\Bigr] \sin\left( \frac{2\pi x}{L}\right)$$ and stick it into the left side of the PDE. This works quite well: $$g_t+kg_{xx} = -\left(\alpha + k\left( \frac{2\pi }{L}\right)^2\right)g$$ Let $\beta=\left(\alpha + k\left( \frac{2\pi }{L}\right)^2\right)$. So, a particular solution is $$ u_P(x,t) = -\beta^{-1}g(x,t) $$ Then the homogeneous solution should satisfy the initial condition $$u_H(x,0)= \beta^{-1} g(x,0)+f(x)$$ and it can be found in the usual way.
I don't know what the remark with $\alpha_0$ is about.