$2^{nd}$ order PDE: Solution I am trying to solve the following equation:
$$\frac{\partial F}{\partial t} = \alpha^2 \, \frac{\partial^2 F}{\partial x^2}-h \, F$$
subject to these conditions:
$$F(x,0) = 0, \hspace{5mm} F(0,t) = F(L,t)=F_{0} \, e^{-ht}.$$
I know that I am suppose to simplify the equation with:
$$F(x,t)=\phi(x,t)e^{-ht}$$
My initial guess is to divide by $$\alpha^2$$
and have this:
$$\frac{d^2F}{dx^2}-\frac{1}{\alpha^2}\frac{dF}{dt}-\frac{h}{\alpha^2} \, F=0.$$
I am having trouble with the next steps. Should I assume a solution of the exponential form?

The following field redefinition, similar to what the other answers have done, simplifies the problem: 
$$F = (F_0 + \phi)e^{-ht} \implies \frac{d\phi}{dt} = \alpha^2\frac{d^2\phi}{dx^2}$$
with $\phi(0,t) = \phi(L,t) = 0$ and $\phi(x,0) = -F_0$. One can now proceed with your favoritte method like separation of variables or expanding $\phi$ in a Fourier series to obtain an equation for the coefficients. 

The other answers have done this so I just want to focus on a problem one encounters when applying the initial condition. No matter what approach you take you will be lead to a Fourier series and to evaluate the coefficients you need to apply the initial condition $\phi(x,0) = -F_0$. This function does not have a non-trivial Fourier-series so a naive application would not give a solution. The usual trick to get around this is by extending the domain of $\phi$ from $[0,L]$ to $[-L,L]$ and then demaning $\phi(x,t)=-\phi(-x,t)$, i.e. we demand that the solution should be odd (since the odd completion do have a nice Fourier series). For the initial condition we therefore consider
$$\phi(x,0) = -F_0\left\{\matrix{-1 & x<0 \\1 & x > 0}\right.$$
This function, the so-called square-wave, has the Fourier series
$$\phi(x,0) = -\frac{4F_0}{\pi}\sum_{n=0}^\infty\frac{1}{2n+1}\sin\left(\frac{(2n+1)\pi x}{L}\right)$$
which can be found by applying the usual method. 

Just for completeness, here is a quick runthough of the solution. We expand $\phi$ in a Fourier $\sin$-series (since we demand that $\phi$ is odd) $\phi(x,t) = \sum_{n=1}^\infty a_n(t) \sin\left(\frac{n\pi x}{L}\right)$ that satisfy the boundary conditions $\phi(0,t)=\phi(L,t) = 0$. By inserting this series into the PDE and equating the left and right hand side we get the equation for the coefficients
$$\frac{da_k}{dt} = -\frac{n^2\pi^2\alpha^2}{L^2} a_k \implies a_k(t) = a_k(0) e^{-\frac{n^2\pi^2\alpha^2}{L^2}t}$$
where $a_k(0)$ is simply the Fourier coefficients of the initial condition which here is just the square-wave given above. The full solution can therefore be written
$$F(x,t) = F_0e^{-ht}\left[1 - \frac{4}{\pi}\sum_{n=0}^\infty\frac{e^{-\frac{(2n+1)^2\pi^2 \alpha^2}{L^2}t}}{2n+1}\sin\left(\frac{(2n+1)\pi x}{L}\right)\right]$$
for all $t>0$ and $0\leq x\leq L$. Note that the formula above is not valid for $t=0$ at the two points $x=0$ and $x=L$ since we get $F(x,0)=F_0$ instead of $F(x,0)=0$.
