Separation of variables in heat equation with decay I just want to see if I completed this problem right. Here is the problem: 
Consider $\frac{\partial T}{\partial t} = k \frac{\partial^{2} T}{\partial x^2} -\alpha T$ where $k,\alpha >0$  are constants and $\partial_x(0,t) = \partial_x (L,t) = 0$ and $T(x,0) = f(x)$. Find the equilibrium temperature and T(x,t). Find the long time asymptotic limit of T and compare to the equilibrium temperature. 
Here is my attempt at the problem: 
Assume the solution is in the form $T(x,t) = F(x)G(t)$. Then, By Separation of Variables, I got 
$(\lambda +\alpha) G \frac{\partial G}{\partial t} = \frac{k}{F} \frac{\partial^{2} T}{\partial x^2}=\lambda $, which lambda is the separation constant.I have examined the cases for $\lambda$, the cases for $\lambda = 0$, and   $\lambda >0$, they would yield the trivial solution for F(x). For $\lambda <0$, I would have                                       $F(x) = A \cos \sqrt{\frac{\lambda}{k} }x +B \sin \sqrt{\frac{\lambda}{k} }x$ which means 
$F'(X) = \sqrt{\frac{\lambda}{k}} (-A \sin (\sqrt{\frac{\lambda}{k}}x) +B\cos (\sqrt{\frac{\lambda}{k}}x))$.
When I initialize the initial conditions, I got the general solution to be a sequence of functions, which is $F_n(x) = A_n \sin (\frac{n \pi x}{L}) $. For G(t), I got the function to be $G(t) = C e ^{-(\lambda -\alpha)t} $. 
Since $\lambda = (\frac{n \pi x}{L})^2 k$. $G(t)$ is also generalized as a sequence of functions which is 
$G_n(t) = C_n e^{((\frac{n \pi x}{L})^2 k - \alpha)t} $ 
Thus the general solution would be a linear sum of sequence of functions: 
$T(x,t) = \sum_{n=1}^{\infty}  a_n \sin (\frac{n \pi x}{L})e^{((\frac{n \pi x}{L})^2 k - \alpha)t}$
From the initial condition, $T(x,0) = F(x)$, this would turn out to be 
$T(x,0) = \sum_{n=1}^{\infty}  a_n \sin (\frac{n \pi x}{L})=F(x) $ which is a Fourier Sine Series representation. 
Am I on the right track on this? Also, for the equilibrium temperature, this is where we set $ \frac{\partial T}{\partial t} = 0$ and solve. What I got for the equilibrium temperature function was 
$F(x) = A e^\sqrt{\frac{\alpha}{k}x} +B e^{-\sqrt{\frac{\alpha}{k}x}}$. From the boundary conditions, I got the solution was the trivial solution. Did I do this correctly? 
Also, for the coefficients, $a_n$, after you are done solving them, do you have to substitute it in into $T(x,t)$? 
Thank you for all of your help.
 A: 
When I initialize the initial conditions, 

Too early to look at initial condition. After you separate the variables, you deal with boundary conditions first. Which brings me to ... 

$F_n(x) = A_n \sin (\frac{n \pi x}{L})$

This is incorrect; you need cosines in order to have zero slope at the endpoints. Sines have zero value but nonzero slope. 
Also, a typo here: 

$G(t) = C e ^{-(\lambda -\alpha)t}$

Sanity check: the coefficients $\alpha$ should contribute to decay, not growth. I think you want $G(t) = C e ^{-(\lambda +\alpha)t}$. Actually, you don't need $C$, since you are multiplying by indefinite coefficient $A_n$ anyway. 
At the very last step, you deal with initial condition: plug $t=0$ into the series you've got, and say what $A_n$ need to be in order for this thing to be $f$. Of course, you can't say much if $f$ is not given: just write down the relation to cosine Fourier coefficients of $f$.
A: Given the pde 
\begin{align}
\partial_{t} u(x,t) = k \partial_{x}^{2} u(x,t) - \alpha u(x,t),
\end{align}
with the conditions $\partial_{x}u(0,t)= \partial_{x}u(L,t) = 0$ and $u(x,0) = f(x)$, then by separation of variables let $u = F(t) G(x)$ to obtain
\begin{align}
\frac{F'}{F} = - \lambda^{2} = k \, \frac{G''}{G} - \alpha.
\end{align}
This yields the two equations
\begin{align}
F' + \lambda^{2} F &= 0 \\
G'' + \left( \frac{\lambda^{2} - \alpha}{k} \right) G &= 0.
\end{align}
The first equation provides the solution
\begin{align}
F(t) = e^{- \lambda^{2} t}.
\end{align}
The equation for $G(x)$ is seen as follows. For the case that $\lambda^{2} = \alpha$ then $G''=0$ for which $G(x) = ax +b$. For the case $\lambda^{2} \neq \alpha$ then
\begin{align}
G(x) = A \cos\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right) + B \sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right)
\end{align} 
and
\begin{align}
G'(x) = - A \sqrt{\frac{\lambda^{2}-\alpha}{k}} \, \sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right) + B \sqrt{\frac{\lambda^{2}-\alpha}{k}} \, \cos\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right).
\end{align} 
Now, given the boundary conditions that $G'(0) = G'(L) = 0$ and $u(x,0) = f(x)$ then for the first case of $G$ then $G'(x) = a$ and $G'(0) = a = 0$. For the second case $G'(0) = 0 \rightarrow B = 0$ and $G'(L) = 0$ leads to
\begin{align}
\sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, L \right) = 0 \rightarrow 
\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, L = n \pi
\end{align}
for $n \geq 1$. This leads to
\begin{align}
\lambda^{2} = \alpha + k \left( \frac{n \pi}{L} \right)^{2}.
\end{align}
Combining the solutions it is seen that
\begin{align}
u(x,t) = A e^{-\alpha t} + \sum_{n=1}^{\infty} B_{n} \sin\left( \frac{n\pi x}{L} \right) \, e^{- \alpha t - k \left( \frac{n \pi}{L}\right)^{2} t}.
\end{align}
For $u(x,0) = f(x)$ then
\begin{align}
f(x) = A + \sum_{n=1}^{\infty} B_{n} \sin\left( \frac{n\pi x}{L} \right).
\end{align}
From Fourier series the coefficients are seen as
\begin{align}
A &= \frac{1}{L} \, \int_{0}^{L} f(x) \, dx \\
B_{n} &= \frac{2}{L} \, \int_{0}^{L} f(x) \, \sin\left( \frac{n \pi x}{L} \right) \, dx.
\end{align}
