Using seperation of variables in Partial Differential Equations I am trying to use separation of variables method to solve 
$$ u_t = 9u_{xx}, \quad 0<x<1,\quad t>0$$
$$ u_x(0,t) = 0$$
$$u(1,t)=u(0,t) $$
$$u(x,0) = \sin\pi x, \quad 0 \leq x \le 1$$
and find an approximation (actual number) for large $t$.
Here are my workings so far but I am running into some problems.
Assume $u(x,t) = F(x)G(t)$ so that 
$$u_x = F'G, u_{xx} = F''G, u_t = FG'$$
Thus our equation PDE becomes,
$$9F''G = FG' $$
Rearranging this equation yields,
$$\frac{F''(x)}{F(x)} = \frac{G'(t)}{9G(t)} = \lambda$$
Thus we form two ODE'S,
$$F''-\lambda F = 0$$
$$G'-9\lambda G = 0 $$
Using the B.C $u_x(0,t) = 0$ gives $F'(0) = 0 $ with $A\ne 0$
Thus we solve $F'' - \lambda F = 0$ using the above B
Then if $\lambda = 0 $ then $F(x) = Ax + B$ and $F'(0) = 0 $ gives $A = 0$. 
But how do I use the other BC to determine $B?$
Then if $\lambda > 0 $ then the characteristic equation is $r^2 - \lambda = 0$ which has real, unequal roots $r = \pm \sqrt{\lambda }$. Then,
$$F(x) = Ae^{\sqrt \lambda x } + B e^{-\sqrt \lambda x }$$
See that $F'(0) = 0 = \sqrt \lambda [Ae^0-Be^0]$ so $A=B$, but how do I use the other BC to determine $B$
Finally if $\lambda = -p^2<0$ then the characteristic eqn $r^2 - \lambda = r^2 - p^2 = 0 $ which has complex roots $r=\pm pi$. Thus $F(x) = A\cos px + B\sin px$. 
Then, $$F'(0) = 0 = -Ap\sin 0 + Bp\cos 0 = Bp$$
Then $B = 0$ But how do I use the other BC to determine $A$
After this I know I must do the same procedure for $G'-9 \lambda G = 0 $
 A: Notes: Going through your attempt, there's a few things you were caught up on. The first is, determining what the last constant is. For aboundary value problem, your general only needs to be correct up to a multiplicative constant. If there's only one constant after applying all the B.C, you don't need to solve for it. The second thing is, use all of the B.C. provided. You correctly noted that $F'(0) = 0$, but you didn't use the other B.Cm which is $F(0) = F(1)$
Solution:
Applying separation of variables to $u(x,t) = F(x)G(t)$ gives
$$ F'' - \lambda F = 0, \quad G' - 9\lambda G = 0 $$
The boundary conditions give
$$ F'(0) = 0 $$ 
$$ F(0) = F(1) $$
Case 1: $\lambda = 0$
$$ F(x) = Ax + B $$
$$ G(t) = C $$
The first B.C. implies $A = 0$, which means $F(x)$ must be a constant, which also satisfies the second B.C. There's no constraint on $G(t)$, so it is also a constant. The solution for this case is a constant $u(x,t) = c$
Case 2: $\lambda > 0$
$$ F(x) = Ae^{\sqrt\lambda x} + Be^{-\sqrt\lambda x} $$
$$ G(t) = e^{9\lambda t} $$
$$F'(0) = 0$$ implies $A-B = 0$ or $A=B$. This reduces to
$$F(x) = A(e^{\sqrt \lambda x} + e^{-\sqrt \lambda x})$$
Then
$$ F(0) = 2A = A(e^{\sqrt\lambda} + e^{-\sqrt\lambda}) = F(1)$$
This could either mean $A=0$, which means $F(x) = 0$, a trivial solution, or $e^{\sqrt\lambda} + e^{-\sqrt\lambda} = 2$, which the only solution is $\lambda = 0$, which contradicts our assumption that $\lambda > 0$
Either way, we conclude that the case $\lambda > 0$ yields no solution
Case 3: $\lambda < 0$ or $\lambda = -p^2$
$$ F(x) = A\cos(px) + B\sin(px)$$
$$ G(t) = \exp(-9p^2t) $$
The first B.C. gives $B = 0$, which reduces to $F(x) = A\cos(px)$, then the second B.C. gives
$$ F(0) = A = A\cos p = F(1) $$
which means either $A = 0$ (giving a trivial solution again), or $\cos p = 1$, giving $p = 2n\pi$, so our solution is
$$F_n(x) = \cos (2n\pi x) $$
$$G_n(t) = \exp(-36n^2\pi^2 t) $$
Combining what we have, and using the law of superposition, the general solution has the form
$$ u(x,t) = A_0 + \sum_{n=1}^{\infty} A_n \cos(2n\pi x) \exp(-36n^2\pi^2 t) $$
We can use our final B.C.
$$ u(x,0) = A_0 + \sum_{n=1}^{\infty} A_n \cos(2n\pi x) = \sin (\pi x) $$
Now you must compute the Fourier series expansion of the RHS to determine the remaining constants
$$ A_0 = \int_0^1 \sin (\pi x) dx $$
$$ A_n = 2\int_0^1 \sin(\pi x) \cos(2n\pi x) dx $$
A: I would not suggest using separation of variables for this. Here is an similar example from a problem I did in PDE, hope this help.
Question: 

Solve the inhomogeneous system:
  $$\begin{cases}
u_t = u_{xx} \ \ \text{for} \ \ 0 < x < 1, t > 0\\
u(0,t) = e^{t}\\
u(1,t) = 0\\
u(x,0) = \sin(\pi x)
\end{cases}$$

Solution:
Using Fourier sine series,
\begin{align*}
u(x,t) &= \sum_{n=1}^{\infty}\hat{u}_n(t)\sin(n\pi x) \ \ \text{where} \ \ \hat{u}_n(t) = 2\int_{0}^{1}u(x,t)\sin(n\pi x)dx\\
u_t(x,t) &= \sum_{n=1}^{\infty}\hat{v}_m(t)\sin(n\pi x) \ \ \text{where} \ \ \hat{v}_m(t) = 2\int_{0}^{1}u_t(x,t)\sin(n\pi x)dx = \frac{d}{dt}(\hat{u}_n(t))\\
u_{xx}(x,t) &= \sum_{n=1}^{\infty}\hat{w}_n(t)\sin(n\pi x) \ \ \text{where} \ \ \hat{w}_n(t) = 2n\pi e^{t} - \lambda_n \hat{u}_n(t), \ \ \text{where} \ \lambda_n = (n\pi)^2\\
\phi(x) &= \sin(\pi x) = \sum_{n=1}^{\infty}\hat{\phi}_n \sin(n\pi x) \Rightarrow \begin{cases}
\hat{\phi}_1 &= 1\\
\hat{\phi}_n &= 0 \ \text{if} \ n\neq 1
\end{cases}
\end{align*}
We have 
$$0 = u_t - u_{xx} = \sum_{n=1}^{\infty}\left[\frac{d}{dt}\hat{u}_n(t) + \lambda_n \hat{u}_n(t) - 2n\pi e^{t} \right]\sin(n\pi x)$$
The inside bracket equals zero. We solve the ODE:
$$\begin{cases}
\frac{d}{dt}\hat{u}_n(t) + \lambda_n \hat{u}_n(t) = 2n\pi e^{t}\\
\hat{u}_n(0) = \hat{\phi}_n
\end{cases}$$
Using the integrating factor $e^{\lambda_n t}$ we have
\begin{align*}
e^{\lambda_n t}\frac{d}{dt}\hat{u}_n(t) + \lambda_n e^{\lambda_n t}\hat{u}_n(t) = 2n\pi e^{(\lambda_n + 1)t}\\
\frac{d}{dt}(e^{\lambda_n t} \hat{u}_n(t) ) = 2n\pi e^{(\lambda_n + 1)t}\\
\Rightarrow e^{\lambda_n t}\hat{u}_n(t) = 2n\pi \frac{1}{1 + \lambda_n}e^{(\lambda_n + 1)t} + 2n\\
\Rightarrow \hat{u}_n(t) = \frac{2n\pi}{1 + \lambda_n}e^{t} + c_n e^{-\lambda_n t}\\
\end{align*}
By the initial condition $\hat{u}_n(0) = \hat{\phi}_n(0)\Rightarrow c_n = \hat{\phi}_n(0) - \frac{2n\pi}{1+\lambda_n} = \begin{cases}
1 - \frac{2n\pi}{1 + \lambda_n} \ \text{if} \ n = 1\\
-\frac{2n\pi}{1 + \lambda_n} \ \text{if} \ n\neq 1
\end{cases}$
Then
$$\hat{u}_n(t) = \frac{2n\pi}{1 + \lambda_n}e^{t} + (\hat{\phi}_n(0) - \frac{2n\pi}{1 + \lambda_n})e^{-\lambda_n t}$$
Therefore
$$u(x,t) = \sum_{n=1}^{\infty}\hat{u}_n(t)\sin(n\pi x) = \left[\frac{2\pi}{1 + \pi^2}e^{t} + \frac{(\pi - 1)^2}{1 + \pi^2}e^{-\pi^2 t} \right]\sin(\pi x) + \sum_{n=2}^{\infty}\left(\frac{2n\pi}{1+n^2 \pi^2}(e^{t} - e^{-n^2 \pi ^2 t})\right)\sin(n\pi x)$$
