How to solve for u(x,t)? Solve the pde

$$(t+1)u_t=u_{xx}$$ 

subject to 

$$\rm{B.C.}:u(0,t)=u(\pi,t)=0,\quad \rm{I.C.}: u(x,0)=\sin x+7\sin 6x.$$ 

I've used separation of variables with $k=(t+1)$ but I did not get the right answer for $u(x,t)$.  I got the right answer for the previous problems, but that was when k equalled a constant, like 5 or 1.  What should I do differently?
I used Paul's formula for the answer here.
 A: $$(t+1)u_t=u_{xx}$$Assuming a separable solution, we let $u(x,t)=F(x)G(t)$ so $u_t=F(x)G'(t),u_{xx}=F''(x)G(t)$ thus:$$(t+1)F(x)G'(t)=F''(x)G(t)\\(t+1)\frac{G'(t)}{G(t)}=\frac{F''(x)}{F(x)}=-\lambda^2\\G'(t)+\frac{\lambda^2}{t+1}G(t)=0\\F''(x)+\lambda^2 F(x)=0$$Clearly $F(x)=C_1\cos\lambda x+C_2\sin\lambda x$. To determine $G'(t)$ observe we can multiply by $\mu=\exp\left(\lambda^2\int\dfrac1{t+1}\,dt\right)=(t+1)^{\lambda^2}$ giving us:$$G(t)=\frac{C_3}{(t+1)^{\lambda^2}}$$Given our boundary conditions:$$u(0,t)=u(\pi,t)=0\\u(x,0)=\sin x+7\sin6x$$hence $F(0)=0$ and therefore $C_1=0$ i.e. $F(x)=C_2\sin\lambda x$.
The key is to consider that since $F(\pi)=0\Leftrightarrow C_2\sin\lambda x=0$ we have either $C_2=0\Leftrightarrow u(x,t)=0$ or $\sin\lambda\pi=0$ hence $\lambda\in\mathbb{Z}$. Given the infinite number of eigenvalues $\lambda$, we write:$$F(x)=\sum_{\lambda=0}^\infty b_\lambda\sin\lambda x$$and so it follows with $a_\lambda=C_2C_3b_\lambda$ $$u(x,t)=F(x)G(t)=\sum_{\lambda=0}^\infty a_{\lambda}\frac1{(t+1)^{\lambda^2}}\sin\lambda x\\u(x,0)=\sum_{\lambda=0}^\infty a_\lambda\sin\lambda x\\\sin x+7\sin6x=\sum_{\lambda=0}^\infty a_\lambda\sin\lambda x$$By the orthogonality of $\sin\lambda x$ we have $a_1=1,a_6=7$ and other $a_\lambda=0$. Thus our solution is given by$$u(x,t)=\frac1{t+1}\sin x+\frac7{(t+1)^{36}}\sin 6x$$
Wolfram agrees. Apologies if I left anything unclear -- let me know and I will try to explain further. I suggest reading up on elementary Sturm-Liouville theory.
A: Solutions which
satisfies ${\rm u}\left(0,t\right) = {\rm u}\left(\pi,t\right) = 0$ are of the form
$$
{\rm u}\left(x,t\right)
=
\sum_{n = 0}^{\infty}{\rm A}_{n}\left(t\right)\sin\left(n\pi x\right)
$$
where
$$
\left(t + 1\right)\,{{\rm d}{\rm A}_{n}\left(t\right) \over {\rm d}t}
=
-\left(n\pi\right)^{2}{\rm A}_{n}\left(t\right)
\quad\Longrightarrow\quad
{{\rm d}%
 \left\lbrack
 \left(t + 1\right)^{\left(n\pi\right)^{2}}{\rm A}_{n}\left(t\right)
 \right\rbrack
 \over {\rm d}t} = 0
$$
with ${\rm A}_{n}\left(0\right) = \delta_{n,1} + 7\delta_{n,6}$. Then
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad{\rm u}\left(x,t\right)
=
{\sin\left(\pi x\right) \over \left(t + 1\right)^{\pi^{2}}}
+
7\,{\sin\left(6\pi x\right) \over \left(t + 1\right)^{36\pi^{2}}}\quad}
\\ \\ \hline
\end{array}
$$
A: You can use separation of variables technique to solve the pde. So, assuming

$$ u(x,t)=F(x)G(t) $$

results in the two ode's

$$ F''(x)=\lambda F(x),\quad G'(t)=\frac{\lambda}{1+t}G(t). $$

Now, solve the first ode using the boundary conditions $u(0)=u(\pi)=0$. You can follow the technique in my answer for the heat equation.  
