PDE Heat Equation with Variable Coefficient {Second ODE Variable Coefficient} Another PDE question:
If I have a non constant coefficients in my heat equation (PDE), how do I solve it? For example we have: 
$\frac {\partial T}{\partial t} =\frac {\partial ^2 T}{\partial r^2} + \frac 2r \frac {\partial T}{\partial r}$
{The original question is actually 
$\frac {\partial T}{\partial t} =\frac {1}{r^2} \frac{\partial}{\partial r} ( r^2 \frac {\partial T}{\partial r})  $ but I simplified it because I dont see any other way that would help me solve the question, or do I?}
We are supposed to make it separable and so now we have: (with ansatz $T(r,t)=F(r)G(t)$) $\frac{G'(t)}{G(t)}=\frac{F''(r)}{F(r)}+\frac 2r \frac{F'(r)}{F(r)}=\lambda $
So we have two separable equations: $F''(r) +\frac 2r F'(r) - \lambda F(r) =0$ and also $G'(t)-\lambda G(t)=0 $
Right. Then how do I proceed with this?
I mean we now have a perfectly fine ODE but I could not solve this, because its either using the reduction order, in which we have to know one of the solution, or using the Greene function, in which I don't know how to even do a complimentary function in this question. 
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Edit:
Boundary Condition is $\frac {\partial T}{\partial r}(1,t) = 0, t>0 $; Initial Condition is $ T(r,0)=r^2, 0<r<1$
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Edit #2:
Boundary condition is $\frac{\partial T}{\partial r}(1,t)=0$. I tried doing the eigenvalue thing but with this being the only boundary condition, I can't do anything to eliminate constants.
 A: Hint:
Let $T=\dfrac{U}{r}$ ,
Then $\dfrac{\partial T}{\partial t}=\dfrac{1}{r}\dfrac{\partial U}{\partial t}$
$\dfrac{\partial T}{\partial r}=\dfrac{1}{r}\dfrac{\partial U}{\partial r}-\dfrac{U}{r^2}$
$\dfrac{\partial^2T}{\partial r^2}=\dfrac{1}{r}\dfrac{\partial^2U}{\partial r^2}-\dfrac{1}{r^2}\dfrac{\partial U}{\partial r}-\dfrac{1}{r^2}\dfrac{\partial U}{\partial r}+\dfrac{2U}{r^3}=\dfrac{1}{r}\dfrac{\partial^2U}{\partial r^2}-\dfrac{2}{r^2}\dfrac{\partial U}{\partial r}+\dfrac{2U}{r^3}$
$\therefore\dfrac{1}{r}\dfrac{\partial U}{\partial t}=\dfrac{1}{r}\dfrac{\partial^2U}{\partial r^2}-\dfrac{2}{r^2}\dfrac{\partial U}{\partial r}+\dfrac{2U}{r^3}+\dfrac{2}{r}\left(\dfrac{1}{r}\dfrac{\partial U}{\partial r}-\dfrac{U}{r^2}\right)$
$\dfrac{1}{r}\dfrac{\partial U}{\partial t}=\dfrac{1}{r}\dfrac{\partial^2U}{\partial r^2}-\dfrac{2}{r^2}\dfrac{\partial U}{\partial r}+\dfrac{2U}{r^3}+\dfrac{2}{r^2}\dfrac{\partial U}{\partial r}-\dfrac{2U}{r^3}$
$\dfrac{1}{r}\dfrac{\partial U}{\partial t}=\dfrac{1}{r}\dfrac{\partial^2U}{\partial r^2}$
$\dfrac{\partial U}{\partial t}=\dfrac{\partial^2U}{\partial r^2}$
Let $U(x,t)=R(r)T(t)$ ,
Then $R(r)T'(t)=R''(r)T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{R''(r)}{R(r)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2\\R''(r)+s^2R(r)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-ts^2}\\R(r)=\begin{cases}c_1(s)\sin((r-1)s)+c_2(s)\cos((r-1)s)&\text{when}~s\neq0\\c_1r+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore T(r,t)=\dfrac{1}{r}\int_0^\infty C_1(s)e^{-ts^2}\sin((r-1)s)~ds+\dfrac{1}{r}\int_0^\infty C_2(s)e^{-ts^2}\cos((r-1)s)~ds$
$\dfrac{\partial T}{\partial r}=\dfrac{1}{r}\int_0^\infty sC_1(s)e^{-ts^2}\cos((r-1)s)~ds-\dfrac{1}{r^2}\int_0^\infty C_1(s)e^{-ts^2}\sin((r-1)s)~ds-\dfrac{1}{r}\int_0^\infty sC_2(s)e^{-ts^2}\sin((r-1)s)~ds-\dfrac{1}{r^2}\int_0^\infty C_2(s)e^{-ts^2}\cos((r-1)s)~ds$
$\dfrac{\partial T}{\partial r}(1,t)=0$ :
$\int_0^\infty sC_1(s)e^{-ts^2}~ds-\int_0^\infty C_2(s)e^{-ts^2}~ds=0$
$\int_0^\infty(sC_1(s)-C_2(s))e^{-ts^2}~ds=0$
$sC_1(s)-C_2(s)=0$
$C_2(s)=sC_1(s)$
$\therefore T(r,t)=\dfrac{1}{r}\int_0^\infty C_1(s)e^{-ts^2}\sin((r-1)s)~ds+\dfrac{1}{r}\int_0^\infty sC_1(s)e^{-ts^2}\cos((r-1)s)~ds$
$T(r,0)=r^2$ :
$\dfrac{1}{r}\int_0^\infty C_1(s)\sin((r-1)s)~ds+\dfrac{1}{r}\int_0^\infty sC_1(s)\cos((r-1)s)~ds=r^2$
$\int_0^\infty C_1(s)\sin((r-1)s)~ds+\int_0^\infty sC_1(s)\cos((r-1)s)~ds=r^3$
$\int_0^\infty C_1(s)\sin rs~ds+\int_0^\infty sC_1(s)\cos rs~ds=(r+1)^3$
$\int_0^\infty C_1(s)\sin rs~ds+\int_0^\infty sC_1(s)\cos rs~ds=r^3+3r^2+3r+1$
