Partial differential equation (heat equation with other terms)? Can some one help me solve the following PDE with the given intial and boundary conditions?
$\gamma t\frac{\partial^{2}f}{\partial x^{2}}=t\frac{\partial f}{\partial t}-\alpha f$
Initial condition: $f(x,t=0)=0$
Outer Boundary Condition: $f(x\rightarrow\infty,t)=0$
Inner Boundary Condition: $\left.\frac{\partial f}{\partial x}\right|_{x=0}=c\left.\frac{\partial f}{\partial t}\right|_{x=0}-1$
I am only interested in the solution at the inner boundary. i.e.
$f(x=0,t>0)=?$
Other information:
$0\leq\alpha\leq0.5$
$\gamma>0$
$c\geq0$
My Skill level:
I have solved similar heat equation problems using Laplace transforms but none for the heat equation with the term $\alpha\frac {f}{t}$
Any help is greatly appreciated!
EDIT I have solved the problem using COMSOL and it is giving me results which are close to my physical problem. However I am interested in getting an analytic solution. 
 A: Assume $\alpha\neq0$ for the key cases:
Of course use separation of variables:
Let $f(x,t)=X(x)T(t)$ ,
Then $\gamma tX''(x)T(t)=tX(x)T'(t)-\alpha X(x)T(t)$
$\gamma tX''(x)T(t)=X(x)(tT'(t)-\alpha T(t))$
$\dfrac{tT'(t)-\alpha T(t)}{\gamma tT(t)}=\dfrac{X''(x)}{X(x)}=-s^2$
$\begin{cases}\dfrac{T'(t)}{T(t)}=\dfrac{\alpha}{t}-\gamma s^2\\X''(x)+s^2X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)t^\alpha e^{-\gamma ts^2}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore f(x,t)=\int_0^\infty C_1(s)t^\alpha e^{-\gamma ts^2}\sin xs~ds+\int_0^\infty C_2'(s)t^\alpha e^{-\gamma ts^2}\cos xs~ds$
Which is automatically satisfied the condition $f(x,0)=0$
$\left.\dfrac{\partial f}{\partial x}\right|_{x=0}=c\left.\dfrac{\partial f}{\partial t}\right|_{x=0}-1$ :
$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds=c\alpha\int_0^\infty C_2'(s)t^{\alpha-1}e^{-\gamma ts^2}~ds-c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-1$
$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-c\alpha\int_0^\infty t^{\alpha-1}e^{-\gamma ts^2}~d(C_2(s))=-1$
$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-c\alpha\left[C_2(s)t^{\alpha-1}e^{-\gamma ts^2}\right]_0^\infty+c\alpha\int_0^\infty C_2(s)~d(t^{\alpha-1}e^{-\gamma ts^2})=-1$
$\int_0^\infty sC_1(s)t^\alpha e^{-\gamma ts^2}~ds+c\gamma\int_0^\infty s^2C_2'(s)t^\alpha e^{-\gamma ts^2}~ds-2c\alpha\gamma\int_0^\infty sC_2(s)t^\alpha e^{-\gamma ts^2}~ds+c\alpha C_2(0)=-1$
$t^\alpha\int_0^\infty(sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s))e^{-\gamma ts^2}~ds=-c\alpha C_2(0)-1$
$\int_0^\infty(sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s))e^{-\gamma ts^2}~ds=-(c\alpha C_2(0)+1)t^{-\alpha}$
$sC_1(s)+c\gamma s^2C_2'(s)-2c\alpha\gamma sC_2(s)=-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)s^{2\alpha-1}}{\Gamma(\alpha)}$ (according to http://en.wikipedia.org/wiki/Gaussian_integral#Integrals_of_similar_form)
$C_1(s)=-c\gamma sC_2'(s)+2c\alpha\gamma C_2(s)-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)s^{2\alpha-2}}{\Gamma(\alpha)}$
$\therefore f(x,t)=\int_0^\infty t^\alpha e^{-\gamma ts^2}(C_2'(s)\cos xs-(c\gamma sC_2'(s)-2c\alpha\gamma C_2(s))\sin xs)~ds-\dfrac{2\gamma^\alpha(c\alpha C_2(0)+1)}{\Gamma(\alpha)}\int_0^\infty t^\alpha s^{2\alpha-2}e^{-\gamma ts^2}\sin xs~ds$
