Second order linear PDE I have the system with the following partial differential equation.
$\\ \frac{\partial u}{\partial t}=\frac{3 a}{4r^ 2}\frac{\partial^ 2 u}{\partial r^ 2}\\$
How can I solve this? 
 A: An infinity of solutions can be found thanks to the separation of variables :
$u(r,t)=f(r)g(t)$
$$\frac{g'}{g}=\frac{3a}{4r^2}\frac{f''}{f}=constant$$
$g(t)=e^{ct}$ with as a consequence $\frac{3a}{4r^2}\frac{f''}{f}=c$
This is a "parabolic cylinder" ODE wich solutions are on the form: $f(r)=D_{-1/2}\left((\frac{8c}{3a})^{1/4}r\right)$ where the function involved is the parabolic cylinder function : http://mathworld.wolfram.com/ParabolicCylinderFunction.html
The solution of the PDE can be expressed as a sum of terms :
$$u(r,t)=\sum_{k} A_k e^{c_k t}D_{-1/2}\left(\left(\frac{8c_k}{3a}\right)^{1/4}r\right)$$
where $A_k$ and $c_k$ are any number of constants than wanted.
A integral form is :
$$u(r,t)=\int A(z) e^{c(z) t}D_{-1/2}\left(\left(\frac{8c(z)}{3a}\right)^{1/4}r\right)dz$$
where $A(z)$ and $c(z)$ are any function, insofar the integral be convergent.
As usual, an infinity of solutions of the PDE can be expressed on that way. This is the easier part of the job. The difficult part of the job is to find the solution which fit with the boundary conditions (not specified in the wording of the question). 
