Solve $\frac {\partial^2 v}{\partial r^2}+\frac 1r \frac {\partial v}{\partial r}-\frac v{r^2} =0$

Question

$$\frac {\partial^2 v}{\partial r^2}+\frac 1r \frac {\partial v}{\partial r}-\frac v{r^2} =0$$ i did $$\frac {1}{r}\frac {\partial }{\partial r}\bigg( r \frac {\partial v}{\partial r} \bigg) =\frac v{r^2} $$ multiply $r $ from each side and then integrate both sides wrt $r $. And then carry on from there. Is what i did correct so far because it seems very dodgy. If it is, can someone give me a solution please.

$v(r=a)=0$ and $v (r=b) = \Omega b $

Answer 1

Hint: Try guessing answers of the form $$ v = r^ \lambda$$ You'll find $$ r^2 v'' + r v ' -v = r^ \lambda P ( \lambda) = 0$$ if $r>0$, then $v$ is a solution if $P( \lambda) =0$.

Answer 2

This is an Euler differential equation. Hint: look for a fundamental set of solutions of the form $v(r) = r^p$.