Here is the problem:
$$ x^2y''-xy+\lambda y = 0,\quad y(1)=0,\quad y(L)=0,\quad L>0 $$
I am asked to find the Eigenvalues and Eigenfunction.
I can't figure out how to get a general equation for y. I tried integrating factors but that was a mess, I'm not sure if there's a better way to do this. It might be a Sturm Liouville equation, but I'm not sure to to solve those. Any help would be appreciated.
Here is the solution given:
$ \lambda_n=1+(n\pi/ln(L))^2,\quad y_n(x)=xsin(n\pi ln(x)/ln(L));\quad n=1,2,3... $
There is likely a typo in the equation. The solution you give solves $$ x^2y''-xy'+\lambda_ny=0. $$ As John mentioned, this is an Euler equation with characteristic equation $m(m-1)-m+\lambda =0$. For $\lambda\leq1$, $m$ is real and there is no solution with $y(L)=0$. For $\lambda>1$ we get solutions $$ y(x)=c_1 x \cos(\sqrt{\lambda-1}\,\log x) + c_2 x \sin(\sqrt{\lambda-1}\,\log x). $$ The condition $y(0)=0$ forces $c_1=0$. The condition $y(L)=0$ forces $$ \sin(\sqrt{\lambda-1}\,\log L)=0, $$so $$\sqrt{\lambda-1}\log L=n\pi,\ \ \ n\in\mathbb N$$ This gives us eigenvalues $$ \lambda_n=1+\left(\frac{n\pi}{\log L}\right)^2 $$ and eigenvectors $$ y_n(x)=x\,\sin\left(\frac{n\pi}{\log L}\,\log x\right). $$