# Sturm-Liouville problem - Partial Differential Equations

I tried looking on the Mathematica forum for a similar solution but unfortunately haven't found one.

I was given the following Sturm-Liouville differential equation problem:

$$x^2y''+5xy'+\lambda y=0$$

$$y(1)=y(e^{\pi}) = 0$$

I displayed the equation in the traditional Sturm-Liouville manner and found that the weight function equals $$x^3$$, but I am having trouble finding the eigenvalues for this problem.

We were told in class that because the coefficients of the ODE are functions of $$x$$ and not constants, we should guess a solution for $$y$$ in the form of $$x^m$$ and continue from there finding the eigenvalues.

Doing that I got a simple quadratic equation involving $$m$$:

$$m^2 -4m +\lambda = 0$$

But because it is a quadratic equation where p and q are both non zero I can't seem to find a way to divide the eigenvalues into the common categories which we usually did in class (for $$\lambda <0, >0, =0$$), like in this case for instance:

$$m^2 + \lambda = 0$$

I believe the solution is very simple but just can't find it, help would be much appreciated!

If you substitute $$x^m$$ into $$x^2y''+5xy'+\lambda y=0$$, you get $$m(m-1)+5m+\lambda = 0 \\ m^2+4m+\lambda = 0 \\ m =\frac{-4\pm\sqrt{16-4\lambda}}{2}=-2\pm\sqrt{4-\lambda}$$ The corresponding solutions are $$\frac{1}{x^2}x^{\sqrt{4-\lambda}}, \frac{1}{x^2}x^{-\sqrt{4-\lambda}}.$$ To put $$x^2y''+5xy'+\lambda y=0$$ into Sturm-Liouville form, multiply by $$x^3$$: $$-(x^5y')'=\lambda x^3y \\ -\frac{1}{x^3}(x^5y')'=\lambda y$$ This problem is considered in $$L^2_{x^3}(1,e^{\pi})$$, where $$x^3$$ is the weight function for the space. The solutions of the above when $$\lambda=0$$ are both in $$L^2_{x^3}(1,e^{\pi})$$ because they are regular on $$(1,e^{\pi})$$. To solve the eigenvalue problem, search for $$A,B$$ such that $$y(x)=A\frac{1}{x^2}x^{\sqrt{4-\lambda}}+B\frac{1}{x^2}x^{-\sqrt{4-\lambda}}$$ satisfies $$y(1)=0,\;\; y(e^{\pi})=0.$$ The first condition is satisfied by setting $$B=-A$$, which gives \begin{align} y(x)&=\frac{A}{x^2}\left[x^{\sqrt{4-\lambda}}-x^{-\sqrt{4-\lambda}}\right] \\ &=\frac{A}{x^2}\left[e^{i\sqrt{\lambda-4}\ln x}-e^{-i\sqrt{\lambda-4}\ln x}\right] \\ &=\frac{A}{x^2}2i\sin(\sqrt{\lambda-4}\ln x) \end{align} The condition $$y(e^{\pi})=0$$ gives the eigenvalue equation $$\sin(\sqrt{\lambda-4}\pi)=0.$$