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!
 A: 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
$$
          \sin(\sqrt{\lambda-4}\pi)=0.
$$
