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!