Help in solving the following Sturm-Liouville problem $-x^2y''-2xy'-\lambda y=0$ I have problems finding the eigenvalues and eigenfunctions of the equation:
$$-x^2y''-2xy'-\lambda y=0$$
The domain is $[1,\pi]$, with conditions $y(1)=y(\pi)=0$.
I have proved the values of $\lambda$ have to be nonnegative.
By making the subsitution $y=x^m$, I get to the condition:
$$m^2+m+\lambda=0$$
Which leads to the solutions:
$$y=ax^{-\frac{1+\sqrt{1-4\lambda}}{2}}+bx^{-\frac{1-\sqrt{1-4\lambda}}{2}}$$
I don't know if I'm going the right way about this or if I'm missing information about how to solve this type of problems.
 A: This is a regular Sturm-Liouville problem because it is non-singular on $[1,\pi]$. Start with a solution $y(1)=0$ and $y'(1)=1$ in order to normalize the problem at $x=1$. This gives
$$
         y_{\lambda}(x) = \frac{1}{\sqrt{1-4\lambda}}\left(-x^{-\frac{1+\sqrt{1-4\lambda}}{2}}+x^{-\frac{1-\sqrt{1-4\lambda}}{2}}\right) \\
         = \frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}\left(x^{\frac{\sqrt{1-4\lambda}}{2}}-x^{-\frac{\sqrt{1-4\lambda}}{2}}\right) \\
%   = \frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}\left(e^{\ln(x)}\right)
%         =\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}x^{-\frac{-\sqrt{1-4\lambda}}{2}}\left(x^{\sqrt{1-4\lambda}}-1\right)
  =\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}2\sinh\left(\frac{\sqrt{1-4\lambda}}{2}\log(x)\right) \\
  = -\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}2\sin\left(i\frac{\sqrt{1-4\lambda}}{2}\log(x)\right)
$$
The eigenvalue equation requires us to find the zeros in $\lambda$ of an entire function:
$$
          y_{\lambda}(\pi)=0.
$$
The eigenvalues satisfy
$$
            i\frac{\sqrt{1-4\lambda}}{2}\log(\pi)=\pm n\pi \\
      \sqrt{1-4\lambda}=\pm 2ni\pi/log(\pi) \\
         1-4\lambda = -4n^2\pi^2/(\log(\pi))^2 \\
          \lambda = \frac{1}{4}+n^2\pi^2/(\log(\pi))^2
$$
