<s>[INST] Find eigenvalues of $x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi = 0$ with boundary conditions $\phi(1) = \phi(b) = 0$ 
Since this is an equidimensional equation, determine all positive eigenvalues.
  \begin{align*}
  x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\
  \phi(1) &= 0 \\
  \phi(b) &= 0 \\
\end{align*}
  The given answer that I need to find the solution for is for $n = 1,2,\ldots$:
  \begin{align*}
  \lambda_n &= \left( \frac{n \pi}{\ln b} \right)^2 \\
\end{align*}

If we multiply by $1/x$ and rearrange we can get this in Sturm-Liouville form with $p(x) = x, q(x) = 0, \sigma(x) = 1/x$:
\begin{align*}
  x \frac{d^2\phi}{dx^2} + \frac{d\phi}{dx} + \frac{1}{x} \lambda \phi &= 0 \\
  \frac{d}{dx} \left( x \frac{d\phi}{dx} \right) + \lambda \frac{1}{x} \phi &= 0 \\  
\end{align*}
If we look at the Rayleigh quotient we can see there are no negative eigenvalues and we can manually verify that zero is not an eigenvalue.
Assume a trial solution of $\phi(x) = x^m, \phi'(x) = m x^{m-1}, \phi''(x) = m(m-1)x^{m-2}$ such that:
\begin{align*}
  x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\
  m (m-1) x^m + m x^m + \lambda x^m &= 0 \\
  \left( m^2 + \lambda \right) x^m &= 0 \\
  \lambda &= -m^2 \\
\end{align*}
Since we know that $\lambda > 0$, then $m^2 < 0$, which means that $m = i r$ where $r \in \mathbb{R}$. However, this can't satisfy the boundary conditions:
\begin{align*}
  \phi(x) &= e^{ir} \\
  \phi(1) = 0 &\neq 1^{ir} = 1 \\
\end{align*}
I'm stuck on what else to try.
 [/INST] You have to take the change of variable $y=\ln x$ and the equation becomes
$$
\frac{d^2\phi}{dy^2}+\lambda\phi=0
$$
that is a standard form having as a solution
$$
\phi(y)=A\cos\sqrt{\lambda} y+B\sin\sqrt{\lambda} y.
$$
I think now you can go on from here.
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