Finding eigenfunctions and eigenvalues to Sturm-Liouville operator I'm struggling to understand how to find the associated eigenfunctions and eigenvalues of a differential operator in Sturm-Liouville form. For instance, one question that I am trying to solve is the following:

By substituting $x=e^{t}$ find the eigenfunctions and eigenvalues of the operator $\hat{\mathcal{L}}$, defined by: $$\hat{\mathcal{L}}[y(x)]=x^{2}y''(x)+2xy'(x)+\frac{1}{4}y(x)$$  With boundary conditions $y(1)=y(e)=0$

I can set up the eigensystem:
$$x^2 y_{n}''(x)+2xy'_{n}(x)+\frac{1}{4}y_{n}(x)=\lambda_{n}y_{n}(x)$$
In Sturm-Lioville form, we have:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}\right)+\frac{1}{4}y_{n}(x)=\lambda_{n}y_{n}(x)$$
However, I'm not sure how to solve this equation as Rodriguez' conditions are not met and if I attempt an ansatz series solution: $y_{n}=\sum_{n=0}^{\infty}\alpha_{n}x^{n}$, I end up with the following equation:
$$\sum_{n=0}^{\infty}\alpha_{n}\cdot n(n-1)x^{n}+\sum_{n=0}^{\infty}2\alpha_{n}\cdot n x^{n}+\sum_{n=0}^{\infty}\frac{\alpha_{n}}{4}x^{n}=\sum_{n=0}^{\infty}\lambda \alpha_{n}x^{n} $$
However, this does not give me a recursion relationship as I would expect, instead I simply get:
$$\lambda = k^{2} + k + \frac{1}{4} \qquad \exists k \in \mathbb{N} \cup \{0\}$$
I tried substituting $x=e^{t}$, giving me:
$$\frac{\mathrm{d}^{2}}{\mathrm{d}t^{2}}\left(y_{n}(e^{t})\right)+\frac{1}{4}y_{n}(e^{t})=\lambda_{n}y_{n}(e^{t})$$
But I wasn't sure how to solve this for the eigenfunctions and eigenvalue either?
I'm missing something big here, so I'd appreciate it if someone could enlighten me! Thanks in advance!
 A: Your eigenfunction equation is Euler's equation, which always has one power of $x$ as a solution. The other solution is a power if the roots are distinct; the case of double roots leads to a logarithmic term that can be gotten by applying variation of parameters starting with the one power solution. The eigenfunction equation to be solved is
$$
                       x^{2}y''+2xy'+\frac{1}{4}y = \lambda y.
$$
Try the solution $y=x^{\alpha}$:
$$
                   \alpha(\alpha-1)+2\alpha +\frac{1}{4}=\lambda \\
                      \left(\alpha+\frac{1}{2}\right)^{2}=\lambda \\
                      \alpha = -\frac{1}{2}\pm \sqrt{\lambda}
$$
There are no repeated roots unless $\lambda = 0$. So, for $\lambda \ne 0$, there are two linearly-independent solutions $x^{(-1/2+\sqrt{\lambda})}$ and $x^{(-1/2-\sqrt{\lambda})}$. At $\lambda =0$, $x^{-1/2}$ is a solution; the second solution is obtained through variation of parameters or through some other means.
It turns out that if we normalize the solution $\phi_{\lambda}(x)=Ax^{(-1/2-\sqrt{\lambda})}+Be^{(-1/2+\sqrt{\lambda})}$ so that $\phi_{\lambda}(1)=0$ and $\phi_{\lambda}'(1)=1$, then the case at $\lambda=0$ works its way out by taking a limit as $\lambda\rightarrow 0$, and nothing else has to be done. This is because of general theories of ODEs that guarantee that $\phi_{\lambda}$ is unique for all $\lambda$ and infinitely differentiable in $\lambda$. The normalization that works for all non-zero $\lambda$ is
$$
     \phi_{\lambda}(x)=\frac{x^{(-1/2+\sqrt{\lambda})}-x^{(-1/2-\sqrt{\lambda})}}{2\sqrt{\lambda}} = \frac{1}{\sqrt{\lambda}}\frac{1}{\sqrt{x}}\left(\frac{e^{\sqrt{\lambda}\ln (x)}-e^{-\sqrt{\lambda}\ln(x)}}{2}\right)
$$
So this is also guaranteed to work in the limit as $\lambda\rightarrow 0$ which, by L'Hopital is
$$
                   \phi_{0}=\frac{\ln(x)}{\sqrt{x}}.
$$
Sure enough, $\phi_{0}(1)=0$ and $\phi_{0}'(1)=1$, and it can be verified that $\phi_{0}$ is a solution of the equation when $\lambda=0$. But I'll leave that to you if you're uncertain. This solution is not relevant anyway because $\phi_{0}(e)\ne 0$, but knowing its form does allow us to exclude $\lambda=0$ as an eigenvalue.
The permissible eigenvalues are found by solving for $\phi_{\lambda}(e)=0$, or, equivalently, by solving
$$
     e^{\sqrt{\lambda}}-e^{-\sqrt{\lambda}} = 0 \iff e^{2\sqrt{\lambda}}=1
    \iff 2\sqrt{\lambda}=\pm 2\pi in,\;\;\; n=1,2,3,\cdots.
$$
(And $n\ne 0$ because $\lambda=0$ has been ruled out.) Therefore the acceptable eigenfunctions are
$$
               \phi_{-n^{2}\pi^{2}}=\frac{1}{n\pi\sqrt{x}}\sin(n\pi\ln(x)),\;\;\; n=1,2,3,\cdots.
$$
The $1/n$ factor is not needed. A factor is needed to normalize the eigenfunctions to have $L^{2}[1,e]$ norm equal to $1$, if you want to have an orthonormal basis. To normalize these,
$$
        \int_{1}^{e}\frac{1}{x}\sin^{2}(n\pi\ln x)\,dx = \int_{0}^{1}\sin^{2}(n\pi y)\,dy
  = \frac{1}{2}.
$$
So the normalized eigenfunctions on $L^{2}[1,e]$ are
$$
      \left\{\sqrt{\frac{2}{x}}\sin(n\pi\ln(x))\right\}_{n=1}^{\infty}
$$
This is guaranteed by the general theory to be a complete orthonormal subset of $L^{2}[1,e]$.
