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!

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 $L^{2}[1,e]$.