Solving the Sturm-Liouville equation $x^2y'' +2xy' +(\lambda +1/4)y = 0$

Solution 1:

Consider the Sturm-Liouville operator $$ Lf= -(x^2f')'-\frac{1}{4}f,\;\;\; 1 \le x \le e, $$ and the eigenfunction problem $$Lf=\lambda f$$ subject to endpoint conditions $$ y(1)=0,\;\; y(e)=0. $$ This is a regular Sturm-Liouville problem because $x^2$ does not vanish on $[1,e]$. If $Lf=\lambda f$ and $Lg=\mu g$, then it is not hard to show that \begin{align} (\lambda-\mu)\langle f,g\rangle&=\langle Lf,g\rangle-\langle f,Lg\rangle \\ &=\int_1^e -(x^2f')'\overline{g}+f(x^2\overline{g}')'dx \\ &=\int_1^e \frac{d}{dx}(-x^2f'\overline{g}+x^2f\overline{g'})dx \\ &=-x^2(f'\overline{g}-f\overline{g}')|_{1}^{e}=0. \end{align} Therefore $f\perp g$ if $\lambda\ne\mu$. Eigenfunctions of $L$ are solutions of $Lf=\lambda f$ subject to $y(1)=0=y(e)$. This is a well-posed regular Sturm-Liouville problem because $x^2$ is non-vanishing on $[1,e]$. If you solve $Lf_{\lambda}=\lambda f_{\lambda}$ subject to $f_{\lambda}(1)=0,\;f_{\lambda}'(1)=1$, then the eigenvalue problem is the following equation in $\lambda$: $$ f_{\lambda}(e)=0. $$