Solving the Sturm-Liouville equation $x^2y'' +2xy' +(\lambda +1/4)y = 0$ I would like to solve the SL equation: $x^2y'' +2xy' +(\lambda +1/4)y = 0$, put it in SL form and show that the eigenfunctions obtained are orthogonal under some inner product.
We have the condition $y(1) = y(e) = 0$
In my textbook, the SL form is $-(p(x)y')' + q(x)y = \mu r(x)y$
So putting it in that form I get $-(x^2y')' - \frac{1}{4}y = \lambda y$. So $p(x) = x^2$, $q(x) = - \frac{1}{4}$ and $r(x) = 1$.
The hint in the question is to "seek solutions of the form $x^{\alpha}$." So I tried to solve the eigenvalue problem for $\lambda = 0$, $\lambda = \mu^2 >0$ and $\lambda = -\mu^2 <0$. In each case I tried solutions of the form $Ax^\alpha$ - I got trivial solutions for $\lambda = 0$ and $\lambda = -\mu^2$. For $\lambda = \mu^2$, i got solutions of the form $Ax^{\frac{-1 \pm -\mu i}{2}}$. When I apply the boundary conditions, the $y(1) = 0$ would imply that $A = 0$ and so only trivial solutions but the $y(e) = 0$ looks like it would get a nice solution in terms of sin and cos.
This makes me think that maybe solutions aren't of the form $Ax^\alpha$.
Would anyone mind helping - this is for revision, not hw. Thanks
 A: 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 $\lambda$:
$$
                      f_{\lambda}(e)=0.
$$
