How to find Eigenfunction of this Sturm-Liouville problem My question what do I have to do to find the eigenfunction in the first case
It is given the following problem
\begin{align} 
xy''(x)+y'(x)+ \frac{\lambda}{x}y(x) &=0, & &x\in(1,e^{2\pi})  \tag1
\end{align}
With $y'(1)=y'(e^{2\pi})=0$
This is a Sturm-Liouville problem in the form
$$(xy'(x))'+\frac{\lambda}{x}y(x)=0 \tag2$$
where $p(x)=x, q(x)=0,r(x)=\frac{1}{x}$.
Now we solve this problem for different values of $\lambda$.

*

*If $\lambda=0$
Then the equation (1) becomes in the form
$$xy''(x)+y'(x)=0$$
This is an Euler-Cauchy equation with general solution
$$y(x)=c_1 \log(x)+c_2 \tag3$$
Now we differentiate the expression $(3)$ and we have
$$y'(x)=c_1 \frac{1}{x}$$
Then we plug in the initial conditions
$$y'(1)=0\Rightarrow c_1=0$$


*If $\lambda>0$ we write $k^2$ with $k>0$
Then equation (1) becomes the form
$$xy''(x)+y'(x)+ \frac{k^2}{x}y(x)=0$$
With general solution $$y(x)=c_1 \cos(k\log(x))+ c_2 \sin(k\log(x)) \tag 4$$
Now we differentiate the expression $(4)$ and we have
$$y'(x)=\frac{c_2 k \cos(k\log(x))-c_1 k \sin(k\log(x))}{x}$$
Then we plug in the initial conditions
$$y'(1)=0\Rightarrow c_2=0$$ so
$$y'(e^{2\pi})=0 \Rightarrow \frac{c_1 k \sin(k \log(e^{2 \pi}))}{e^{2 \pi}} =0$$
Where $k \neq 0$ and $c_1\neq 0$ because we are looking for no trivial solutions.
Thus,
$\sin(2\pi k)=0$ so $k=\frac{n}{2}, \ n \in \mathbb{Z}$
Then the eigenfunction is $$y(x)=c_1 \cos(\frac{n}{2} \log(x))$$


*If $\lambda<0$ we write $-k^2$ with $k>0$
Then the equation (1) becomes in the form
$$xy''(x)+y'(x)- \frac{k^2}{x}y(x)=0$$
With general solution $$y(x)=c_1 \cosh(k\log(x))+i c_2 \sinh(k\log(x)) \tag 5$$
Now we differentiate the expression $(5)$ and we have
$$y'(x)=\frac{c_2 k \cosh(k\log(x))+ c_1 k \sinh(k\log(x))}{x}$$
Then we plug in the initial conditions
$$y'(1)=0\Rightarrow i \ c_2 \ k=0 \Rightarrow c_2=0$$ so
$$y'(e^{2\pi})=0 \Rightarrow \frac{c_1 k \sinh(k \log(e^{2 \pi}))}{e^{2 \pi}} =0$$
Where $k \neq 0$ and $c_1\neq 0$, because we are looking for no trivial solutions.
Thus,
$\sinh(2\pi k)=0$ so $k=\frac{in}{2}, \ n \in \mathbb{Z}$.
Then the eigenfunction is $$y(x)=c_1 \cosh(\frac{in}{2} \log(x))$$
 A: Start by solving the following
$$
      xy''(x)+y'(x)+\frac{\lambda}{x}y(x)=0 \\
       y'(1)=0,\; y(1)=1.
$$
The reason for the added condition $y(1)=1$ is that it is always possible to multiply any solution without that equation by a non-zero factor that will result in $y(1)=1$, which this problem a unique solution. Furthermore, there is no loss in generality because, if $y(1)=0$, then $y\equiv 0$, and, otherwise, multiplying $y$ by a non-zero scale factor will result in a solution where $y(1)=1$.
As was noted, the following equation is an Euler equation:
$$
         x^2y''(x)+xy'(x)+\lambda y(x)=0
$$
So there is at least one solution of the form $u^{\alpha}$ where
$$
           \alpha(\alpha-1)+\alpha+\lambda = 0 \\
       \alpha^2+\lambda= 0 \\
         \alpha = \pm\sqrt{\lambda}.
$$
Assuming a solution $y(x)=Ax^{-\alpha}+Bx^{\alpha}$, then the requirements that $y(1)=1$ and $y'(1)=0$ gives
$$
                   A+B=1 \\
                  (-A+B)\alpha=0.
$$
That gives $A=B=1/2$ and corresponding solution
$$
       y(x)=\frac{1}{2}(x^{\alpha}+x^{-\alpha})=\frac{1}{2}(x^{\sqrt{\lambda}}+x^{-\sqrt{\lambda}})
$$
This solution is an eigenfunction iff $y'(e^{2\pi})=0$, which gives the eigenvalue equation for $\lambda$.
