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 $$y(x)=c_1 \cosh(\frac{in}{2} \log(x))$$
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 $$ \alpha(\alpha-1)+\alpha+\lambda = 0 \\ \alpha^2+\lambda= 0 \\ \alpha = \pm i\sqrt{\lambda}. $$ 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 $$ y(x)=\frac{1}{2}(x^{\alpha}+x^{-\alpha})=\frac{1}{2}(x^{i\sqrt{\lambda}}+x^{-i\sqrt{\lambda}})=\cos(\sqrt{\lambda}\ln x) $$ This solution is an eigenfunction iff $y'(e^{2\pi})=0$, which results in the eigenvalue equation for $\lambda$: $$ \frac{\sqrt{\lambda}\sin(2\pi\sqrt{\lambda})}{e^{2\pi}}=0 \implies \sqrt{\lambda}=0,1/2,1,3/2,2,\cdots. $$