$(xy')'+{\lambda \over x}y=0, \forall 1The boundary value problem 
$$(xy')'+{\lambda \over x}y=0, \forall 1<x<e^{2\pi}$$ such that $y'(1)=0=y'(e^{2\pi})$
then 
$(a)$ every eigen function of boundary value problem is bounded 
$(b)$ There exist eigen function of boundary value of boundary value problem which is periodic 
$(c)$ There exists non-periodic eigen function of boundary value problem 
$(d)$ $ \forall \lambda >0$ eigen value, corresponding eigenfunction is bounded. 
How should I approach this problem?
 A: Hint: $x^2y''+xy'+\lambda y=0$ is Cauchy-Euler equation which on substitution $x=e^t$ reduces to $(D^2+\lambda )y=0$ where$D\equiv \frac{d}{dt}$.
A: The given equation is 
$(xy')'+{\lambda \over x}y=0$
$\implies x^2 y'' + x y' + \lambda y = 0$. . . . $(1)$
Put $x=e^z$, then the given equation becomes $\{D(D-1)+D+\lambda\}y=0 $, where $D\equiv \frac{d}{dz}$
i.e., $(D^2+\lambda)y=0$ . . . . $(2)$
For $\lambda=0$,
From $(2)$ we have $y=az+b\implies y=a \log x+b$, where $a, b$ are arbitrary constants.
Given that $y'(1)=0\implies a= 0$, so $y=b$. 
Again $y=b$ also satisfies $y'(e^{2\pi})=0$.
Hence $\lambda=0$  is an eigenvalue for this BVP and the eigenfunctions corresponding to this eigenvalue is $y=b$ where $b \neq 0$. 
For $\lambda \lt0$, let $\lambda = -m^2, m\neq0$
From $(2)$ we have $y= a e^{mz} +  b e^{-mz}\implies y= a e^{m \log x} +  b e^{-m \log x}$
Given that $y'(1)=0\implies ma -mb= 0$ . . . $(3)$ 
$y'(e^{2\pi})=0 \implies mae^{-2\pi} e^{2\pi m}-mb e^{-2\pi} e^{-2\pi m}=0$ . . . . $(4)$
Solving equation $(3)$ & $(4)$, we have $a=b=0$, as $m \neq 0$
So, for this case we only get the trivial solution and so there are no negative eigenvalues.
For $\lambda \gt 0$, let $\lambda= m^2, m\neq0$
From $(2)$ we have $y= a \cos \left( {mz} \right) +  b \sin \left( {mz} \right)\implies y= a \cos \left( {m \log x} \right) +  b \sin \left( {m \log x} \right)$
Given that $y'(1)=0\implies b= 0$, 
$y'(e^{2\pi})=0 \implies mae^{-2\pi} \sin (2\pi m)=0\implies a\neq0$ (since we want non-trivial solutions).
So $\sin (2\pi m)=0\implies 2\pi m=n\pi \implies m= \frac{n}{2}\hspace{0.25in}n = 1,2,3, \ldots$. 
So, ${\lambda _{\,n}} = {\left( {\frac{n}{2}} \right)^2} = \frac{{{n^2}}}{4}\hspace{0.25in}n = 1,2,3, \ldots$
The eigenfunctions that correspond to these eigenvalues are $$y_n(x)= a_n \cos \left( {\frac{n}{2} \log x} \right)$$

Clearly we have the following eigenvalues/eigenfunctions for the given BVP,
\begin{align*}{\lambda _{\,n}} & = \frac{{{n^2}}}{4} & {y_n}\left( x \right) & = \cos \left( {\frac{{n\,\log x}}{2}} \right)\hspace{0.25in}n = 1,2,3, \ldots \\ {\lambda _{\,0}} & = 0 & {y_0}\left( x \right) & = 1\end{align*}
Since we are taking eigen functions so we dropped the constant term (also taking $b=1$).

Now come to the points:
$(a).$ Since $\cos$ function is bounded and also $1<x<e^{2\pi}$, so every eigen function of boundary value problem is bounded.
$(b).$ Since $\cos(\frac{n \log x}{2})$ function is not periodic, so there exists non-periodic eigen function of boundary value problem. 
$(c).$ Since the eigen function for $\lambda = 0$ is  periodic (as constant function is always periodic ),  so there exist eigen function of boundary value of boundary value problem which is periodic. 
$(d).$ Also $\forall \lambda >0$ eigen value, corresponding eigenfunction is bounded.
A: If you're familiar with Cauchy-Euler equations, you can directly solve
$$ x^2y'' + xy' + \lambda y = 0 $$
The characteristic polynomial is
$$ r(r-1) + r + \lambda = 0 \implies r^2 + \lambda = 0 $$


*

*If $\lambda < 0$, there is no non-trivial solution.

*If $\lambda = 0$, the only non-trivial solution is a constant $y(x) \equiv c$. This is the only periodic solution.

*If $\lambda > 0$, there exists solutions of the form $y(x) = c_1\cos(\sqrt{\lambda}\ln x) + c_2 \sin(\sqrt{\lambda}\ln x)$. These solutions are all non-periodic.
Additionally, all solutions are bounded in the interval $(1,e^{2\pi})$
