Let
$$x^2y''+xy'+\lambda y=0,\,\,\,\,\,y(1)=0,\,\,y(b)=0\tag1$$
Using the fact this equation is equidimensional. Calculate all associated eigenvalues.
Is $\lambda=0$ an eigenvalue?
Prove there exist infinite eigenvalues with one minimum, but there doesn't exist a maximum eigenvalue.
My attempt:
Note the equation $(1)$ is a Cauchy-Euler equation.
Suppose $y=x^r$ then $y'=r x^{r-1}$, $y''=r(r-1)x^{r-2}$.
Then
$$x^2r(r-1)x^{r-2}+xr x^{r-1}+r x^r=0\implies(r^2-r)x^r=0\iff r_1=0,\,\,\, r_2=-1$$
Then the solution for the ODE is: $$y(x)=C_1+C_2x^{-1}$$
Here I'm a little confused, to solve my problem. Can someone help me?
Considering
$$ \left(x^2\frac{d^2}{dt^2}+x\frac{d}{dt}+\lambda\right)y = 0 $$
after the substitution $y = x^r$ we get
$$ x^r\left(r(r-1)+r +\lambda\right)=0 $$
then
$$ r = \pm i \sqrt\lambda $$
then for $\lambda > 0$ we have
$$ y = C_1\sin(\sqrt\lambda \ln x) + C_2\cos(\sqrt\lambda \ln x) $$
and now from the boundary conditions
$$ y(1) = C_2 = 0\\ y(b) = C_1\sin(\sqrt\lambda \ln b) = 0 $$
then we conclude
$$ \sqrt\lambda\ln b = k \pi\Rightarrow \lambda_ k = \frac{\pi^2}{\ln^2 b}k^2 $$