The problem: Show the eigenvalue problem has no negative eigenvalues
$-y(\rho)-\frac{2}{\rho} y'(\rho)=\lambda y(\rho)$
$y(0)$ bounded, $y(\pi)=0$
Solution:
Making the transformation
$Y = \rho y$ we get $−Y=-\lambda Y$.
If $\lambda = −k^{2}$, then
$Y = a \sinh(k\rho) + b \cosh (k\rho)$
or
$y = \frac{1}{\rho}a \sinh(k\rho) + b \cosh (k\rho)$
For boundedness at $\rho = 0$ we set $b = 0$. Then $y(\pi) = 0$ forces $\sinh kπ = 0$. Thus
$k = 0$. Consequently, there are no negative eigenvalues.
My question is, I do not understand why we set $b = 0$ for boundedness? Why is $\cosh(k\rho)$ unbounded at $y(0)$?
Because $$ y(\rho)=\frac1\rho(a\sinh(k\rho)+b\cosh(k\rho)) $$ and the limit $\displaystyle \lim_{\rho\to0}\frac{\sinh(k\rho)}{\rho}=\lim_{\rho\to0}\frac{e^{k\rho}-e^{-k\rho}}{2\rho} $ exists. But if $b\ne 0$, the term $\frac{b\cosh(k\rho)}{\rho}$ would blow up as $\rho\to 0$.
It is not that $\cosh(k\rho)$ is unbounded at $\rho=0$ but that $\cosh(k\rho)/\rho$ is unbounded near $\rho=0$.