Solving Linear ODE Can someone help me prove that given
$$
   y'' + \frac{2}{x}y' + \lambda y = 0
$$
and boundary conditions
$$
  {\lim_{x \to 0} xy = 0 , \hspace{5mm} y(1) =0 }
$$
that
$$
    y(x) = \frac{1}{x}\left[ A \sin{(\sqrt{\lambda}x)} + B \cos{(\sqrt\lambda x)} \right]
$$
When trying to solve it the usual way I get
$$ m^2 + \frac{2}{x}m + \lambda = 0 $$
$$ m = -\frac{1}{x} \pm j\sqrt{\lambda - \frac{1}{x^2}}$$
which obviously won't me to the solution because I have the $\sqrt{\lambda - \frac{1}{x^2}}$ term instead of a $\sqrt{\lambda x}$ term.
 A: Set $u(x) = x y(x)$.  You may then show that
$$u'' + \lambda u = 0$$
with $u(0)=0$ and $u(1)=0$.  Now solve as usual.  When you get the solution for $u(x)$, then $y(x)=u(x)/x$.
EDIT
A few more details about where the above equation comes from.  Note that
$$ y' = \frac{x u'-u}{x^2} $$
$$ y'' = \frac{x u'' - u'}{x^2} - \frac{x^2 u' - 2 x u}{x^4} $$
Plugging this back into the original equation above results in 
$$\frac{u''}{x} + \lambda  \frac{u}{x} = 0$$
which results in the derived equation for $u$ above.
A: Here is a related problem. Already, you have given the solution $ y(x) $
$$ y(x) = \frac{1}{x}\left[ A \sin{(\sqrt{\lambda}x)} + B \cos{(\sqrt\lambda x)} \right]. $$
and you need to find the eigenvalues $\lambda$ and the corresponding eigenfunctions. Exploiting the boundary conditions we get the two equations
$$ 0 = B. $$
$$ 0 = A \sin( \sqrt{\lambda} ) + B\cos(\sqrt{\lambda} ). $$
From the above two equations, we have
$$ \sin(\sqrt{\lambda}) = 0  \implies \lambda_n = n^2\pi^2 . $$
Now, you can find the eigenfunctions corresponding to $\lambda_n$. 
