Second Order Differential Equation $u''-\frac{l(l+1)u}{p^{2}}=0$ I am following the derivation of the equation for the hydrogen atom (this is NOT a physics question) and a second order differential equation is produced, considering when radius $r\to 0$, that looks like:
$\frac{d^2u}{dp^2}=\frac{l(l+1)}{p^2}u$ 
To solve this, I let:
$u\propto p^k$
And return:
$k(k-1)p^{k-2}=\frac{l(l+1)}{p^2}p^k$
Therefore, I'm left with:
$k(k-1)=l(l+1)$
Which I don't know how to solve. I have a feeling that I may have gone wrong somewhere as I know the answer to be:
$u=A p^{l+1}+Bp^{-l}$
Could anyone tell me where I am going wrong or how to finish this derivation?
 A: From $$k(k-1)=l(l+1),$$ we have $$4k(k-1)=4l(l+1).$$ Thus
$$(2k-1)^2=4k(k-1)+1=4l(l+1)+1=(2l+1)^2.$$
That is
$$2k-1=\pm(2l+1).$$
That is $k=l+1$ or $k=-l$, which seems to work well with the given general form of $u$, but I think you have a typo, that is, you should have
$$u=Ap^{l+1}+Bp^{-l}.$$
Alternatively, you can just factor
$$k(k-1)-l(l+1)=0.$$ 
This is equivalent to
$$k^2+\big((-l-1)+l\big)k+(-l-1)l=0.$$
So
$$\big(k+(-l-1)\big)(k+l)=0.$$
Here is yet another solution.  From $k(k-1)=l(l+1)$, we can write
$$k^2-l^2=k+l.$$
But $k^2-l^2=(k+l)(k-l)$.  That is,
$$(k+l)(k-l)=k+l.$$
Hence, $k+l=0$ or $k-l=1$.  The same conclusion follows.

Btw, the case $l=-1/2$ is special.  You can show that the solutions are $u=Ap^{1/2}+Bp^{1/2}\ln p$, where $A$ and $B$ are constants.
A: To start withe you second order ODE is Euler's homogeneous equation whose solution is correctly otained by you by putting $u(p)=p^k$ where $k^2-k-l(l+1)=0 \implies k=\frac{1\pm \sqrt{1+4l(l+1)}}{2} \implies k=l+1,-l.$ We get two linearly independent solutions so 
$$u(p_=A p^{l+1} + B p^{=l}.$$ Further the condition og regularity $u(0)=0$ allows only one solution. Finally, $$u(p)= A p^{l+1}.$$
