General Solution for a second order ODE So we have this problem in mathematical physics.
We are asked to find the general solution for the ODE
$$
(x^2)y''+(x)y'-(n^2)y=0
$$
for integer $n$.
But this is what I tried to do. I let 
$$
y=x^p\\
y'=px^{p-1}\\
y''=p(p-1)x^{p-2}
$$
Then substituted it back to the original equation,
$$
(x^2)p(p-1)x^{p-2}+xpx^{p-1}-n^2x^p=0
$$
So  I have
$$
x^p(p^2-n^2)=0
$$
And there, i'm stuck.
My answer is
$$
C_1x^{p_1}+C_2x^{p_2}=y(x)
$$
But is this correct? :'(
 A: You are almost there.
You have $$p^2=n^2 \implies p=\pm n$$
Thus your general solution is $$y=C_1 x^n+C_2 x^{-n}$$  
A: Here's another possible approach:
Let us consider your differential equation
$$x^2\cdot\frac{d^2y}{dx^2}+x\cdot\frac{dy}{dx}-n^2\cdot y=0$$
Let $t=\ln(x)\rightarrow x=e^t$.
Then,
$$\begin{align}
\ \frac{dy}{dx} & =\frac{dt}{dx}\cdot\frac{dy}{dt} \\
 & = \frac{d}{dx}\left(\ln(x)\right)\cdot\frac{dy}{dt} \\ 
 & = \frac{1}{x}\cdot\frac{dy}{dt} \\
 & = e^{-t}\cdot\frac{dy}{dt} \\ 
\end{align}$$
and
$$\begin{align}
\ \frac{d^2y}{dx^2} & =\frac{d}{dx}\left(\frac{dt}{dx}\cdot\frac{dy}{dt}\right) \\
 & = \frac{d}{dx}\left(\frac{1}{x}\cdot\frac{dy}{dt}\right) \\ 
 & = \frac{d}{dx}\left(\frac{1}{x}\right)\cdot\frac{dy}{dt}+\left(\frac{1}{x}\right)^2\cdot\frac{d^2y}{dt^2} \\
 & = \frac{1}{x^2}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right) \\ 
 & = e^{-2\cdot t}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)
\end{align}$$
Substitution into the differential equation gives
$$\begin{aligned} e^{2\cdot t}\cdot e^{-2\cdot t}\cdot\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)+e^t\cdot e^{-t}\cdot\frac{dy}{dt}-n^2\cdot y&=0 \\ 
\frac{d^2y}{dt^2}-n^2\cdot y&=0 \\ 
\end{aligned}$$
Assume that $y\propto e^{\lambda\cdot t}$ for some constant $\lambda$. 
Substitution into the differential equation gives
$$\begin{aligned} \frac{d^2}{dt^2}(e^{\lambda\cdot t})-n^2\cdot e^{\lambda\cdot t}&=0 \\ 
(\lambda^2-n^2)\cdot e^{\lambda\cdot t}&=0 \\ 
\end{aligned}$$
Because $e^{\lambda\cdot t}\ne 0$ for any finite $\lambda$, $\lambda^2-n^2=0\rightarrow \lambda_{1,2}=\pm n$.
Therefore,
$$y(t)=\text{c}_1\cdot e^{-n\cdot t}+\text{c}_2\cdot e^{n\cdot t}$$
Let $t=\ln(x)$:
$$\begin{align}
\ y(x) & =\text{c}_1\cdot e^{-n\cdot\ln(x)}+\text{c}_2\cdot e^{n\cdot\ln(x)} \\
 & = \text{c}_1\cdot e^{\ln(x^{-n})}+\text{c}_2\cdot e^{\ln(x^{n})} \\ 
 & = \text{c}_1\cdot x^{-n}+\text{c}_2\cdot x^{n} \\
\end{align}$$
where $n\in\mathbb{Z}$.
