Answer
$$ n=4,\ \ \ n= -1 $$
Work Step by Step
Given $$x^{2} y^{\prime \prime}-2 x y^{\prime}=4 y$$
Since
\begin{align*}
y&=x^n \\
y'&=nx^{n-1}\\
y''&=n(n-1)x^{n-2}
\end{align*}
Then
\begin{align*}
x^{2} y^{\prime \prime}-2 x y^{\prime}&=4 y\\
n(n-1)x^{n-2}x^2-2 xnx^{n-1}&=4x^n \\
[n(n-1) -2n-4]x^{n} &=0\\
(n-4)(n+1)&=0
\end{align*}
Hence $$ n=4,\ \ \ n= -1 $$
