The general solution of the differential equation $\frac{d y}{d x}+\frac{2}{x} y=x^2$, is

$y=c x^{-2}+\frac{x^3}{5}$

The given differential equation is of the form $\frac{d y}{d x}+P y=Q$, where $P=\frac{2}{x}$ and $Q=x^2$.

Integrating factor = $e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^2$

Multiplying both sides of the differential equation by integrating factor $=x^2$ and integrating with respect to $x$, we get

$y x^2=\frac{x^5}{5}+c$ or, $y=\frac{x^3}{5}+c x^{-2}$