Ine solution of the differential equation

$\frac{d y}{d x}+\frac{2}{x} y=x$ and $y(1)=1$ (given)

Since, the above differential equation is the linear

differential equation, then $I . F=e^{\int \frac{2}{x} d x}=x^{2}$

Now, the solution of the linear differential equation

$y \times x^{2}=\int x^{3} d x$

$\Rightarrow y x^{2}=\frac{x^{4}}{4}+C$

$\because y(1)=1$

$\therefore 1 \times 1=\frac{1}{4}+C \Rightarrow C=\frac{3}{4}$

$\therefore$ solution becomes

$y=\frac{x^{2}}{4}+\frac{3}{4 x^{2}}$