Answer
\[y(x)=x^2+\frac{1}{x^2}\]
Work Step by Step
$\large y'+\frac{2}{x}y=4x$ ____(1)
(1) is Linear differential equation
Integrating Factor:-
\[I(x)=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2\]
Multiply (1) by $I(x)$
\[x^2\frac{dy}{dx}+2xy=4x^3\]
\[\frac{d}{dx}(yx^2)=4x^3\]
Integrating,
\[x^2y=\int 4x^3 dx+C\]
$C$ is constant of integration
$x^2y=x^4+C$ ____(2)
Using initial condition $y(1)=2$
\[2=1+C\Rightarrow C=1\]
From (2)
\[y(x)=x^2+\frac{1}{x^2}\]
