Answer
\[y(x)=(1-x^2)[-\ln (1-x^2)^2+C]\]
Work Step by Step
$\frac{dy}{dx}+\frac{2x}{1-x^2}y=4x$ ___(1)
(1) is Linear differential equation
Integrating Factor:-
\[I(x)=e^{\int\frac{2x}{1-x^2}dx}=e^{-\ln|1-x^2|}=\frac{1}{1-x^2}\]
Multiply (1) by $I(x)$
\[\frac{1}{(1-x^2)}\frac{dy}{dx}+\frac{2x}{(1-x^2)^2}y=\frac{4x}{1-x^2}\]
\[\frac{d}{dx}\left[\frac{y}{1-x^2}\right]=\frac{4x}{1-x^2}\]
Integrating,
\[\frac{y}{1-x^2}=4\int\frac{x}{1-x^2}dx+C\]
$C$ is constant of integration
\[\frac{y}{1-x^2}=-2\ln|1-x^2|+C\]
\[y(x)=(1-x^2)[-\ln (1-x^2)^2+C]\]
Hence General solution of (1) is $y(x)=(1-x^2)[-\ln (1-x^2)^2+C]$
