Answer
$|\ln y | -\dfrac{y^2}{2}=\dfrac{x^2}{2} +C$
Work Step by Step
We have the differential equation:
$\dfrac{dy}{dx}=\dfrac{xy}{(1-y^2)x}$
$\implies \dfrac{1-y^2}{y} \ dy = x \ dx$
Now, we will integrate it.
$\int \dfrac{1-y^2}{y} \ dy = \int x \ dx$
$\implies \int \dfrac{1}{y} \ dy -\int y \ dy =\int x dx$
$\implies |\ln y | -\dfrac{y^2}{2}=\dfrac{x^2}{2} +C$
