Solution of the differential equation (x + xy)dy – y(1 – x²)dx = 0 is

$y=\log \frac{x}{y}-\frac{x^2}{2}+C$

$(x+x y) d y-y\left(1-x^2\right) d x=0$

$\Rightarrow(x+x y) d y=y\left(1-x^2\right) d x$

so $\int \frac{1+y}{y} d y=\int \frac{1-x^2}{x} d x$

$\Rightarrow \int 1+\frac{1}{y} d y=\int \frac{1}{x}-x d x \Rightarrow y+\log y=\log x-\frac{x^2}{2}+C$

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

Option: 1