$ \left(x^2-y x^2\right) d y+\left(y^2+x y^2\right) d x=0$
$\therefore x^2(1-y) d y+y^2(1+x) d x=0$
$\therefore x^2(1-y) d y=-y^2(1+x) d x$
$\therefore\left(\frac{1-y}{y^2}\right) d y=-\left(\frac{1+x}{x^2}\right) d x $
Integrating on both sides, we get
$ \int\left(\frac{1-y}{y^2}\right) d y=-\int\left(\frac{1+x}{x^2}\right) d x$
$\therefore \int \frac{1}{y^2} d y-\int \frac{1}{y} d y=-\int \frac{1}{x^2} d x-\int \frac{1}{x} d x$
$\therefore \frac{y^{-1}}{-1}-\log |y|=\left(\frac{x^{-1}}{-1}\right)-\log |x|+ c$
$\therefore-\frac{1}{y}-\log |y|=\frac{1}{x}-\log |x|+ c $
$\therefore \log |x|-\log |y|=\frac{1}{x}+\frac{1}{y}+c$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.