Given $ydx + (x - y^3) dy = 0$
$\Rightarrow y \frac{dx}{dy}+x-y^{3}=0 \Rightarrow \frac{dx}{dy}+\frac{1}{y}x=y^{2}$
compare this equation to general equation
i.e. $ \frac{dx}{dy}+Px=Q \Rightarrow P=\frac{1}{y}, Q = y^{2}$
$\therefore $ I.f .=$e^{\int Pdy}=e^{\int\frac{1}{y}dy}=e^{log y}=y$
x × I.f. =$\int Q×I.f.dy +C_{1}$
$\Rightarrow x.y = \int y^{2}.y dy+C_{1}\Rightarrow xy =\frac{y^{4}}{4}+C_{1}$
$\Rightarrow 4xy = y ^{4}+ 4C_{1}$
$\Rightarrow y ^{4}= 4xy-4C_{1} \Rightarrow y ^{4}= 4xy+C,$
where C =-4C$_{1}$