The solution of differential equation $\frac{dy}{dx}\left(x^2{y}^3+xy\right)=1$ is :

$\frac{dy}{dx}\left(x^2{y}^3+xy\right)=1$

$\frac{dx}{dy}-xy=x^2{y}^3\therefore \frac{1}{x^2}\frac{dx}{dy}-\frac{1}{x}y=y^3$

Put $-\frac{1}{x}=z\therefore \frac{1}{x^2}\frac{dx}{dy}=\frac{dz}{dy}$

$\begin{array}{l}\therefore \frac{dz}{dy}+z\cdot y=y^3\text{ linear }\\ \text{ I.F. }=e^{\int Pdy}=e^{\int ydy}=e^{y^2/2}\end{array}$

Multiply both sides by I.F. and integrate

$z\cdot e^{y^2/2}=\int e^{y^2/2}\cdot y\cdot y^{2}dy$

Put $y^2/2=t\therefore ydy=dt$

$\therefore z{e}^t=2\int l{e}^{t}dt=2e^t(t-1)+c$

$\begin{array}{l}z=2(t-1)+c{e}^{-t}\\ \text{ or }-\frac{1}{x}=2\left(\frac{y^2}{2}-1\right)+c{e}^{-y^2/2}\\ \text{ or }\frac{1}{x}=2-y^2-c{e}^{-y^2/2}.\text{ Choose }-c=+k.\end{array}$