$(y – xy^2) dx – (x + x^2y)dy = 0$

Solution:

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

Comparing the given eqn with $M d x+N d y=0$

$\begin{array}{cc}{\therefore \mathrm{M}=\left(y-x y^{2}\right)} & {\therefore \mathrm{N}=-\left(x+x^{2} y\right)} \\ {\frac{\partial M}{\partial y}=1-2 x y} & {\frac{\partial N}{\partial x}=-(1+2 x y)} \\ {\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}}\end{array}$

The given differential eqn is not exact diff. eqn.

But the given diff. eqn is in the form of $y . f(x y) d x+x f(x y) d y=0$

Integrating factor $=I . \mathrm{F} .=\frac{1}{M x-N y}=\frac{1}{x y-x^{2} y^{2}+x y+x^{2} y^{2}}=\frac{1}{2 x y}$

Multiply the I.F.to eqn $(1)$ $\left(\frac{1}{2 x}-\frac{y}{2}\right) d x-\left(\frac{1}{2 y}+\frac{x}{2}\right) d y=0$

$\therefore M_{1}=\left(\frac{1}{2 x}-\frac{y}{2}\right) \quad N_{1}=-\left(\frac{1}{2 y}+\frac{x}{2}\right)$

$\int M_{1} d x=\int\left(\frac{1}{2 x}-\frac{y}{2}\right) d x=\frac{1}{2}(\log x)-\frac{x y}{2}$

$\frac{\partial}{\partial y} \int M_{1} d x=\frac{-x}{2}$

$\int\left[N_{1}-\frac{\partial}{\partial y} \int M_{1} d x\right] d y=\int \frac{-1}{2 y} d y=\frac{-1}{2}(\log y)$

The solution of given diff. eqn is given by,

$\int M_{1} d x+\int\left[N_{1}-\frac{\partial}{\partial y} \int M_{1} d x\right] d y=c$

$\therefore \frac{1}{2}(\log x)-\frac{x y}{2}-\frac{1}{2}(\log y)=c$

$\therefore \log \left(\frac{\sqrt{x}}{\sqrt{y}}\right)-\frac{x y}{2}=c$