$y \ dx + x (1 – 3 x^2 y^2) dy = 0$

Solution:

$y d x+x\left(1-3 x^{2} y^{2}\right) d y=0----(1)$

$\begin{array}{ll}{\text { Compare the given eqn with } \mathrm{Mdx}+\mathrm{Ndy}=0} \\ {\therefore \mathrm{M}=\mathrm{y}} & {\therefore \mathrm{N}=x\left(1-3 x^{2} y^{2}\right)} \\ {\frac{\partial M}{\partial y}=1} & {\frac{\partial N}{\partial x}=1-9 x^{2} y^{2}}\end{array}$

$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$

Hence the given diff. eqn is not exact.

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 $=\frac{1}{M x-N y}=\frac{1}{x y-x y+3 x^{3} y^{3}}=\frac{1}{3 x^{3} y^{3}}$

Multiply the $I .$ F. to eqn $(1)$

$\frac{1}{3 x^{3} y^{2}} d x+\left[\frac{1}{3 x^{2} y^{3}}-\frac{1}{y}\right] d y=0$

$\therefore M_{1}=\frac{1}{3 x^{3} y^{2}} \quad N_{1}=\left[\frac{1}{3 x^{2} y^{3}}-\frac{1}{y}\right]$

Now this diff. eqn is exact.

The solution of given diff. eqn is given by,

$\int M d x+\int\left[N-\frac{\partial}{\partial y} M d x\right] d y=c$

$\int M_{1} d x=\int \frac{1}{3 x^{3} y^{2}} d x=\frac{-1}{6 y^{2} x^{2}}$

$\frac{\partial}{\partial y} \int M_{1} d x=\frac{1}{3 x^{2} y^{3}}$

$\begin{aligned} \int\left[N_{1}-\frac{\partial}{\partial y} \int M_{1} d x\right] d y &=\int\left[\frac{1}{3 x^{2} y^{3}}-\frac{1}{y}-\frac{1}{3 x^{2} y^{3}}\right] d y \\ &=\int \frac{-1}{y} d y==-\log y \\ \therefore \frac{-1}{6 y^{2} x^{2}}-\log y &=c \end{aligned}$