Solution:

Compare the given diff. eqn with $\mathrm{M} \mathrm{d} x+\mathrm{Ndy}=0$

$\therefore \mathrm{M}=\left(y+\frac{1}{3} y^{3}+\frac{1}{2} x^{2}\right) \quad \therefore \mathrm{N}=\left(x+x y^{2}\right)$

$\frac{\partial M}{\partial y}=1+y^{2} \quad \frac{\partial N}{\partial x}=1+y^{2}$

$\therefore \quad \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

The given differential eqn is exact.

The solution of exact differential eqn is given by,

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

$\int M d x=\int\left(y+\frac{1}{3} y^{3}+\frac{1}{2} x^{2}\right) d x=x y+\frac{x}{3} y^{3}+\frac{x^{3}}{6}$

$\frac{\partial}{\partial y} \int M d x=x+x y^{2}$

$\int\left[N-\frac{\partial}{\partial y} M d x\right] d y=\int\left[x+x y^{2}-\left(x+x y^{2}\right)\right] d y=0$

$\therefore \quad x y+\frac{x}{3} y^{3}+\frac{x^{3}}{6}=c$