Problem: Show that the given equation becomes exact when multiplied by the integrating factor and solve the equation.
$x^{2}y^{3}+x(1+y^{2})y'=0$
$My=x^{2}3y^{2} \neq Nx=\frac{1}{2}x^{2}$
Therefore it is not exact.
Multiply $\mu(\frac{1}{xy^{3}})$ to both side:
$\frac{x^{2}y^{3}}{xy^{3}}+(\frac{x+xy^{2}}{xy^{3}})y'=0$
$=x+(\frac{x}{xy^{3}}+\frac{xy^{2}}{xy^{3}})y'=0$
$=x+(\frac{1}{y^{3}}+\frac{1}{y})y'=0$
$M=x$
$N=\frac{1}{y^{3}}+\frac{1}{y}$
$My=0 = Nx=0$
Now it is exact.
Integrate M with respect to $x$ we get:
$\phi(x,y)=\frac{1}{2}x^{2}+h(y)$
Take derivative with respect to $y$ on both side:
$\phi y=N=0+h'(y)$
So $h'(y)=(\frac{1}{y^{3}}+\frac{1}{y})$
$h(y)=\int y^{-3}+y^{-1}dy$
$h(y)=-\frac{1}{4}y^{-4}+lny+C$
General Solution: $\frac{1}{2}x^{2}-\frac{1}{4}y^{-4}+lny=C$
