Answer
$C=\frac{x^2(xy^2-1)}{y}$
Work Step by Step
We are given
$(3xy-2y^{-1})dx+x(x+y^{-2})dy=0$ ___(1)
Here, $M(x,y)=3xy-2y^{-1}$
$N(x,y)=x(x+y^{-2})$
$(M)_y=3x+\frac{2}{y^2}$
$(N)_x=2x+\frac{1}{y^2}$
$(M)_y \ne (N)_x$
$\Rightarrow $ (2) is not Exact
We have:
$\frac{M_y-N_x}{N}=\frac{1}{x}=f(x)$
The integrating factor is:
$I(x)=e^{\int f(x)dx}=e^{\int \frac{1}{x}dx}=e^{\ln x }=x$
Multiplying equation (1) by $I(x)=x$
$x(3xy-2y^{-1})dx+x^2(x+y^{-2})dy=0$
Then:
$\frac{\partial \phi}{\partial x}=x(3xy-\frac{2}{y})$
Integrating both sides:
$\phi (x,y)=g(y)+\int x(3xy -\frac{2}{y})dx$
We have:
$\phi(x,y)=g(y)+\frac{x^2(xy^2-1)}{y}$
For
$\frac{\partial phi}{\partial y}=x^2(x+y^{-2})$
$\rightarrow g'(y)+x^2(x+\frac{1}{y^2})=x^2(x+y^{-2})$
Since $g'(y)=0 \rightarrow g(y)=C$
C is a constant integration
The solution is:
$\phi (x,y)=\frac{x^2(xy^2-1)}{y}+C=0$
or
$C=\frac{x^2(xy^2-1)}{y}$
