Answer
$C=xy+\ln(\frac{y}{x})$
Work Step by Step
We are given:
$$x^{-1}(xy-1)dx+y^{-1}(xy+1)dy=0 (1)$$
Here,
$M(x,y)=\frac{\partial}{\partial y}(x^{-1}(xy-1))=1$
$N(x,y)=\frac{\partial}{\partial x}(y^{-1}(xy+1))=1$
$M_y=1 \;\;\;\;,\;\;\;\;N_x=1$
$M_y=N_x$
$\Rightarrow$ (1) is exact differential equation
Therefore there exists a potential function $\phi$ such that
$\frac{\partial\phi}{\partial x}=x^{-1}(xy-1)$ ___(2)
Integrating (2) with respect to $x$ holding $y$ fixed
$\phi =xy-\ln x+f(y)$ ___(3)
Where $f(y)$ is arbitrary function of $y$
Hence we get:
$\frac{\partial\phi}{\partial y}=x+f'(y)=x+\frac{1}{y}$ (4)
We have
$f'(y)= \frac{1}{y}\rightarrow f(y)=\ln y$
$C$ is constant
The solution is:
$\phi(x,y)= xy-\ln x+ \ln y+C$
and $\phi(x,y)=0$
so:
$C=xy+ln(\frac{y}{x})$
