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