Solution of $(1+xy)ydx+(1-xy)xdy=0$ is

The correct option is C

$\log \left(\frac{x}{y}\right)-\left(\frac{1}{xy}\right)=c^{}$

Differentiating the given expression:

$(1+xy)ydx+(1-xy)xdy=0$

This expression can be written as

$ydx+xy\left(ydx\right)+xdy-xy\left(xdy\right)=0$

Solving further we get,

$$\begin{gathered} \left(ydx+xdy\right)+xy\left(ydx-xdy\right)=0 \\ \Rightarrow d\left(xy\right)+{\left(xy\right)}^2\left(\frac{ydx-xdy}{xy}\right)=0\left[d\left(xy\right)=\left(ydx+xdy\right)\right] \\ \Rightarrow d\left(xy\right)+{\left(xy\right)}^2\left(\frac{dx}{x}-\frac{dy}{y}\right)=0 \\ \text{Dividing by}{\left(xy\right)}^2 \\ \Rightarrow \frac{1}{{\left(xy\right)}^2}d\left(xy\right)+\frac{dx}{x}-\frac{dy}{y}=0 \\ \Rightarrow {\left(xy\right)}^{-2}d\left(xy\right)+\frac{dx}{x}-\frac{dy}{y}=0 \\ \Rightarrow \frac{{\left(xy\right)}^{-1}}{-1}+\log x-\log y=c\left[\because {\left(xy\right)}^{-2}d\left(xy\right)=\frac{{\left(xy\right)}^{-1}}{-1}\right] \\ \Rightarrow \log \left(\frac{x}{y}\right)-\frac{1}{xy}=c \end{gathered}$$

Therefore, option (C) is the correct answer.