The solution of the differential equation $(1 + x y) x d y + (1 - x y) y d x = 0,$ is

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$\frac{1}{x y} + \log (\frac{y}{x}) = C$

$- \frac{1}{x y} + \log (\frac{y}{x}) = C$

$- x y + \log (\frac{y}{x}) = C$

$- \frac{1}{x y} + \log (\frac{x}{y}) = C$

answer is C.

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Detailed Solution

The given equation can be written as

$\begin{array}{l} (x d y + y d x) + x y (x d y - y d x) = 0 \\ \Rightarrow d (x y) + x y (x d y - y d x) = 0 \\ \Rightarrow \frac{d (x y)}{(x y)^{2}} + \frac{x d y - y d x}{x y} = 0 \\ \Rightarrow \frac{d (x y)}{(x y)^{2}} + \frac{d y}{y} - \frac{d x}{x} = 0 \end{array}$

$\begin{array}{l} \Rightarrow \frac{d (x y)}{(x y)^{2}} + d (\log y) - d (\log x) = 0 \\ \Rightarrow \frac{d (x y)}{(x y)^{2}} + d (\log y - \log x) = 0 \\ \Rightarrow \frac{d (x y)}{(x y)^{2}} + d \log (\frac{y}{x}) = 0 \end{array}$