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The solution of $(1 + x y) y d x + (1 - x y) x d y = 0$ is:

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$\frac{x}{y} + \frac{1}{x y} = k$

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

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

answer is B.

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Detailed Solution

$y d x + x d y + x y^{2} d x - x^{2} y d y = 0 \frac{y d x + x d y}{x^{2} y^{2}} + \frac{d x}{x} - \frac{d y}{y} = 0 .$
On integrating, we get $- \frac{1}{x y} + \log x - \log y = k$

$\Rightarrow \log \frac{x}{y} = \frac{1}{x y} + k .$

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