Solution of $(1 + x y) y d x + (1 - x y) x d y = 0$ is

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

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$\log (\frac{x}{y}) = c$

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

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None of these

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Solution

The correct option is C
$\log (\frac{x}{y}) - (\frac{1}{x y}) = c^{}$

Differentiating the given expression:
$(1 + x y) y d x + (1 - x y) x d y = 0$
This expression can be written as
$y d x + x y (y d x) + x d y - x y (x d y) = 0$
Solving further we get,
$(y d x + x d y) + x y (y d x - x d y) = 0 \Rightarrow d (x y) + {(x y)}^{2} (\frac{y d x - x d y}{x y}) = 0 [d (x y) = (y d x + x d y)] \Rightarrow d (x y) + {(x y)}^{2} (\frac{d x}{x} - \frac{d y}{y}) = 0 Dividingby {(x y)}^{2} \Rightarrow \frac{1}{{(x y)}^{2}} d (x y) + \frac{d x}{x} - \frac{d y}{y} = 0 \Rightarrow {(x y)}^{- 2} d (x y) + \frac{d x}{x} - \frac{d y}{y} = 0 \Rightarrow \frac{{(x y)}^{- 1}}{- 1} + \log x - \log y = c [∵ {(x y)}^{- 2} d (x y) = \frac{{(x y)}^{- 1}}{- 1}] \Rightarrow \log (\frac{x}{y}) - \frac{1}{x y} = c$
Therefore, option (C) is the correct answer.

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