Find the solution of $(e^{x} + 1) y d y + (y + 1) d x = 0$

e^{x + y} = k (y + 1) (1 + e^{x}) .

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e^{x - y} = k (y - 1) (1 + e^{x}) .

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e^{x + y} = k (y - 1) (1 + e^{x}) .

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e^{x - y} = k (y + 1) (1 + e^{x}) .

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Solution

The correct option is A $e^{x + y} = k (y + 1) (1 + e^{x}) .$
$(e^{x} + 1) y d y + (y + 1) d x = 0$
$\frac{y}{y + 1} d y + \frac{1}{e^{- x} + 1} d x = 0$
or $(1 - \frac{1}{y + 1}) d y + \frac{e^{- x}}{1 + e^{- x}} d x = 0$
Integrating,
$y - log (y + 1) - log (1 + e^{- x}) = log k$
$y = log k (y + 1) (1 + e^{- x})$
$e^{y} = k (y + 1) (\frac{1 + e^{x}}{e^{x}})$
$e^{x + y} = k (y + 1) (1 + e^{x}) .$

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