The solution of differential equation $(e^{x} + 1) y d y = (y + 1) e^{x} d x$ is

(e^{x} + 1) (y + 1) = C e^{y}

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(e^{x} + 1) | (y + 1) | = C e^{- y}

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(e^{x} + 1) (y + 1) = \pm C e^{y}

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None of these

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Solution

The correct option is A $(e^{x} + 1) (y + 1) = C e^{y}$
The given differential equation is
$(e^{x} + 1) \cdot y d y = (y + 1) e^{x} d x$
$\Rightarrow \frac{y d y}{(y + 1)} = \frac{e^{x}}{(e^{x} + 1)} d x$
$\Rightarrow (\frac{y + 1 - 1}{y + 1}) d y = \frac{e^{x}}{e^{x} + 1} d x$
$\Rightarrow d y - \frac{1}{y + 1} d y = \frac{e^{x}}{e^{x} + 1} d x$
On integrating, we get
$\Rightarrow y - log | y + 1 | = log (e^{x} + 1) + log k$
$\Rightarrow y = log | (y + 1) (e^{x} + 1) | k$
$\Rightarrow (y + 1) (e^{x} + 1) = e^{y} C$ .

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