Solution of $(e^{x} + 1) y d y + (y + 1) d x = 0$ is:

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

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

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

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(e^{x} + 1) x = c

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Solution

The correct option is B $(y + 1) (1 + e^{- x}) = c e^{y}$
$(e^{x} + 1) y d y + (y + 1) d x = 0$
$\Rightarrow (\frac{y}{y + 1}) d y + (\frac{e^{- x}}{1 + e^{- x}}) d x = 0$
$\Rightarrow \int (1 - \frac{1}{y + 1}) d y - \int (\frac{- e^{- x}}{1 + e^{- x}}) d x = - c$
$\Rightarrow y - l o g (y + 1) - l o g (1 + e^{- x}) = - c$
$y + c = l o g (y + 1) (1 + e^{- x})$
$\Rightarrow (y + 1) (1 + e^{- x}) = c \cdot e^{y}$

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Similar questions

(e^{x} + 1) y d y + (y + 1) d x = 0

Let $c$ be the arbitrary constant, then the solution of the differential equation $e^{x} cot y d x + (1 - e^{x}) {cosec}^{2} y d y = 0$ is

(e^{x} + 1) y d y + (y + 1) d x = 0

(e^{x} + 1) y d y + (y + 1) d x = 0

\frac{d y}{d x} = e^{x + y}

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