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

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

$y + 1 = e^{x} + 1 + k$

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

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

answer is C.

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Detailed Solution

Option (C) is correct.

Given differential equation.

$(e^{x} + 1) y d y = (y + 1) e^{x} d x \begin{array}{ll} \Rightarrow \frac{d y}{d x} = \frac{e^{x} (1 + y)}{(e^{x} + 1) y} \\ \Rightarrow \frac{d x}{d y} = \frac{(e^{x} + 1) y}{e^{x} (1 + y)} \\ \Rightarrow \frac{d x}{d y} = \frac{e^{x} y}{e^{x} (1 + y)} + \frac{y}{e^{x} (1 + y)} \\ \Rightarrow \frac{d x}{d y} = \frac{y}{1 + y} + \frac{y}{(1 + y) e^{x}} \\ \Rightarrow \frac{d x}{d y} = \frac{y}{1 + y} (1 + \frac{1}{e^{x}}) \end{array}$

$\begin{aligned} \Rightarrow \frac{d x}{d y} = \frac{y}{1 + y} (\frac{e^{x} + 1}{e^{x}}) \\ \Rightarrow (\frac{y}{1 + y}) d y = (\frac{e^{x}}{e^{x} + 1}) d x \end{aligned}$

On integrating both sides, we get

$\int \frac{y}{1 + y} d y = \int \frac{e^{x}}{1 + e^{x}} d x \begin{array}{l} \Rightarrow \int \frac{1 + y - 1}{1 + y} d y = \int \frac{e^{x}}{1 + e^{x}} d x \\ \Rightarrow \int 1 d y - \int \frac{1}{1 + y} d y = \int \frac{e^{x}}{1 + e^{x}} d x \\ \Rightarrow y - \log | (1 + y) = \log | (1 + e^{x}) + \log k \\ \Rightarrow y = \log (1 + y) + \log (1 + e^{x}) + \\ \Rightarrow y = \log \{k (1 + y) (1 + e^{x})\} \end{array}$