The general solution of differential equation $(e^x + 1) y dy = (y + 1) e^x dx$ is

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

The correct answer is Option (3) → $y = \log \{k(y + 1)(e^x + 1)\}$ ##

Given differential equation is

$(e^x + 1) y dy = (y + 1) e^x dx$$

$\Rightarrow \frac{dy}{dx} = \frac{e^x (1 + y)}{(e^x + 1) y} \Rightarrow \frac{dx}{dy} = \frac{(e^x + 1) y}{e^x (1 + y)}$

$\Rightarrow \frac{dx}{dy} = \frac{e^x y}{e^x (1 + y)} + \frac{y}{e^x (1 + y)}$

$\Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} + \frac{y}{(1 + y) e^x}$

$\Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} \left( 1 + \frac{1}{e^x} \right)$

$\Rightarrow \frac{dx}{dy} = \frac{y}{1 + y} \left( \frac{e^x + 1}{e^x} \right)$

$\Rightarrow \left( \frac{y}{1 + y} \right) dy = \left( \frac{e^x}{e^x + 1} \right) dx \quad \text{[using variable separable method]}$

On integrating both sides, we get

$\int \frac{y}{1 + y} dy = \int \frac{e^x}{1 + e^x} dx$

$\Rightarrow \int \frac{1 + y - 1}{1 + y} dy = \int \frac{e^x}{1 + e^x} dx$

$\Rightarrow \int 1 \, dy - \int \frac{1}{1 + y} \, dy = \int \frac{e^x}{1 + e^x} \, dx$

$\Rightarrow y - \log |(1 + y)| = \log |(1 + e^x)| + \log k$

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

$\Rightarrow y = \log \{ k(1 + y)(1 + e^x) \} \quad [\log a + \log b = \log ab]$