Find the general solution of the differential equation: $xdy = (e^y - 1)dx$.

$1 - e^{-y} = Cx$

The correct answer is Option (2) → $1 - e^{-y} = Cx$ ##

Given differential equation can be written as

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

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

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

$\Rightarrow \log |1 - e^{-y}| = \log |x| + \log C$

$\Rightarrow 1 - e^{-y} = Cx$