The general solution of $e^x \cos y \, dx - e^x \sin y \, dy = 0$ is

$e^x \cos y = k$

The correct answer is Option (1) → $e^x \cos y = k$ ##

Given that, $e^x \cos y \, dx - e^x \sin y \, dy = 0$

$\Rightarrow e^x \cos y \, dx = e^x \sin y \, dy$

$\Rightarrow \frac{dx}{dy} = \tan y$

$\Rightarrow dx = \tan y \, dy \quad \text{[applying variable separable method]}$

On integrating both sides, we get

$x = \log \sec y + C$

$\Rightarrow x - C = \log \sec y$

$\Rightarrow \sec y = e^{x - C}$

$\Rightarrow \sec y = e^x \cdot e^{-C}$

$\Rightarrow \frac{1}{\cos y} = \frac{e^x}{e^C}$

$\Rightarrow e^x \cos y = e^C$

$\Rightarrow e^x \cos y = K \quad [\text{where, } K = e^C]$