The general solution of $\frac{dy}{dx} = 2x e^{x^2 - y}$ is

$e^y = e^{x^2} + C$

The correct answer is Option (3) → $e^y = e^{x^2} + C$ ##

Given that, $\frac{dy}{dx} = 2x e^{x^2 - y} = 2x e^{x^2} \cdot e^{-y}$

$\Rightarrow e^y \frac{dy}{dx} = 2x e^{x^2}$

$\Rightarrow e^y \, dy = 2x e^{x^2} \, dx \quad \text{[using variable separable method]}$

On integrating both sides, we get

$\int e^y \, dy = 2 \int x e^{x^2} \, dx$

Put $x^2 = t$ in RHS integral, we get

$2x \, dx = dt$

$\int e^y \, dy = \int e^t \, dt$

$\Rightarrow e^y = e^t + C$

$\Rightarrow e^y = e^{x^2} + C \quad [∵t = x^2]$