The general solution of the differential equation $\frac{dy}{dx}e^{x-y}+x^2e^{-y}$ is equal to:

$e^y=e^x+\frac{x^3}{3}+c$, where c is an arbitrary constant.

The correct answer is Option (1) → $e^y=e^x+\frac{x^3}{3}+c$, where c is an arbitrary constant.

$e^{x-y}\frac{dy}{dx}+x^{2}e^{-y}=0$

$\text{Multiply by }e^{y}:\quad e^{x}\frac{dy}{dx}+x^{2}=0$

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

$\text{Let }u=e^{y}\;\Rightarrow\;\frac{du}{dx}=e^{y}\frac{dy}{dx}$

$e^{y}\frac{dy}{dx}=\frac{du}{dx}$

$\frac{du}{dx}=e^{y}\frac{dy}{dx}=e^{y}(-x^{2}e^{-x})=-x^{2}e^{y-x}$

$\text{But from }e^{x}\frac{dy}{dx}+x^{2}=0:\; e^{y}\frac{dy}{dx}=e^{y-x}(-x^{2})$

$\Rightarrow\; \frac{du}{dx}=e^{x}+x^{2}$

$\int du=\int (e^{x}+x^{2})dx$

$u=e^{x}+\frac{x^{3}}{3}+C$

$e^{y}=e^{x}+\frac{x^{3}}{3}+C$

The general solution is

$e^{y}=e^{x}+\frac{x^{3}}{3}+C$