Question

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

Answer 1

We have,

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

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

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

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

$dy=(1-e^x)dx$

On integrating both sides, we get

$\int dy=\int (1-e^x)dx$

$y=x-e^x+C$

Hence, this is the answer.