Answer
$y=\frac{\ln (C-2e^{-x})}{2}$
Work Step by Step
We are given:
$$e^{2x+y}dy-e^{x-y}dx=0$$
We can write it as:
$$\frac{dy}{dx}=\frac{e^{x-y}}{e^{2x+y}}$$
$$\frac{dy}{dx}=e^{-x-2y}$$
Integrating both sides:
$$\int \frac{dy}{dx}=\int e^{-x-2y}$$
$$\frac{1}{2}e^{2y}=-e^{-x}+C$$
$$e^{2y}=-2e^{-x}+C$$
$$2y=\ln(-2e^{-x}+C)$$
Hence general solution is $y=\frac{\ln (C-2e^{-x})}{2}$
