Answer
$(\dfrac{1}{2})e^{2y}-e^x=c$
Work Step by Step
Given: $\dfrac{dy}{dx}=\dfrac{e^{2x-y}}{e^{x+y}}$
Re-arrange the given equation and integrate as follows:.
Then $\dfrac{dy}{dx}= \int e^{2x-y-x-y} \implies \int \dfrac{dy}{dx}= \int\dfrac{e^x}{e^{2y}} $
or, $\int e^{2y} dy =\int e^x dx$ ....(1)
As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$
Equation (1) becomes: $ \dfrac{1}{2} e^{2y}=e^x+c$
Hence, $(\dfrac{1}{2})e^{2y}-e^x=c$
