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

Question

how can I solve the given first order differential equation: $ \frac{dy}{dx} - e^{x-y} + e^y = 0 $

Answer 1

How about multiplying the equation by $e^y$, then call $z=e^y$. You get: $$e^y\frac{dy}{dx}+(e^y)^2=e^x$$ then:$$\frac{dz}{dx}=e^y\frac{dy}{dx}$$ so:$$\frac{dz}{dx}+z^2=e^x$$ Is this something you can solve?