Solution of the equation (dy)/(dx) = e^ x - y (e^ x - e^ y ) is

We have,

$\dfrac{dy}{dx} = e^{x - y} \left(e^{x} - e^{y}\right)$

$⇒$e^{y}$\dfrac{dy}{dx} + e^{y}e^{x} = e^{2x}$

Put $e^{y} = v$

$∴$e^{y}$\dfrac{dy}{dx} = \dfrac{dv}{dx}$

$∴ \dfrac{dv}{dx} + ve^{x} = e^{2x}$

This is linear equation

$I. F = = e^{∫ e^{x}dx} = e^{ex}$

$∴$ Required solution is

$v \cdote^{ex} = ∫ e^{ex} \cdote^{2x}dx + C$

$v \cdote^{ex} = e^{ex}\left(e^{x} - 1\right) + C$

$⇒ e^{y} \cdote^{ex} = e^{ex}\left(e^{x} - 1\right) + C$

$⇒ e^{y} = e^{x} - 1 + ce^{- e^{x}}$