# How will you solve:(dy)/(dx)=e^(x-y)[e^x-e^y]?

Mar 21, 2018

$y = \ln \left({C}_{0} {e}^{- {e}^{x}} + {e}^{x} - 1\right)$

#### Explanation:

Making the substitution

$y = \ln u$ we have the transformed differential equation

$\frac{u ' - {e}^{2 x} + {e}^{x} u}{u} = 0$ or assuming $u \ne 0$

$u ' + {e}^{x} u - {e}^{2 x} = 0$

This is a linear non homogeneous differential equation easily soluble giving

$u = {C}_{0} {e}^{- {e}^{x}} + {e}^{x} - 1$ and finally

$y = \ln \left({C}_{0} {e}^{- {e}^{x}} + {e}^{x} - 1\right)$