# How do you find the general solution to dy/dx=e^(x−y)?

How do you find the general solution to $\frac{dy}{dx}={e}^{x-y}$?
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Jeremy Mayo
$y\prime ={e}^{x-y}={e}^{x}{e}^{-y}$
so this is separable
${e}^{y}y\prime ={e}^{x}$

${e}^{y}={e}^{x}+C$
$y=\mathrm{ln}\left({e}^{x}+C\right)$