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

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Answer 1 · 2016-08-01T21:59:23

$y = ln (e^x + C)$

$y' = e^(x-y) = e^x e^(-y)$

so this is separable

$e^y y' = e^x$

$int \ e^y y' \ dx =int\ e^x \ dx$

$int \ d/dx(e^y) \ dx =int\ e^x \ dx$

$e^y = e^x + C$

$y = ln (e^x + C)$

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