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)#