What is a solution to the differential equation #e^x(y'+1)=1#?

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Answer 1 · 2016-07-10T17:06:45

#y = - e^-x - x+ C#

you can tell just by looking that this is separable, so we separate it out

#e^x(y'+1)=1#

#y'+1=e^-x#

#y'=e^-x - 1#

integrate both sides

#int \ y' \ dx=int \ e^-x - 1 \ dx#

#y = - e^-x - x+ C#