#e^x (y'+1)=1# ? using Separation of Variables

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Answer 1 · 2018-04-25T21:59:36

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

We have:

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

Which we can write as:

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

# :. dy/dx = e^(-x) - 1 #

This is a separable DE, so we can "separate the variables":

# int \ dy = int \ (e^(-x) - 1 \) dx #

Both integrals are standard function, so we can immediately integrate:

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

Which is the General Solution.