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

Question

1812 views around the world

You can reuse this answer
Creative Commons License

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.

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