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

Question

Samantha Adkison · Accepted Answer

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

We have: # e^x (y'+1)=1# which we could type as: # y'+1 = e^(-x) # # :. dy/dx = e^(-x) - 1 # We can "separate the variables" because this is a separable DE: # int \ dy = int \ (e^(-x) - 1 \) dx # We can integrate right away because both integrals are standard functions: # y = -e^(-x) - x + C # which is the overall resolution.

Emily Ammerman · Answer

undefined To solve the differential equation $ e^x (y' + 1) = 1 $ using separation of variables, first isolate $ y' $:

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

Now, divide both sides by $ e^x $:

$$ y' = \frac{1 - e^x}{e^x} $$

Next, separate variables:

$$ \frac{dy}{dx} = \frac{1 - e^x}{e^x} $$

Multiply both sides by $ dx $:

$$ dy = \frac{1 - e^x}{e^x} dx $$

Now, integrate both sides:

$$ \int dy = \int \frac{1 - e^x}{e^x} dx $$

$$ y = \int \left( \frac{1}{e^x} - 1 \right) dx $$

$$ y = \int e^{-x} - \int 1 \, dx $$

$$ y = -e^{-x} - x + C $$

Where $ C $ is the constant of integration.