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

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

We have:

which we could type as:

We can "separate the variables" because this is a separable DE:

We can integrate right away because both integrals are standard functions:

which is the overall resolution.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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