# What is a solution to the differential equation #e^xdy/dx+2e^xy=1#?

this is non-separable, we use an integrating factor (IF)

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this linear non-homogeneus differential equation has as solution

such that

and

For the homogeneus we obtain easily ( variables grouping)

so the solution is

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To solve the differential equation ( e^x \frac{dy}{dx} + 2e^x y = 1 ), we can use an integrating factor method.

First, rearrange the equation to get ( \frac{dy}{dx} + 2y = e^{-x} ).

The integrating factor ( \mu(x) ) is given by ( e^{\int 2 , dx} = e^{2x} ).

Multiply both sides of the differential equation by ( e^{2x} ) to get ( e^{2x} \frac{dy}{dx} + 2e^{3x}y = e^x ).

This equation can be rewritten as ( \frac{d}{dx}(e^{2x}y) = e^x ).

Now, integrate both sides with respect to ( x ) to get ( e^{2x}y = \int e^x , dx ).

Integrate ( e^x ) with respect to ( x ) to get ( e^{2x}y = e^x + C ), where ( C ) is the constant of integration.

Finally, solve for ( y ) to get ( y = \frac{e^x + C}{e^{2x}} ).

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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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