# How do you find the general solution to #dy/dx=2y-1#?

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To find the general solution to the differential equation ( \frac{dy}{dx} = 2y - 1 ):

Separate variables:

[ \frac{dy}{2y - 1} = dx ]

Integrate both sides:

[ \int \frac{dy}{2y - 1} = \int dx ]

For the left integral, let ( u = 2y - 1 ) and ( du = 2dy ):

[ \frac{1}{2} \int \frac{du}{u} = \int dx ]

[ \frac{1}{2} \ln|u| = x + C_1 ] [ \ln|2y - 1| = 2x + C_1 ]

Exponentiate both sides:

[ 2y - 1 = e^{2x + C_1} ]

[ 2y - 1 = e^{C_1}e^{2x} ]

Let ( C = e^{C_1} ):

[ 2y - 1 = Ce^{2x} ]

[ 2y = Ce^{2x} + 1 ]

[ y = \frac{Ce^{2x} + 1}{2} ]

This is the general solution to the differential equation.

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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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