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

Aug 1, 2016

$y = C {e}^{2 x} + \frac{1}{2}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 y - 1$

separating the variables
$\frac{1}{2 y - 1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

integrating
$\int \setminus \frac{1}{2 y - 1} \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus \mathrm{dx}$

$\int \setminus \frac{1}{2 y - 1} \setminus \mathrm{dy} = \int \setminus \mathrm{dx}$

$\frac{1}{2} \ln \left(2 y - 1\right) = x + C$

$\ln \left(2 y - 1\right) = 2 x + C$

$2 y - 1 = {e}^{2 x + C} = C {e}^{2 x}$

$y - \frac{1}{2} = C {e}^{2 x}$

$y = C {e}^{2 x} + \frac{1}{2}$