# What is a general solution to the differential equation y'-3y=5?

Jun 26, 2016

$= \gamma {e}^{3 x} - \frac{5}{3}$

#### Explanation:

$y ' - 3 y = 5$

$y ' = 3 y + 5$

$\frac{y '}{3 y + 5} = 1$

$\int \setminus \frac{\mathrm{dy}}{3 y + 5} = \int \setminus \mathrm{dx}$

$\frac{1}{3} \ln \left(3 y + 5\right) = x + \alpha$

$\ln \left(3 y + 5\right) = 3 \left(x + \alpha\right)$

$3 y + 5 = {e}^{3 \left(x + \alpha\right)} = {e}^{3 x} {e}^{3 \alpha} = \beta {e}^{3 x}$

$y = \frac{1}{3} \left(\beta {e}^{3 x} - 5\right)$

$= \gamma {e}^{3 x} - \frac{5}{3}$