# What is a general solution to the differential equation dy/dx=10-2y?

Jul 19, 2016

$y = \frac{{e}^{- 2 \left(x + C\right)} + 10}{2}$

OR

$y = C {e}^{- 2 x} + 5$

#### Explanation:

This differential equation is separable.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 10 - 2 y$

$\mathrm{dy} = \left(10 - 2 y\right) \mathrm{dx}$

$\frac{1}{10 - 2 y} \mathrm{dy} = \mathrm{dx}$

$\int \frac{\mathrm{dy}}{10 - 2 y} = \int \mathrm{dx}$

Let $u = 10 - 2 y \to \mathrm{du} = - 2 \mathrm{dy} \to - \frac{1}{2} \mathrm{du} = \mathrm{dy}$

$- \frac{1}{2} \int \frac{\mathrm{du}}{u} = \int x \mathrm{dx}$

$- \frac{1}{2} \ln \left(10 - 2 y\right) = x + C$

$\ln \left(10 - 2 y\right) = - 2 \left(x + C\right)$

$10 - 2 y = {e}^{- 2 \left(x + C\right)}$

$- 2 y + 10 = {e}^{- 2 \left(x + C\right)}$

$- 2 y = {e}^{- 2 \left(x + C\right)} - 10$

$\textcolor{red}{- 2 y = {e}^{- 2 x} {e}^{- 2 C} - 10}$

$\textcolor{red}{- 2 y = C {e}^{- 2 x} - 10}$

$\textcolor{red}{y = \frac{C}{- 2} {e}^{- 2 x} - \frac{10}{- 2}}$

$\textcolor{red}{y = C {e}^{- 2 x} + 5}$

$y = - \frac{{e}^{- 2 \left(x + C\right)} - 10}{2}$

$y = \frac{{e}^{- 2 \left(x + C\right)} + 10}{2}$