# What is a solution to the differential equation dy/dx = 20ycos(-5x)?

Jul 10, 2016

$y = C {e}^{4 \sin 5 x}$

#### Explanation:

it's separable, so we separate

$\frac{\mathrm{dy}}{\mathrm{dx}} = 20 y \cos \left(- 5 x\right)$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 20 \cos \left(- 5 x\right)$

we integrate

$\int \setminus \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus 20 \cos \left(- 5 x\right) \setminus \mathrm{dx}$

or

$\int \setminus \frac{1}{y} \mathrm{dy} = 20 \int \setminus \cos \left(- 5 x\right) \setminus \mathrm{dx}$

$\int \setminus \frac{1}{y} \mathrm{dy} = 20 \int \setminus \cos 5 x \setminus \mathrm{dx}$ $q \quad q \quad$ as $\cos \left(- \varphi\right) = \cos \varphi$

$\ln y = 20 \frac{1}{5} \sin 5 x + C$

$\ln y = 4 \sin 5 x + C$

$y = {e}^{4 \sin 5 x + C} = C {e}^{4 \sin 5 x}$