# What is a solution to the differential equation y'=1/2sin(2x)?

Mar 26, 2018

Question makes no sense

#### Explanation:

You can't solve or simply such an equation do you mean differentiate the equation?

Then you have to chain rule it

$f \left(u\right) = \frac{1}{2} \sin \left(u\right)$
$u = 2 x$

$f ' \left(u\right) = \frac{1}{2} \cos \left(u\right)$
$u ' = 2$

So now you multiply the two derivatives together $y ' ' = f ' \left(u\right) \cdot u '$

$y ' ' = \cos \left(u\right) = \cos \left(2 x\right)$

Mar 26, 2018

$y = - \frac{1}{4} \cos \left(2 x\right) + C$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \sin \left(2 x\right)$

This is a First Order Separable ODE, so we can "separate the variables" .

$\int \setminus \mathrm{dy} = \int \setminus \frac{1}{2} \sin \left(2 x\right) \setminus \mathrm{dx}$

Both integrals have well known trivial result so we can immediately integrate to get:

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

$\therefore y = - \frac{1}{4} \cos \left(2 x\right) + C$