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

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Answer 1 · 2018-03-26T11:45:48

Question makes no sense

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

Then you have to chain rule it

$f(u)=1/2sin(u)$
$u=2x$

$f'(u)=1/2cos(u)$
$u'=2$

So now you multiply the two derivatives together $y''=f'(u)*u'$

$y''=cos(u)=cos(2x)$

Answer 2 · 2018-03-26T11:47:36

$y = -1/4cos(2x) + C$

We have:

$dy/dx = 1/2sin(2x)$

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

$int \ dy = int \ 1/2sin(2x) \ dx$

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

$y = 1/2(-cos(2x)/2) + C$

$:. y = -1/4cos(2x) + C$

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