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 #