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

Answer 1

Charles Anctil

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)#

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Answer 2

Ezekiel Alexander

# 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 #

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Answer 3

Charles Arocha

The solution to the differential equation ( y' = \frac{1}{2}\sin(2x) ) is ( y = -\frac{1}{4}\cos(2x) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.