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

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

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

We have:

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

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

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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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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.

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.

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.

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