- #1

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**Hard intergral!!**

**If y = ke**

^{-x}(sin2x), prove that [tex]\frac{d^2y}{dx^2}[/tex] + 2[tex]\frac{dy}{dx}[/tex] + 5y = 0Need help asap, No clue how to figure it out. teacher is away until next term, test next week.

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- Thread starter Sirsh
- Start date

- #1

- 267

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Need help asap, No clue how to figure it out. teacher is away until next term, test next week.

- #2

rock.freak667

Homework Helper

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you have y, you can find d

- #3

- 267

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y'= sin2x(-ke

y''= ke

I've got my first and second intergral, can i get a confirmation? thanks alot.

- #4

Dick

Science Advisor

Homework Helper

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No, not right. All of your terms in y' and y'' should have a k and exp(-x) in them. Do you see why? Just use the product rule and the chain rule.

- #5

- 267

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I dont see why there would be a exp(-x)?? i thought the derivative of e^x was e^x

- #6

cronxeh

Gold Member

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I dont see why there would be a exp(-x)?? i thought the derivative of e^x was e^x

Derivative of exp(-x) is -exp(-x)

But more importantly you have 2 functions that you are taking a derivative of. Also, stop saying 'integral' - that is not what you are doing.

And finally,

(f*g)' = f' * g + f * g'

y=k*exp(-x)*sin(2*x)

f=k*exp(-x)

g=k*sin(2*x)

y=k*(f*g)

Find y' = (f*g)' = f' * g + f * g'

when f' = -exp(-x)

and g'=2*cos(2*x)

Last edited:

- #7

Dick

Science Advisor

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Because the problem you started out with has exp(-x) in it, not exp(x). What's the derivative of exp(-x)?

- #8

- 267

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lol sorry i put integral.. i am also doing integrals at the moment, kind of side tracked.

alright so the frist derivative is -e^(-x)sin2x+ke^(-x)2cos2x.

And second derivative is (e^(-x)2cos2x-e^(-x)sin2x)+(-ke^(-x)2cos2x-ke^(-x)4sin2x)

- #9

Dick

Science Advisor

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That's getting better. But why no k in the first term of y'=-e^(-x)sin2x+ke^(-x)2cos2x? I told you there should be one.

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