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Hard intergral

  • Thread starter Sirsh
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  • #1
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Hard intergral!!

:cry:If y = ke-x(sin2x), prove that [tex]\frac{d^2y}{dx^2}[/tex] + 2[tex]\frac{dy}{dx}[/tex] + 5y = 0:cry:


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

Answers and Replies

  • #2
rock.freak667
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you have y, you can find d2y/dx2 and dy/dx. Just find them and sub them into the equation and see if the left side is the same as the right side.
 
  • #3
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y'= sin2x(-ke-x)-(sin2x*cos2x)

y''= ke-xcos2x+sin2x(ke-x)+1

I've got my first and second intergral, can i get a confirmation? thanks alot.
 
  • #4
Dick
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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
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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
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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
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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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