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Homework Help: Hard intergral

  1. Mar 2, 2010 #1
    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.
     
  2. jcsd
  3. Mar 2, 2010 #2

    rock.freak667

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    Re: Hard intergral!!

    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.
     
  4. Mar 2, 2010 #3
    Re: Hard intergral!!

    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.
     
  5. Mar 2, 2010 #4

    Dick

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    Re: Hard intergral!!

    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.
     
  6. Mar 2, 2010 #5
    Re: Hard intergral!!

    I dont see why there would be a exp(-x)?? i thought the derivative of e^x was e^x
     
  7. Mar 2, 2010 #6

    cronxeh

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    Re: Hard intergral!!

    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: Mar 2, 2010
  8. Mar 2, 2010 #7

    Dick

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    Re: Hard intergral!!

    Because the problem you started out with has exp(-x) in it, not exp(x). What's the derivative of exp(-x)?
     
  9. Mar 2, 2010 #8
    Re: Hard intergral!!

    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)
     
  10. Mar 2, 2010 #9

    Dick

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    Re: Hard intergral!!

    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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