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Math Help - show that y'' − y = F(t) can be written...

  1. #1
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    show that y'' − y = F(t) can be written...

    Show that y'' − y = F(t) can be written as y= Ae^t+Be^-t + \int sinh(t-\tau )F\tau d\tau

    Obviously, the first is the complementary solution which is easily obtained. The second part I have no idea about. Tried using a variation of parameters but how to get that integral I don't know.
    Last edited by mr fantastic; April 27th 2011 at 03:59 AM.
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  2. #2
    A Plied Mathematician
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    Why don't you just differentiate the y there twice, plug into the DE, and see if you get equality?
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    Ok I've tried that but I'm having difficulty integrating with respect to tau as I use parts but that just creates another difficult integral and then i will have to differentiate all that twice
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    My suggestion was to differentiate, not integrate. That is, you have

    y(t)=Ae^{t}+Be^{-t}+\int_{0}^{t}\sinh(t-\tau)F(\tau)\,d\tau\implies

    \dot{y}(t)=Ae^{t}-Be^{-t}+\overbrace{\int_{0}^{t}\cosh(t-\tau)F(\tau)\,d\tau+\underbrace{\sinh(t-t)F(t)\cdot 1}_{=0}-\underbrace{\sinh(t-0)F(0)\cdot 0}_{=0}}^{\text{Use Leibniz's Integral Rule here}}

    =Ae^{t}-Be^{-t}+\int_{0}^{t}\cosh(t-\tau)F(\tau)\,d\tau.

    Here's a link to Leibniz's Integral Rule. You'll need to differentiate again.
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    when I differentiate again I just get y so my final answer would be 0. partial d/dt(cosh(t-tau)) = sinh(t-tau)
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    No, you should not get just y. The second time around, the two boundary terms from the Leibniz integral rule are not both zero. Check the rule a little more carefully.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ackbeet View Post
    No, you should not get just y. The second time around, the two boundary terms from the Leibniz integral rule are not both zero. Check the rule a little more carefully.
    A quick check-in, then I'll get out of the way of the main discussion. The Leibniz rule. That is essentially the Fundamental Theorem mixed in with the chain rule, right? (I know that's only a brief description, I can fill in the details on my own if that is where it comes from.)

    Thanks!
    -Dan
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    Quote Originally Posted by topsquark View Post
    A quick check-in, then I'll get out of the way of the main discussion. The Leibniz rule. That is essentially the Fundamental Theorem mixed in with the chain rule, right? (I know that's only a brief description, I can fill in the details on my own if that is where it comes from.)

    Thanks!
    -Dan
    You are essentially correct, yes.
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    I have got the answer. Thanks a lot for your help. Just one question: when you have multiplied the boundary terms by 1 and 0 respectively, does this represent d/dt(t) and d/dt(0) respectively.
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    Quote Originally Posted by poirot View Post
    I have got the answers. Thanks a lot for your help. Just one question: when you have multiplied the boundary terms by 1 and 0 respectively, does this represent d/dt(t) and d/dt(0) respectively.
    Exactly right. You're welcome!
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by poirot View Post
    Show that y'' − y = F(t) can be written as y= Ae^t+Be^-t + \int sinh(t-\tau )F\tau d\tau

    Obviously, the first is the complementary solution which is easily obtained. The second part I have no idea about. Tried using a variation of parameters but how to get that integral I don't know.
    In terms of Laplace Transform, supposing y(0)=y'(0)=0, the DE is written as...

    (1)

    ... so that y(t) is derived as follows...



    Kind regards

    \chi \sigma
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