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Math Help - Differential equations using Laplace transforms

  1. #1
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    Differential equations using Laplace transforms

    y''+9y=r(t)=\left\{\begin{matrix} 8sin(t),t\in (0,\pi) \\ 0,t \in (\pi,+\infty) \end{matrix}\right. ,y(0)=0,y'(0)=4

    How can I express r(t) using Dirac delta function or Heaviside step function?
    I know that Heaviside is the antiderivative of Dirac.

    H(x)=\left\{\begin{matrix} 0, x<0 \\ 1,x\geqslant 0 \end{matrix}\right.

    \delta(x)=\left\{\begin{matrix} +\infty, x=0 \\ 0,x\neq 0 \end{matrix}\right.

    EDIT:

    r(t)=8sin(t)*(1-u(t-\pi))
    L(r(t))(s)=L(8sin(t))(s)-L(8sin(t)*u(t-\pi))
    L(8sin(t)*u(t-\pi))(s)=8*e^{-\pi s}*L(sin(t+\pi))=-8e^{-\pi s}\frac{1}{s^{2}+1} =>
    L(r(t))(s)=8\frac{1}{s^{2}+1}+8e^{-\pi s}\frac{1}{s^{2}+1}
    then i got
    Y(s)=\frac{8e^{-\pi s}+4s^{2}+12}{(s^{2}+1)(s^{2}+9)}

    I believe that I have to use the Frequency shift property to get rid of e^{-\pi s}, but how?
    L(e^{at}f(t))(s)=F(s-a)
    Last edited by cristi92; May 9th 2012 at 07:23 AM.
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  2. #2
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    Re: Differential equations using Laplace transforms

    Quote Originally Posted by cristi92 View Post
    y''+9y=r(t)=\left\{\begin{matrix} 8sin(t),t\in (0,\pi) \\ 0,t \in (\pi,+\infty) \end{matrix}\right. ,y(0)=0,y'(0)=4

    How can I express r(t) using Dirac delta function or Heaviside step function?
    I know that Heaviside is the antiderivative of Dirac.
    The Heaviside function, H(x), has the value 0 for x< 0, 1 for x> 0. Here, you have to move that "cut point" to x= 1 and reverse ">" and "<". We can do that using
    H(1- x). When x< 1, 1- x> 1 so H(1- x)= 1. When x> 1, 1- x< 0 so H(1- x)= 0. You want 8sin(t) for t< 1, 0 for t> 1 so you want 8sin(t)H(1- t).

    H(x)=\left\{\begin{matrix} 0, x<0 \\ 1,x\geqslant 0 \end{matrix}\right.

    \delta(x)=\left\{\begin{matrix} +\infty, x=0 \\ 0,x\neq 0 \end{matrix}\right.

    EDIT:

    r(t)=8sin(t)*(1-u(t-\pi))
    L(r(t))(s)=L(8sin(t))(s)-L(8sin(t)*u(t-\pi))
    L(8sin(t)*u(t-\pi))(s)=8*e^{-\pi s}*L(sin(t+\pi))=-8e^{-\pi s}\frac{1}{s^{2}+1} =>
    L(r(t))(s)=8\frac{1}{s^{2}+1}+8e^{-\pi s}\frac{1}{s^{2}+1}
    then i got
    Y(s)=\frac{8e^{-\pi s}+4s^{2}+12}{(s^{2}+1)(s^{2}+9)}

    I believe that I have to use the Frequency shift property to get rid of e^{-\pi s}, but how?
    L(e^{at}f(t))(s)=F(s-a)
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  3. #3
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    Re: Differential equations using Laplace transforms

    Thank you!
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