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Math Help - Laplace shifting problem

  1. #1
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    Laplace shifting problem

    Hi,

    I am trying to solve

    y''+4y=0, 0<t<a; exp(-t), t>a

    using Laplace transforms.

    I end up with
    H(t-a)(exp(-(t-a))-cos(2(t-a))+1/2sin(2(t-a))

    where H(t-a) is the Heaviside/unit step function. Is this correct? If not, where have I gone wrong?

    Cheers,
    Ben
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  2. #2
    A Plied Mathematician
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    Hmm. That isn't what I get. You're missing the complimentary function, I think.

    1. How can you write the original DE in one equation using the Heaviside step function?
    2. What do you get when you LT the DE?
    3. What are the initial conditions?
    4. So what do you get when you solve for the LT of y?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by weedingb View Post
    Hi,

    I am trying to solve

    y''+4y=0, 0<t<a; exp(-t), t>a

    using Laplace transforms.

    I end up with
    H(t-a)(exp(-(t-a))-cos(2(t-a))+1/2sin(2(t-a))

    where H(t-a) is the Heaviside/unit step function. Is this correct? If not, where have I gone wrong?

    Cheers,
    Ben
    You have proposed a second order linear DE for the solution of which it is necessary to know the values y(0) and y^{'}(0) at the moment not yet specified...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor chisigma's Avatar
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    The DE is [in my opinion]...

    y^{''} + 4\ y= \left\{\begin{array}{ll} 0,\,\,0 < x < a{}\\e^{-t} ,\,\, x\ge a\end{array}\right. (1)

    ... and if we suppose y(0)= y^{'}(0)=0 in terms of Laplace Transform is written...

    \displaystyle s^{2}\ Y(s) + 4\ Y(s) = \frac{e^{-a\ (s+1)}}{s+1} \implies Y(s)= \frac{e^{-a\ (s+1)}}{(s+1)\ (s^{2}+4)} (2)

    Now if You consider that is...

    \displaystyle \frac{1}{(s+1)\ (s^{2}+4)} = \frac{1}{5}\ \{\frac{1}{s+1}+ \frac{1-s}{s^{2}+4} \} (3)

    ... You obtain...

    \displaystyle y(t)= \mathcal{L}^{-1} \{Y(s)\} = \frac{e^{-a}}{5}\ \{e^{-(t-a)} + \cos 2(t-a) - \sin 2(t-a)\}\ \mathcal{H} (t-a) (4)

    Kind regards

    \chi \sigma
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