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Thread: Nontrivial Solution/Laplace Transform

  1. #1
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    Nontrivial Solution/Laplace Transform

    So on this practice exam paper problem #5 (page 7) http://www.math.lsa.umich.edu/course...inalW10sol.pdf

    I am told to find the Laplace Transform of:
    tx''-3x'+x=0 where x(0)=0

    Now I've always transformed tx'' in the following way:

    Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
    So I would get -[(2*s*X(s))+((s^2)*X'(s)))]

    But the solution gives tx'' -> -(s*X(s)-x(0))'
    Which gives -[(s*X'(s))+x(s)]

    Why?

    Even on my textbook: Differential Equations: Computing and Modelling by Edwards & Penney they are using the method that I have always used.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Nontrivial Solution/Laplace Transform

    Quote Originally Posted by chaosier View Post
    Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
    So I would get -[(2*s*X(s))+((s^2)*X'(s)))]
    You are right. Indeed:

    $\displaystyle \mathcal{L}\{tx''(t)\}=-\frac{d}{ds}\mathcal{L}\{x''(t)\}=-\frac{d}{ds}(s^2X(s)-sx(0)-x''(0))=$

    $\displaystyle -\frac{d}{ds}(s^2X(s)-x''(0))=-(2sX(s)+s^2X'(s))$
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: Nontrivial Solution/Laplace Transform

    Quote Originally Posted by chaosier View Post
    So on this practice exam paper problem #5 (page 7) http://www.math.lsa.umich.edu/course...inalW10sol.pdf

    I am told to find the Laplace Transform of:
    tx''-3x'+x=0 where x(0)=0

    Now I've always transformed tx'' in the following way:

    Transform of tx'' -> -[(s^2)*X(s)-x'(0)]'
    So I would get -[(2*s*X(s))+((s^2)*X'(s)))]

    But the solution gives tx'' -> -(s*X(s)-x(0))'
    Which gives -[(s*X'(s))+x(s)]

    Why?

    Even on my textbook: Differential Equations: Computing and Modelling by Edwards & Penney they are using the method that I have always used.
    Starting from the second order DE...

    $\displaystyle t\ x^{''} -3\ x^{'} + x=0,\ x(0)=0$ (1)

    ... taking into account that is...

    $\displaystyle \mathcal{L}\{t\ x^{''}\}= - \frac{d}{d s} \{s^{2}\ X(s) -s\ x(0) - x^{'}(0)\} = -s^{2}\ X^{'} (s)-2\ s\ X(s)$ (2)

    ... You arrive to write...

    $\displaystyle s^{2}\ X^{'} +(5\ s -1)\ X=0$ (3)

    Now You can find X(s) solving the first order DE (3), which is linear and relatively 'comfortable'...




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