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Homework Help: Differential equation, Laplace transform

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    solve the differential equation by using Laplace transform

    [tex]x(0)=0, x^{'}(0)=0[/tex]

    2. Relevant equations
    then i get
    [tex]X(s) = \frac{3}{s^2+9}\cdot\frac{1}{(s+5)(s+2)}[/tex]

    \begin{cases}A+C+D=0\\7A+2C+B+5D=0\\10A+7B+9C+9D=0\\10B+18C+45D=3\end{cases}\Rightarrow \begin{cases}A=-\frac{21}{442}\\B=\frac{3}{442}\\C=-\frac{1}{34}\\D=\frac{1}{13}\end{cases}[/tex]

    [tex]x(t) = =-\frac{21}{442}cos3t+\frac{3}{442}sin3t-\frac{1}{34}e^{-2t}+\frac{1}{13}e^{-5t}[/tex]
    1) well i made a quck check (used other method to calculate A, B) to se if the are correct
    and this is what i get

    and from this i obtain

    can someone please show me where did I make a mistake??

    3. The attempt at a solution
    Last edited: Jun 2, 2010
  2. jcsd
  3. Jun 2, 2010 #2


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    X(s) = \frac{-3(7s-1)}{442(s^2+9)}\ + \frac{1}{13(s+2)}\ - \frac{1}{34(s+5)}\ }


    x(t) = \frac{1}{442}(sin(3t)-21cos(3t))\ + \frac{e^{-2t}}{13}\ - \frac{e^{-5t}}{34}\
    Last edited: Jun 2, 2010
  4. Jun 2, 2010 #3


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    You forgot about the factor of 3 on the top of the Laplace transform of sine

    [tex]\frac{B}{s^2+9} = \frac{3/442}{s^2+9} = \frac{1}{442}\left(\frac{3}{s^2+9}\right) \rightarrow \frac{1}{442} \sin 3t[/tex]

    As for the sign of that term, I'll let you track that down yourself. :tongue:
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