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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^{''}(t)+7x^{'}(t)+10x(t)=sin3t[/tex]
    [tex]x(0)=0, x^{'}(0)=0[/tex]







    2. Relevant equations
    [tex]X(s)s^2+X(s)7s+X(s)10=\frac{3}{s^2+9}[/tex]
    then i get
    [tex]X(s) = \frac{3}{s^2+9}\cdot\frac{1}{(s+5)(s+2)}[/tex]

    [tex]\frac{3}{(s^2+9)(s+5)(s+2)}=\frac{As+B}{s^2+9}+\frac{C}{s+5}+\frac{D}{s+2}\\
    \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
    [tex]x=A\cos3t+B\sin3t[/tex]
    [tex]x'=-3A\sin3t+3B\cos3t[/tex]
    [tex]x''=-9A\cos3t-9B\sin3t[/tex]

    and from this i obtain
    [tex]A=-\frac{21}{442}[/tex]
    [tex]B=-\frac{1}{442}[/tex]

    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

    cronxeh

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    Gold Member

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

    \begin{cases}A=-\frac{21}{442}\\B=\frac{3}{442}\\C=-\frac{1}{34}\\D=\frac{1}{13}\end{cases}
    correct
    [/tex]

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

    vela

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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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