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Simultaneous Laplace transforms

  1. Feb 15, 2012 #1
    I have to try and solve the following simultaneous Laplace transform problem and don't really know which path to take can someone give me a nudge in the right direction please.

    dx/dt=4x-2y & dy/dt=5x+2y given that x(0)=2, y(0)=-2
    this is what i have so far for dx/dt=4x-2y
    sx-x(0)=4x-2y
    sx-2=4x-2y
    (s-4)x+2y=2

    And for dy/dt=5x+2y
    sy-y(0)=5x+2y
    sy+2=5x+2y
    (s-2)y-5x=-2
    Not really sure where to go from here, or even if this is correct.
     
  2. jcsd
  3. Feb 16, 2012 #2
    [tex](s-4)X+2Y=2,5X-(s-2)Y=2.[/tex]
    [tex]X=\frac{2s}{s^2-6s+18},Y=-\frac{2s-18}{s^2-6s+18}.[/tex]
     
  4. Feb 16, 2012 #3
    Can you ellaborate a little please.
    Where did this all come from?
     
  5. Feb 16, 2012 #4
    ##\begin{cases}(s-4)X+2Y=2&...(1)\\5X-(s-2)Y=2&...(2)\end{cases}##
    ##(s-2)\times(1)+2\times(2):((s-4)(s-2)+10)X=2(s-2)+4,##[tex]X=\frac{2s}{s^2-6s+18}.[/tex]
    ##5\times(1)-(s-4)\times(2):(10+(s-2)(s-4))Y=10-2(s-4),##[tex]Y=\frac{18-2s}{s^2-6s+18}.[/tex]
     
  6. Feb 16, 2012 #5
    Thats great, thanks alot.
    Just out of interest where has the 2s in X come from and the 18 - 2s in Y come from, i can work out the bottom lines. sorry if i appear stupid but it is 5.20am.
    From there i can use partial fractions to determine the inverse Laplace transform (I think anyway).
     
  7. Feb 18, 2012 #6
    I get the 2s & the 18-2s.
    Am i correct in thinking these sre complex roots and by definition are quite complex to solve especially the 18-2s one.?
    any help is appreciated
     
  8. Feb 20, 2012 #7
    I need to then try and find the inverse Laplace transform of X & Y can anyone assist me in telling me if i am close with:-
    X=2e^(-3t)*cosh3t
    Y=e^(-18t)-2e^(-3t)*cosh3t

    This forum has been more than helpful so far and is highy recommended
     
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