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Simultaneous Laplace transforms |
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| Feb15-12, 10:03 PM | #1 |
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Simultaneous Laplace transforms
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. |
| Feb16-12, 05:42 AM | #2 |
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[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] |
| Feb16-12, 08:56 AM | #3 |
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Can you ellaborate a little please.
Where did this all come from? |
| Feb16-12, 10:10 PM | #4 |
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Simultaneous Laplace transforms
##\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] |
| Feb16-12, 11:20 PM | #5 |
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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). |
| Feb18-12, 04:05 PM | #6 |
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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 |
| Feb20-12, 03:24 PM | #7 |
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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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