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Simultaneous Laplace transforms

by hurcw
Tags: laplace, simultaneous, transforms
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hurcw
#1
Feb15-12, 10:03 PM
P: 20
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.
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Some Pig
#2
Feb16-12, 05:42 AM
P: 38
[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]
hurcw
#3
Feb16-12, 08:56 AM
P: 20
Can you ellaborate a little please.
Where did this all come from?

Some Pig
#4
Feb16-12, 10:10 PM
P: 38
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]
hurcw
#5
Feb16-12, 11:20 PM
P: 20
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).
hurcw
#6
Feb18-12, 04:05 PM
P: 20
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
hurcw
#7
Feb20-12, 03:24 PM
P: 20
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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