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Laplace transform of a function squared, help with this system

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Laplace transform to the system:

    [itex] \frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt} [/itex]

    [itex] x(0) = 2 ; y(0) = 3 [/itex]
    3. The attempt at a solution

    I've tried everything on this one. I first solved [itex] \frac{dy}{dt} + 6y = 2\frac{dy}{dt} [/itex] and I got [itex] y = 3e^{6t} [/itex].

    Next I tried writing it:

    [itex] 36e^{6t} = 3 \frac{d}{dt}(\frac{x^2}{2}) - \frac{dx}{dt} [/itex] so that I could use the identity of the laplace transform of derivatives. That still leaves me with trying to find the transform of x2(t)...

    So then I tried

    [itex] 36e^{6t} dt = 3x - 1 dx [/itex]

    and integrating, but this brings me to the same problem.

    I can't either figure out how to solve it without using laplace transform, so I'm really stuck. What am I doing wrong???
     
    Last edited: Nov 24, 2012
  2. jcsd
  3. Nov 24, 2012 #2

    LCKurtz

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    Everything except what you were asked to do. Start by taking the Laplace transforms of the original equations to get equations involving ##X(s)## and ##Y(s)##.
     
  4. Nov 24, 2012 #3
    Ok, I'm still not sure how that changes anything...

    [itex](s+6)Y(s) - 3 = 3L(\frac{dx}{dt}x) - sX(s) +2 = 2sY(s) -6 [/itex]

    I could solve for that middle transform, but replacing it into another equation will just give me 0=0...

    What now?
     
  5. Nov 24, 2012 #4

    LCKurtz

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    You have the "system" as$$
    \frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt}$$
    I apparently don't know what "system" you are thinking of because that isn't how you normally write one.

    I read that to mean this pair of equations:$$
    \frac{dy}{dt} + 6y = \frac{dx}{dt}$$ $$
    3x - \frac{dx}{dt} = 2\frac{dy}{dt}$$
     
  6. Nov 24, 2012 #5
    Ok well that would be much easier to solve. I guess I'll assume there's a typo in the question because I was asking myself the same thing.
    Glad to know I'm not insane after all.
     
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