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Differential Equations - First Order Systems

  1. Jun 27, 2007 #1

    dx / dt = 3*x + y

    dy / dt = -y

    As for solving this, here is what I've got so far:

    Since dy/dt is separable, I found that dy / y = -dt, integrated, and solved explicitly for y:

    y = Ce^-t

    I then plugged Ce^-t for y in the dx / dt portion of the system and found that:

    dx/dt - (3*x) = Ce^-t

    Then used Integrating Factor method, found u(t) = e^(-3t), multiplied to both sides and integrated and solved explicitly for x.

    x = (-1/3)C + Ce^(3t)

    For the second portion of the question, it asks to verify that the solution is correct.

    To do that, I plugged in the x and y solutions from part 1, as well as took derivatives of x and y and plugged them in as well.

    dy / dt = -y, -Ce^-t = -Ce^-t

    However, the problem I am having is that when I plug in for the dx / dt equation, I get down to t = ln1 = 0.

    I don't understand if this is wrong or what, as the y solution works but the x one doesn't seem to come to any reasonable conclusion.

    Any help in understanding these results, or in finding mistakes would be much appreciated.

  2. jcsd
  3. Jun 27, 2007 #2


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    so far so good
    I don't get this. I get [itex] x(t) = - \frac{1}{4C} e^{-Ct} [/itex].
    You have done a mistake either in integrating or in solving for x.
  4. Jun 27, 2007 #3
    hrm, I've re-done the integration multiple times and keep getting the same thing.

    I am integrating this:

    (x*e^(-3t))' = C*e^(-4t)

    After integrating I get:

    x*e^(-3t) = ( -C / [3e^(3t)] ) + C

    Divide out the u(t) = e^(-3t) and end up with this:

    x = (-C / 3) + C*e^(3t)

    I've also done this integration in both Maple and Mathematica and got the same thing as what I got by hand.

    So...I'm not sure its the integration or solving for x, I'm baffled.
  5. Jun 27, 2007 #4


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    Sorry, I had made a few mistakes .

    I did forget the constant of integration (sorry) and I made typos by putting the C in the denominator instead of the numerator and putting it in the exponential (it's getting too late for me to do simple maths!).
    here is my work

    [itex] d/dt(x e^{-3t}) = C e^{-4t} \rightarrow x e^{-3t} = -\frac{C}{4} e^{-4t} + K [/itex]

    wher I use K for the constant of integration to distinguish it from C. Solving, we get

    [itex] x = -\frac{C}{4} e^{-t} + K e^{3t} [/itex]
    Last edited: Jun 27, 2007
  6. Jun 27, 2007 #5


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    It's your integration of e^(-4t) which is incorrect.
    You may check that my answer satisfies the DE for x(t)


  7. Jun 27, 2007 #6
    Haha, I just caught my error. Ima give it a whirl now with the CORRECT integral, hehe.

    Thank you nrged

    *edit* - got it, much thanks to you nrged
    Last edited: Jun 27, 2007
  8. Jun 27, 2007 #7


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    :biggrin:Good! And you are welcome. Sorry for my mistakes.

    Best luck!
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