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System of differential equations

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of

    [tex]\dot{x}[/tex] = x + e^{2t}p [/tex]
    [tex]\dot{p} = 2e^{-2t}x - p [/tex]

    2. Relevant equations

    3. The attempt at a solution
    I know how to solve systems with constant coefficients using eigenvalues and eigenvectors, but what should I do in this case?
  2. jcsd
  3. Aug 10, 2010 #2
    Hello Max. Here's a challenging way: try it with power series. I did for the IVP:

    [tex]x'=x+e^{2t}p,\quad x(0)=1[/tex]

    [tex]p'=2e^{-2t}x-p,\quad p(0)=1[/tex]

    and obtained the plot below which agrees nicely with the numerical results cus' I actually plotted them both and they're right on top of one another. Also,I'm no expert so may be an easier way to solve this. :)

    Attached Files:

    Last edited: Aug 10, 2010
  4. Aug 12, 2010 #3
    Thanks, but we have not learned power series, I looked up the method, but I can't solve the system. Did you find the solution?
    The book says
    [tex] x = Ae^{(1+\sqrt{2})t} + Be^{(1-\sqrt{2})t} [/tex]
    [tex] p = A\sqrt(2)e^{(\sqrt{2}-1)t} -B\sqrt{2}e^{(-\sqrt{2}-1)t} [/tex]
    Last edited: Aug 12, 2010
  5. Aug 12, 2010 #4
    Well not that solution but a purist would say the power series is the solution although not in closed form like the above. I don't know how to arrive at your solution.
  6. Aug 12, 2010 #5


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    Find the eigenvalues and eigenvectors of [[1,exp(2t)],[2*exp(-2t),-1]]. The eigenvectors will be functions of t, but that shouldn't stop you from treating this the same way you would if the matrix were numerical.
  7. Aug 12, 2010 #6
    Ok. I see now. Thank you. I assume Max got it too. Personally I think Max should work out all the details in nice pretty latex and post it but I don't want to bug him about it. :)
  8. Aug 15, 2010 #7
    [tex] (1-\lambda)(-1-\lambda) - e^{2t}2e^{-2t} = 0 [/tex]
    [tex] \lambda = \pm- \sqrt{3} [/tex]

    for [tex] \lambda = \sqrt(3) [/tex]

    [tex] (1-\sqrt{3})x + e^{2t}y = 0 [/tex]

    [tex] v_1 = \begin{array}{c} 1\\ (\sqrt{3}-1})e^{-2t}\\ [/tex]

    for [tex] \lambda = -\sqrt{3} [/tex]
    [tex] (1+ \sqrt{3})x + e^{2t}y = 0 [/tex]
    [tex] v_2 = \begin{array}{c} 1\\ (-\sqrt{3}-1})e^{-2t}\\ [/tex]

    [tex] x = Ae^{\sqrt{3t}} + Be^{-\sqrt{3t} [/tex]
    [tex] p = A(\sqrt{3}-1)e^{(-2+\sqrt{3})t} - B(\sqrt{3}+1)e^{(-2-\sqrt{3})t} [/tex]

    Not sure what I did wrong, but tried calculating the eigenvalue several times, but that is not important. Thanks again to both of you for all the help.
  9. Aug 15, 2010 #8


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    A somewhat more primitive way: differentiate the first equation again to get
    [tex]x''= x'+ 2e^{2t}p+ e^{2t}p'[/tex]
    From the second equation,
    [tex]p'= 2e^{-2t}x- p[/tex]
    Putting that in,
    [tex]x''= x'+ 2e^{2t}p+ 2x- e^{2t}p= x' + 2x[/tex]

    [tex]x''- x' - 2x= 0[/tex]
    for x(t).
    Again, from the first equation,
    [tex]e^{2t}p= x'- x[/tex]
    so that
    [tex]p= e^{-2t}(x'- x)[/tex].
  10. Aug 15, 2010 #9


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    Your solution is correct, the book is wrong. Try to plug in back into the original equations
  11. Aug 15, 2010 #10


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    A term e2tp is missing.

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