1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Homework Helper

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


    User Avatar
    Homework Helper

    A term e2tp is missing.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook