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Particular solution DE

  1. Jul 29, 2006 #1
    Hi, I'm having trouble finding the particular solution of the following system.

    {\mathop x\limits^ \bullet } \\
    {\mathop y\limits^ \bullet } \\
    1 & { - 1} \\
    { - 1} & 1 \\
    x \\
    y \\
    \end{array}} \right] + \left[ {\begin{array}{*{20}c}
    2 \\
    { - 5} \\
    \end{array}} \right]

    I found the complimentary function, it had some sines and cosines in it but I don't think that matters in terms of finding the particular solution. The independent variable t isn't present anywhere in the equation so I don't need to do any differentiation. If I set x' = y' = 0 (the prime denotes differentiation wrt t) then I end up with an 'inconsistent' set of equations, namely, x - y = 2 and x - y = 5. Can someone tell me how I can find the particular solution? Thanks.
  2. jcsd
  3. Jul 30, 2006 #2
    I would be leary of your complimentary solution as a first order system should usually be exponential not trigonometric. I will help you to find the complete solution, (complimentary + particular)

    you'll want to use eigenvalues and vectors to "decouple" the equations

    the way I solved it was to let:
    {x;y}={phi}e^nt (where phis is the matrix of eigenvectors) and find the eigenvalues of the matrix [1 -1;-1 1] (the variable n is the eigenvalue)

    do this by subsituting into the original equation.

    at this point I will apologize for my unwillingness to use latex.

    the eigenvalues are n= 0 and 2

    the eigenvectors are then [1;1] and [1;-1]

    let psi be the fundamental matrix = [1 e^2t;1 -e^2t]

    now your equation is of the form:


    where f(t)=(2;5)
    at this point I will introduce the missing function u(t) needed to solve the equation.

    the solution is x=psi*u(t)
    where psi*u'(t)=f(t)

    solve for u1' and u2' and integrate to find u1 and u2

    then [x;y]=psi*[u1;u2]

    where c1 and c2 are determined by y'(0) and x'(0)

    this solution is correct, substitute it back into the original equation if you are unsure (it satisfies the heck out of it)

    hope this helps.
  4. Jul 30, 2006 #3
    by the way

    x=xc +xp

    where xp=7/2*t+0.75
    and xc=c1+c2e^2t

    y of course is separated similarly.
  5. Jul 30, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Let me clarify that: A linear system can have either exponential or trigonometric (or polynomial or some combination) solutions. In this case, since the coefficient matrix is symmetric and so has real eigenvalues, the complementary solution is exponential, not trigonometric.

    Benny, recalculate the eigenvalues! Here, the problem is that one of the eigenvalues is 0 so one of the solutions to the associated homogeneous equation is the constant solution. Try a solution of the form
    [tex]\left[ \begin{array}{c}At \\ Bt \end{array} \right][/tex]
  6. Jul 30, 2006 #5
    Yep, the complimentary function should be in terms of exponentials, I mixed signs when I calculated the determinant.

    Thanks for the suggestion HallsofIvy. Does the method of undetermined coefficients for single ODEs carry over to systems?

    Edit: Hmm...I still get down to the problem of having an inconsistent system of equations. If I assume a solution of the form (At, Bt)^T and I plug it back into the matrix equation I end up with two equations.

    A = At - Bt + 2
    B = -At + Bt - 5

    Rewriting this:

    (A-B)t + (2-A) = 0...(1)
    (B-A)t + (-5-B) = 0...(2)

    Equating coefficients I get A - B = 0 so that A = B. But I also need A = 2 meaning that B = 2. But then -5-B wouldn't be zero.
    Last edited: Jul 30, 2006
  7. Jul 30, 2006 #6


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    Staff Emeritus
    Science Advisor

    Sorry about that: you will need a particular solution of the form
    [tex]\left[ \begin{array}{c}At \\ Bt+ C \end{array} \right][/tex]
  8. Jul 31, 2006 #7
    Ok thanks I'll try that.

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