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Repeated eigenvalues

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    I want to solve this system

    x' = [itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex]x + [itex]\left( \begin{array}\\ t \\ 2t \end{array} \right)[/itex]

    2. Relevant equations

    3. The attempt at a solution

    i found the eigenvalues to both be 5. The eigenvector is (1,-2) and the generalized eigenvector i found to be (0,1)

    I'm confused on how to solve the non-homogeneous part, since I got repeated eigenvalues. Is the procedure the same? Can I use, say, variation of parameters to solve this?
    Last edited: May 7, 2016
  2. jcsd
  3. May 7, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I apologize for being dense, but I don't understand this notation. Is the expression on the right a 2x2 matrix? If so, it would make it more readable if you used Tex notation:

    [itex]\left( \begin{array}\\ 7 & 1 \\ -4 & 3 \end{array} \right)[/itex]
  4. May 7, 2016 #3
    Yes, that's right. I'm not really familiar with Latex, I'll have to read up on it. I've edited my post, thanks
  5. May 7, 2016 #4


    Staff: Mentor

    As written, this makes no sense. You can't add a 2 x 2 array to a 2 x1 array (column matrix).

    This would make more sense if it were something like this:
    ##\vec{x'} = \begin{bmatrix} 7 & 1 \\ -4 & 3 \end{bmatrix}\vec{x} + \begin{bmatrix} 1 \\ 2 \end{bmatrix}##

    Here ##\vec{x}## means ##\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}## and similar for its derivative.
  6. May 7, 2016 #5
    Yes that's that I meant, sorry for the confusion. So could i set up a fundamental matrix with the homogeneous solutions and solve the non-homogeneous system with, say, variation of parameters? I tried and this is what I get:

    x(t) = c1e5t[itex]\left( \begin{array}\\ -1 \\ 2 \end{array} \right)[/itex] + c2e5t[itex]\left( \begin{array}\\ -t-1/2 \\ 2t \end{array} \right)[/itex] - (t/25)[itex]\left( \begin{array}\\ 1 \\ 18 \end{array} \right)[/itex] + (1/125)[itex]\left( \begin{array}\\ 3 \\ -26 \end{array} \right)[/itex]

    I checked my solution on wolfram and it's slightly different, which annoys me.
  7. May 7, 2016 #6


    Staff: Mentor

    Just check that your solution satisfies the system of diff. equations.
  8. May 7, 2016 #7

    Ray Vickson

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    Science Advisor
    Homework Helper

    Isn't there a factor ##x## missing on the right? Should you not be dealing with the system
    [tex] \pmatrix{x_1'\\x_2'} = \pmatrix{7 & 1 \\ -4 & 3} \pmatrix{x_1\\x_2} + \pmatrix{t \\2t} ? [/tex]
    You can either plug in the matrix exponential in the solution
    [tex] {\mathbf{x}} = e^{At} \int_0^t e^{-A \tau} {\mathbf{f}}(\tau) \, d \tau [/tex]
    to your equation ##{\mathbf{x}}'(t) = A {\mathbf{x}}(t) + {\mathbf{f}}(t)##, or else use the Laplace-transform method.

    As for the matrix exponential: you have ##A = P J P^{-1}##, where ##J## is the Jordan canonical form of ##A##:
    [tex] J = \pmatrix{5 & 1 \\0 & 5} [/tex]
    Furthermore, for any scalar ##x## we have ##e^{Ax} = P e^{Jx} P^{-1}##, and ##e^{Jx}## is easy to determine; see
    webpages on matrix exponentials.
  9. May 7, 2016 #8
    I just checked, and its satisfied. Thanks!
  10. May 7, 2016 #9
    Yes, I forgot the vector 'x' next to the coefficient matrix. Thanks for your tips, I actually did the variation of parameters method and everything worked out after I checked my solution as Mark44 suggested.
  11. May 7, 2016 #10
    Thanks everyone, I have no further questions.
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