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Homework Help: Solution to Matrix Differential Equation

  1. Oct 11, 2008 #1
    How do I write the form of the solution to this equation:

    [tex]

    \dot{\vec{x}}(t) =
    \left [ \begin{array}{cc}
    a_{11}(t) & a_{12}(t) \\
    a_{21}(t) & a_{22}(t)
    \end{array} \right ] \vec{x}(t)

    [/tex]


    I just need to be able to write x1(t) and x2(t) so I can do the rest of the problem I'm working on. Getting this would just be a small step in my solution, but I am very rusty with my differential equations!! :(

    Initially, I thought to write:

    [tex]
    x_1(t) = \int_{t_0}^{t}x_1(\tau)a_{11}(\tau) + x_2(\tau)a_{12}(\tau)d\tau
    [/tex]


    [tex]
    x_2(t) = \int_{t_0}^{t}x_1(\tau)a_{21}(\tau) + x_2(\tau)a_{22}(\tau)d\tau
    [/tex]

    But that has the solutions with dependence on x1(t) and x2(t). That's not the way to write it, is it?

    Thanks.
     
  2. jcsd
  3. Oct 11, 2008 #2

    Mark44

    Staff: Mentor

    Let's look at your system of DEs this way:
    [tex]
    \dot{\vec{x}}(t) = A \vec{x}(t)
    [/tex]

    What you'd like is a system that looks like this:
    [tex]
    \dot{\vec{x}}(t) = D \vec{x}(t)
    [/tex]
    where D is a diagonal matrix.

    This will untangle things so that you have x1'(t) = d11 x1(t) and x2'(t) = d22 x2(t).

    These are easy to solve, since each one involves only a single variable.

    Getting the matrix D is the hard part, though, since doing this involves changing to a different basis (for R2 in your case). Without going into too many details, you'll want to find a matrix D that is similar to your original matrix A, which I'm assuming is invertible. Similarity is precisely defined this way: If A ~ D, then for some invertible matrix P, AP = PD.

    Equivalently, P[tex]^{-1}[/tex]AP = P[tex]^{-1}[/tex]PD = D.

    You will need to come up with a matrix P whose columns are the new basis, and a matrix P[tex]^{-1}[/tex], the inverse of P.

    To wind this up, the columns of P are the eigenvectors of matrix A, and it turns out that the diagonal entries of D are the eigenvalues of A.

    I hope I've given you enough to at least get you started searching for the things to learn more about. Solving a system of DEs, even the simplest possible system in two variables requires a significant amount of understanding in linear algebra.

    Mark
     
  4. Oct 12, 2008 #3
    Yes, my linear algebra is a few years out of service, but it is not non-existent. You have refreshed my memory. Thanks.
     
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