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Solving differential equations through matrix

  1. May 9, 2007 #1

    I'd like to know how to solve the following equation with matrix, if possible at all:

    d ( x^2 ) / dt^2 + w^2 x = 0

    I know how to solve it without having to use a matrix, but I heard it is possible to do it with matrix. How about doing it? How is this method called?

  2. jcsd
  3. May 9, 2007 #2


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    Is that d ( x^2 ) / dt^2 + w^2 x = 0
    or d^2 x/dt^2+ w^2x= 0?

    I'm going to assume it is the latter. Define y= dx/dt so dy/dt= d^2x/dt^2 and the equation becomes dy/dt= -w^2x. You now have the two equations dx/dt= y and dy/dt= -w^2x. If you write
    [tex]X= \left(\begin{array}{c} x \\ y\end{array}\right)[/tex]
    Then the two equations become the single matrix equation
    [tex]\frac{dX}{dt}= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)X[/tex]
    To solve that, find the eigenvalues of the coefficient array (they are [itex]\pm w i[/itex]). The general solution then can be written as exponentials of those eigenvalues times t or, since they are imaginary, sine and cosine.
  4. May 9, 2007 #3
    it's the latter one, as you thought. thanks by the response!
  5. May 10, 2007 #4
    Hi again

    Sorry but I could not follow the step X=(x, y) and why then it becomes the next equation.
  6. May 10, 2007 #5


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    [tex]\left(\begin{array}{c}\frac{dx}{dt} \\ \frac{dy}{dt}\end{array}\right)= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)[/tex]

    Do you see how the matrix multiplication on the right works out?
  7. May 10, 2007 #6

    D H

    Staff: Mentor

    Taking this one step further, define [itex]\mathbf X[/itex] and [itex]\mathbf A[/itex] as
    \mathbf X &\equiv \bmatrix x\\y\endbmatrix \\[12pt]
    \mathbf A &\equiv \bmatrix 0&&1\\-w^2&&0\endbmatrix


    [tex]\frac {d\mathbf X}{dt}= \mathbf A\mathbf X[/tex]

    If [itex]\mathbf X[/itex] and [itex]\mathbf A[/itex] were scalars, the solution to the above would be the exponential

    [tex]\mathbf X = e^{\mathbf A t}\mathbf X|_{t=0}[/tex]

    The series expansion of the exponential function works for matrices as well as scalars (for example, see http://mathworld.wolfram.com/MatrixExponential.html or http://www.sosmath.com/matrix/expo/expo.html).

    In this case,

    [tex]\mathbf A^2 = -w^2 \mathbf I[/tex]

    where [itex]\mathbf I[/itex] is the identity matrix. Thus

    (\mathbf A t)^{2n} &= (-1)^n (wt)^{2n} \mathbf I \\
    (\mathbf A t)^{2n+1} &= (-1)^n \frac 1 w (wt)^{2n+1} \mathbf A\\

    The matrix exponential is thus

    e^{\mathbf A t} &=\sum_{n=0}^{\infty} \frac {(\mathbf At)^n}{n!}
    &= \sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n}}{(2n)!} +
    \sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n+1}}{(2n+1)!}
    &= \sum_{n=0}^{\infty} (-1)^n\frac {(wt)^{2n}}{(2n)!} \mathbf I +
    \frac 1 w\sum_{n=0}^{\infty}(-1)^n\frac{(wt)^{2n+1}}{(2n+1)!} \mathbf A
    &= \cos(wt) \mathbf I +\frac 1 w\sin(wt) \mathbf A
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