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Solving a system of ODEs

  1. Nov 20, 2013 #1
    Given the matrix [itex]b=\begin{pmatrix}-1&0&-1\\-4&3&-1\\0&0&-2\end{pmatrix}[/itex] decide if the system of ODEs, [itex]\frac{dx}{dt}=Bx[/itex] is decoupled. If yes find the general solution x=xh(t)

    2. Relevant equations



    3. The attempt at a solution
    I would say the matrix is decoupled since the second equation involving [itex]2x[/itex]2(t) can be solved without the other two equations. Then the third equation can be solved without knowing [itex]x[/itex]1(t). We have:
    [itex]
    x'_1 = -x_1 - x_3 \\
    x'_2 = -4x_1 + 3x_2 - x_3 \\
    x'_3 = -2x_3
    [/itex]
    Im not sure where to go from here.
     
  2. jcsd
  3. Nov 20, 2013 #2

    Filip Larsen

    User Avatar
    Gold Member

    You may want to look into matrix diagonalization [1]. If B can written as A D A-1, where D is a diagonal matrix, can you then use this to rewrite you ODE system to a new uncoupled variable basis?

    [1] http://en.wikipedia.org/wiki/Diagonalizable_matrix
     
  4. Nov 20, 2013 #3
    I have found the diagonal matrix,
    [tex]D=
    \begin{pmatrix}
    -1 & 0 & 0\\
    0 & 3 & 0\\
    0 & 0 & -2
    \end{pmatrix}
    [/tex]
    I thought the matrix B was already uncoupled though. Is this not the case?
     
  5. Nov 20, 2013 #4

    Mark44

    Staff: Mentor

    ???
    Do you mean the third equation? It involves only x3' and x3.
    The system of equations was not uncoupled. The purpose of finding a diagonal matrix that is similar to B gives you a system that is uncoupled. In an uncoupled system, each equation involves only a single variable and its derivative.
     
  6. Nov 20, 2013 #5
    That was an error on my part, what I meant was the matrix is decoupled since the second equation involving [itex]-2x_3(t)[/itex] can be solved without the other two equations and then we can solve for [itex]x_1(t)[/itex] without knowing [itex] x_2(t)[/itex]
     
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