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Variation of parameters

  1. Oct 12, 2015 #1
    Given a ODE like this:

    y''(t) - (a + b) y'(t) + (a b) y(t) = x(t)

    The general solution is: y(t) = A exp(a t) + B exp(b t) + u(t) exp(a t) + v(t) exp(b t)

    So, for determine u(t) and v(t), is used the method of variation of parameters:
    [tex]
    \begin{bmatrix}
    u'(t)\\
    v'(t)\\
    \end{bmatrix}
    =
    \begin{bmatrix}
    y_1(t) & y_2(t) \\
    y_1'(t) & y_2'(t) \\
    \end{bmatrix}^{-1}
    \begin{bmatrix}
    0\\
    x(t)\\
    \end{bmatrix}[/tex] Where:

    y1(t) = exp(a t)
    y2(t) = exp(b t)

    So, my question is: AND IF the matrix equation above woud be like this:
    [tex]\begin{bmatrix}
    u'(t)\\
    v'(t)\\
    \end{bmatrix}
    =
    \begin{bmatrix}
    y_1(t) & y_2(t) \\
    y_1'(t) & y_2'(t) \\
    \end{bmatrix}^{-1}
    \begin{bmatrix}
    x_1(t)\\
    x_2(t)\\
    \end{bmatrix}[/tex]

    How would be the right side of the ODE for matrix equation above?

    Would be like this:
    y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t) + x2(t)

    Or like this:
    y''(t) - (a + b) y'(t) + (a b) y(t) = x1(t)
    y''(t) - (a + b) y'(t) + (a b) y(t) = x2(t)

    Or other form?
     
  2. jcsd
  3. Oct 13, 2015 #2

    Mark44

    Staff: Mentor

    I don't see how you could get the matrix equation you show, with ##
    \begin{bmatrix}
    x_1(t)\\
    x_2(t)\\
    \end{bmatrix}## on the right. The zero term in the
    ##\begin{bmatrix}
    0\\
    x(t)\\
    \end{bmatrix}##
    vector comes from the homogeneous equation, which in this context is y'' -(a + b)y' + (ab)y = 0. The x(t) term in that vector comes from the related nonhomogeneous equation, which is y'' -(a + b)y' + (ab)y = x(t).
     
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