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Math Help - ODE order change

  1. #1
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    ODE order change

    Hi, I would to check if my transformation from an ODE 2nd degree to ODE 1st degree is fine:

    X''(t)+X(t)=0

    I set:
    X1(t)=X(t)
    X2(t)=X'(t)
    this implies that:
    X1'(t)=X2(t)
    so the original equation becomes:
    X2'(t)=-X1(t)

    PLease tell me if I am right?

    B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy
    Hi, I would to check if my transformation from an ODE 2nd degree to ODE 1st degree is fine:

    X''(t)+X(t)=0

    I set:
    X1(t)=X(t)
    X2(t)=X'(t)
    this implies that:
    X1'(t)=X2(t)
    so the original equation becomes:
    X2'(t)=-X1(t)

    PLease tell me if I am right?

    B
    Yes this is the standard way of reducing a second order ODE
    to a first order system. The only comment that I would make on
    this is that the original equation becomes the system:

    \left[ \begin{array}{cc}x_1(t) \\ x_2(t)\end{array}  \right]'\ =\ \left[{ \begin{array}{cc}x_2(t) \\ -x_1(t)\end{array}  \right]

    RonL
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