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Transformation from an ODE 2nd degree to ODE

  1. Dec 2, 2005 #1
    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
     
  2. jcsd
  3. Dec 2, 2005 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct. So you have two first order equations:
    [tex]\frac{dX_1}{dt}= X_2[/tex]
    and
    [tex]\frac{dX_2}{dt}= -X_1[/tex]
    instead of the single second order equation
    [tex]\frac{d^2X}{dt^2}+ X= 0[/tex]
     
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