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From a System of 1st ODE to a 2nd ODE and back to the system of 1st ODEs

  1. Apr 17, 2010 #1
    I have a somewhat theoretical question regarding Differential Equations:

    How can we reconcile the fact that if I go from let's say this system of 1st ODE

    x' = 2y-x
    y' = -x+y

    to a 2nd ODE "using x(t) instead of y(t)" we get: x" + x =0

    then back to a system of 1st ODE:

    letting y=x' and then from x"=-x we get x"=-y' so y'=-x.

    So now our new 1st ODE system has the following equations instead


    which is different from our original system:

    x' = 2y-x
    y' = -x+y

    Is there a good Mathematical explanation to this?
  2. jcsd
  3. Apr 18, 2010 #2


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    I don't know what you mean by "using x(t) instead of y(t)". And the x satisfying the system of equations definitely does NOT satisfy x"= x= 0. What we can do is differentiate the first equation again to get x"= 2y'- x. Then, since y'= -x+ y, x"= 2(-x+ y)+-x= -3x+ 2y. Since, from the first equatioj, 2y= x'+ x, x"= -3x+ (x'+ x)= x'- 2x or x"- x'+ x= 0.

    Yes, you made a mistake!
    However, your question is still valid.

    If, in the differential equation x"- x'+ x= 0, you let y= x', then x"= y' and so you have y'- y+ x= 0 or y'= y- x.

    Now the system is

    x'= y,
    y'= y- x.

    Yes, that is different from the original system because this "y" is different from the y in the original system.
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