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Differential Equation stability at critical point

  1. Mar 8, 2009 #1

    JJBladester

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    Gold Member

    1. The problem statement, all variables and given/known data

    This is the autonomous differential equation: x" - 2x' + 37x = 0

    Solve the above DEQ and state whether the critical point (0,0) is stable, unstable, or semi-stable.

    2. Relevant equations

    Solution to the above DEQ is x = c1excos6x + c2exsin6x

    3. The attempt at a solution

    I worked out the solution using the quadratic formula and got roots 1[tex]\pm[/tex]6i. This gives you an [tex]\alpha[/tex] of 1 and a [tex]\beta[/tex] of 6, which yields the equation I put in part 2 above.

    From there, I read that when you get a general solution in the form x = e[tex]\alpha[/tex]t(c1cos[tex]\beta[/tex]t + c2sin[tex]\beta[/tex]t) with [tex]\alpha[/tex] < 0 and [tex]\beta[/tex] [tex]\neq[/tex]0, then you have a spiral point.

    My problem is I'm not sure how to classify the stability of the critical point (0,0).
     
  2. jcsd
  3. Mar 8, 2009 #2

    HallsofIvy

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    It's easy. Your characteristic roots have positive real part so (0, 0) is an unstable critical point. You can see why from your general solution. (0,0) (in the phase plane) is a critical point because x(t)= 0 for all t (so that x'(t)= 0 also) is itself a solution. It is an unstable critical point because if x is just slightly different from 0, that exponential means it gets larger and larger, moving rapidly away from (0,0).
     
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