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Quadratic Formula (Equilibrium Points)

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data

    The following is part of a solution to a problem about finding equiblirum points of a differential equation and sketching its bifurcation diagram

    http://img32.imageshack.us/img32/9194/63226925.jpg [Broken]

    How did they get [itex]y= -1 \pm \sqrt{1- \mu}[/itex]?

    2. Relevant equations

    The quadratic equation:

    [itex]\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

    3. The attempt at a solution

    Setting the DE equalt to 0:

    [itex]\frac{dy}{dt} = \mu +2y+y^2 = 0[/itex]

    Using the quadratic formula:

    [itex]y = \frac{-2 \pm \sqrt{4-4 \mu}}{2}[/itex]

    [itex]= -1 \pm \frac{\sqrt{4-4 \mu}}{2}[/itex]

    So how did they get that answer? :confused: Any explanation is appreciated.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 21, 2012 #2

    ehild

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    [itex]\sqrt{4-4\mu}=\sqrt{4(1-\mu)}=2\sqrt{(1-\mu)}[/itex]

    ehild
     
  4. Apr 21, 2012 #3

    Ray Vickson

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    This type of thing happens over and over again, so you should learn it once and for all: [tex] \frac{\sqrt{X}}{a} = \sqrt{\frac{X}{a^2}}[/tex] if [itex] a > 0.[/itex]

    RGV
     
    Last edited by a moderator: May 5, 2017
  5. Apr 21, 2012 #4
    Ah, I didn't see that :redface:

    Thank you very much guys.
     
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