Quadratic Formula (Equilibrium Points)

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  • #1
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Homework Statement



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]?

Homework Equations



The quadratic equation:

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

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.
 
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Answers and Replies

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

ehild
 
  • #3
Ray Vickson
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Homework Statement



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]?

Homework Equations



The quadratic equation:

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

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.

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
 
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  • #4
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Ah, I didn't see that :redface:

Thank you very much guys.
 

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