1. Apr 21, 2012

### roam

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 $y= -1 \pm \sqrt{1- \mu}$?

2. Relevant equations

$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

3. The attempt at a solution

Setting the DE equalt to 0:

$\frac{dy}{dt} = \mu +2y+y^2 = 0$

$y = \frac{-2 \pm \sqrt{4-4 \mu}}{2}$

$= -1 \pm \frac{\sqrt{4-4 \mu}}{2}$

So how did they get that answer? Any explanation is appreciated.

Last edited by a moderator: May 5, 2017
2. Apr 21, 2012

### ehild

$\sqrt{4-4\mu}=\sqrt{4(1-\mu)}=2\sqrt{(1-\mu)}$

ehild

3. Apr 21, 2012

### Ray Vickson

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

RGV

Last edited by a moderator: May 5, 2017
4. Apr 21, 2012

### roam

Ah, I didn't see that

Thank you very much guys.