# Quadratic Formula (Equilibrium Points)

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

## Homework Equations

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

## The Attempt at a Solution

Setting the DE equalt to 0:

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

Using the quadratic formula:

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

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

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

Last edited by a moderator:

## Answers and Replies

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

ehild

Ray Vickson
Homework Helper
Dearly Missed

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

## Homework Equations

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

## The Attempt at a Solution

Setting the DE equalt to 0:

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

Using the quadratic formula:

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

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

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

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:
Ah, I didn't see that

Thank you very much guys.