# Differential equation bifurcation, how to find equilibrium points?

## Homework Statement

What are the bifurcation values for the equation:

dy/dt = y^3 +ay^2

## The Attempt at a Solution

Equilibrium solutions:
y^3 + ay^2 = 0
==> y^2 (y + a) = 0
==> y = 0 (double root), or y = -a.

a = 0 is the sole bifurcation point, since

a < 0 ==> two equilibrium solutions
a = 0 ==> one equilibrium solution
a > 0 ==> two equilibrium solutions.

my question is how can you tell that a < 0 has two equilibrium solutions and a=0 has one and a>0 has two again?

epenguin
Homework Helper
Gold Member

## Homework Statement

What are the bifurcation values for the equation:

dy/dt = y^3 +ay^2

## The Attempt at a Solution

Equilibrium solutions:
y^3 + ay^2 = 0
==> y^2 (y + a) = 0
==> y = 0 (double root), or y = -a.

a = 0 is the sole bifurcation point, since

a < 0 ==> two equilibrium solutions
a = 0 ==> one equilibrium solution
a > 0 ==> two equilibrium solutions.

my question is how can you tell that a < 0 has two equilibrium solutions and a=0 has one and a>0 has two again?

What is an equilibrium solution? It is a point where dy/dt = 0. When you are exactly at such a point the system will no longer move from there, that is called equilibrium. Your algebra tells you there is only one such point (I.e. value of y) when a = 0, and two otherwise.

You should also look at in what direction the y moves when it is not at an equilibrium point to start to understand what this is for.