Question

How does one find all of the equilibria of an ODE?

Answer 1

so the system in question is \[[\begin{matrix} x' \\ y'\end{matrix}]=[\begin{matrix} x-y \\ 2-x^2-y^2\end{matrix}]\]

Answer 2

do I just find eigenvalues?

Answer 3

yes u do.

Answer 4

ok, and do I find the eigenvalues of this matrix? \[[\begin{matrix} 1&-1\\-2x&-2y\end{matrix}]\]

Answer 5

siths is good at this :)

Answer 6

i know :/

Answer 7

i haven't done much on ODE :/ heheh

Answer 8

@perl

Answer 9

@wio any talent here?

Answer 10

@SithsAndGiggles !!!!!!

Answer 11

The equilibrium points are the points for which \({\bf x}'=\vec{0}\), as in both \(x'\) and \(y'\) are \(0\).

Answer 12

This means you're solving the system
\[\begin{cases}x-y=0\\2-x^2-y^2=0\end{cases}\]

Answer 13

Can't help have weird headache, but @rational might help

Answer 14

ok, I got it! Thanks guys!

Answer 15

yw