# Equilibria in Nonlinear Systems

1. Nov 8, 2014

### BrainHurts

So I'm reading the Example on page 161 of Differential Equations, Dynamical Systems and an Introduction to Chaos by Hirsh, Smale, and Devaney.

I'm not understanding everything.

So given the system

$x' = x + y^2$
$y' = -y$

we see this is non-linear. I get it that near the origin, y^2 tends to zero "fast". So we can consider the system

x' = x

y' = -y

I see that the solution to this system is $X = c_1 e^t (1,0) + c_2e^{-t}(0,1) = (c_1e^{t},c_2e^{-2})$and we have that the y-axis is the stable line and the x-axis is the unstable line.

Here's where I'm unsure of where things are going on since we can solve $y' = -y \rightarrow y = y_0e^{-t}$, we solve $x' = x \rightarrow x = x_0e^t$

Since we have the solution to the homogenous eig buation and we'll say that the particular solution $x_p(t) = Ce^{-2t}, x_p'(t) = -2Ce^{-2t},$ we'll get that $C = -\dfrac{1}{3}y_0^2$, so we'll get that $x(t) = ce^t - \dfrac{1}{3}y_0^2e^{-2t}$

Not seeing how they got that
$x(t) = \left( x_0 + \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t}$

$y(t) = y_0e^{-t}$

why isn't it $x(t) = \left( x_0 - \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t}$?

This seems like a big deal because we're getting ready to do a change of coordinates.

2. Nov 9, 2014

### pasmith

You have $x(t) = ce^t - \frac13 y_0^2 e^{-2t}$ and you need $x(0) = x_0$. Hence $c - \frac13 y_0^2 = x_0$ so that $c = x_ 0 + \frac13 y_0^2$.

3. Nov 10, 2014

### BrainHurts

Thanks so much! I know we want to do that in order to make the linearization work correct?