Why is the Particular Solution Different in Nonlinear System Equilibria?

[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.

 

[pasmith]

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.

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 $c = x_ 0 + \frac13 y_0^2$.

 

[BrainHurts]

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