- #1
BrainHurts
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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
[itex]x' = x + y^2 [/itex]
[itex]y' = -y[/itex]
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 [itex]X = c_1 e^t (1,0) + c_2e^{-t}(0,1) = (c_1e^{t},c_2e^{-2}) [/itex]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 [itex]y' = -y \rightarrow y = y_0e^{-t}[/itex], we solve [itex]x' = x \rightarrow x = x_0e^t[/itex]
Since we have the solution to the homogenous eig buation and we'll say that the particular solution [itex] x_p(t) = Ce^{-2t}, x_p'(t) = -2Ce^{-2t},[/itex] we'll get that [itex]C = -\dfrac{1}{3}y_0^2[/itex], so we'll get that [itex]x(t) = ce^t - \dfrac{1}{3}y_0^2e^{-2t}[/itex]
Not seeing how they got that
[itex]x(t) = \left( x_0 + \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]
[itex]y(t) = y_0e^{-t} [/itex]
why isn't it [itex]x(t) = \left( x_0 - \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]?
This seems like a big deal because we're getting ready to do a change of coordinates.
I'm not understanding everything.
So given the system
[itex]x' = x + y^2 [/itex]
[itex]y' = -y[/itex]
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 [itex]X = c_1 e^t (1,0) + c_2e^{-t}(0,1) = (c_1e^{t},c_2e^{-2}) [/itex]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 [itex]y' = -y \rightarrow y = y_0e^{-t}[/itex], we solve [itex]x' = x \rightarrow x = x_0e^t[/itex]
Since we have the solution to the homogenous eig buation and we'll say that the particular solution [itex] x_p(t) = Ce^{-2t}, x_p'(t) = -2Ce^{-2t},[/itex] we'll get that [itex]C = -\dfrac{1}{3}y_0^2[/itex], so we'll get that [itex]x(t) = ce^t - \dfrac{1}{3}y_0^2e^{-2t}[/itex]
Not seeing how they got that
[itex]x(t) = \left( x_0 + \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]
[itex]y(t) = y_0e^{-t} [/itex]
why isn't it [itex]x(t) = \left( x_0 - \dfrac{1}{3}y_0^2 \right) e^t - \dfrac{1}{3}y_0^2e^{-2t} [/itex]?
This seems like a big deal because we're getting ready to do a change of coordinates.