[6] An equation dy dt = f(y) has the following phase portrait. 2 y (a) Find all equilibrium solutions. (b) Determine whether each of the equilibrium solutions is stable, asymptotically stable or unstable. (c) Graph the solutions y(t) vs t, for the initial values y(1.4) = 0, y(0) = 0.5, y(0) = 1, y(0) = 1.1, y(0) = 1.5, y(-0.5) = 1.5, y(0) = 2, y(0) = 2.5, y(0) = 3, y(0) = 3.5, y(0) = 4, y(0) = 4.5, y(-1) = 4.5. (Without further quantitative information about the equation and the solution formula, it's clearly impossible to draw accurate graphs of y(t) vs t. Here, try to sketch graphs qualitatively to show the correct dynamic properties. The point is that a great deal of info about solution dynamics can be read off from one simple figure of phase portrait.)

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter14: Discrete Dynamical Systems
Section14.3: Determining Stability
Problem 1YT
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[6] An equation
dt
=
f(y) has the following phase portrait.
2
Y
(a) Find all equilibrium solutions.
(b) Determine whether each of the equilibrium solutions is stable, asymptotically stable
or unstable.
(c) Graph the solutions y(t) vs t, for the initial values y(1.4) = 0, y(0) = 0.5, y(0) = 1,
y (0) = 1.1, y(0) = 1.5, y(-0.5) = 1.5, y(0) = 2, y(0) = 2.5, y(0) = 3, y(0) = 3.5,
y(0) = 4, y(0) = 4.5, y(-1) = 4.5. (Without further quantitative information about
the equation and the solution formula, it's clearly impossible to draw accurate graphs
of y(t) vs t. Here, try to sketch graphs qualitatively to show the correct dynamic
properties. The point is that a great deal of info about solution dynamics can be
read off from one simple figure of phase portrait.)
Transcribed Image Text:[6] An equation dt = f(y) has the following phase portrait. 2 Y (a) Find all equilibrium solutions. (b) Determine whether each of the equilibrium solutions is stable, asymptotically stable or unstable. (c) Graph the solutions y(t) vs t, for the initial values y(1.4) = 0, y(0) = 0.5, y(0) = 1, y (0) = 1.1, y(0) = 1.5, y(-0.5) = 1.5, y(0) = 2, y(0) = 2.5, y(0) = 3, y(0) = 3.5, y(0) = 4, y(0) = 4.5, y(-1) = 4.5. (Without further quantitative information about the equation and the solution formula, it's clearly impossible to draw accurate graphs of y(t) vs t. Here, try to sketch graphs qualitatively to show the correct dynamic properties. The point is that a great deal of info about solution dynamics can be read off from one simple figure of phase portrait.)
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