a. Consider the equation dy = k(1 – y)? where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution dt) = 1. b. Sketch ly) versus y. Show that y is increasing as a function of t for y< 1 and also for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, ɖt) = 1 is semistable. c. Solve the equation subject to the initial condition y(0) = yo and confirm the conclusions reached in part b.
a. Consider the equation dy = k(1 – y)? where k is a positive constant. Show that y = 1 is the only critical point, with the corresponding equilibrium solution dt) = 1. b. Sketch ly) versus y. Show that y is increasing as a function of t for y< 1 and also for y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, ɖt) = 1 is semistable. c. Solve the equation subject to the initial condition y(0) = yo and confirm the conclusions reached in part b.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Problem 6
a. Consider the equation
dy
%3D
dt
= k(1 – y)²
where k is a positive constant. Show that y = 1 is the only critical point, with the
corresponding equilibrium solution (1) = 1.
b. Sketch fly) versus y. Show that y is increasing as a function of t for y < 1 and also for
y > 1. The phase line has upward-pointing arrows both below and above y = 1. Thus
solutions below the equilibrium solution approach it, and those above it grow farther
away. Therefore, (t) = 1 is semistable.
c. Solve the equation subject to the initial condition y(0)
yo and confirm the
conclusions reached in part b.
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