6. Without solving for y(t), find the steady states of the logistic equation 4y - y², and determine whether each steady state is stable, unstable, or semi-stable. Based on this information, make approximate graphs of solutions y(t) starting from y(0) = −1, y(0) = 2, and y(0) = 5. You can graph all the solutions on the same axes (but label them clearly). 7. Same question for and y(0) = 5. dy dt 1 dy = y(y-2)(y-4), and starting points y(0) = −1, y(0) = 1, y(0) = 3, dt
6. Without solving for y(t), find the steady states of the logistic equation 4y - y², and determine whether each steady state is stable, unstable, or semi-stable. Based on this information, make approximate graphs of solutions y(t) starting from y(0) = −1, y(0) = 2, and y(0) = 5. You can graph all the solutions on the same axes (but label them clearly). 7. Same question for and y(0) = 5. dy dt 1 dy = y(y-2)(y-4), and starting points y(0) = −1, y(0) = 1, y(0) = 3, dt
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
dolve question 7 please

Transcribed Image Text:dy
4y - y²,
=
6. Without solving for y(t), find the steady states of the logistic equation
dt
and determine whether each steady state is stable, unstable, or semi-stable. Based on
this information, make approximate graphs of solutions y(t) starting from y(0) = −1,
y(0) = 2, and y(0) = 5. You can graph all the solutions on the same axes (but label them
clearly).
7. Same question for
and y(0) = 5.
1
-
dy
= y(y− 2)(y—4), and starting points y(0) = −1, y(0) = 1, y(0) = 3,
dt
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