Slope Fields: Answer to Example 2

Example: Consider the autonomous differential equation with the initial condition

displaymath29

where the graph of f(y) is given below


1.
Draw the phase line for this differential equation, and classify the equilibrium points (critical points) as sinks, sources, or nodes.
2.
Give a rough sketch of the slope field for this differential equation, and draw a few solutions into the slope field.
3.
Consider the solution to the differential equation which satisfies the initial condition y(1)=2. Find

displaymath35

4.
Same as in 3., if y(2)=1, that is find

displaymath35


Solution:

1.
Here is a picture of the phase line (with the slope field):

The equilibrium points are y=0, y=1 and y=3.

The equilibrium point y=0 is a source, y=1 is a node, and y=3 is a sink.

2.
Below you can see the same picture with some solutions (in blue):

3.
Since y(1)=2, the solution will increase over time and eventually approach the sink at y=3. Thus displaymath35 will be equal to 3.

3.
Since y=1 is an equilibrium point, the solution will be constant, in particular displaymath35 will be equal to 1.

[Differential Equations] [Slope Fields] [Previous Example]
[Geometry] [Algebra] [Trigonometry ]
[Calculus] [Complex Variables] [Matrix Algebra]

S.O.S. MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Author: Helmut Knaust
Last Update 6-22-98

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour