Qualitative Techniques: Slope Fields
A differentiable function--and the solutions to differential equations better be differentiable--has tangent lines at
every point. Let's draw small pieces of some of these tangent lines of the function
:
Slope fields (also called vector fields or direction fields) are a tool to graphically obtain the solutions to a first
order differential equation. Consider the following example:
The slope, y'(x), of the solutions y(x), is determined once we know the values for x and y , e.g., if x=1 and y=-1,
then the slope of the solution y(x) passing through the point (1,-1) will be
. If we graph
y(x) in the x-y plane, it will have slope 2, given x=1 and y=-1. We indicate this graphically by inserting a small
line segment at the point (1,-1) of slope 2.
1
Thus, the solution of the differential equation with the initial condition y(1)=-1 will look similar to this line
segment as long as we stay close to x=-1.
Of course, doing this at just one point does not give much information about the solutions. We want to do this
simultaneously at many points in the x-y plane.
We can get an idea as to the form of the differential equation's solutions by " connecting the dots." So far, we
have graphed little pieces of the tangent lines of our solutions. The " true" solutions should not differ very much
from those tangent line pieces!
Let's consider the following differential equation:
Here, the right-hand side of the differential equation depends only on the dependent variable y, not on the
independent variable x. Such a differential equation is called autonomous. Autonomous differential equations
are always separable.
2
Autonomous differential equations have a very special property; their slope fields are horizontal-shiftinvariant, i.e. along a horizontal line the slope does not vary.
Try it yourself!
What is special about the solutions to an autonomous differential equation?
Here is an example of the logistic equation which describes growth with a natural population ceiling:
Note that this equation is also autonomous!
3
The solutions of this logistic equation have the following form:
As a last example, we consider the non-autonomous differential equation
Now the slope field looks slightly more complicated.
4
Here is the same slope field again. What is special about the points on the red parabola?
Slope Fields: Answer to Example 1
Example: Consider the autonomous differential equation
,
5
where the graph of f(y) is given by
1.
Sketch the Slope Fields of this differential equation.
(Hint: the graph of the solutions and the graph of f(y) are two different entities.)
2.
Sketch the graph of the solution to the IVP
Find the limit
.
3.
Sketch the graph of the solution to the IVP
Find the limit
.
Answer:
1.
Since we do not know the function f(y), we will only be able to sketch the slope fields. This will give us
an idea about the behavior of the solutions. Therefore, we should be looking for the critical solutions
(given by the roots of f(y)=0), and the sign of f(y) which will give the variation of the solutions. Note
that we should be careful not to mix between the graph of f(y) and the graphs of the solutions y(t).
So, according to the graph of f(y), the critical solutions are y= -1, y=0, and y=1. Using the sign of f(y),
we conclude that

the solutions located in the region y<-1 are decreasing,
6



the solutions located in the region -1 < y<0 are increasing,
the solutions located in the region 0<y<1 are decreasing,
the solutions located in the region 1 < y are increasing.
The sketch of the slope fields is given below.
2.
Using the slope fields, we sketch the graph of the solution satisfying the initial condition y(0) = 0.5.
Clearly, we have
3.
7
Using the slope fields, we sketch the graph of the solution satisfying the initial condition y(0) = -0.5.
Clearly, we have
Slope Fields: Answer to Example 2
Example: Consider the autonomous differential equation with the initial condition
where the graph of f(y) is given below
8
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
4.
Same as in 3., if y(2)=1, that is find
Solution:
1.
Here is a picture of the phase line (with the slope field):
9
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
will be equal to 3.
3.
Since y=1 is an equilibrium point, the solution will be constant, in particular
will be equal to 1.
Source: sosmath.com
10