Slope FieldsWe will consider differential equations of theform

dydt

= f(t,y).

Example

dydt

= y − t.

A solution of this differential equation is

y = t + 1 − et.

Check: Let y be the function y = t + 1 − et.Then

dydt

= 1 − et

andy − t = (t + 1 − et ) − t = 1 − et

so dy/dt = y − t, which shows that the functiony = 1 + t − et is a solution of dy/dt = y − t.

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Here is a graph of the solution y = 1 + t − et:

-3-2-1

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y

-3 -2 -1 1 2 3t

Recall that the value of dy/dt at any givenpoint (t1,y1) on the graph of y is the slope ofthe tangent line to the graph of y at that point.For example,

• At t = −1, we have

y = −1 + 1 − e−1 = −e−1

so the point(−1,−e−1) lies on the graph ofy.At this point, we have

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dydt

= y − t

= −e−1 − (−1)= 1 − e−1

≈ 0.63212

so the slope of the tangent line to the graph ofy at the point(−1,−e−1) is approximately0.63212.

• At t = 0, we have

y = 0 + 1 − e0 = 0

so the point(0,0) lies on the graph ofy. Atthis point, we have

dydt

= y − t

= 0 − 0

= 0

so the slope of the tangent line to the graph ofy at the point(0,0) is 0.

• At t = 1, we have

y = 1 + 1 − e1 = 2 − e

so the point(1,2− e) lies on the graph ofy.

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At this point, we havedydt

= y − t

= (2 − e) − (1)= 1 − e

≈ −1.7183

so the slope of the tangent line to the graph ofy at the point(1,2− e) is approximately−1.7183.

The following picture shows the graph of yalong with graphs of the three tangent lineswhose slopes were just computed.

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-1

1

2

3

y

-3 -2 -1 1 2 3t

The main thing to observe here is that if wepick any point (t1,y1) in the plane, we caneasily compute the value of y1 − t1. Thus,without knowing the solution of the differentialequation

dydt

= y − t

whose graph passes through the point (t1,y1),we do know that the slope of the tangent lineto this graph at the point (t1,y1) is y1 − t1.

For example, suppose that we pick the point(0,3) in the (t,y) plane. At this point, we have

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y − t = 3 − 0 = 3

which tells us that 3 is the slope of thetangent line to the graph of the solution of thedifferential equation

dydt

= y − t

whose graph passes through the point (0,3).We can draw a “minitangent” line with slope 3at the point (0,3). This minitangent line givesus visual information about the graph of thesolution of the initial value problem

dydt

= y − t

y(0) = 3

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-2

-1

1

2

3

4

y

-4 -3 -2 -1 1 2 3 4t

The actual solution of the initial value problemdydt

= y − t

y(0) = 3

is

y = t + 1 + 2et.The graph of this solution is shown below.

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-2

-1

1

2

3

4

y

-4 -3 -2 -1 1 2 3 4t

If we pick a large number of points in the (t,y)plane, compute the value of y − t, and drawminitangents at these points, we get a visualpicture of what the entire family of solutions of

dydt

= y − t

looks like. What is nice about this is that it isvery easy. We can get a good picture of whatsolutions look like without actually knowing(explicitly) what the solutions are. Below is atable with slopes of minitangents computed atselected points and a picture showing theseminitangents. This type of picture is called a

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slope field.

(t,y) y − t

(−2,2) 4

(−2,0) 2

(−2,−2) 0

(t,y) y − t

(0,2) 2

(0,0) 0

(0,−2) −2

(t,y) y − t

(2,2) 0

(2,0) −2

(2,−2) −4

-3

-2

-1

0

1

2

3

y

-3 -2 -1 1 2 3t

Of course, to get a really good picture of thebehavior of solutions, we need to computey − t at many points. This is easily handled bya computer with a program written tocompute and draw slope fields. Here is a

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computer–generated picture of the slope fieldof

dydt

= y − t

using a much larger sampling of points in the(t,y) plane.

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-2

-1

0

1

2

3

y

-3 -2 -1 1 2 3t

By looking at the above computer–generatedslope field, we can visualize the family ofsolutions of dy/dy = y − t. In particular, notethat there appears to be a solution whosegraph is a line with slope 1 passing throughthe point (0,1). This line is, of course,y = t + 1. It is easy to check that the function

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y = t + 1 is indeed a solution of the differentialequation dy/dt = y − t.

Furthermore, it appears from the slope fieldthat solutions whose graphs lie above the liney = t + 1 “blow up” as t → ∞ (meaning thatlim t→∞ y = ∞) and approach the line y = t + 1as t → −∞ (meaning thatlim t→−∞(y − (t + 1)) = 0) and it appears thatsolutions whose graphs lie below the liney = t + 1 “blow up” as t → ∞ (in this case, inthe sense that lim t→∞ y = −∞) and approachthe line y = t + 1 as t → −∞.

In fact, the family of solutions of dy/dt = y − tconsists of all functions of the form

y = t + 1 + cet

(where c can be any constant). The valuec = 0 gives us the solution y = t + 1. Solutionswhose graphs lie above the line y = t + 1correspond to positive values of c andsolutions whose graphs lie below the liney = t + 1 correspond to negative values of c.The figure below shows graphs of somemembers of this family (corresponding to the

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choices c = 0, c = 1, c = 1.5, c = −1, andc = −1.5).

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1

2

3

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y

-4 -3 -2 -1 1 2 3 4t

Summary of the GeneralIdea About Slope FieldsGiven a differential equation of the form

dydt

= f(t,y),

we can choose any point, (t1,y1), in the (t,y)plane and compute the value of f(t1,y1). Thevalue of f(t1,y1) is the slope of the line tangentto the graph of the function, y, that satisfies

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dydt

= f(t,y)

y(t1) = y1.If we compute the value of f(t,y) at manydifferent points in the (t,y) plane and drawminitangents with these slopes at thesepoints, then we will have obtained a slopefield for the differential equation

dydt

= f(t,y).

The slope field gives us a visual picture ofwhat the graphs of solutions look like.

A special case: Differentialequations whose right handside does not depend on yIf the right hand side of

dydt

= f(t,y)

does not depend on y, then the minitangentswhich make up the slope field of thisdifferential equation have the same slope

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along any given vertical line. Hence, theseslope fields are somewhat easier to createand visualize.

ExampleLet us create a slope field for the differentialequation

dydt

= 12

t2.

The right hand side of this differentialequation is f(t,y) = 1

2 t2 (which does notdepend on y). Observe that

f(−2,0) = 12

(−2)2 = 2

f(−2,12) = 12

(−2)2 = 2

f −2,− 17 = 12

(−2)2 = 2

etc.meaning that the slope of any minitangentthat lies along the vertical line t = −2 hasslope 2. Hence, we only really need to do onecomputation using t = −2:

f(−2,y) = 2 for all y.

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-2

2

4

y

-4 -2 2 4t

A similar result holds for any chosen value oft. For example,

f(0,y) = 0 for all y.

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-2

2

4

y

-4 -2 2 4t

A computer–generated graph of the slopefield of

dydt

= 12

t2

is shown below.

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-3

-2

-1

1

2

3

y

-3 -2 -1 1 2 3t

Of course, we can easily deduce (byantidifferentiation) that the family of solutionsof the differential equation

dydt

= 12

t2

consists of all functions of the form

y = 16

t3 + C

(where C can be any constant). The fact thatthe graphs of solutions are all verticaltranslations of each other is evident from theslope field. Some members of the family ofsolutions are pictured below. (What values of

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C were used to obtain the family members inthe picture?)

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1

2

3

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y

-4 -3 -2 -1 1 2 3 4t

A special case: Differentialequations whose right handside does not depend on tIf the right hand side of

dydt

= f(t,y)

does not depend on t, then the minitangentswhich make up the slope field of this

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differential equation have the same slopealong any given horizontal line. Differentialequations whose right hand sides do notdepend on t are called autonomousdifferential equations. The word autonomousmeans “not depending on time”. Autonomousdifferential equations are used to modelsystems for which the rules (hypotheses)governing the system do not depend on whattime it is. For example, the Malthusianpopulation model,

dPdt

= kP

is autonomous because the rule governingthe system is simply that a population alwaysgrows at a rate proportional to itself. Themodel predicts, for example, that the growthhistory of a population that starts with 3million individuals in the year 1790 would bethe same as the growth history of apopulation that starts with 3 million individualsin the year 1985. Both would, for instance,have the same number of individuals after 20years.

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ExampleLet us create a slope field for the autonomousdifferential equation

dydt

= 0.2y 1 − y6

.

The right hand side of this differentialequation is f(t,y) = 0.2y(1 − y/6) (which doesnot depend on t). After doing a fewcomputations:

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f(t,−2) = − 815

f(t,−1) = − 730

f(t,0) = 0

f(t,1) = 16

f(t,2) = 415

f(t,3) = 310

f(t,4) = 415

f(t,5) = 16

f(t,6) = 0

f(t,7) = − 730

f(t,8) = − 815

we observe that the slope field of thisdifferential equation has the appearanceshown in the following figure.

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-2

0

2

4

8

y

-2 2 4 6 8t

Note that the slopes of minitangents are thesame along any given horizontal line. Inparticular, note that since f(t,0) = 0 andf(t,6) = 0 for any t, there are constantsolutions that lie along the lines y = 0 andy = 6. These solutions are called equilibriumsolutions and are a common feature ofautonomous differential equations.

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