1. Calculus: Concepts and Contexts 3rd ed.
James Stewart

2. Differential Equations
Chapter Seven
Differential Equations

3. Section 7.1 Modeling with Differential Equations
Goals
Introduce differential equations and their use in modeling physical phenomena
Discuss initial value problems

4. Models of Population Growth
One possible model for the growth of a population is that the population grows at a rate proportional to the size of the population.
For this model we can define
t to be the time
P to be the size of the population

5. Population Growth (cont’d)
The rate of growth is dP/dt, so our assumption is written as the equation
where k is the proportionality constant.
This is a differential equation because it has
an unknown function P and
its derivative dP/dt.

6. Population Growth (cont’d)
Consequences of this model:
The equation shows that P (t) > 0, so the population is always increasing.
Assuming we rule out a population of zero, that is!
In fact, as P(t) increases, dP/dt becomes larger.
So the growth rate increases with the population.
Direct substitution shows that P(t) = Cekt is a solution of the differential equation.
Later we will show there is no other.

7. Population Growth (cont’d)
Consequences (cont’d)
Allowing C to vary leads to the family of solutions shown on the next slide.
The only physically meaningful solutions are those with C > 0, for t ≥ 0.
Putting t = 0 gives P(0) = Cek(0) = C, so C turns out to be the initial population.

8. Population Growth (cont’d)

9. The Logistic Equation
The preceding model is appropriate if resources are unlimited.
A more realistic model is based on the assumption that the population levels off as it reaches its carrying capacity K :

10. The Logistic Equation (cont’d)
We can incorporate both assumptions using the equation
Note that if
P is small compared with K, then dP/dt ≈ kP ;
P > K, then dP/dt < 0.

11. The Logistic Equation (cont’d)
The preceding equation is called the logistic equation.
Later we will solve this equation explicitly, but for now we can simply look at the sign of dP/dt to see how the solutions approach the equilibrium solutions P = 0 and P = K :

12. The Logistic Equation (cont’d)

13. The Motion of a Spring
We consider the motion of a mass m at the end of a vertical spring, as shown on the next slide.
If the spring is stretched (or compressed) x units from its natural length, then by Hooke’s Law the spring exerts a force proportional to x :
restoring force = –kx

14. Motion of a Spring (cont’d)

15. Motion of a Spring (cont’d)
Here the spring constant k is positive.
Ignoring external forces, then Newton’s Second Law gives
This a second-order differential equation, because it involves a second derivative.

16. Motion of a Spring (cont’d)
What can we guess about the form of the solutions of this equation?
We see that the second derivative of x is proportional to x but has the opposite sign.
Two functions with this property are the sine and cosine functions.
In fact, all solutions of this equation can be written in terms of sines and cosines.
Not surprising: The spring will oscillate!

17. General Differential Equations
In general, a differential equation is an equation that contains
an unknown function and
one or more of its derivatives.
The order of a differential equation is the order of the highest derivative that occurs in the equation.

18. General Equations (cont’d)
A function f is a solution of a differential equation if the equation is satisfied when y = f(x) and its derivatives are substituted into the equation.
To solve a differential equation is to find all possible solutions of the equation.
Thus, the general solution of the equation y  = x3 is given by y = x4/4 + C.

19. Example
Show that every member of the family of functions y = (1 + cet)/(1 – cet) is a solution of the differential equation y  = ½(y2 – 1).
Solution We apply the Quotient Rule to y:

20. Solution (cont’d) The right side of the differential equation becomes
So for every value of c, the given function is a solution of the differential equation.

21. Initial-Value Problems
In many physical problems we need to find the particular solution that satisfies a condition of the form y(t0) = y0 .
This is called an initial condition, and
the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem.

22. Initial-Value Problems (cont’d)
Geometrically, when we impose an initial condition, we
look at the family of solution curves and
pick the one that passes through the point (t0, y0).
Physically, this corresponds to
measuring the state of a system at time t0 and
using the solution of the initial-value problem to predict the future behavior of the system.

23. Example
Find a solution of the differential equation y  = ½(y2 – 1) that satisfies the initial condition y(0) = 2.
Solution Substituting t = 0 and y = 2 into the formula y = (1 + cet)/(1 – cet) from our preceding example gives

24. Solution (cont’d) Solving this equation for c gives c = ⅓.
So the solution of the initial-value problem is

25. Review Models for General differential equations
population growth
the motion of a spring
General differential equations
Initial-value problems

26. Section 7.2 Direction Fields and Euler’s Method
Goals
Gain insight about first-order differential equations from their direction fields
Introduce Euler's method for finding approximate solutions to differential equations.

27. Direction Fields
A direction field can give a picture of a differential equation even when no solution can be found.
For example, suppose we want the graph of the solution of the initial-value problem
y  = x + y, y(0) = 1
Since y  gives the slope of the solution curve, as the next slide illustrates…

28. Direction Fields (cont’d)

29. Direction Fields (cont’d)
…the slope at the point (0, 1) is = 1.
Thus, the solution curve near the point (0, 1) looks like a short line segment with slope 1.
In the same way we can draw short line segments at a number of points (x, y) with slope x + y , resulting in a direction field.
This allows us to visualize the solution curve:

30. Direction Fields (cont’d)

31. Direction Fields (cont’d)
In general, suppose we have a first-order equation of the form y  = F(x, y).
Here F(x, y) is an expression in x and y.
The differential equation says that the slope of a solution curve at a point (x, y) on the curve is F(x, y).
If we draw segments with slope F(x, y) at several points (x, y), we get a direction field.

32. Example
Sketch the direction field for the differential equation y  = x2 + y2 – 1.
Use the field to sketch the solution curve that passes through the origin.
Solution We start by computing the slope at several points, as shown in the chart on the next slide:

33. Solution (cont’d)

34. Solution (cont’d)
Now we draw short line segments with these slopes at these points.
To sketch the desired solution curve, we
start at the origin and move to the right in the direction of the line segment;
then continue to draw the curve so that it moves parallel to the nearby line segments.
Both are shown on the next two slides:

35. Solution (cont’d)

36. Solution (cont’d)

37. An Application
The next slide shows a simple electric circuit with an electromotive force (battery or generator) that produces at time t…
a voltage of E(t) volts (V) and
a current of I(t) amperes (A).
The circuit also contains a
resistor with a resistance of R ohms (Ω) and
an inductor with an inductance of L henries (H):

38. Application (cont’d)

39. Application (cont’d)
Ohm’s Law gives the drop in voltage due to the resistor as RI.
The voltage drop due to the inductor is L(dI/dt).
One of Kirchoff’s Laws says that the sum of the voltage drops is equal to the supplied voltage E(t).

40. Application (cont’d) Thus, we have
a first–order differential equation that models the current I at time t.
Studying this equation can reveal important information about the current flow.

41. Example Suppose that in the circuit of Fig. 9,
the resistance is 12 Ω,
the inductance is 4 H, and
a battery gives a constant voltage of 60 V.
Draw the corresponding direction field.
What is the limiting value of the current?
Identify any equilibrium solutions.

42. Example (cont’d)
If the switch is closed when t = 0 so the current starts with I(0) = 0, use the direction field to sketch the solution curve.
Solution If we put L = 4, R = 12, and E(t) = 60 in our earlier equation, we get

43. Solution (cont’d)
The direction field for this differential equation is shown on the next slide.
It appears from the direction field that all solutions approach the value 5 A, that is,
We can verify from the equation that the constant I(t) = 5 is an equilibrium solution.

44. Solution (cont’d)

45. Remarks
Notice from Fig. 11 that the line segments along any horizontal line are parallel.
That is because the variable t does not occur on the right side of the equation .
In general, a differential equation of the form y  = f(y) in which the independent variable is missing from the right side, is called autonomous.

46. Remarks (cont’d)
For such an equation, the slopes corresponding to two different points with the same y-coordinate must be equal.
This means that if we know one solution to an autonomous differential equation, then we can obtain infinitely many others just by shifting the graph of the known solution to the right or left.

47. Remarks (cont’d)
In Figure 11 we have shown the solutions that result from shifting the solution curve of the example one and two seconds to the right.
They correspond to closing the switch when t = 1 or t = 2.

48. Euler’s Method
The basic idea behind direction fields can be used to find numerical approximations to solutions of differential equations.
We illustrate the method on the initial-value problem that we used to introduce direction fields:
y  = x + y, y(0) = 1

49. Euler’s Method (cont’d)
As we know, the solution curve has slope 1 at the point (0, 1).
As a first approximation to the solution we could use the linear approximation L(x) = x + 1.
In other words, we could use the tangent line at (0, 1) as a rough approximation to the solution curve, as the next slide illustrates:

50. Euler’s Method (cont’d)

51. Euler’s Method (cont’d)
Euler’s idea was to improve on this approximation by…
proceeding only a short distance along this tangent line and then
making a midcourse correction according to the direction field.
The next slides illustrate this using step sizes of first 0.5, then 0.25:

52. Euler’s Method (cont’d)

53. Euler’s Method (cont’d)

54. Euler’s Method (cont’d)
In general, Euler’s method says to…
start at the point given by the initial value and
proceed in the direction indicated by the direction field.
Stop after a short time, look at the slope at the new location, and proceed in that direction.
Keep stopping and changing direction according to the direction field.

55. Euler’s Method (cont’d)
Euler’s method produces only approximations to the exact solution of an initial-value problem.
But by decreasing the step size…
and therefore increasing the number of midcourse corrections…
we obtain successively better approximations to the exact solution.

56. Euler’s Method (cont’d)
For the general first-order initial-value problem
y  = F(x, y), y(x0) = y0 ,
we want approximate values for the solution at equally spaced numbers
x0 , x1 = x0 + h, x2 = x1 + h, …,
where h is the step size.

57. Euler’s Method (cont’d)
The figure on the next slide shows that the approximate value of the solution when x = x1 is
y1 = y0 + hF(x0, y0)
Similarly,
y2 = y1 + hF(x1, y1)
In general,
yn = yn-1 + hF(xn-1, yn-1)

58. Euler’s Method (cont’d)

59. Example
Use Euler’s method with step size 0.1 to construct a table of approximate values for the solution of the initial-value problem
y  = x + y, y(0) = 1
Solution Here h = 0.1, x0 = 0, y0 = 1, and F(x, y) = x + y. On the next slide we begin to apply Euler’s method:

60. Solution (cont’d)
This means that if y(x) is the exact solution, then y(0.3) ≈
We can proceed with similar calculations to arrive at the table on the next slide:

61. Solution (cont’d)

62. Decreasing Step Size
For a more accurate table of values in the preceding example we could decrease the step size.
But for a large number of steps we need to program a calculator or computer to carry out the calculations.
The table on the next slide shows the results of applying Euler’s method with decreasing step-size to this problem:

63. Step Size (cont’d)

64. Step Size (cont’d)
Notice that the Euler estimates seem to be approaching limits, namely, the true values of y(0.5) and y(1).
The figure on the next slide shows the Euler approximations with step sizes 0.5, 0.25, 0.1, 0.05, 0.02, 0.01, and
They are approaching the exact solution curve as the step size h approaches 0:

65. Step Size (cont’d)

66. Review Direction fields Euler’s method
Autonomous differential equations
Euler’s method

67. Section 7.3 Separable Equations
Goals
Learn to solve separable differential equations
Discuss two applications
Orthogonal trajectories
Mixing problems

68. Introduction
We have examined differential equations so far from two points of view:
Geometric (direction fields), and
Numerical (Euler’s method).
We would also like to approach differential equations symbolically.
That is, we would like to have an explicit formula for a solution.

69. Introduction (cont’d)
This is not always possible, but here we examine a type of equation that we can solve explicitly.
A separable differential equation can be written in the form

70. h(y) = g(x)dx, where h(y) = 1/f(y).
Separable Equations
To solve this equation we write it in the differential form
h(y) = g(x)dx, where h(y) = 1/f(y).
In this form the variables have been separated:
All the y’s are on one side of the equation, and
all the x’s are on the other.

71. Separable Equations (cont’d)
Next we integrate both sides:
∫ h(y) = ∫ g(x)dx
This equation implicitly defines y as a function of x.
In some cases we may be able to solve for y in terms of x.
On the next slide we justify this procedure using the Chain Rule:

72. Separable Equations (cont’d)
If h and g satisfy the preceding equation, then they satisfy the original equation:

73. Example Solve the differential equation
Find the solution of the equation that satisfies the initial condition y(0) = 2.
Solution We write the equation in terms of differentials and integrate both sides:

74. Solution (cont’d)
where C is an arbitrary constant.

75. Solution (cont’d) Solving for y, we get
We could also write this as where K = 3C.
Putting x = 0 in this general solution gives
y(0) = 2, we must have

76. Solution (cont’d) Thus, the solution of the initial-value problem is
The next slide shows several members of the family of solutions of the given differential equation.
The graph of our particular solution above is shown in red.

77. Solution (cont’d)

78. Example Solve the differential equation
Solution Writing the equation in differential form and integrating gives
∫ (2y + cos y) dy = ∫ 6x2 dx, or

79. Solution (cont’d) y2 + sin y = 2x3 + C where C is a constant.
In this case it’s impossible to solve the equation explicitly for y.
The next slide shows the above solution for the following values of C (from left to right): C = 3, 2, 1, 0, –1, –2, and –3 :

80. Solution (cont’d)

81. Orthogonal Trajectories
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family orthogonally, that is, at right angles.
This is illustrated on the next slide:

82. Orthogonal Trajectories (cont’d)

83. Orthogonal Trajectories (cont’d)
For instance, as the next slide illustrates,
each member of the family y = mx of straight lines through the origin is an orthogonal trajectory of
the family x2 + y2 = r2 of concentric circles with center the origin.
We say that the two families are orthogonal trajectories of each other:

84. Orthogonal Trajectories (cont’d)

85. Example
Find the orthogonal trajectories of the family of curves x = ky2, where k is an arbitrary constant.
Solution The curves x = ky2 form a family of parabolas whose axis of symmetry is the x-axis.
Differentiating x = ky2 gives

86. Solution (cont’d)
We need an equation that is valid for all values of k at the same time.
To eliminate k we note from the given general parabola that k = x/y2.
So the differential equation can be written as

87. Solution (cont’d) or

88. Solution (cont’d)
This means that the slope of the tangent line at any point (x, y) on one of the parabolas is y  = y/(2x).
On an orthogonal trajectory the slope of the tangent line must be the negative reciprocal of this slope.
Therefore, the orthogonal trajectories must satisfy the differential equation

89. Solution (cont’d)
We solve this separable equation:

90. Solution (cont’d) Here C is an arbitrary positive constant.
Thus, the orthogonal trajectories are a family of ellipses centered at the origin.
Both the family of parabolas and the family of ellipses are shown on the next slide:

91. Solution (cont’d)

92. Mixing Problems
A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed solution of some substance, such as salt.
A solution of a given concentration enters the tank at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which may differ from the entering rate.

93. Mixing Problems
If y(t) denotes the amount of substance in the tank at time t, then y (t) is the rate at which the substance is being added minus the rate at which it is being removed.
The mathematical description of this situation often leads to a first-order separable differential equation.

94. Example
A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
How much salt remains in the tank after half an hour?

95. Solution Let y(t) be the amount of salt (in kg) after t minutes. We…
are given that y(0) = 20 and
want to find y(30).
We do this by finding a differential equation satisfied by y(t).
Note that dy/dt is the rate of change of the amount of salt, so

96. Solution (cont’d) We have
Here “rate in” is the rate at which salt enters the tank and
“rate out” is the rate at which it leaves.
We have

97. Solution (cont’d)
The tank always contains 5000 L of liquid, so the salt concentration at time t is y(t)/5000 (kg/L).
Since the brine flows out at a rate of 25 L/min, we have

98. Solution (cont’d) Thus
Solving this separable differential equation leads to

99. Solution (cont’d) Since y(0) = 20, we have –ln 130 = C, so Therefore
Since y(t) is continuous and y(0) = 20 and the right side is never 0, 150 – y(t) is always positive.

100. Solution (cont’d) Thus, |150 – y| = 150 – y and so
y(t) = 150 – 130e–t/200
The amount of salt after 30 min is
y(30) = 150 – 130e–30/200 ≈ 38.1 kg.

101. Review Separable equations and their solution Two applications:
Orthogonal trajectories
Mixing problems

102. Section 7.4 Exponential Growth and Decay
Goals
Study the law of natural growth (or decay)
Discuss applications
Population growth
Radioactive decay
Newton’s Law of Cooling

103. Law of Natural Growth/Decay
Earlier we assumed that a population grew at a rate proportional to the size of the population.
This assumption applies in many other situations as well.
For example, the mass of a radioactive substance decays at a rate proportional to the mass.

104. Natural Growth/Decay (cont’d)
In general, if…
y(t) is the value of a quantity y at time t and
the rate of change of y with respect to t is proportional to its size y(t) at any time,
then
where k is a constant.

105. Natural Growth/Decay (cont’d)
This is called the
law of natural growth (if k > 0) or the
law of natural decay (if k < 0)
We can solve it as a separable differential equation:

106. Natural Growth/Decay (cont’d)
This leads to
ln y = kt + C, or
y = ekt + C = eCekt, or
y = Aekt,
where A (= ±eC or 0) is a constant.
Actually, since y(0) = Aek∙0 = A, A is the initial value of the function.

107. Population Growth
What is the significance of the proportionality constant k?
In the context of population growth,
The quantity
relative growth rate.

108. Population Growth (cont’d)
Notice that the relative growth rate appears as the coefficient of t in the exponential function y0ekt.
For example, assuming that the growth rate is proportional to population size,
use the data on the next slide to model the population of the world in the 20th century.
What is the relative growth rate? How well does the model fit the data?

109. Example (cont’d) Solution We…
measure the time in years;
let t = 0 in the year 1990.
The population P(t) is measured in millions of people.
The initial condition is P(0) = 1650.

110. Solution (cont’d) The initial-value problem is
By our previous work, the solution is
P(t) = 1650ekt
One way to estimate the relative growth rate is to use the fact that P(50) = 2560.

111. Solution (cont’d) This means that 1650ek(50) = 2560, which leads to
Thus the relative growth rate is about 88% per year and the model becomes
P(t) = 1650e t

112. Comparison
The table at right and the figure on the next slide allow us to compare the predictions of this model with the actual data.

113. Comparison (cont’d)

114. Comparison (cont’d)
You can see that the predictions become quite inaccurate after about 60 years.
We might try to improve the model by using the given population for 1970, instead of 1950, to estimate k.
This leads to the estimate k ≈
This second model is shown on the next slide; it is more accurate after 1970 but less accurate before 1950:

115. Comparison (cont’d)

116. Radioactive Decay
Radioactive substances have been found experimentally to decay at a rate proportional to the remaining mass.
Thus the amount m(t) can be expressed in terms of the initial amount m0 by
m(t) = m0ekt

117. Radioactive Decay (cont’d)
We can express the rate of decay in terms of half-life, the time required for half of any given quantity to decay.
For example, the half-life of radium-226 is 1590 years:
A sample of this isotope has a mass of 100 mg. Find a formula for the amount remaining after t years.

118. Radioactive Decay (cont’d)
Radium-226 (cont’d)
Find the mass after 1000 years correct to the nearest milligram.
When will the mass be reduced to 30 mg?
Solution Let m(t) be the mass of radium-226 (in mg) remaining after t years.
Them dm/dt = km and y(0) = 100, so
m(t) = m(0)ekt = 100ekt

119. Solution (cont’d) The fact that y(1590) = ½(100) gives Therefore
The mass after 1000 years is

120. 100e–(ln 2/1590)t = 30 or e–(ln 2/1590)t = 0.3
Solution (cont’d)
We want to find the value of t such that m(t) = 30, that is,
100e–(ln 2/1590)t = 30 or e–(ln 2/1590)t = 0.3
Solving this equation for t gives t ≈ 2762 years.
We can check this result by graphing, as the next slide shows:

121. Solution (cont’d)

122. Newton’s Law of Cooling
This states that the rate of cooling (or warming) of an object is proportional to the temperature difference between the object and its surroundings.
If we let
T(t) be the temperature of the object at time t
TS be the temperature of the surroundings,

123. Newton’s Law of Cooling (cont’d)
then we can formulate Newton’s Law of Cooling as a differential equation:
where k is a constant.
We could solve this as a separable differential equation…

124. Newton’s Law of Cooling (cont’d)
…but it is easier to make the change of variable y(t) = T(t) – TS :
Because TS is constant, y (t) = T (t), and so the equation becomes
We can solve this for y, and then find T.

125. Example
A bottle of soda pop at room temperature ( 72°F) is placed in a refrigerator where the temperature is 44°F. After half an hour the soda pop has cooled to 61°F.
What is the temperature of the soda pop after another half hour?
How long does it take for the soda pop to cool to 50°F?

126. Solution Let T(t) be the temperature of the soda after t minutes.
Here TS = 44°F, so Newton’s Law of Cooling gives
If we let y = T – 44, then
y(0) = T(0) – 44 = 72 – 44 = 28

127. Solution (cont’d) Thus y is a solution of the initial-value problem
As before, y(t) = y(0)ekt = 28ekt.
We are given that T(30) = 61, so y(30) = 61 – 44 =17, which leads to

128. Solution (cont’d) Thus T(60) = 44 + 28e–0.01663(60) ≈ 54.3°F.
So after another half-hour the pop has cooled to about 54°F.
Next, T(t) = 50 when e– t = 50, which leads to

129. Solution (cont’d) Notice that as we would expect.
The pop cools to 50°F after about 1 hour 33 minutes.
Notice that
as we would expect.
The temperature function is graphed on the next slide:

130. Solution (cont’d)

131. Review Law of natural growth/decay Applications: Population growth
Relative growth rate
Radioactive decay
Newton’s Law of Cooling

132. Section 7.5 The Logistic Equation
Goals
Introduce the logistic differential equation and study its direction field
Solve the logistic equation analytically
Mention other growth models

133. Introduction
As we discussed in Section 7.1, a population often increases exponentially in its early stages…
…but levels off eventually and approaches its carrying capacity because of limited resources.
Another way to say this is that…

134. Introduction (cont’d)
…the relative growth rate
is almost constant when the population is small, but
decreases as the population P increases, and
becomes negative if P ever exceeds its carrying capacity K, the maximum population that the environment is capable of sustaining in the long run.

135. The Logistic Model
The simplest expression for relative growth rate that incorporates these assumption is
Multiplying by P, we obtain the logistic differential equation:

136. The Logistic Model (cont’d)
Notice that if
P is small compared with K, then P/K is close to 0 and so dP/dt ≈ kP;
P  K, then P/K  1, so dP/dt  0.
Also, if
0 < P < K, then dP/dt > 0 and the population is increasing, whereas
dP/dt < 0 if P > K.

137. Direction Field
We can learn quite a bit about the logistic model by looking at its direction field.
For example, on the next slide we draw a direction field for the logistic equation with k = 0.08 and carrying capacity K = 1000.
So the model is

138. Direction Field (cont’d)

139. Direction Field (cont’d)
We note that the slopes are
positive for 0 < P < 1000 and negative for P > 1000;
small when P is close to 0 or 1000 (= K)
Also the solutions move
away from the equilibrium solution P = 0 and
toward the equilibrium solution P = 1000.

140. Direction Field (cont’d)
On the next slide we sketch solution curves with initial populations P(0) = 100, 400, and 1300.
Note that the solution curves
that start below P = 1000 are increasing;
that start above P = 1000 are decreasing;
that start below P = 500 have inflection points at P = 500:

141. Direction Field (cont’d)

142. Euler’s Method
Next we use Euler’s method with step sizes 20, 10, 5, 1, and 0.1 to estimate the population sizes P(40) and P(80), where P is the solution of the initial-value problem

143. Solution
We apply Euler’s method as in Section 7.2, using the above values for the step size h. Also we use a calculator or computer as needed.
The next slide shows a table with the results.
Our estimates for the population sizes at times t = 40 and t = 80 are
P(40) ≈ 539 and P(80) ≈ 1032

144. Solution (cont’d)
The following slide shows a graph of the Euler approximations using h = 10 and h = 1:

145. Solution (cont’d)

146. The Analytic Solution
Since the logistic equation is separable, we can solve it explicitly using the method of Section From
we have

147. Analytic Solution (cont’d)
Applying partial fractions to the integral on the left leads to
We continue on the next slide:

148. Analytic Solution (cont’d)
where A = ±e–C . Solving for P gives

149. Analytic Solution (cont’d)
If t = 0, then P = P0, so

150. Analytic Solution (cont’d)
So the solution to the logistic equation is
As we would expect,

151. Example Write the solution of the initial-value problem
and use it to find the population sizes P(40) and P(80).
When does the population reach 900?

152. Solution We apply the preceding formulas with
k = 0.08,
carrying capacity K = 1000, and
initial population P0 = 100.
So the population at time t is

153. Solution (cont’d) Thus
Calculation shows that P(40) = and P(80) = 985.3, respectively.
Further, the population reaches 900 when

154. Solution (cont’d)
As a check on our work, here is the population curve.
It does indeed intersect the line P = 900 when t ≈ 55:

155. Other Growth Models
There are at least two other modifications of the logistic model:
The differential equation
has been used to model populations that are subject to “harvesting”.

156. Other Growth Models (cont’d)
An example might be a population of fish being caught at a constant rate.
Also, there may be a minimum population level m below which the species tends to become extinct.
Adults may not be able to find suitable mates, for example.

157. Other Growth Models (cont’d)
Such populations have been modeled by the differential equation
where the extra factor 1 – m/P takes into account the consequences of a sparse population.

158. Review The logistic model Analysis using The analytic solution
Direction fields
Euler’s method
The analytic solution
Other models for population growth

159. Section 7.6 Predator-Prey Systems
Goals
Use differential equations to study the interaction of two species that share the same habitat.
Introduce the use of the phase plane and phase portraits.

160. Introduction We first consider the situation in which
one species, called the prey, has an ample food supply and
the second species, called the predators, feeds on the prey.
Examples of prey and predators include
rabbits and wolves in an isolated forest,
food fish and sharks, and
aphids and ladybugs.

161. Introduction (cont’d)
Our model will have two dependent variables and both are functions of time.
We let…
R(t) be the number of prey (using R for rabbits) and
W(t) be the number of predators (with W for wolves)
at time t.

162. Introduction (cont’d)
In the absence of predators, the ample food supply would support exponential growth of the prey, that is,
In the absence of prey, we assume that a predator population would decline at a rate proportional to itself, that is,

163. Introduction (cont’d)
With both species present, however, we assume that
the principal cause of death among the prey is being eaten by a predator, and
the birth and survival rates of the predators depend on their available food supply, namely, the prey.

164. Introduction (cont’d)
We also assume that the two species encounter each other at a rate that is proportional to both populations…
…and is therefore proportional to the product RW.
The more there are of either population, the more encounters there are likely to be.

165. Introduction (cont’d)
A system of two differential equations that incorporates these assumptions is as follows:
Here k, r, a, and b are positive constants.

166. Predator-Prey Equations
Notice that…
the term aRW decreases the natural growth rate of the prey and
the term bRW increases the natural growth rate of the predators.
The equations above are known as the predator-prey equations, or the Lotka-Volterra equations.

167. Predator-Prey (cont’d)
A solution of this system of equations is a pair of functions R(t) and W(t) that describe the populations of prey and predator as functions of time.
Because the system is coupled (R and W occur in both equations), we can’t solve one equation and then the other; we have to solve them simultaneously.

168. Predator-Prey (cont’d)
Unfortunately, it is usually impossible to find explicit formulas for R and W as functions of t.
We can, however, use graphical methods to analyze the equations.

169. Example
Suppose that populations of rabbits and wolves are described by the Lotka-Volterra equations with k = 0.08, a = 0.001, r = 0.02, and b =
The time is measured in months.
Find the constant solutions (called the equilibrium solutions) and interpret the answer.

170. Example (cont’d)
Use the system of differential equations to find an expression for dW/dR.
Draw a direction field for the resulting differential equation in the RW-plane.
Then use that direction field to sketch some solution curves.
Suppose that at some point there are 1000 rabbits and 40 wolves.
Draw the corresponding solution curve and use it to describe the changes in both population levels.

171. Example (cont’d)
Use the preceding part to make sketches of R and W as functions of t.
Solution With the given values of k, a, r, and b, the Lotka-Volterra equations become

172. Solution (cont’d)
Both R and W will be constant if both derivatives are 0, that is,
R  = R(0.08 – 0.001W) = 0
W  = W(– R) = 0
One solution is given by R = 0 and W = 0.
This makes sense: If there are no rabbits or wolves, the populations are certainly not going to increase!

173. Solution (cont’d) The other constant solution is
So the equilibrium populations consist of 80 wolves and 1000 rabbits.
This means that 1000 rabbits are just enough to support a constant wolf population of 80.

174. Solution (cont’d)
Next we can use the Chain Rule to eliminate t: So

175. Solution (cont’d)
If we think of W as a function of R, we have the differential equation
The next slide shows the direction field for this differential equation:

176. Solution (cont’d)

177. Solution (cont’d)
On the next slide we use this direction field to sketch several solution curves.
If we move along a solution curve, we observe how the relationship between R and W changes as time passes:

178. Solution (cont’d)

179. Solution (cont’d) Notice that
the curves appear to be closed in the sense that if we travel along a curve, we always return to the same point.
the point (1000, 80) is inside all the solution curves.
That point is called an equilibrium point because it corresponds to the equilibrium solution R = 1000, W = 80.

180. Phase Plane
When we represent solutions of a system of differential equations as in Fig. 2, we call the RW-plane the phase plane, and the solution curves phase trajectories.
So a phase trajectory is a path traced out by solutions as time goes by.
A phase portrait consists of equilibrium points and typical phase trajectories.

181. Solution (cont’d)
Starting with 1000 rabbits and 40 wolves corresponds to drawing the solution curve through the point P0(1000, 40).
The next slide shows this phase trajectory with the direction field removed.
Starting at the point at time t = 0 and letting t increase, do we move clockwise or counterclockwise around the phase trajectory?

182. Solution (cont’d)

183. Solution (cont’d)
If we put R = 1000 and W = 40 in the first differential equation, we get
Since dR/dt > 0, R is increasing at P0 and so we move counterclockwise around the phase trajectory.

184. Solution (cont’d)
We see that at P0 there aren’t enough wolves to maintain a balance between the populations, so the rabbit population increases.
That results in more wolves.
Eventually there are so many wolves that the rabbits have a hard time avoiding them.

185. Solution (cont’d) So the number of rabbits begins to decline…
…at P1 , where we estimate that R reaches its maximum population of about 2800.
This means that at some later time the wolf population starts to fall…
…at P2 , where R = 1000 and W ≈ 140.

186. Solution (cont’d)
But this benefits the rabbits, so their population later starts to increase…
at P3 , where W = 80 and R ≈ 210.
As a consequence, the wolf population eventually starts to increase as well.
This happens when the populations return to their initial values of R = 1000 and W = 40, and the entire cycle begins again.

187. Solution (cont’d)
From this description of how the rabbit and wolf populations rise and fall, we can sketch the graphs of R(t) and W(t).
These graphs are shown on the next two slides.
We denote by t1, t2, t3 the times at which the points P1, P2, P3 are reached:

188. Solution (cont’d)

189. Solution (cont’d)

190. Solution (cont’d)
To make the graphs easier to compare, we draw the graphs
on the same axes, but
with different scales for R and W, as shown on the next slide.
Notice that the rabbits reach their maximum populations about a quarter of a cycle before the wolves:

191. Solution (cont’d)

192. Review
Predator-prey equations
Phase plane and phase trajectories