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Appendix F.1 ■ Solutions of Differential Equations F
■ Find general solutions of differential equations. ■ Find particular solutions of differential equations.
General Solution of a Differential Equation A differential equation is an equation involving a differentiable function and one or more of its derivatives. For instance, Differential equation is a differential equation. A function is a solution of a differential equation if the equation is satisfied when and its derivatives are replaced by and its derivatives. For example, Solution of differential equation is a solution of the differential equation shown above. To see this, substitute for and in the original equation. Substitute for y and
In the same way, you can show that and are also solutions of the differential equation. In fact, each function given by General solution where is a real number, is a solution of the equation. This family of solutions is called the general solution of the differential equation.
Example 1 Verifying Solutions
Determine whether each function is a solution of the differential equation a. b. SOLUTION a. Because and it follows that
So, is a solution. b. Because and it follows that
So, is also a solution.
Checkpoint 1
Determine whether y Ce^4 x is a solution of the differential equation y y. ■
y Ce x
y y Ce x^ Ce x^ 0.
y Ce x y Ce x ,
y Ce x
y y Ce x^ Ce x^ 0.
y Ce x y Ce x ,
y Ce x y Ce x
y y 0.
C
y Ce ^2 x
y 2 e ^2 x , y 3 e ^2 x , y 12 e ^2 x
y 2 y 2 e ^2 x^ 2 e ^2 x y .
y 2 e ^2 x
y
y e ^2 x
y f x
y f x
y 2 y 0
F Differential Equations
F.1 Solutions of Differential Equations
Particular Solutions and Initial Conditions A particular solution of a differential equation is any solution that is obtained by assigning specific values to the arbitrary constant(s) in the general solution.* Geometrically, the general solution of a differential equation represents a family of curves known as solution curves. For instance, the general solution of the differential equation is General solution Figure F.1 shows several solution curves corresponding to different values of Particular solutions of a differential equation are obtained from initial conditions placed on the unknown function and its derivatives. For instance, in Figure F.1, suppose you want to find the particular solution whose graph passes through the point This initial condition can be written as when Initial condition Substituting these values into the general solution produces which implies that So, the particular solution is Particular solution
Example 2 Finding a Particular Solution
For the differential equation
verify that is a solution. Then find the particular solution determined by the initial condition when SOLUTION You know that is a solution because and
Furthermore, the initial condition when yields General solution Substitute initial condition.
Solve for C.
and you can conclude that the particular solution is
Particular solution
Try checking this solution by substituting for and in the original differential equation.
Checkpoint 2
For the differential equation verify that is a solution. Then find the particular solution determined by the initial condition when ■
*Some differential equations have solutions other than those given by their general solutions. These are called singular solutions. In this brief discussion of differential equations, singular solutions will not be discussed.
y 1 x 4.
xy 2 y 0, y Cx^2
y y
y 2 x^3 27
C
2 C 3 ^3
y Cx^3
y 2 x 3
3 Cx^3 3 Cx^3
xy 3 y x 3 Cx^2 3 Cx^3
y Cx^3 y 3 Cx^2
y 2 x 3.
y Cx^3
xy 3 y 0
y 3 x^2.
C 3.
3 C 1 ^2 ,
y 3 x 1.
C.
y Cx^2.
xy 2 y 0
F2 Appendix F ■ Differential Equations
STUDY TIP
To determine a particular solution, the number of initial conditions must match the number of constants in the general solution.
− 3
− 2
3 2 1
− 3 − 2 2 3
y
x
General solution: y = Cx^2 (1, 3)
Solution Curves for FIGURE F.
xy 2 y 0
F4 Appendix F ■ Differential Equations
In the first three examples in this section, each solution was given in explicit form, such as Sometimes you will encounter solutions for which it is more convenient to write the solution in implicit form, as shown in Example 4.
Example 4 Sketching Graphs of Solutions
Given that General solution is the general solution of the differential equation
sketch the particular solutions represented by and SOLUTION The particular solutions represented by and are shown in Figure F.3.
Graphs of Five Particular Solutions FIGURE F.
Checkpoint 4
Given that
is the general solution of
sketch the particular solutions represented by C 1, C 2,and C 4. ■
xy 2 y 0
y Cx^2
2 2
2
3
2
2
2
2
3
1
y
x
x
y
x
y
x
x
C = 0
C = 1
C = − 1 C = − 4
y (^) y
C = 4
C 0,±1, ± 4
C 0, ±1, ±4.
2 yy x 0
2 y^2 x^2 C
y f x .
SUMMARIZE (Section F.1)
1. Explain how to verify a solution of a differential equation (page F1). For an example of verifying a solution, see Example 1. 2. Describe the difference between a general solution of a differential equation and a particular solution (pages F1 and F2). For an example of a general solution of a differential equation and a particular solution, see Example 2. 3. Describe a real-life example of how a differential equation can be used to model the sales of a company’s product (page F3, Example 3).
Appendix F.1 ■ Solutions of Differential Equations F
Exercises F.
Verifying Solutions In Exercises 1–12, verify that the function is a solution of the differential equation. See Example 1.
Solution Differential Equation 1. 2.
3.
Determining Solutions In Exercises 13–16, determine whether the function is a solution of the differential equation
**13. 14.
- 16.**
Determining Solutions In Exercises 17–20, determine whether the function is a solution of the differential equation
17.
18.
19.
20.
Finding a Particular Solution In Exercises 21–24, verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition. See Example 2.
21. General solution: Differential equation: Initial condition: when 22. General solution: Differential equation: Initial condition: when 23. General solution: Differential equation: Initial condition: and when 24. General solution: Differential equation: Initial condition: and when
Sketching Graphs of Solutions In Exercises 25 and 26, the general solution of the differential equation is given. Sketch the particular solutions that correspond to the indicated values of C. See Example 4. General Solution Differential Equation C-Values 25. 26.
Finding General Solutions In Exercises 27–34, use integration to find the general solution of the differential equation.
27. 28.
dy dx 4 sin x
dy dx cos 4 x
dy dx
x 1 x^2
dy dx
x x^2 6
dy dx
x 2 x
dy dx
1 x
dy dx
2 x^3 3 x dy dx
3 x^2
y Ce x y y 0 0, ±1, ± 4
y C x 2 ^2 x 2 y 2 y 0 0, ±1, ± 2
y 5 y 6 x 0
y y 12 y 0
y C 1 e^4 x^ C 2 e ^3 x
y 5 y 0.5 x 1
xy y 0
y C 1 C 2 ln x
y 2 x 1
2 x 3 yy 0
2 x^2 3 y^2 C
y 3 x 0
y 2 y 0
y Ce ^2 x
y x ln x
y xe x
y cos x
y 29 xe ^2 x
y 3 y 2 y 0.
y y 3 sin 2 x
x
y e ^2 x y 5 ln x
y ^4 ^ 16 y 0.
y C 1 e^4 x^ C 2 e x y 3 y 4 y 0
y C 1 sin x C 2 cos x y y 0
y e x^3 y 3 x^2 y 6 xy 0
y x^2 x^2 y 2 y 0
y Ce x x^2 y 2 x 1 y 0
y x ln x Cx 4 x y 1 y 4 0
y x^2 2 x xy y x 3 x 4
C
x
y Cx^2 3 x xy 3 x 2 y 0
y
x y 4 x^2 y 0
y
x
y 2 x^3 y 0
y e ^2 x y 2 y 0
y Ce^4 x y 4 y
The following warm-up exercises involve skills that were covered in earlier sections. You will use these skills in the exercise set for this section. For additional help, review Sections 2.2, 2.6, 4.3, and 4.4.
In Exercises 1–4, find the first and second derivatives of the function.
1. 2. 3. 4.
In Exercises 5 and 6, solve for k.
5. 0.5 9 9 e k 6. 14.75 25 25 e ^2 k
y 3 x^2 2 x 1 y 2 x^3 8 x 4 y 3 e^2 x y 3 e x^2
SKILLS WARM U P F.
Appendix F.2 ■ Separation of Variables F
F.2 Separation of Variables
■ Use separation of variables to solve differential equations. ■ Use differential equations to model and solve real-life problems.
Separation of Variables The simplest type of differential equation is one of the form You know that this type of equation can be solved by integration to obtain
In this section, you will learn how to use integration to solve another important family of differential equations—those in which the variables can be separated. This technique is called separation of variables.
Essentially, the technique of separation of variables is just what its name implies. For a differential equation involving and you separate the variables by grouping the variables on one side and the variables on the other. After separating variables, integrate each side to obtain the general solution.
Example 1 Solving a Differential Equation
Find the general solution of
SOLUTION Begin by separating variables, then integrate each side.
Differential equation
Separate variables.
Integrate each side.
General solution
Checkpoint 1
Find the general solution of ■
dy dx
x^2 y
y^3 3
y x^2 2
C
y^2 ^1 ^ dy^ ^ x^ dx
(y^2 1 dy x dx
dy dx
x y^2 1
dy dx
x y^2 1
x y
x y,
y (^) f x dx.
y f x.
Separation of Variables If and are continuous functions, then the differential equation
has a general solution of
gy
dy (^) f x dx C.
dy dx
f xgy
f g
TECH TUTOR
You can use a symbolic integration utility to solve a differential equation that has separable variables. Use a symbolic integration utility to solve the differential equation in Example 1.
F8 Appendix F ■ Differential Equations
Example 2 Solving a Differential Equation
Find the general solution of
SOLUTION Begin by separating variables, then integrate each side.
Differential equation
Separate variables.
Integrate each side.
Find antiderivative of each side.
Multiply each side by 2. So, the general solution is Note that is used as a temporary constant of integration in anticipation of multiplying each side of the equation by 2 to produce the constant C.
Checkpoint 2
Find the general solution of
■
Example 3 Solving a Differential Equation
Find the general solution of Use a graphing utility to graph several solutions.
SOLUTION Begin by separating variables, then integrate each side.
Differential equation
Separate variables.
Integrate each side.
Find antiderivative of each side. By taking the natural logarithm of each side, you can write the general solution as General solution The graphs of the particular solutions given by 5, 10, and 15 are shown in Figure F.4.
Checkpoint 3
Find the general solution of
Use a graphing utility to graph the particular solutions given by C 1,2, and 4. ■
2 y dy dx
2 x.
C 0,
y lnx^2 C.
e y^ x^2 C
e^ y^ dy^ ^ 2 x^ dx
e y^ dy 2 x dx
e y^ dy dx
2 x
e y^ dy dx
2 x.
dy dx
x 1 y
y^2 x^2 C. C 1
y^2 x^2 C
y^2 2
x^2 2
C 1
y^ dy^ ^ x^ dx
y dy x dx
dy dx
x y
dy dx
x y
− 5
− 6 6
5
C = 0
C = 5 (^) C = 10
C = 15
FIGURE F.
F10 Appendix F ■ Differential Equations
Application
Example 6 Corporate Investing
A corporation invests part of its receipts at a rate of dollars per year in a fund for future corporate expansion. The fund earns percent interest per year compounded continuously. The rate of growth of the amount in the fund is
where is the time (in years). Solve the differential equation for as a function of where when SOLUTION You can solve the differential equation using separation of variables.
Differential equation
Differential form
Separate variables.
Integrate.
Assume and multiply each side by r. Exponentiate each side.
Solve for A.
General solution
Using when you find the value of
So, the differential equation for as a function of can be written as
Checkpoint 6
Use the result of Example 6 to find when and t 25 years. ■
A P $550,000, r 5.9%,
A
P
r
e rt^ 1 .
A t
C
P
r
0 Ce r^0 ^
P
r
A 0 t 0, C.
A Ce rt^
P
r
A
C 3 e rt^ P r
rA P ertC^2
lnrA P rt C 2 rA P > 0
r
lnrA P t C 1
dA rA P
dt
dA rA P dt
dA dt
rA P
A 0 t 0.
t A t,
dA dt
rA P
A
r
P
SUMMARIZE (Section F.2)
1. Explain how to use separation of variables to solve a differential equation (page F7). For examples of solving a differential equation using separation of variables, see Examples 1, 2, 3, 4, and 5. 2. Describe a real-life example of how separation of variables can be used to solve a differential equation that models corporate investing (page F10, Example 6).
Appendix F.2 ■ Separation of Variables F
Exercises F.
Separation of Variables In Exercises 1–6, decide whether the variables in the differential equation can be separated.
1. 2.
Solving a Differential Equation In Exercises 7–26, use separation of variables to find the general solution of the differential equation. See Examples 1 and 2.
7. 8.
Solving a Differential Equation In Exercises 27–30, (a) find the general solution of the differential equation and (b) use a graphing utility to graph the particular solutions given by 2, and 4. See Example 3.
27. 28.
Finding a Particular Solution In Exercises 31–38, use the initial condition to find the particular solution of the differential equation. See Example 4. Differential Equation Initial Condition **31.
33.**
34.
35. 36.
37.
Finding an Equation In Exercises 39 and 40, find an equation of the graph that passes through the point and has the specified slope. Then graph the equation.
39. Point: 40. Point:
Slope: Slope: y
2 y 3 x y
9 x 16 y
y 1 when x 0
dy dx 2 xy sin x^2
y 1 when x 0 dy dx
y cos x
y e x^2 y y 0 when x 0
x^2 16 y 5 x y 2 when x 5
y 3 when x 0
dy dx
x^2 1 y
xy 4 y 0 y 5 when x 0
x y y 0 y 4 when x 1
yy e x^ 0 y 4 when x 0
dy dx
0.25x 4 y
dy dx y 3
dy dx
2 x y
dy dx
x
C 1,
y
dy dx
y 6 cos x
dy dx sin x
2 xy 2 y y yx 1 0
y x y
x 1 y
dy dx
1 y
e y^ e xy 1 1
dy dt 3 t^2 1
y xy 0 y y 5
1 y
dy dx
x^2 4 y 4 x 0 dy dx
dy dx 3 y^2 x^2 y
dy dx
dy dx
x^2 3 y^2
dy dx
x 1 y^3
dr ds 0.05s
dr ds 0.05r
dy dx
x
dy dx
2 x
x dy dx
y
dy dx
x y
dy dx
x x y
dy dx
x
dy dx
x 1 x
dy dx
x y 3
The following warm-up exercises involve skills that were covered in earlier sections. You will use these skills in the exercise set for this section. For additional help, review Sections 4.4, 5.2, and 5.3.
In Exercises 1–6, find the indefinite integral and check your result by differentiating.
1. 2. 3.
In Exercises 7–10, solve the equation for C or k.
7. 8.
9. 10 2 e^2 k 10. 6 ^2 3 6 ek
3 ^2 6 3 1 C 1 ^2 2 ^2 C
xe^1 x
(^2) dx e^2 y^ dy
y 2 y^2 1
dy
x 5 x^3 ^2 dx t^3 ^ t^1 ^3 ^ dt dx
SKILLS WARM U P F.
■ Quiz Yourself F
QUIZ YOURSELF
Take this quiz as you would take a quiz in class. When you are done, check your work against the answers given in the back of the book.
In Exercises 1–4, verify that the function is a solution of the differential equation. Solution Differential Equation 1. 2.
3.
In Exercises 5 and 6, verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition.
5. General solution: Differential equation: Initial condition: and when 6. General solution: Differential equation: Initial condition: and when
In Exercises 7–10, use separation of variables to find the general solution of the differential equation.
7. 8.
In Exercises 11 and 12, (a) find the general solution of the differential equation and (b) use a graphing utility to graph the particular solutions given by and
In Exercises 13 and 14, use the initial condition to find the particular solution of the differential equation. Differential Equation Initial Condition 13.
14.
15. Find an equation of the graph that passes through the point and has a slope of Then graph the equation. 16. Ignoring resistance, a sailboat starting from rest accelerates at a rate proportional to the difference between the velocities of the wind and the boat. With a 20-knot wind, this acceleration is described by the differential equation where is the velocity of the boat (in knots) and is the time (in hours). After half an hour, the boat is moving at 10 knots. Write the velocity as a function of time.
v t
dvdt k 20 v,
y 3 x^2 y.
y 3 when x
dy dx y sin x
y 2 y 1 0 y 1 when x 0
dy dx
y x 3
dy dx
x^2 1 2 y
C ± 1.
C 0
dy dx
x 3 y^2 1 y
dy dx
2 x 1
y x 2 y 1
dy dx
4 x 4
y 0 y 4 x 2
x^2 y 3 xy 3 y 0
y C 1 x C 2 x^3
y 2 y 1 x 6
y 9 y 0
y C 1 sin 3x C 2 cos 3x
y 2 xy y x^3 x
x^3 5 x Cx
y xy 2 y 0
x
y C 1 cos x C 2 sin x y y 0
y Cex^22 y y 0
F14 Appendix F ■ Differential Equations
F.3 First-Order Linear Differential Equations
■ Solve first-order linear differential equations. ■ Use first-order linear differential equations to model and solve real-life problems.
First-Order Linear Differential Equations In this section, you will see how to solve a very important class of differential equations—first-order linear differential equations. The term “first-order” refers to the fact that the highest-order derivative of in the equation is the first derivative.
To solve a linear differential equation, write it in standard form to identify the functions and Then integrate and form the expression Integrating factor which is called an integrating factor. The general solution of the equation is
General solution
Example 1 Solving a Linear Differential Equation
Find the general solution of SOLUTION For this equation, and So, the integrating factor is
Integrating factor This implies that the general solution is
General solution
Checkpoint 1
Find the general solution of y y 10. ■
e x^ Ce x.
e x
e^2 x^ C
e x
e x e x dx
y
u x
Q x u x dx
e x.
e dx
u x e P x dx
P x 1 Q x e x.
y y e x.
y
u x
Q x u x dx.
u x e P x ^ dx
P x Q x . P x
y
Definition of a First-Order Linear Differential Equation A first-order linear differential equation is an equation of the form
where and are functions of An equation that is written in this form is said to be in standard form.
P Q x.
y P x y Q x
F16 Appendix F ■ Differential Equations
Application
Example 3 Intravenous Feeding
Glucose is added intravenously to the bloodstream at the rate of units per minute, and the body removes glucose from the bloodstream at a rate proportional to the amount present. Assume that is the amount of glucose in the bloodstream at time and that the rate of change of the amount of glucose is
where is a constant. Find the general solution of the differential equation. SOLUTION In standard form, this linear differential equation is
Standard form
which implies that and So, the integrating factor is
Integrating factor and the general solution is
General solution
Checkpoint 3
Use the general solution
from Example 3 to find the particular solution determined by the initial condition A 0 when t 0.(Assume k 0.05and q 0.05.) ■
A
q k Ce kt
q k
Ce kt.
e kt q k
e kt^ C
e^ ^ kt qe ktdt
A
u t
Q t u t dt
e kt
e k^ dt
u t e P t ^ dt
P t k Q t q.
dA dt
kA q
k
dA dt
q kA
A t
q
SUMMARIZE (Section F.3)
1. State the definition of a first-order linear differential equation (page F14). For examples of solving a first-order linear differential equation, see Examples 1 and 2. 2. State the guidelines for solving a first-order linear differential equation (page F15). For examples of solving a first-order linear differential equation, see Examples 1 and 2. 3. Describe a real-life example of how a first-order linear differential equation can be used to analyze intravenous feeding (page F16, Example 3).
Appendix F.3 ■ First-Order Linear Differential Equations F
Exercises F.
The following warm-up exercises involve skills that were covered in earlier sections. You will use these skills in the exercise set for this section. For additional help, review Sections 4.2, 4.4, and 5.1–5.3.
In Exercises 1–4, simplify the expression.
1. 2. 3. 4.
In Exercises 5–10, find the indefinite integral.
5. 6. 7. 8. (^) 9. (^) 4 x 3 ^2 dx 10. (^) x 1 x^2 ^2 dx
x 1 x^2 2 x 3 dx
2 x 5 4 e^2 x^ dx xe^3 x^2^ dx dx
e ln^ x e 2 ln^ x x
e x^
e x e^2 x^ e x e x^ e^2 x
SKILLS WARM U P F.
Writing in Standard Form In Exercises 1–6, write the first-order linear differential equation in standard form.
**1. 2.
- 6.**
Solving a Linear Differential Equation In Exercises 7–18, find the general solution of the first-order linear differential equation. See Examples 1 and 2.
7. 8.
Using Two Methods In Exercises 19–22, solve for y in two ways.
19.
20.
21.
22.
Matching In Exercises 23–26, match the differential equation with its solution without solving the differential equation. Explain your reasoning. Differential Equation Solution
23. (a) 24. (b) 25. (c) 26. (d)
Finding a Particular Solution In Exercises 27–34, find the particular solution that satisfies the initial condition. Differential Equation Initial Condition **27.
34.**
35. Sales The rate of change (in thousands of units) in sales of a biomedical syringe is modeled by
where is the time (in years). Solve this differential equation and use the result to complete the table.
t 0 1 2 3 4 5 6 7 8 9 10 S 0
t
dS dt
0.2 100 S 0.2 t
S
2 xy y x^3 x y 2 when x 4
xy 2 y 3 x^2 5 x y 3 when x 1
y 2 x 1 y 0 y 2 when x 1
y 3 x^2 y 3 x^2 y 6 when x 0
y y x y 4 when x 0
xy y 0 y 2 when x 2
y 2 y e ^2 x y 4 when x 1
y y 6 e x y 3 when x 0
y 2 xy x y Ce^2 x
y 2 xy 0 y x^2 C
y 2 y 0 y ^12 Ce x^2
y 2 x 0 y Ce x^2
y 4 xy x
y 2 xy 2 x
y 3 y 2
y y 4
xy y x^2 ln x
x^3 y 2 y e^1 x^2
xy y x^2 1
x 1 y y x^2 1
y 5 y e^5 x
y 2 xy 10 x
dy dx
e ^2 x 1 e ^2 x
dy dx
x^2 3 x
dy dx 3 y e ^3 x
dy dx y e^4 x
dy dx
5 y 15 dy dx
3 y 6
y 1 x 1 y x x^2 y y
xy y xe x xy y x^3 y
x^3 2 x^2 y 3 y 0 y 5 2 x y 0
Appendix F.4 ■ Applications of Differential Equations F
F.4 Applications of Differential Equations
■ Use differential equations to model and solve real-life problems.
Applications of Differential Equations
Example 1 Modeling a Chemical Reaction
During a chemical reaction, substance A is converted into substance B at a rate that is proportional to the square of the amount of substance A. When 60 grams of A are present, and after 1 hour only 10 grams of A remain unconverted. How much of A is present after 2 hours? SOLUTION Let be the amount of unconverted substance A at any time From the given assumption about the conversion rate, you can write the differential equation as shown.
Using separation of variables or a symbolic integration utility, you can find the general solution to be
General solution
To solve for the constants and use the initial conditions. That is, because when you can determine that Similarly, because when it follows that
which implies that So, the particular solution is
Substitute for k and C.
Particular solution
Using the model, you can determine that the amount of unconverted substance A after 2 hours is
In Figure F.5, note that the chemical conversion is occurring rapidly during the first hour. Then, as more and more of substance A is converted, the conversion rate slows down.
Checkpoint 1
Use the chemical reaction model in Example 1 to find the amount of substance A (in grams) as a function of (in hours) given that grams when and y 5 grams when t 2. ■
t y 40 t 0
y
5.45 grams.
y
5 t 1
y
1 12 t 1 60
k 121.
k 1 60
t 0, C 601. y 10 t 1,
C k , y 60
y
kt C
dy dt ky^2
y t.
t 1 ,
t 0,
Rate of change of y
is propor- tional to
the square of y.
y
t
60 50 40 30 20 10
(0, 60)
(1, 10) (2, 5.45)
y = 60 5 t + 1
1 2 3 Time (in hours)
Amo
unt (in
g rams)
Chemical Reaction
FIGURE F.
F20 Appendix F ■ Differential Equations
Example 2 Modeling Advertising Awareness
The new cereal product from Example 3 in Section F.1 is introduced through an advertising campaign to a population of 1 million potential customers. The rate at which the population hears about the product is assumed to be proportional to the number of people who are not yet aware of the product. By the end of 1 year, half of the population has heard of the product. How many will have heard of it by the end of 2 years? SOLUTION Let be the number (in millions) of people at time who have heard of the product. This means that is the number (in millions) of people who have not heard of it, and is the rate at which the population hears about the product. From the given assumption, you can write the differential equation as shown.
Using separation of variables or a symbolic integration utility, you can find the general solution to be General solution To solve for the constants and use the initial conditions. That is, because when you can determine that Similarly, because when it follows that which implies that
So, the particular solution is Particular solution This model is shown graphically in Figure F.6. Using the model, you can determine that the number of people who have heard of the product after 2 years is
FIGURE F.
Checkpoint 2
Repeat Example 2 given that by the end of 1 year, only one-fourth of the population have heard of the product. ■
y
t
(2, 0.75) (1, 0.50)
(0, 0) 1 2 3 4 5 Time (in years)
Potential c
ustomers (in millions)
Advertising Awareness
y = 1 − e −0.693 t
0.75 or 750,000 people.
y 1 e 0.693^2
y 1 e 0.693 t.
k ln 0.5 0.693.
0.5 1 e k ,
t 0, C 1. y 0.5 t 1,
C k , y 0
y 1 Ce kt.
dy dt
k 1 y
dy dt
1 y
y t
Rate of change of y
is propor- tional to
the difference between 1 and y.
