Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 103 pdf

For each differential equation below, sketch the slope field and find the general solution.. 31.3 QUALITATIVE ANALYSIS OF SOLUTIONS TO AUTONOMOUS DIFFERENTIAL EQUATIONS One way of gettin

19 Let M = M(t) be the amount of money in a trust fund earning interest at an annual interest rate of r compounded continuously Suppose money is withdrawn at a rate of

wdollars per year Assuming the withdrawals are being made continuously, we can use a differential equation to model the situation Analysis of this differential equation shows that the rate of change of money in the account at t = 0 could be either positive or negative, depending on the size of M(0) = M0 Find the threshold value of M0 Then,

by analyzing the sign of the first and second derivatives, argue that if M0is less than this threshold value, then M(t) is decreasing and concave down, and if M0is greater than this threshold value, then M(t) is increasing and concave up

20 A miser spends money at a rate proportional to the amount he has Suppose that right now he has $100,000 stowed under his mattress; he does not pay any taxes and does not earn any return on his money Assume that this is all the money he has and that he has no other source of income At the moment he is spending the money at a rate of

$10,000 per year

(a) At what rate will he be spending money when he has $50,000?

(b) At what time will the amount of money be down to $10,000?

21 A drosophila colony (a colony of fruit flies) is being kept in a laboratory for study It is being provided with essentially unlimited resources, so if left to grow, the colony will grow at a rate proportional to its size If we let N (t) be the number of drosophila in the colony at time t, t given in weeks, then the proportionality constant is k

(a) Write a differential equation reflecting the situation

(b) Solve the differential equation using N0to represent N (0)

(c) Suppose the drosophila are being cultivated to provide a source for genetic study, and therefore drosophila are being siphoned off at a rate of S drosophila per week Modify the differential equation given in part (a) to reflect the siphoning off (d) One of your classmates is convinced that the solution to the differential equation

in part (c) is given by

N (t ) = N0ekt− St

Show him that this is not a solution to the differential equation

(e) Your classmate is having a hard time giving up the solution he brought up in part (d) He sees that it does not satisfy the differential equation, but he still has a strong gut feeling that it ought to be right Convince him that it is wrong by using

a more intuitive argument Use words and talk about fruit flies

22 Solve the differential equations below Find the general solution

(a) dydt = sin 3t (b) dydt = 5 · 2t (c)dxdt =t +1t (d)dxdt =t +1t2

23 Solve the differential equations below Find the general solution

(a)dydt = 3t + 5 (b) dydt = 3y (c)dydt = −y

(d) dy= 0 (e)dy = 3y − 6

24 For each differential equation below, sketch the slope field and find the general solution (a) dydt = −y (b) dydt = −t (c)dydt = e−t

25 Each function below is a solution to one of the second order differential equations listed

To each function match the appropriate differential equation C1and C2are constants

Differential Equations

I ddt2x2 − 9x = 0 II.ddt2x2 + 9x = 0 III.ddt2x2 = 3x

Solution Functions

(a) x(t) = 5e3t (b) x(t) = −2e√3t (c) x(t) = 7 sin 3t (d) x(t) = C1sin 3t + C2cos 3t (e) x(t) = C1e

√ 3t + C2e−

√ 3t

26 For what value(s) of β, if any, is (a) y = C1sin βt a solution to y= 16y?

(b) y = C2cos βt a solution to y= 16y?

(c) y = C3eβta solution to y= 16y?

27 For what value(s) of β, if any, is (a) y = C1sin βt a solution to y= −16y?

(b) y = C2cos βt a solution to y= −16y?

(c) y = C3eβta solution to y= −16y?

28 (a) There are two values of λ such that y = eλt is a solution to y+ 7y+ 12y = 0 Find them and label them λ, and λ2

(b) Let y = C1eλ1 t

+ C2eλ2 t, where C1and C2are arbitrary constants Verify that y(t)

is a solution to y+ 7y+ 12y = 0

29 (a) Find λ such that y = eλtis a solution to y+ 4y+ 4y = 0

(b) Verify that y = teλt is also a solution to y+ 4y+ 4y + 0

31.3 QUALITATIVE ANALYSIS OF SOLUTIONS TO

AUTONOMOUS DIFFERENTIAL EQUATIONS

One way of getting information about the behavior of solutions to a differential equation

is to use the differential equation itself to sketch a picture of the solution curves This will not give us a formula for the solutions; however, a graphical approach will often give us enough qualitative information about the solutions to answer some important questions

In this section we will focus exclusively on autonomous differential equations, differ-ential equations of the formdydt = f (y)

EXAMPLE 31.14 Newton’s law of cooling, revisited. Suppose a hot or cold beverage is put in a room that is

kept at 65 degrees Then the rate of change of the temperature of the beverage is proportional

to the difference between the temperature of the room and the temperature of the drink.

dT

dt = k(65 − T ), where k is a positive constant and T = T (t) is the temperature of the beverage at time t Answer the following questions by taking a graphical perspective

(a) What must the temperature of the beverage be in order for its temperature to remain constant?

(b) For what temperatures is the beverage cooling down? In other words, where is dTdt negative?

(c) Sketch representative solution curves corresponding to a variety of initial conditions

SOLUTION Solving a differential equation involving dT

dt means finding temperature, T , as a function

of time, t, so we label the vertical axis T and the horizontal axis t

(a) If a quantity is not changing, then its derivative must be zero dT

dt = k(65 − T ) is zero only if T = 65 The beverage’s initial temperature must be 65◦ in order to remain constant This agrees perfectly with our intuition

(b) Because the equation dTdt = k(65 − T ) is a continuous function of T , the sign of dTdt can only change around the zeros ofdT

dt; dT

dt = 0 only at T = 65

We draw the T number line vertically because temperature, T , is plotted on the vertical axis

When T > 65, dTdt is negative; this makes sense because a hot beverage will cool off When T < 65,dTdt is positive; this makes sense because a cold beverage will warm up (c) We use the information contained in Figure 31.3 dTdt = k(65 − T ), so the further T is from 65 the greater the magnitude of the slope and the closer T is to 65 the more gentle the slope

65 65

T (t) is

decreasing

T (t) is

increasing

dT

dt < 0 dT

dt = 0 dT

dt > 0

T T

t

Figure 31.3

65 65°

T

t

Figure 31.4

The nonconstant solutions are asymptotic to the constant solution, T = 65, meaning that the temperature of the drink will approach the temperature of the room The constant solution T (t) = 65 is called the equilibrium solution

D e f i n i t i o n

An equilibrium solution to a differential equation is a solution that is constant for all

values of the independent variable (often t = time) Ifdydt = f (y), then the equilibrium solutions can be found by settingdydt = 0 and solving for y Equilibrium solutions are

also referred to as constant solutions.

The next example will serve as a case study of differential equations of the formdxdt = f (y), where f (y) is continuous

EXAMPLE 31.15 Do a qualitative analysis of the solutions to the differential equation

dy

dt = (y − 1)(y − 3)

Sketch representatives of the family of solutions

SOLUTION Solving the differential equation means finding y as a function of t, so we label the vertical

axis y and the horizontal axis t

First we identify the equilibrium, or constant solutions, solutions for whichdydt = 0 These correspond to horizontal lines in the ty-plane The constant solutions are y = 1 and y = 3 Becausedydt is continuous, the sign ofdydt can only change around the zeros ofdydt

For y > 3, dy

dt >0 ⇒ y(t) is increasing

For 1 < y < 3, dy

dt <0 ⇒ y(t) is decreasing

For y < 1, dy

dt >0 ⇒ y(t) is increasing

t

y = 3

y = 1 3

1

2

y

t

y = 3

y = 1 3

1 2

y

3 1

(+) (–) (+)

sign of dy

dt graph of y

Figure 31.5

The equilibrium solutions ofdydt = f (y) divide the plane into horizontal strips f (y) is continuous so two distinct solutions to the differential equation cannot intersect Therefore, each nonconstant solution lies completely within one strip We check the sign ofdydt in each

of these strips Because f (y) is a continuous function, the sign of dydt within each strip does not change Therefore each nonconstant solution is either strictly increasing or strictly decreasing

Representative solution curves are given in Figure 31.6 Note that within each horizontal band the solutions are of the same form (In each band one arbitrary solution has been highlighted; the other solutions within the band can be obtained by shifting the highlighted solution horizontally.)

A t

3

1 2

Figure 31.6

Each solution in a bounded strip is asymptotic to a constant solution (Why? The solutions are horizontal translates and there will not be an “empty” horizontal band.) Each solution

in an unbounded strip is either asymptotic to a constant solution or increases or decreases without bound

In summary, given a differential equation of the form dydt = f (y), where f (y) is continuous we know the following:

The solution curves will not intersect

The constant solutions (equilibrium solutions) partition the ty-plane into horizontal strips in which each solution is of the same type (i.e., the solutions are horizontal translates of one another)

Within each horizontal strip the solution curve is either strictly increasing or strictly decreasing because the sign ofdy can only change around the zeros of f (y)

Every solution curve is either asymptotic to a constant solution or increases or decreases without bound

Equilibria and Stability

Equilibrium solutions to differential equations can be classified as stable, unstable, or semistable In Example 31.5, y = 1 and y = 3 are the equilibrium solutions The equilibrium

at y = 1 is referred to as a stable equilibrium; under slight perturbation the system will

tend back toward the equilibrium For instance, if y is slightly less than 1, as t → ∞, y increases toward 1 Similarly, if y is a bit more than 1 (i.e., anything under 3), then as

t → ∞, y decreases toward 1 The equilibrium at y = 3 is an unstable equilibrium; under

slight perturbation the system does not return to this equilibrium If y is slightly greater than

3, then as t → ∞, y increases without bound On the other hand, if y is slightly less than

3, then as t → ∞, y tends away from 3 and toward the stable equilibrium of 1

As illustration, consider two stationary coins, one lying flat and the other balancing

on its edge Since their positions are not changing with time, both coins are at equilibrium The former configuration is stable under slight perturbations The latter, however, is unstable because the coin balancing on its edge will topple under small perturbations

In terms of modeling real-world situations, stable equilibrium solutions are states that systems would naturally gravitate toward, while unstable equilibrium solutions are thresholds between two qualitatively different types of outcomes In any modeling situation knowing whether an equilibrium solution is stable or unstable is of tremendous importance, because real life is chock full of perturbations

y

y = c semistable

y = b stable

y = a

y

y = c

y = b

y = a

unstable

Figure 31.7

EXAMPLE 31.16 Find and classify the equilibrium solutions of

dx

dt = x2− x

SOLUTION To find the equilibrium solutions, setdxdt = 0

dx

dt = x(x − 1) = 0

x = 0 or x = 1 Now look at the sign ofdx on either side of the equilibrium values

t

x = 1

x

t

x = 1 1

0

x

(+)

(+) (–)

sign of dx

dt

Figure 31.8

x(t ) = 0 is a stable equilibrium solution, while x(t) = 1 is an unstable equilibrium solution

EXERCISE 31.4 Do a qualitative analysis of the solutions of the differential equation

dS

dt = 0.01S(300 − S)

(This is the differential equation that arose in modeling the spread of a flu in a college dormitory.) Sketch some representative solution curves and any constant solutions Classify the equilibria After analyzing the solutions in the abstract, determine what this says about the spread of the flu

EXERCISE 31.5 Write a differential equation whose solution curves look like those sketched below (Begin

by constructing a differential equation with equilibrium solutions at y = 3 and y = −2.)

y

t

y = 3

y = –2 –2

3

Figure 31.9

EXAMPLE 31.17 An industrial plant produces radioactive material at a constant rate of 4 kilograms per year

The radioactive material decays at a rate proportional to the amount present and has a half-life of 20 years

(a) Write a differential equation whose solution is R(t), the amount of material present t years after this practice begins

(b) Sketch some representative solutions corresponding to different initial values of R Include the equilibrium solution Can we predict the level of radioactive material in the long run?

SOLUTION (a) The rate of production is constant at 4 kilograms per year regardless of the time and

regardless of the amount present

The rate out is proportional to the amount present, so it is kR, where k is a constant.

rate of change = rate of production − rate of decay dR

We must find the value of k using the half-life information We know that if a substance decays at a rate proportional to the amount present (and there are no further additions

to the amount), then it can be modeled by dSdt = kS The solution to this differential equation is S(t) = Cekt We know the half-life to be 20 years, so we can compute k

S(20) =12Ce0t

Ce20k= 0.5C

e20k= 0.5 20k = ln 0.5

k =− ln 2

20 ≈ −0.0347

In our case, sincedRdt = 4 − kR already incorporates the negative sign indicating decay,

k =ln 220 ≈ 0.0347

The differential equation is dRdt = 4 −ln 220R

(b) The equilibrium solution occurs where dRdt = 0 4 = ln 220R ⇒ R = ln 280 ≈ 115.42 kg This is a stable equilibrium

R (t)

R (t) =

t

80

ln 2

Figure 31.10

In the long run, the amount of radioactive material present approaches ln 280 kg, or approximately 115.42 kg

Logistic Population Growth

Because unlimited resources are not observed in the real world, population growth is not exponential indefinitely At a certain point members of the population begin to compete with one another for limited resources and the growth rate slows The larger the population, the more prominent the role of competition Ecologists and population biologists call the

number of animals a particular environment can support the carrying capacity of that

environment for that animal Suppose a Tanzanian savannah has the carrying capacity L for lions, where L is a fixed constant Let P = P (t) be the population of lions at time t

We expect the graph of P versus t to look something like the graph drawn in Figure 31.11 below

P = L

t

Figure 31.11

For P small the graph should be concave up As P gets closer to the carrying capacity we expect the population growth to slow down and hence the graph of P to be concave down The most basic observation is that there are constant solutions at P = 0 and P = L; if there are no lions we don’t expect lions to be spontaneously generated, and if the population

is at the carrying capacity we expect it to stay there (barring natural or unnatural disasters)

To formulate a mathematical model for the situation we’ll write a differential equation that has equilibrium solutions at P = 0 and P = L and is increasing for P ∈ (0, L)

dP

dt = kP (L − P ) will work, provided k is a positive constant

NOTE This equation,dP

dt = kLP − kP2, can be thought of as dP

dt = k1P − k2P2, where k1and k2are constant

Without the −k2P2 term this would just be the familiar exponential growth differential equation But because lions are competing with one another for limited resources we need

to incorporate a braking factor This braking factor cannot be of the form kP , as this would still give exponential growth; only the proportionality constant would have changed A braking factor of the form k(P · P ) is reasonable because the competition is proportional

to “interactions” between lions and lions

A population growth model of the form

dP

dt = kP (L − P )

is known as a logistic growth model The logistic model was first used in the early 1800s

by the Belgian demographer Pierre-Franc¸ois Verthust to model human world population The population figures he predicted for 100 years into the future were off by less than 1%

EXAMPLE 31.18 Suppose the number of fish in a lake grows according to the equation

dP

dt = 0.45P − 0.0005P2 (a) What is the lake’s carrying capacity for fish? Is it a stable equilibrium?

(b) What size is the fish population when it is growing most rapidly?

SOLUTION (a) Let’s sketch representative solution curves We begin by identifying the equilibrium

solutions

dP

dt = P (0.45 − 0.0005P ) = 0

P = 0 or 0.0005P = 0.45

P = 0.45 0.0005= 900

P P

t

stable 900

900

0

P

t

900 (+)

(–)

sign of dp

dt

Figure 31.12

From this analysis we see that the carrying capacity is 900 fish and that this is a stable equilibrium

(b) The fish population is growing most rapidly at the point of inflection of the curves lying

in the strip between P = 0 and P = 900 Our analysis so far tells us that there ought to

be at least one such point for each curve because the curve is asymptotic to both P = 0 and P = 900 We will show that for each solution curve in this interval there is only one such point of inflection and determine for what value of P this occurs

There are two different approaches we can take to this problem

Approach 1.We must determine the value of P in the interval (0, 900) such thatdPdt is maximum We want to know where

0.45P − 0.0005P2

is maximum But the expression 0.45P − 0.0005P2is quadratic, corresponding to a parabola with P -intercepts of 0 and 900, a parabola opening downward; hence its maximum value is at P = −0.45

2(−0.0005)=0.450.01= 450 The point of inflection is at P = 450

Approach 2. dPdt will be maximum at the point at which the solution curve changes concavity The point of inflection of P (t) is the point at which the sign of d2P

dt 2 changes dP

dt = 0.45P − 0.0005P2

Differentiate both sides of the equation above with respect to t keeping in mind that P

itself is a function of t