Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 102 pdf

At any point P in the ty-plane we can use the differential equation to find the slope of the tangent line to the solution curve through P.. In particular the Existence and Uniqueness The

6 Let’s suppose that the population in a certain country has a growth rate of 2% and

a population of 9 million at a time we’ll designate as t= 0 Due to the political and economic situation, there is a massive rearrangement of populations in the region The immigration and emigration rates are both constant, with people entering the country

at a rate of 100,000 per year and leaving at a rate of 300,000 per year Let P= P (t) be the population in millions at time t

(a) Write a differential equation reflecting the situation Keep in mind that P is in millions.

(b) If this situation goes on indefinitely, what will happen to the country’s population? (c) What initial population would support a net emigration of 200,000 per year?

7 In the beginning of a chemical reaction there are 600 moles of substance A and none

of substance B Over the course of the reaction, the 600 moles of substance A are converted to 600 moles of substance B (Each molecule of A is converted to a molecule

of B via the reaction.) Suppose the rate at which A is turning into B is proportional to the product of the number of moles of A and the number of moles of B

(a) Let N= N(t) be the number of moles of substance A at time t Translate the

statement above into mathematical language (Note: The number of moles of

substance B should be expressed in terms of the number of moles of substance A.)

(b) Using your answer to part (a), findd2N

dt 2 Your answer will involve the proportion-ality constant used in part (a)

(c) N (t) is a decreasing function The rate at which N is changing is a function of N , the number of moles of substance A When the rate at which A is being converted

to B is highest, how many moles are there of substance A?

8 There are many places in the world where populations are changing and immigration and/or emigration play a big role People may move to find food, or to find jobs, or

to flee political or religious persecution Pick a situation that interests you You could look at the number of Tibetans in Tibet, or the number of Tibetans in India, or the number of lions in the Serengeti, or the number of tourists in Nepal Get some data and try to model the population dynamics using a differential equation What simplifying assumptions have you made?

AN INTRODUCTION

Although knowing about the rate of change of a quantity is useful, often what we really

want to know about is the actual amount of that quantity After all, it’s nice to know that

the money in your bank account is growing at an instantaneous rate of 5%, but what you really want to know is when you’ll finally have enough money to buy that new motorcycle

or whatever it is that you’re saving for In this section we turn our attention to the solutions

of differential equations

What Does It Mean to Be a Solution to a Differential Equation?

We have said that a function f is a solution to a differential equation if it satisfies the

differential equation By this we mean that when the function and its derivative(s) are substituted in the appropriate places in the differential equation, the two sides of the equation are equal We review work presented in Section 15.2

EXAMPLE 31.10 Is y= x3a solution to the differential equationdydx =3yx?

SOLUTION To determine whether y= x3is a solution, we need to substitute it into the differential

equation

dy

dx =3yx Replace y by x3wherever it appears

d

dxx

3 ?

= 3x

3

x 3x2 ?

= 3x

3

x 3x2= 3x2 True

y= x3is a solution todxdy=3yx because it satisfies the equation

EXAMPLE 31.11 Is y= xe3xa solution to the differential equationdxdy=3yx?

SOLUTION

dy

dx =3yx Replace y by xe3xwherever it appears

d

dx[xe3x] ?

= 3xe

3x

x 3xe3x+ e3x ?

= 3e3x

e3x[3x+ 1] = e3x(3) The equation is not satisfied, so y= xe3xis not a solution to the differential equation.
Differential equations have families of solutions In Section 15.2 we looked at the family

of solutions to each of the three differential equations given below The discussion is summarized here

i dydt = 2 ii dydt = 2t iii dydt = 2y Graphical perspective: Solving a differential equation that involvesdydt means finding y as

a function of t; therefore, on our graph we will label the vertical axis y and the horizontal axis t

At any point P in the ty-plane we can use the differential equation to find the slope of the tangent line to the solution curve through P We’ll draw a short line segment through

P indicating the slope of the solution curve there The resulting diagram is called a slope

field (See Figure 31.1 on the following page.)

Observations

In part (i), dy= 2, the slope is independent of the point P chosen

In part (ii),dydt = 2t, the slope depends only on the t-coordinate of P

This is true whenever we have a differential equation of the form dydt = f (t)

In part (iii), dydt = 2y, the slope depends only on the y-coordinate of P

This is true whenever we have a differential equation of the form dydt = f (y)

y

t

y

t

y

t

(i) dy

dy

dy

dt = 2y

Figure 31.1

In Figure 31.1 (ii) the slope is positive whenever t is positive, negative whenever t is negative, and zero at t= 0 At (2, 3) the slope is 4; at (1, 5) the slope is 2

In Figure 31.1 (iii) the slope is positive whenever y is positive, negative whenever y is negative, and zero at y= 0 At (2, 3) the slope is 6; at (1, 5) the slope is 10

By following the slope fields, we can get a rough idea of the shapes of the solution curves (See Figure 31.2.)

y

t

y

t

y

t C>O

C<O

(i) dy

dy

dy

dt = 2y

y = Ce2t

y = t2 + C

y = 2 t + C

Figure 31.2

We state, without proof, the following fact (known as the Existence and Uniqueness Theorem).4

4

E x i s t e n c e a n d U n i q u e n e s s T h e o r e m ( W e a k f o r m )

Let (a, b) be a point in the plane Any differential equation of the form dydt = g(t) where g is continuous, or of the form dydt = f (y) where f and fare continuous, has

a solution passing through (a, b) The solution exists, and it is unique

In particular the Existence and Uniqueness Theorem tells us given any point P= (a, b) in the plane, a differential equation of the formdydt = k,dydt = kt, ordydt = ky where k is constant has exactly one solution passing through the point P This makes sense in our examples above; because the slope at P is completely determined by the coordinates of P , no two solution curves can cross We can combine this graphical analysis with our knowledge of analytic solutions to these differential equations.5

y= 2t + C is a solution to part (i) for any constant C

y= t2+ C is a solution to part (ii) for any constant C

y= Ce2t is a solution to part (iii) for any constant C

The Existence and Uniqueness Theorem tells us that we have written general solutions to each of these differential equations In other words, any solution todydt = 2 can be expressed

in the form y= 2t + C; any solution to dydt = 2t can be expressed in the form y = t2+ C; any solution todydt = 2y can be expressed in the form y = Ce2t

EXERCISE 31.1 Let k be an arbitrary constant Show that for any point P in the plane there is a unique value

of C such that the curve y= Cektpasses through P

EXERCISE 31.2 Let C be an arbitrary constant Show that for any point P in the plane there is a unique value

of C such that the curve y= t2+ C passes through P Suppose we know the general solution of a first order differential equation We can determine

a particular solution if an initial condition is specified Geometrically this is equivalent to knowing one point through which the solution curve passes

The conclusions drawn from these specific examples can be generalized

1 Suppose a differential equations is of the form dydt = f (t).

. The slope of the solution curve at P is determined completely by the t-coordinate

. If F is an antiderivative of f , then F (t)+ C is the general solution todydt = f (t) Solvingdydt = f (t) is equivalent to finding f (t ) dt

. The solutions todydt = f (t) are vertical translates of one another (They differ from

one another by an additive constant.)

2 Suppose a differential equation is of the form dydt = f (y).

. The slope of the solution curve at P is determined completely by the y-coordinate

. Solutions todydt = f (y) are horizontal translates of one another.

5 See Section 15.2 for a review of solutions to differential equations of the formdy= ky.

Solving Differential Equations: Analytic Solutions

Solving a differential equation can be quite difficult However, there are several types of differential equations that we can already solve In this section we will look at differential equations of the form dydt = f (t) and dydt = ky In Section 31.3 we’ll look qualitatively at solutions to differential equations of the form dydt = f (y) Then in Section 31.4 we will look at solutions to a larger class of differential equations, differential equations of the form

dy

dt = f (y)g(t) The type of differential equations we look at here are special cases of those

we will look at in Section 31.4

Differential Equations of the Form dydt = f (t)

Differential equations of the form dydt = f (t) can be solved by integration To solve such

an equation is to find a function y whose derivative is f (t), i.e., to find an antiderivative of

f (t ) If dydt = f (t), then y = f (t ) dt

EXAMPLE 31.12 The differential equation governing the path of a projectile can be solved by finding

antiderivatives In Example 31.7 we considered an object falling through the air Ignoring air resistance, we say that the object undergoes a constant downward acceleration of 32 feet per second due to the force of gravity The object’s initial vertical velocity, v(0), is denoted by the constant v0and its initial height, s(0), by s0 Solve the differential equations modeling this situation

SOLUTION Let v(t) be the vertical component of the velocity of the object at time t

dv

dt = −32 v(t )=

−32 dt v(t )= −32t + C1

We solve for C1using v(0)= v0to get

v0= −32(0) + C1, so C1= v0

v(t )= −32t + v0

We know that v(t)=dsdt, where s(t) is the height (vertical position) of the object at time t

ds

dt = −32t + v0

s(t )=

(−32t + v0) dt

Then s(t)= −32t

2

2 + v0t+ C2 Using s(0)= s0, we can solve for C2to get

s0= −32 ·0

2 + v0· 0 + C2, so C2= s0 s(t )= −16t2+ v0t+ s0

Review: Differential Equations of the Form dydt = ky

In Section 15.2 we looked at the differential equationdydt = ky, an equation that models any situation in which a quantity grows or decays at a rate proportional to the amount of the quantity itself and conclude the following

The general solution to dy

dt = ky is y(t) = Cekt, where C is an arbitrary constant

The general solution todydt = ky, together with an initial condition, enables us to determine

a particular solution Using substitution, we are able to get quite a bit of mileage out of knowing how to find a solution to differential equations of this form

EXAMPLE 31.13 Solve the following differential equations

(a)dPdt = −2P (b) dtd(700− F ) = −0.01(700 − F ) (c)dTdt = −k(T − 65)

SOLUTION Our strategy is to put each of these equations in the formdydt = ky Once in this form we

know y(t)= Cekt (a)dP

dt = −2P The general solution is P (t) = Ce−2t (b) d

dt(700− F ) = −0.01(700 − F )

700− F (t) plays the role of the dependent variable, y; the constant of proportionality

is−0.01

The general solution is 700− F (t) = Ce−0.01t

We can write F (t)= 700 − Ce0.01t (c)dTdt = −k(T − 65) We need to rewrite this to get it into the formdydt = ky

Let y= T − 65 Then dydt =dTdt and the original equation becomesdydt = −ky

y= Cekt The general solution is T − 65 = Ce−kt, or T (t)= 65 + Ce−kt

We see that knowing how to solve any differential equation of the form

dy

dt = ky allows us to solve dTdt = kT − b because we can use substitution to transform it into the form dydt = ky Using substitution, we can solve several of the other differential equations that arose from examples in Section 31.1, namely,

(a)dCdt = k(C − N), where k and N are constants (Example 31.2);

(b) dAdt = 5 − kA, where k is a constant (Example 31.5); and (c)dMdt = 0.05M − 4000 (Example 31.6)

We solve by using substitution to alter the form to bedy= ky

(a) dCdt = k(C − N) can be solved using the substitution y = C − N, where N is constant (b) dAdt = 5 − kA can be solved either by writingdAdt = −kA + 5 = −k A−5k

and using the substitution y= A −5k orsimply by using the substitution y= 5 − kA

(c) dMdt = 0.05M − 4000 can be solved by writing dMdt = 0.05 M−40000.05and using the substitution y= M −40000.05 orby letting y= 0.05M − 4000

We can solve any differential equation of the formdPdt = aP + b either by rewriting it first as

dP

dt = a P +b

a

and then using the substitution y= P +b

ato get it into the formdydt = ay

or by using the substitution u(t)= aP + b, P =1a[u(t)− b]

EXERCISE 31.3 Solve the differential equation dMdt = 0.05M − 4000 with the initial condition M(0) =

10,000 (This is the differential equation from Example 31.6.) Check your answer to make sure it works

The differential equation dPdt = aP + b is a special case of differential equations of the formdydt = f (y) Differential equations of the formdydt = f (y) are called autonomous

differential equations.The word autonomous means “not controlled by outside forces”;

in an autonomous differential equation the rate of change of y is controlled (determined) only by y, not by other outside “forces.” In other words, dydt can be described in terms of

yonly, without reference to t or any other variables In the next section we will study the behavior of solutions to autonomous differential equations (such as dS

dt = kS(300 − S)) from a qualitative perspective

P R O B L E M S F O R S E C T I O N 3 1 2

1 Which one of the following is a solution to the differential equation y(t )= −5y?

2 Which one of the following is a solution to the differential equation y (t )= −25y?

3 (a) Verify that y(t)= Cektis a solution to the differential equationdydt = ky (b) Verify that y= ket is not a solution todydt = ky

(c) Verify that y= ekt+ C is not a solution to dydt = ky

4 Which of the following is a solution todydx =−yx +(ln x)y2 2? (a) y= x + 1 (b) y= 1 + 1/x (c) y=ln xx (d) y= ln x

5 Determine which of the following functions are solutions to each of the differential equations below (A given differential equation may have more than one solution.) Differential Equations:

i dydt = t ii dydt = y iii dydt = et iv.d2y2 = 4y

Solution choices:

(a) y=t2

6 Which of the following is a solution to the differential equation

y − y− 6y = 0?

(a) y= Cet (b) y= sin 2t (c) y= 5e3t+ e−2t (d) y= e3t− 2

7 Which of the following is a solution to the differential equation

y + 9y = 0?

(a) y= e3t+ e−3t (b) y= Cet− t (c) y= C(t2+ t) (d) y= sin 3t + 6 (e) y= 5 cos 3t

8 Which of the following is a solution to the differential equation

dy

dt = y + 1?

(a) y= Cet (b) y= Cet− t (c) y= C(t2+ t) (d) y= Cet− 1 (e) y= Ce−t+ 1

9 Is y=xe2x +3xex a solution to the differential equation xdydx+ (1 − x)y = xex? Justify your answer

10 Solve the following differential equations by using the method of substitution to put them into the form dydt = ky

(a) dP

dt = 0.3(1000 − P ) (b) dM

dt = 0.4M − 2000

11 Solve the following

(a) dxdt = 6 − 2x Do this using substitution in two ways

i Factor out a−2 from the right-hand side and let u = x − 3 Then solve

ii Let v= 6 − 2x Express dx

dt in terms of v and then convert the equation

dx

dt = 6 − 2x to an equation in v and dvdt and solve

(b) dx

dt = 3x − 7 (c) dydt = ky + B

12 When a population has unlimited resources and is free from disease and strife, the rate at which the population grows in often modeled as being proportional to the population Assume that both the bee and the mosquito populations described below behave according to this model

In both scenarios described below you are given enough information to find the proportionality constant k In one case the information allows you to find k solely using the differential equation, without requiring that you solve it In the other scenario you must actually solve the differential equation in order to find k

(a) Let M= M(t) be the mosquito population at time t, t in weeks At t = 0 there are

1000 mosquitoes Suppose that when there are 5000 mosquitoes the population

is growing at a rate of 250 mosquitoes per week Write a differential equation reflecting the situation Include a value for k, the proportionality constant (b) Let B= B(t) be the bee population at time t, t in weeks At t = 0 there are 600 bees When t= 10 there are 800 bees Write a differential equation reflecting the situation Include a value for k, the proportionality constant

13 The population in a certain country grows at a rate proportional to the population at time t, with a proportionality constant of 0.03 Due to political turmoil, people are leaving the country at a constant rate of 6000 people per year Assume that there is no immigration into the country Let P= P (t) denote the population at time t

(a) Write a differential equation reflecting the situation

(b) Solve the differential equation for P (t) given the information that at time t= 0 there are 3 million people in the country In other words, find P (t), the number of people in the country at time t

14 A population of otters is declining New otters are born at a rate proportional to the population with constant of proportionality 0.04, but otters die at a rate proportional

to the population with constant 0.09 Today, the population is 1000 A group of people wants to try to prevent the otter population from dying out, so they plan to bring in otters from elsewhere at a rate of 40 otters per year We’ll model the situation with continuous functions Let P (t) be the population of the otters t years after today (a) Write a differential equation whose solution is P (t)

(b) Solve this differential equation Your answer should include no unknown con-stants

(c) According to this model, will the attempt to save the otter population work? Explain your answer If it won’t work, at what rate must otters be brought in to ensure the population’s survival? If it will work, for how many years must the importation of otters continue?

15 (a) Suppose a hot object is placed in a room whose temperature is kept fixed at F degrees Let T (t) be the temperature of the object Newton’s law of cooling says that the hot object will cool at a rate proportional to the difference in temperature between the object and its environment Write a differential equation reflecting this statement and involving T Explain why this differential equation is a special case of the differential equation in Example 31.3 part (a)

(b) What is the sign of the constant of proportionality in the equation you wrote above? Explain

(c) Suppose that we are interested in the temperature of a cold cup of lemonade as it warms up to room temperature Let L(t) represent the temperature of the lemonade

at time t and assume that it sits in a room that is kept at 65 degrees At time t= 0 the lemonade is at 40 degrees Fifteen minutes later it has warmed to 50 degrees

i Sketch a graph of L(t) using your intuition and the information given

ii Is L(t) increasing at an increasing rate or a decreasing rate?

iii Write a differential equation reflecting the situation Indicate the sign of the proportionality constant

iv Find L(t) Your final answer should have no undetermined constants and should be consistent with the answer to part (i)

v How long will it take the lemonade to reach a temperature of 55 degrees?

16 Money in a bank account is earning interest at a nominal rate of 4% per year com-pounded continuously Withdrawals are made at a rate of $8000 per year Assume that withdrawals are made continuously

(a) Write a differential equation modeling the situation

(b) Depending on the initial deposit, the amount of money in the account will either in-crease, dein-crease, or remain constant Explain this in words; refer to the differential equation

(c) Suppose the money in the account remains constant What was the initial deposit? For what initial deposits will the amount of money in the account actually continue

to grow?

(d) Show that M(t)= M0e0.04t− 8000t is not a solution to the differential equation

you got in part (a)

17 Suppose a population is changing according to the equationdP

dt = kP − E, where E is the rate at which people are emigrating from the country As established in part (d) of the previous problem, P (t)= P0ekt− Et is not a solution to this differential equation.

(a) Use substitution to solve dPdt = kP − E (Your answer ought to agree with that given in part (b).)

(b) Verify that P (t)= Cekt+E

k, where C is a constant, is a solution to the differential equation dPdt = kP − E

18 P (t)= Cekt+E

k, where C is a constant, is the general solution to the differential equation dPdt = kP − E Below is the slope field fordPdt = 2P − 6

P

t

5 4 3 2 1

–1 –2 –3

(a) i Find the particular solution that corresponds to the initial condition P (0)= 2

ii Sketch the solution curve through (0, 2)

(b) i Find the particular solution that corresponds to the initial condition P (0)= 3

ii Sketch the solution curve through (0, 3)

(c) i Find the particular solution that corresponds to the initial condition P (0)= 4

ii Sketch the solution curve through (0, 4)

y= 2t + C is a solution to part (i) for any constant...

y= t2+ C is a solution to part (ii) for any constant C

y= Ce2t is a solution to part (iii) for any constant C

The Existence and Uniqueness Theorem tells us that... ky, an equation that models any situation in which a quantity grows or decays at a rate proportional to the amount of the quantity itself and conclude the following

The general solution to