Ebook Advanced engineering mathematics (7th edition) Part 1

(BQ) Part 1 book Advanced engineering mathematics has contents: Linear second order equations, first order differential equations, the laplace transform, series solutions, approximation of solutions, vectors and vector spaces, matrices and linear systems,... and other contents.

Guide to Notation

L[ f ] Laplace transform of f

L[ f ](s) Laplace transform of f evaluated at s

L−1[F ] inverse Laplace transform of F

H (t) Heaviside function

f ∗ g often denotes a convolution with respect to an integral transform, such as the Laplacetransform or the Fourier transform

δ(t) delta function

< a, b, c > vector with components a, b, c

F · G dot product of vectors F and G

F × G cross product of F and G

R n n-space, consisting of n-vectors < x1, x2, · · · , x n >

[a i j] matrix whose i , j-element is a i j If the matrix is denoted A, this i , j element may also be

denoted A i j

At transpose of A

AR reduced (row echelon) form of A

rank(A) rank of a matrix A [A B] augmented matrix

A−1 inverse of the matrix A

|A| or det(A) determinant of A

pA(λ) characteristic polynomial of A

 often denotes the fundamental matrix of a system X= AX

T often denotes a tangent vector

N often denotes a normal vector

n often denotes a unit normal vector

∂( f, g)

∂(u, v) Jacobian of f and g with respect to u and v

 

 f (x, y, z) dσ surface integral of f over 

f (x0−), f (x0+) left and right limits, respectively, of f (x) at x0

F[ f ] or ˆf Fourier transform of f

F[ f ](ω) or ˆF(ω) Fourier transform of f evaluated at ω

F−1 inverse Fourier transform

F C [ f ] or ˆ f C Fourier cosine transform of f

F−1

C or ˆf C−1 inverse Fourier cosine transform

F S [ f ] or ˆ f S Fourier sine transform of f

F−1

S or ˆf S−1 inverse Fourier sine transform

D[u] discrete N - point Fourier transform (DFT) of a sequence u j

ˆfwin windowed Fourier transform

χ I often denotes the characteristic function of an interval I

σ N (t) often denotes the N th Cesàro sum of a Fourier series

Z (t) in the context of filtering, denotes a filter function

P n (x) nth Legendre polynomial

(x) gamma function

B (x, y) beta function

J ν Bessel function of the first kind of orderν

γ depending on context, may denote Euler’s constant

Y ν Bessel function of the second kind of orderν

I0, K0 modified Bessel functions of the first and second kinds, respectively, of order zero

∇2u Laplacian of u

Re(z) real part of a complex number z

Im(z) imaginary part of a complex number z

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Preface xi

PART 1 Ordinary Differential Equations 1

CHAPTER 1 First-Order Differential Equations 3

CHAPTER 2 Linear Second-Order Equations 43

CHAPTER 3 The Laplace Transform 77

3.3.1 The First Shifting Theorem 84

CHAPTER 4 Series Solutions 121

CHAPTER 5 Approximation of Solutions 137

PART 2 Vectors, Linear Algebra, and Systems of Linear Differential Equations 145

CHAPTER 6 Vectors and Vector Spaces 147

CHAPTER 7 Matrices and Linear Systems 187

CHAPTER 9 Eigenvalues, Diagonalization, and Special Matrices 267

CHAPTER 11 Vector Differential Calculus 345

CHAPTER 12 Vector Integral Calculus 367

PART 4 Fourier Analysis, Special Functions, and Eigenfunction Expansions 425

CHAPTER 13 Fourier Series 427

CHAPTER 14 The Fourier Integral and Transforms 465

14.3.3 The Shannon Sampling Theorem 485

CHAPTER 15 Special Functions and Eigenfunction Expansions 505

CHAPTER 16 The Wave Equation 565

16.3 Wave Motion in an Infinite Medium 579

CHAPTER 17 The Heat Equation 611

CHAPTER 18 The Potential Equation 641

PART 6 Complex Functions 667

CHAPTER 19 Complex Numbers and Functions 669

19.2.1 Limits, Continuity, and Differentiability 677

CHAPTER 20 Complex Integration 695

CHAPTER 21 Series Representations of Functions 715

CHAPTER 22 Singularities and the Residue Theorem 729

CHAPTER 23 Conformal Mappings and Applications 751

23.3 Conformal Mapping Solutions of Dirichlet Problems 776

APPENDIX A MAPLE Primer 789

Answers to Selected Problems 801

Index 867

Preface

This seventh edition of Advanced Engineering Mathematics differs from the sixth in four ways.

First, based on reviews and user comments, new material has been added, including thefollowing

• Orthogonal projections and least squares approximations of vectors and functions This vides a unifying theme in recognizing partial sums of eigenfunction expansions as projectionsonto subspaces, as well as understanding lines of best fit to data points

pro-• Orthogonalization and the production of orthogonal bases

• LU factorization of matrices

• Linear transformations and matrix representations

• Application of the Laplace transform to the solution of Bessel’s equation and to problemsinvolving wave motion and diffusion

• Expanded treatment of properties and applications of Legendre polynomials and Besselfunctions, including a solution of Kepler’s problem and a model of alternating current flow

• Heaviside’s formula for the computation of inverse Laplace transforms

• A complex integral formula for the inverse Laplace transform, including an application to heatdiffusion in a slab

• Vector operations in orthogonal curvilinear coordinates

• Application of vector integral theorems to the development of Maxwell’s equations

• An application of the Laplace transform convolution to a replacement scheduling problem

Third, there is an added emphasis on constructing and analyzing models, using ordinary andpartial differential equations, integral transforms, special functions, eigenfunction expansions,and matrix and complex function methods

Finally, the answer section in the back of the book has been expanded to provide moreinformation to the student

This edition is also shorter and more convenient to use than preceding editions The chapterscomprising Part 8 of the Sixth Edition, Counting and Probability, and Statistics, are now available

on the 7e book website for instructors and students

Supplements for Instructors:

• A detailed and completely revised Instructor’s Solutions Manual and

• PowerPoint Slidesare available through the Instructor’s Resource site at login.cengage.com

Supplements for Students:

CourseMate from Cengage Learning offers students book-specific interactive learningtools at an incredible value Each CourseMate website includes an e-book and interactivelearning tools To access additional course materials (including CourseMate), please visitwww.cengagebrain.com At the cengagebrain.com home page, search for the ISBN of your title(from the back cover of your book) using the search box at the top of the page This will take you

to the product page where these resources can be found

In preparing this edition, the author is indebted to many individuals, including:

Charles S Campbell, University of Southern CaliforniaDavid Y Gao, Virginia Tech

Donald Hartig, California Polytechnic State University, San Luis ObispoKonstantin A Lurie, Worcester Polytechnic Institute

Allen Plotkin, San Diego State UniversityMehdi Pourazady, University of ToledoCarl Prather, Virginia Tech

Scott Short, Northern Illinois University

PETER V O’NEIL

P A R T

1

Ordinary Differential Equations

CHAPTER 1

First-Order Differential Equations

CHAPTER 2

Linear Second-Order Equations

C H A P T E R 1

First-Order Differential Equations

L I N E A R E Q U AT I O N S E X A C T E Q U AT I O N S

H O M O G E N E O U S B E R N O U L L I A N D R I C C AT I

E Q U AT I O N S E X I S T E N C E A N D U N I Q U E N E S S

Part 1 of this book deals with ordinary differential equations, which are equations that contain

one or more derivatives of a function of a single variable Such equations can be used to model arich variety of phenomena of interest in the sciences, engineering, economics, ecological studies,and other areas

We begin in this chapter with first-order differential equations, in which only the firstderivative of the unknown function appears As an example,

y+ xy = 0

any function satisfying the equation It is routine to check by substitution that y = ce −x2/2is a

solution of y+ xy = 0 for any constant c.

We will develop techniques for solving several kinds of first-order equations which arise inimportant contexts, beginning with separable equations

A differential equation is separable if it can be written (perhaps after some algebraic

manipulation) as

d y

in which the derivative equals a product of a function just of x and a function just of y.

This suggests a method of solution

Step 1 For y such that G (y) = 0, write the differential form

G(y) d y=



F (x) dx.

Step 3 Attempt to solve the resulting equation for y in terms of x If this is possible, we have

an explicit solution (as in Examples 1.1 through 1.3) If this is not possible, the solution

is implicitly defined by an equation involving x and y (as in Example 1.4).

Step 4 Following this, go back and check the differential equation for any values of y such that

G(y) = 0 Such values of y were excluded in writing 1/G(y) in step (1) and may lead

to additional solutions beyond those found in step (3) This happens in Example 1.1

Observe that y= 0 by itself satisfies the differential equation, hence it provides another solution

(called a singular solution).

In summary, we have the general solution

y (x) = 1

e −x − k

This expression for y (x) is called the general solution of this differential equation because

it contains an arbitrary constant We obtain particular solutions by making specific choices for

k In Example 1.1,

0 0 –0.2

integral curves of the differential equation Graphs of these integral curves are shown in

an equation for the solution corresponding to that k, but not yet an explicit expression for this

sides of the equation to get

|1 + y| = e k

e −1/x = ae −1/x ,

Eliminate the absolute value symbol by writing

1+ y = ±ae −1/x = be −1/x ,

y = −1 + be −1/x with b= 0

–2 0

FIGURE 1.2 Some integral curves from Example 1.2.

Example 1.1, we can include this singular solution in the solution by separation of variables by

in which b can be any real number, including zero This expression contains all solutions.

Each of these examples has infinitely many solutions because of the arbitrary constant in the

general solution If we specify that the solution is to satisfy a condition y (x0) = y0 with x0 and

y0 given numbers, then we pick out the particular integral curve passing through(x0, y0) The

differential equation, together with a condition y (x0) = x0, is called an initial value problem The condition y (x0) = y0is called an initial condition.

One way to solve an initial value problem is to find the general solution and then solve forthe constant to find the particular solution satisfying the initial condition

case, we must be satisfied with an equation implicitly defining the general solution or the solution

of an initial value problem

–0.8 –0.4

–1.2

–2 –1.6

x

3 2

1

FIGURE 1.3 Graph of the solution of Example 1.4.

Some Applications of Separable Equations

Separable differential equations arise in many contexts We will discuss three of these

EXAMPLE 1.5 Estimated Time of Death

A homicide victim is discovered and a lieutenant from the forensics laboratory is summoned toestimate the time of death

can be used to estimate the last time at which the victim was alive and had a “normal” bodytemperature This last time was the time of death

To find T (t), some information is needed First, the lieutenant finds that the body is located

in a room that is kept at a constant 68◦Fahrenheit For some time after death, the body will loseheat into the cooler room Assume, for want of better information, that the victim’s temperaturewas 98.6◦at the time of death

By Newton’s law of cooling, heat energy is transferred from the body into the room at a rate

temperature at time t, then Newton’s law says that, for some constant of proportionality k,

To solve for T , take the exponential of both sides of this equation to get

Now the constants k and B must be determined Since there are two constants, we will need two

pieces of information Suppose the lieutenant arrived at 9:40 p.m and immediately measured thebody temperature, obtaining 94.4◦ It is convenient to let 9:40 p.m be time zero in carrying outmeasurements Then

T (0) = 94.4 = 68 + B,

T (t) = 68 + 26.4e kt

To determine k, we need another measurement The lieutenant takes the body temperature again

at 11:00 p.m and finds it to be 89.2◦ Since 11:00 p.m is 80 minutes after 9:40 p.m., this meansthat

The time of death was the last time at which the body temperature was 98.6◦(just before it began

to cool) Solve for the time t at which

(because of the negative sign) the first measurement at 9:40 p.m., which was chosen as timezero in the model This puts the murder at about 8:46 p.m

This is an estimate, because an educated guess was made of the body’s temperature before death.

that small changes in the body’s normal temperature and in the constant temperature of the roomyield small changes in the estimated time of death This can be verified by trying a slightlydifferent normal temperature for the body, say 99.3◦, to see how much this changes the estimated

EXAMPLE 1.6 Radioactive Decay and Carbon Dating

In radioactive decay, mass is lost by its conversion to energy which is radiated away It has

is proportional to the mass itself This means that, for some constant of proportionality k that is

unique to the element,

dm

Here k must be negative, because the mass is decreasing with time.

This differential equation for m is separable Write it as

in which A can be any positive number Any radioactive element has its mass decrease according

to a rule of this form, and this reveals an important characteristic of radioactive decay Suppose

at some timeτ there are M grams Look for h so that, at the later time τ + h, exactly half of this

mass has radiated away This would mean that



kln(2).

This is positive because k < 0.

Notice that h, the time it takes for half of the mass to convert to energy, depends only

on the number k, and not on the mass itself or the time at which we started measuring the loss If we measure the mass of a radioactive element at any time (say in years), then h years later exactly half of this mass will have radiated away This number h is called the half-life

of the element The constants h and k are both uniquely tied to the particular element and

half-life

k is tied to the element’s half-life The meaning of A is made clear by observing that

m(0) = Ae0= A.

A is the mass that is present at some time designated for convenience as time zero (think of this

as starting the clock when the first measurement is made) A is called the initial mass, usually denoted m0 Then

m(t) = m0e kt

It is sometimes convenient to write this expression in terms of the half-life h Since

m (t) = m0e kt = m0 e − ln(2)t/h (1.1)This expression is the basis for an important technique used to estimate the ages of certain ancientartifacts The Earth’s upper atmosphere is bombarded by high-energy cosmic rays, producinglarge numbers of neutrons which collide with nitrogen, converting some of it into radioactive

C This has a half-life h = 5, 730 years Over the geologically short time in

which life has evolved on Earth, the ratio of14

C to regular carbon in the atmosphere has remained

amount of this decay and hence the time it took, giving an estimate of the last time the organism

lived This method of estimating the age of an artifact is called carbon dating Since an artifact

may have been contaminated by exposure to other living organisms, this is a sensitive process.However, when applied rigorously and combined with other tests and information, carbon datinghas proved a valuable tool in historical and archeological studies

m (t) = e − ln(2)t/5730 ≈ e −0.000120968t

As a specific example, suppose we have a piece of fossilized wood Measurements show that theratio of14

C to carbon is 37 of the current ratio To calibrate our clock, say the wood died at time

gram, then T satisfies the equation

0.37 = e −0.000120968T

from which we obtain

T= − ln(0.37)

0.000120968 ≈ 8, 219

EXAMPLE 1.7 Draining a Container

Suppose we have a container or tank that is at least partially filled with a fluid The container

is drained through an opening How long will it take the container to empty? This is a simpleenough problem for something like a soda can, but it is not so easy with a large storage tank(such as the gasoline tank at a service station)

We will derive a differential equation to model this problem We need two principles fromphysics The first is that the rate of discharge of a fluid flowing through an opening at the bottom

of a container is given by

d V

in which V (t) is the volume of fluid remaining in the container at time t; v(t) is the velocity of

the discharge of fluid through the opening; A is the constant cross sectional area of the opening; and k is a constant determined by the viscosity of the fluid, the shape of the opening, and the

fact that the cross-sectional area of fluid pouring out of the opening is in reality slightly lessthan the area of the opening itself Molasses will flow at a different rate than gasoline, andthe shape of the opening will obviously play some role in how the fluid empties through thisopening

velocity of a free-falling body released from a height equal to the depth of the fluid at time t (Free-falling means influenced by gravity only.) In practice, k must be determined for the

particular fluid, container, and opening and is a number between 0 and 1

Equation (1.2) contains two unknown functions, so we must eliminate one To do this, let

r (t) be the radius of the surface of the fluid at time t, and consider an interval of time from t0

1/4 feet), so its area is A = π/16 For water and an opening of this shape and size, experiment

60h3/2 − h5/2 = −t + C with C arbitrary For the problem under consideration, the radius of the hemisphere is 18 feet, so

h (0) = 18 Therefore,

60(18)3/2 − (18)5/2 = C.

60h3/2 − h5/2= 2268√2− t.

28 seconds This is time it takes for the tank to drain 

These last three examples illustrate an important point A differential equation or initialvalue problem may be used to model and describe a process of interest However, the processusually occurs as something we observe and want to understand, not as a differential equation.This must be derived, using whatever information and fundamental principles may apply (such

as laws of physics, chemistry, or economics), as well as the measurements we may take Wesaw this in Examples 1.5, 1.6, and 1.7 The solution of the differential equation or initial valueproblem gives us a function that quantifies some part of the process and enables us to understandits behavior in the hope of being able to predict future behavior or perhaps design a process

that better suits our purpose This approach to the analysis of phenomena is called mathematical

modeling We see it today in studies of global warming, ecological and financial systems, and

physical and biological processes

SECTION 1.1 PROBLEMS

In each of Problems 1 through 6, determine whether

y = ϕ(x) is a solution of the differential equation C is

constant wherever it appears

In each of Problems 7 through 16, determine if the

dif-ferential equation is separable If it is, find the general

solution (perhaps implicitly defined) and also any

singu-lar solutions the equation might have If it is not separable,

do not attempt a solution

16 [cos(x + y) + sin(x − y)]y= cos(2x)

In each of Problems 17 through 21, solve the initial value

22 An object having a temperature of 90◦ Fahrenheit is

placed in an environment kept at 60◦ Ten minutes

later the object has cooled to 88◦ What will be the

temperature of the object after it has been in this

envi-ronment for 20 minutes? How long will it take for the

object to cool to 65◦?

23 A thermometer is carried outside a house whose ent temperature is 70◦Fahrenheit After five minutes,the thermometer reads 60◦, and fifteen minutes afterthis, it reads 50.4◦ What is the outside temperature(which is assumed to be constant)?

ambi-24 A radioactive element has a half-life of ln(2) weeks.

If e3tons are present at a given time, how much will

be left three weeks later?

25 The half-life of Uranium-238 is approximately

4.5(109) years How much of a 10 kilogram block of

U− 238 will be present one billion years from now?

26 Given that 12 grams of a radioactive element decays

to 9.1 grams in 4 minutes, what is the half-life of this

Calculate I(x) and find a differential equation for

I (x) Use the standard integral∞

0 e −t2dt=√π/2 to

determine I (0), and use this initial condition to solve

for I (x) Finally, evaluate I (3).

28 (Draining a Hot Tub) Consider a cylindrical hot tubwith a 5-foot radius and a height of 4 feet placed onone of its circular ends Water is draining from the tubthrough a circular hole 5/8 inches in diameter in the

base of the tub

(a) With k = 0.6, determine the rate at which the

depth of the water is changing Here it is useful

(b) Calculate the time T required to drain the hot tub

if it is initially full Hint: One way to do this is

(c) Determine how much longer it takes to drain the

lower half than the upper half of the tub Hint:

Use the integral of part (b) with different limitsfor each half

29 Calculate the time required to empty the cal tank of Example 1.7 if the tank is inverted to lie

hemispheri-on a flat cap across the open part of the hemisphere

The drain hole is in this cap Take k = 0.8 as in the

example

30 Determine the time it takes to drain a spherical tank

with a radius of 18 feet if it is initially full of water,

which drains through a circular hole with a radius of 3

inches in the bottom of the tank Use k = 0.8.

31 A tank shaped like a right circular cone, vertex down,

is 9 feet high and has a diameter of 8 feet It is initially

full of water

(a) Determine the time required to drain the tank

through a circular hole with a diameter of 2 inches

at the vertex Take k = 0.6.

(b) Determine the time it takes to drain the tank if it is

inverted and the drain hole is of the same size and

shape as in (a), but now located in the new (flat)

base

32 Determine the rate of change of the depth of water

in the tank of Problem 31 (vertex at the bottom) if

the drain hole is located in the side of the cone 2

feet above the bottom of the tank What is the rate of

change in the depth of the water when the drain hole

is located in the bottom of the tank? Is it possible to

determine the location of the drain hole if we are told

the rate of change of the depth and the depth of the

water in the tank? Can this be done without knowing

the size of the drain opening?

33 (Logistic Model of Population Growth) In 1837,

the Dutch biologist Verhulst developed a differential

equation to model changes in a population (he was

studying fish populations in the Adriatic Sea)

Ver-hulst reasoned that the rate of change of a population

P (t) with respect to time should be influenced by

growth factors (for example, current population) andalso factors tending to retard the population (such aslimitations on food and space) He formed a model

by assuming that growth factors can be incorporated

into a term a P (t) and retarding factors into a term

−bP(t)2with a and b as positive constants whose

val-ues depend on the particular population This led to his

This is the logistic model of population growth Show

that, unlike exponential growth, the logistic model

produces a population function P (t) that is bounded

above and increases asymptotically toward a /b as

t→ ∞ Thus, a logistic model produces a populationfunction that never grows beyond a certain value

34 Continuing Problem 33, a 1920 study by Pearl and

Reed (appearing in the Proceedings of the National

Academy of Sciences) suggested the values

a = 0.03134, b = (1.5887)10−10

for the population of the United States Table 1.1gives the census data for the United States in ten year

Year Population P (t) Percent error Q (t) Percent error

increments from 1790 through 1980 Taking 1790 as

year zero to determine p0, show that the logistic model

for the United States population is

P (t) = 123, 141.5668

0.03072 + 000062e0.03134t e0.03134t

Calculate P (t) in ten year increments from 1790 to

fill in the P (t) column in the table Remember that

(with 1790 as the base year) 1800 is year t= 10 in the

model, 1810 is t= 20, and so on Also, calculate the

percentage error in the model and fill in this column

Plot the census figures and the numbers predicted by

the logistic model on the same set of axes You should

observe that the model is fairly accurate for a long

period of time, then diverges from the actual census

numbers Show that the limit of the population in this

model is about 197, 300, 000, which the United States

actually exceeded in 1970

Sometimes an exponential model Q(t) = k Q(t)

is used for population growth Use the census data

(again with 1790 as year zero) to solve for Q (t).

Compute Q (t) for the years of the census data and

the percentage error in this exponential prediction ofpopulation Plot the census data and the exponentialmodel predicted data on the same set of axes It should

be clear that Q (t) diverges rapidly from the actual

census figures Exponential models are useful for verysimple populations (such as bacteria in a dish) butare not sophisticated enough for human or (in gen-eral) animal populations, despite occasional claims byexperts that the population of the world is increasingexponentially

A first-order differential equation is linear if it has the form

for some functions p and q.

There is a general approach to solving a linear equation Let

This is the general solution with the arbitrary constant c.

procedure

Step 1 If the differential equation is linear, y+ p(x)y = q(x) First compute

ep (x) dx

This is called an integrating factor for the linear equation.

Step 2 Multiply the differential equation by the integrating factor

Step 3 Write the left side of the resulting equation as the derivative of the product of y and the

integrating factor The integrating factor is designed to make this possible The right side

is a function of just x.

Step 4 Integrate both sides of this equation and solve the resulting equation for y,

obtain-ing the general solution The resultobtain-ing general solution may involve integrals (such as

cos(x2) dx) which cannot be evaluated in elementary form.

x y+ y = 3x3or

(xy)= 3x3.

As suggested previously, solving a linear differential equation may lead to integrals we

e x2 /2 d x in elementary terms (as a finite algebraic combination of

power series about 0, we could integrate this series term by term This would yield an infiniteseries expression for the solution

Here is an application of linear equations to a mixing problem

EXAMPLE 1.10 A Mixing Problem

We want to determine how much of a given substance is present in a container in which

var-ious substances are being added, mixed, and drained out This is a mixing problem, and it is

encountered in the chemical industry, manufacturing processes, swimming pools and (on a moresophisticated level) in ocean currents and atmospheric activity

As a specific example, suppose a tank contains 200 gallons of brine (salt mixed with water)

3 gal/min

1/8 lb/gal

3 gal/min

FIGURE 1.5 Storage tank in Example 1.10.

is pumped into the tank at a rate of 3 gallons per minute, and the mixture is continuously stirred.Brine also is allowed to empty out of the tank at the same rate of 3 gallons per minute (seeFigure 1.5) How much salt is in the tank at any time?

Let Q (t) be the amount of salt in the tank at time t The rate of change of Q(t) with respect

to time must equal the rate at which salt is pumped in minus the rate at which it is pumped out:

d Q

dt = (rate in) − (rate out)

=

18

poundsgallon

 

3gallonsminute

 

3gallonsminute

called the transient part of the solution, and it decays to zero as t increases Q (t) is the sum of a

steady-state part and a transient part This type of decomposition of a solution is found in many

settings For example, the current in a circuit is often written as a sum of a steady-state term and

a transient term

The initial ratio of salt to brine in the tank is 100 pounds per 200 gallons or 1/2 pound per

brine mixture to dilute toward the incoming ratio with a terminal amount of salt in the tank of

1/8 pound per gallon times 200 gallons This leads to the expectation (in the long term) that the

11 Find all functions with the property that the y intercept

of the tangent to the graph at(x, y) is 2x2

12 A 500 gallon tank initially contains 50 gallons of

brine solution in which 28 pounds of salt have been

dissolved Beginning at time zero, brine containing 2pounds of salt per gallon is added at the rate of 3 gal-lons per minute, and the mixture is poured out of thetank at the rate of 2 gallons per minute How much salt

is in the tank when it contains 100 gallons of brine?

Hint: The amount of brine in the tank at time t is

50+ t.

13 Two tanks are connected as in Figure 1.6 Tank 1initially contains 20 pounds of salt dissolved in 100gallons of brine Tank 2 initially contains 150 gallons

of brine in which 90 pounds of salt are dissolved Attime zero, a brine solution containing 1/2 pound of

salt per gallon is added to tank 1 at the rate of 5 gallonsper minute Tank 1 has an output that discharges brineinto tank 2 at the rate of 5 gallons per minute, and tank

2 also has an output of 5 gallons per minute mine the amount of salt in each tank at any time Also,determine when the concentration of salt in tank 2 is aminimum and how much salt is in the tank at that time

Deter-Hint: Solve for the amount of salt in tank 1 at time t

and use this solution to help determine the amount intank 2

Tank 2 Tank 1

5 gal/min; 1/2 lb/gal 5 gal/min

5 gal/min

FIGURE 1.6 Storage tank in Problem 13, Section 1.2.

CHAPTER 17 The Heat Equation 611

CHAPTER 18 The Potential Equation 6 41< /h3>

PART. .. class="text_page_counter">Trang 12

3.3 .1 The First Shifting Theorem 84

CHAPTER Series Solutions 12 1

CHAPTER Approximation of Solutions 13 7... class="text_page_counter">Trang 11

Preface xi

PART Ordinary Differential Equations 1< /b>

CHAPTER