ADVANCED ENGINEERING MATHEMATICS 7e CUSTOM EDITION PDF

Preview

Summary

I NSTRUCTOR ’ S S OLUTIONS M ANUAL TO ACCOMPANY ADVANCED ENGINEERING MATHEMATICS 7e CUSTOM EDITION PETER V. O’NEIL © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, co...

Description

I NSTRUCTOR ’ S S OLUTIONS M ANUAL TO

ACCOMPANY

ADVANCED ENGINEERING MATHEMATICS 7e CUSTOM EDITION

PETER V. O’NEIL

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Contents 1

First-Order Differential Equations 1.1 Terminology and Separable Equations 1.2 Linear Equations 1.3 Exact Equations 1.4 Integrating Factors 1.5 Homogeneous, Bernoulli and Riccati Equations 1.6 Additional Applications 1.7 Existence and Uniqueness Questions

1 1 13 17 24 26 29 37

2

Linear Second-Order Equations 2.1 The Linear Second-Order Equation 2.2 Reduction of Order 2.3 The Constant Coefficient Case 2.4 The Nonhomogeneous Equation 2.5 Spring Motion 2.6 Euler’s Differential Equation

41 41 44 45 48 53 61

3

The Laplace Transform 3.1 Definition and Notation 3.2 Solution of Initial Value Problems 3.3 Shifting and the Heaviside Function 3.4 Convolution 3.5 Impulses and the Delta Function 3.6 Solution of Systems 3.7 Polynomial Coefficients

64 64 68 71 78 85 87 96

4

Series Solutions 4.1 Power Series Solutions 4.2 Frobenius Solutions

99 99 103

5

Approximation of Solutions 5.1 Direction Fields 5.2 Euler’s Method 5.3 Taylor and Modified Euler Methods

108 108 111 114

6

Vectors and Vector Spaces 6.1 Vectors in the Plane and 3-Space 6.2 The Dot Product 6.3 The Cross Product 6.4 The Vector Space Rn 6.5 Orthogonalization 6.6 Orthogonal Complements and Projections 6.7 The Function Space C[a, b]

117 117 118 119 120 125 127 128

iii © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

iv

CONTENTS

7

Matrices and Linear Systems 7.1 Matrices 7.2 Elementary Row Operations 7.3 Reduced Row Echelon Form 7.4 Row and Column Spaces 7.5 Homogeneous Systems 7.6 Nonhomogeneous Systems 7.7 Matrix Inverses 7.8 Least Squares Vectors and Data Fitting 7.9 LU Factorization 7.10 Linear Transformations

133 133 137 139 141 143 149 155 157 160 164

8

Determinants 8.1 Definition of the Determinant 8.2 Evaluation of Determinants I 8.3 Evaluation of Determinants II 8.4 A Determinant Formula for A−1 8.5 Cramer’s Rule 8.6 The Matrix Tree Theorem

166 166 167 168 170 171 172

9

Eigenvalues, Diagonalization, and Special Matrices 9.1 Eigenvalues and Eigenvectors 9.2 Diagonalization 9.3 Some Special Types of Matrices

174 174 178 183

10

Systems of Linear Differential Equations 10.1 Linear Systems 10.2 Solution of X0 = AX for Constant A 10.3 Solution of X0 = AX + G 10.4 Exponential Matrix Solutions 10.5 Applications and Illustrations of Techniques 10.6 Phase Portraits

190 190 192 197 205 207 216

11

Vector Differential Calculus 11.1 Vector Functions of One Variable 11.2 Velocity and Curvature 11.3 Vector Fields and Streamlines 11.4 The Gradient Field 11.5 Divergence and Curl

225 225 228 232 234 237

12

Vector Integral Calculus 12.1 Line Integrals 12.2 Green’s Theorem 12.3 An Extension of Green’s Theorem 12.4 Independence of Path and Potential Theory 12.5 Surface Integrals 12.6 Applications of Surface Integrals 12.7 Lifting Green’s Theorem to R3 12.8 The Divergence Theorem of Gauss 12.9 Stokes’s Theorem 12.10 Curvilinear Coordinates

240 240 242 245 247 253 255 258 259 260 263

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

CONTENTS

v

13

Fourier Series 13.1 Why Fourier Series? 13.2 The Fourier Series of a Function 13.3 Sine and Cosine Series 13.4 Integration and Differentiation of Fourier Series 13.5 Phase Angle Form 13.6 Complex Fourier Series 13.7 Filtering of Signals

267 267 268 277 289 292 295 297

14

The Fourier Integral and Transforms 14.1 The Fourier Integral 14.2 Fourier Cosine and Sine Integrals 14.3 The Fourier Transform 14.4 Fourier Cosine and Sine Transforms 14.5 The Discrete Fourier Transform 14.6 Sampled Fourier Series 14.7 DFT Approximation of the Fourier Transform

310 310 314 318 328 329 335 339

15

Special Functions and Eigenfunction Expansions 15.1 Eigenfunction Expansions 15.2 Legendre Polynomials 15.3 Bessel Functions

341 341 351 359

16

The Wave Equation 16.1 Derivation of the Wave Equation 16.2 Wave Motion on an Interval 16.3 Wave Motion in an Infinite Medium 16.4 Wave Motion in a Semi-Infinite Medium 16.5 Laplace Transform Techniques 16.6 Characteristics and d’Alembert’s Solution 16.7 Vibrations in a Circular Membrane I 16.8 Vibrations in a Circular Membrane II 16.9 Vibrations in a Rectangular Membrane

380 380 381 398 401 404 407 422 424 425

17

The Heat Equation 17.1 Initial and Boundary Conditions 17.2 The Heat Equation on [0, L] 17.3 Solutions in an Infinite Medium 17.4 Laplace Transform Techniques 17.5 Heat Conduction in an Infinite Cylinder 17.6 Heat Conduction in a Rectangular Plate

428 428 429 454 458 461 462

18

The Potential Equation 18.1 Laplace’s Equation 18.2 Dirichlet Problem for a Rectangle 18.3 Dirichlet Problem for a Disk 18.4 Poisson’s Integral Formula 18.5 Dirichlet Problem for Unbounded Regions 18.6 A Dirichlet Problem for a Cube 18.7 Steady-State Equation for a Sphere 18.8 The Neumann Problem

465 465 466 471 474 475 478 480 483

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

vi

CONTENTS

19

Complex Numbers and Functions 19.1 Geometry and Arithmetic of Complex Numbers 19.2 Complex Functions 19.3 The Exponential and Trigonometric Functions 19.4 The Complex Logarithm 19.5 Powers

489 489 492 497 502 503

20

Complex Integration 20.1 The Integral of a Complex Function 20.2 Cauchy’s Theorem 20.3 Consequences of Cauchy’s Theorem Series Representations of Functions 21.1 Power Series 21.2 The Laurent Expansion

507 507 510 512 517 517 523

22

Singularities and the Residue Theorem 22.1 Singularities 22.2 The Residue Theorem 22.3 Evaluation of Real Integrals 22.4 Residues and the Inverse Laplace Transform

527 527 529 534 543

23

Conformal Mappings and Applications 23.1 Conformal Mappings 23.2 Construction of Conformal Mappings 23.3 Conformal Mapping Solutions of Dirichlet Problems 23.4 Models of Plane Fluid Flow

546 546 561 564 568

21

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

Chapter 1

First-Order Differential Equations 1.1

Terminology and Separable Equations

1. For x > 0, rewrite the equation as 2xy 0 + 2y = ex . With y = ϕ(x) = 21 x−1 (C − ex ), compute y0 =


1 −x−2 (C − ex ) − x−1 ex . 2

Then
2xy 0 + 2y = x −x−2 (C − ex ) − x−1 ex + x−1 (C − ex ) = ex . Therefore ϕ(x) is a solution. 2. For all x, ϕ0 + ϕ = −Ce−x + (1 + Ce−x ) = 1 so ϕ(x) = 1 + Ce−x is a solution. 3. On any interval not containing x = 0 we have      2  1 3 3 x x −3 xϕ0 = x + 2 =x+ − =x− = x − ϕ, 2 2x 2x 2 2x so ϕ is a solution. 4. With ϕ(x) = Ce−x , ϕ0 + ϕ = −Ce−x + Ce−x = 0, so ϕ is a solution. 5. For x > 1,

√ 1 2ϕϕ0 = 2 x − 1 √ = 1, 2 x−1

so ϕ is a solution. 1 © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

2

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS √ 6. For x 6= ± 2, ϕ0 =

−2cx = (x2 − 2)2



2x 2 − x2



c x2 − 2

 =

2xϕ , 2 − x2

so ϕ is a solution. 7. This equation is separable since we can write it as sin(y) 1 dy = dx cos(y) x if cos(y) 6= 0 and x 6= 0. A routine integration gives the implicitly defined general solution sec(y) = kx. Now cos(y) = 0 if y = (2n + 1)π/2 for n any integer. y = (2n + 1)π/2 also satisfies the original differential equation and is a singular solution. 8. Substitute sin(x − y) = sin(x) cos(y) − cos(x) sin(y), cos(x + y) = cos(x) cos(y) − sin(x) sin(y), and cos(2x) = cos2 (x) − sin2 (x) into the differential equation to obtain the separated equation (cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx. Upon integrating we obtain the implicitly defined solution cos(y) + sin(y) = cos(x) + sin(x) + c. 9. This differential equation is not separable. 10. The differential equation itself assumes that y 6= 0 and x 6= −1. Write 2y 2 + 1 x dy = , y dx x+1 which separates as 1 1 dy = dx. y(2y 2 + 1) x(x + 1) Use a partial fractions decomposition to write     2y 1 1 1 − − dx. dy = y 1 + 2y 2 x 1+x Integration this equation to obtain ln |y| −

1 ln(1 + 2y 2 ) = ln |x| − ln |x + 1| + c. 2

Then, ln

y p

1 + 2y 2

!

 = ln

x x+1

 + c,

in which we have taken the case that y > 0 and x > 0 to drop the absolute values. Finally, take the exponential of both sides of this equation to obtain the implicitly defined solution   y x p =k . x+1 1 + 2y 2 Since y = 0 satisfies the original differential equation, y = 0 is a singular solution. © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS 11. Write 3

3

dy 4x = 2 dx y

and separate variables: 3y 2 dy = 4x dx. Integrate to obtain y 3 = 2x2 + k, which implicitly defines the general solution. We can also write y = 2x2 + k


1/3

.

12. This equation is not separable. 13. Write the differential equation as sin(x + y) dy = dx cos(y) sin(x) cos(y) + cos(x) sin(y) = cos(y) sin(y) . = sin(x) + cos(x) cos(y) There is no way to separate the variables in this equation, so the differential equation is not separable. 14. Since ex+y = ex ey , we can write the differential equation as ex ey

dy = 3x dx

or, in separated form, ey dy = 3xe−x dx. Integration gives us the implicitly defined general solution ey = −3e−x (x + 1) + c. 15. Write the differential equation as dy = y(y − 1). dx This is separable. If y = 6 0 and y = 6 1, we can write x

1 1 dx = dy. x y(y − 1) Use partial fractions to write this as 1 1 1 dx = dy − dy. x y−1 y Integrate to obtain ln |x| = ln |y − 1| − ln |y| + c, © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

4

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS or

y − 1 ln |x| = ln + c. y This can be solved for x to obtain the general solution y=

1 . 1 − kx

The trivial solution y(x) = 0 is a singular solution, as is the constant solution y(x) = 1. We assumed that y 6= 0, 1 in the algebra of separating the variables. 16. Write the differential equation as x

dy = −y dx

and separate the variables: 1 1 dy = − dx. y x This separation requires that x 6= 0 and y 6= 0. Integration gives us ln |y| = − ln |x| + c. Then ln |y| + ln |x| = c so ln |xy| = c. Then xy = ec = k, in which k can be any positive constant. Notice now that y = 0 is also a solution of the original differential equation. Therefore, if we allow k to be any constant (positive, negative or zero), we can omit the absolute values and write the general solution in the implicit form xy = k. 17. Write ln(y x ) = x ln(y) and separate the variables to write ln(y) dy = 3x dx. y Integrate to obtain (ln(y))2 = 3x2 + c. Substitute the initial condition to obtain c = −3, so the solution is implicitly defined by (ln(y))2 = 3x2 − 3. 2

2

18. Write ex−y = ex e−y and Separate the variables to obtain 2

2yey dy = ex dx. 2

Integrate to get ey = ex + c. The condition y(4) = −2 requires that c = 0, so the solution is √ 2 defined implicitly by ey = ex , or x = y 2 . Since y(4) = −2, the explicit solution is y = − x. 19. Separate the variables to obtain y cos(3y) dy = 2x dx, with solution given implicitly by 1 1 y sin(3y) + cos(3y) = x2 + c. 3 9 The initial condition requires that π 1 4 sin(π) + cos(π) = + c, 9 9 9 so c = −5/9. The solution is implicitly defined by 3y sin(3y) + cos(3y) = 9x2 − 5. © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

5

20. Integrate 1 dy = 3x2 dx y+2 to obtain ln |2 + y| = x3 + c. Substitute the initial condition to obtain c = ln(10) − 8. The solution is defined by   2+y = x3 − 8. ln 10 21. If y 6= −1 and x 6= 0, we obtain the separated equation y2 1 dy = dx. y+1 x Write this as

 y−1+

1 1+y

 dy =

1 dx. x

Integrate to obtain 1 2 y − y + ln |1 + y| = ln |x| + c. 2 Now use the initial condition y(3e2 ) = 2 to obtain 2 − 2 + ln(3) = ln(3) + 2 + c so c = −2 and the solution is implicitly defined by 1 2 y − y + ln(1 + y) = ln(x) − 2, 2 in which the absolute values have been removed because the initial condition puts the solution in a part of the x, y− plane where x > 0 and y > −1. 22. By Newton’s law of cooling the temperature function T (t) satisfies T 0 (t) = k(T − 60), with k a constant of proportionality to be determined, and with T (0) = 90 and T (10) = 88. This is based on the object being placed in the environment at time zero. This differential equation is separable (as in the text) and we solve it subject to T (0) = 90 to obtain T (t) = 60 + 30ekt . Now T (10) = 88 = 60 + 30e10k gives us e10k = 14/15. Then 1 k= ln 10



14 15



≈ −6.899287(10−3 ).

Since e10k = 14/15, we can write T (t) = 60 + 30(e10k )t/10 = 60 + 30 Now  T (20) = 60 + 30

14 15



14 15

t/10 .

2 ≈ 86.13

degrees Fahrenheit. To reach 65 degrees, solve  65 = 60 + 30

14 15

t/10

© 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

6

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS to obtain t=

10 ln(1/6) ≈ 259.7 ln(14/15)

minutes. 23. Compute I 0 (x) = −

Z

∞

0

2x −(t2 +(x/t)2 ) e dt. t

Let u = x/t to obtain I 0 (x) = 2

Z

0

2

e−((x/u)

+u2 )

du

∞ ∞

Z

e−(u

= −2

2

+(x/u)2 )

du = −2I(x).

0

This is the separable equation I 0 = −2I. Write this as 1 dI = −2 dx I and integrate to obtain I(x) = ce−2x . Now √ Z ∞ π −t2 , I(0) = e dt = 2 0 a standard result often used in statistics. Then √ I(x) = Put x = 3 to obtain

Z 0

π −2x e . 2 √

∞

e−t

2

−(9/t2 )

dt =

π −6 e . 2

24. (a) For water h feet deep in the cylindrical hot tub, V = 25πh, so  2 √ 5 dh = −0.6π 25π 64h, dt 16 with h(0) = 4. Thus

√ dh 3 h =− . dt 160 (b) The time it will take to drain the tank is Z 0  dt T = dh dh 4 Z 0 640 160 = − √ dh = 3 3 h 4

seconds. (c) To drain the upper half will require Z 2 √ 160 320 (2 − 2) T1 = − √ dh = 3 3 h 4 © 2012 Cengage Learning. All Rights Reserved. This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. May not be scanned, copied, duplicated, or posted to a publicly accessible website, in whole or in part.

Landsburg-1120953

book

May 3, 2012

2:0

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

7

r(t) h(t)

18

Figure 1.1: Problem 25, Section 1.1.

seconds, approximately 62.5 seconds. The lower half requires Z 0 160 320 √ − √ dh = 2 T2 = 3 3 h 2 seconds, about 150.8 seconds. 25. Model the problem using Torricelli’s law and the geometry of the hemispherical tank. Let h(t) be the depth of the liquid at time t, r(t) the radius of the top surface of the draining liquid, and V (t) the volume in the container (See Figure 1.1). Then p dV dV dh = −kA 2gh and = πr2 . dt dt dt Here r2 +h2 = 182 , since the radius of the tub is 18. We are given k = 0.8 and A = π(1/4)2 = π/16 is the area of the drain hole. With g = 32 feet per second per second, we obtain the initial value problem √ dh π(324 − h2 ) = 0.4π h; h(0) = 18. dt This is a separable differential equation with the general solution √ 1620 h − h5/2 = −t + k. √ Then h(0) = 18 yields k = 3888 2, so √ √ 1620 h − h5/2 = 3888 2 − t. √ The hemisphere is emptied at the instant that h = 0, hence at t = 3888 2 seconds, about 91 minutes, 39 seconds. √ 26. From the geometry of the sphere (Figure 1.2), dV /dt = −kA 2gh becomes  2 1 √ 2 dh π(324 − (h − 18) ) = −0.8π 64h, dt 4 with h(0) = 36. Here h(t) is the height of the upper surface of the fluid above the bottom of the sphere. This equation simplifies to √ (36 h − h3/2 ) dh = −0.4 dt, √ a separated equation with general solution h ...