Solution manual for advanced engineering mathematics 8th edition by ONeil

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ADVANCED ENGINEERING MATHEMATICS,

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INSTRUCTOR'S SOLUTIONS MANUAL

TO ACCOMPANY

ADVANCED ENGINEERNG

MATHEMATICS

8 th EDITION

P ETER V O’N EIL

U NIVERSITY OF A LABAMA AT B IRMINGHAM

.

1 First-Order Differential Equations 11.1 Terminology and Separable Equations 11.2 The Linear First-Order Equation 121.3 Exact Equations 191.4 Homogeneous, Bernoulli and Riccati Equations 28

2 Second-Order Differential Equations 372.1 The Linear Second-Order Equation 372.2 The Constant Coefficient Homogeneous Equation 412.3 Particular Solutions of the Nonhomogeneous Equation 462.4 The Euler Differential Equation 532.5 Series Solutions 58

3 The Laplace Transform 693.1 Definition and Notation 693.2 Solution of Initial Value Problems 723.3 The Heaviside Function and Shifting Theorems 773.4 Convolution 863.5 Impulses and the Dirac Delta Function 923.6 Systems of Linear Differential Equations 93

iv CONTENTS

4 Sturm-Liouville Problems and Eigenfunction Expansions 1014.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems 1014.2 Eigenfunction Expansions 1074.3 Fourier Series 114

5 The Heat Equation 1375.1 Diffusion Problems on a Bounded Medium 1375.2 The Heat Equation With a Forcing Term F (x, t) 1475.3 The Heat Equation on the Real Line 1505.4 The Heat Equation on a Half-Line 1535.5 The Two-Dimensional Heat Equation 155

6 The Wave Equation 1576.1 Wave Motion on a Bounded Interval 1576.2 Wave Motion in an Unbounded Medium 1676.3 d’Alembert’s Solution and Characteristics 1736.4 The Wave Equation With a Forcing Term K(x, t) 1906.5 The Wave Equation in Higher Dimensions 192

7 Laplace’s Equation 1977.1 The Dirichlet Problem for a Rectangle 1977.2 The Dirichlet Problem for a Disk 2027.3 The Poisson Integral Formula 2057.4 The Dirichlet Problem for Unbounded Regions 2057.5 A Dirichlet Problem in 3 Dimensions 2087.6 The Neumann Problem 2117.7 Poisson’s Equation 217

8 Special Functions and Applications 2218.1 Legendre Polynomials 2218.2 Bessel Functions 2358.3 Some Applications of Bessel Functions 251

9 Transform Methods of Solution 2639.1 Laplace Transform Methods 2639.2 Fourier Transform Methods 2689.3 Fourier Sine and Cosine Transforms 271

10 Vectors and the Vector Space Rn 27510.1 Vectors in the Plane and 3− Space 27510.2 The Dot Product 27710.3 The Cross Product 27810.4 n− Vectors and the Algebraic Structure of Rn 28010.5 Orthogonal Sets and Orthogonalization 28410.6 Orthogonal Complements and Projections 287

11 Matrices, Determinants and Linear Systems 29111.1 Matrices and Matrix Algebra 29111.2 Row Operations and Reduced Matrices 29511.3 Solution of Homogeneous Linear Systems 29911.4 Nonhomogeneous Systems 30611.5 Matrix Inverses 31311.6 Determinants 31511.7 Cramer’s Rule 31811.8 The Matrix Tree Theorem 320

.

12 Eigenvalues, Diagonalization and Special Matrices 32312.1 Eigenvalues and Eigenvectors 32312.2 Diagonalization 32712.3 Special Matrices and Their Eigenvalues and Eigenvectors 33212.4 Quadratic Forms 336

13 Systems of Linear Differential Equations 33913.1 Linear Systems 33913.2 Solution of X0= AX When A Is Constant 34113.3 Exponential Matrix Solutions 34813.4 Solution of X0= AX + G for Constant A 35013.5 Solution by Diagonalization 353

14 Nonlinear Systems and Qualitative Analysis 35914.1 Nonlinear Systems and Phase Portraits 35914.2 Critical Points and Stability 36314.3 Almost Linear Systems 36414.4 Linearization 369

15 Vector Differential Calculus 37315.1 Vector Functions of One Variable 37315.2 Velocity, Acceleration and Curvature 37615.3 The Gradient Field 38115.4 Divergence and Curl 38515.5 Streamlines of a Vector Field 387

16 Vector Integral Calculus 39116.1 Line Integrals 39116.2 Green’s Theorem 39316.3 Independence of Path and Potential Theory 39816.4 Surface Integrals 40516.5 Applications of Surface Integrals 40816.6 Gauss’s Divergence Theorem 41216.7 Stokes’s Theorem 414

17.1 Fourier Series on [−L, L] 41917.2 Sine and Cosine Series 42317.3 Integration and Differentiation of Fourier Series 42817.4 Properties of Fourier Coefficients 43017.5 Phase Angle Form 43217.6 Complex Fourier Series 43517.7 Filtering of Signals 438

vi CONTENTS

18 Fourier Transforms 44118.1 The Fourier Transform 44118.2 Fourier sine and Cosine Transforms 448

19 Complex Numbers and Functions 45119.1 Geometry and Arithmetic of Complex Numbers 45119.2 Complex Functions 45519.3 The Exponential and Trigonometric Functions 46119.4 The Complex Logarithm 467

20 Complex Integration 47320.1 The Integral of a Complex Function 47320.2 Cauchy’s Theorem 47720.3 Consequences of Cauchy’s Theorem 479

21 Series Representations of Functions 48521.1 Power Series 48521.2 The Laurent Expansion 492

22 Singularities and the Residue Theorem 49722.1 Classification of Singularities 49722.2 The Residue Theorem 49922.3 Evaluation of Real Integrals 505

23 Conformal Mappings 51523.1 The Idea of a Conformal Mapping 51523.2 Construction of Conformal Mappings 533

.

Chapter 1

First-Order Differential Equations

1.1 Terminology and Separable Equations

1 The differential equation is separable because it can be written

2 Write the differential equation as

xdy

dx = −y,which separates as

2 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

with c constant (c = ek) y = 0 is a singular solution, satisfying theoriginal differential equation

3 If cos(y) 6= 0, the differential equation is

y

dx =

sin(x + y)cos(y)

= sin(x) cos(y) + cos(x) sin(y)

cos(y)

= sin(x) + cos(x) tan(y)

There is no way to separate the variables in this equation, so the tial equation is not separable

differen-4 Write the differential equation as

exeydy

dx = 3x,which separates in differential form as

eydy = 3xe−xdx

Integrate to get

ey= −3e−x(x + 1) + c,with c constant This implicitly defines a general solution

5 The differential equation can be written

xdy

dx = y

2− y,or

1y(y − 1)dy =

1

xdx,and is therefore separable Separating the variables assumes that y 6= 0and y 6= 1 We can further write

1

y − 1xy = k

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.

y − 1

xy = c,with c = ek constant Solve this for y to obtain the general solution

y = 1

1 − cx.

y = 0 and y = 1 are singular solutions because these satisfy the differentialequation, but were excluded in the algebra of separating the variables

6 The differential equation is not separable

7 The equation is separable because it can be written in differential form as

sin(y)cos(y)dy =

The algebra of separating the variables required that cos(y) 6= 0 Nowcos(y) = 0 if y = (2n+1)π/2, with n any integer Now y = (2n+1)π/2 alsosatisfies the original differential equation, so these are singular solutions

8 The differential equation itself requires that y 6= 0 and x 6= −1 Write theequation as

xy

dy

dx =

2y2+ 1xand separate the variables to get

1y(2y2+ 1)dy =

1x(x + 1)dx.

Use a partial fractions decomposition to write this as

 1

y − 2y2y2+ 1

Integrate to obtain

ln |y| −1

2ln(1 + 2y

2) = ln |x| − ln |x + 1| + c

4 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

with c constant This implicitly defines a general solution We can go astep further by writing this equation as

x + 1

+ c

and take the exponential of both sides to get

yp

1 + 2y2 = k

 x

x + 1

,

which also defines a general solution

9 The differential equation is

dy

dx = e

x− y + sin(y),and this is not separable It is not possible to separate all terms involving

x on one side of the equation and all terms involving y on the other

10 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 get the separated differential form

(cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx

Integrate to obtain the implicitly defined general solution

cos(y) + sin(y) = cos(x) + sin(x) + c

11 If y 6= −1 and x 6= 0, we obtain the separated equation

This implicitly defines a general solution The initial condition is y(3e2) =

2, so put y = 2 and x = 3e2 to obtain

2 − 2 + ln(3) = ln(3e2) + c

Now

ln(3e2) = ln(3) + ln(e2) = ln(3) + 2,so

6 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

or 9 = 12 + c Then c = −3 and the solution of the initial value problem

x = y2.Because y(4) = −2, the explicit solution is y = −√

13

3y sin(3y) + cos(3y) = 9x2− 1

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.

16 Let T (t) be the temperature function By Newton’s law of cooling, T0(t) =k(T −60) for some constant k to be determined This equation is separableand is easily solved to obtain:

t

10ln(14/15) = ln(1/6) = − ln(6).

The object reaches 65 degrees at time

t = − 10 ln(6)ln(14/15) ≈ 259.7minutes

8 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

17 Suppose the thermometer was removed from the house at time t = 0, andlet T (t) be the temperature function Let A be the ambient temperatureoutside the house (assumed constant) By Newton’s law,

T (0) = 70 = A + c,

so c = 70 − A and

T (t) = A + (70 − A)ekt.Now use the other two conditions:

T (5) = A + (70 − A)e5k= 15.5 and T (15) = A + (70 − A)e15k = 50.4

From the equation for T (5), solve for e5k to get

e5k =60 − A

70 − A.Then

10.4A2− 1156A + 30960 = 0,with roots 45 and (approximately) 66.15385 Clearly the outside temper-ature must be less than 50, and must therefore equal 45 degree

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.

18 The amount A(t) of radioactive material at time t is modeled by

A0(t) = kA, A(0) = e3together with the given half-life of the material,

.The half-life of this element is the time t∗ it will take for there to be 6grams, so

A(t∗) = 6 = 12eln(9.1/12)t∗/4.Solve this to get

t∗= 4 ln(1/2)ln(9.1/12)≈ 10.02 minutes

21 Let

I(x) =

Z ∞ 0

e−t2−(x/t)2dt

The integral we want is I(3) Compute

I0(x) = −2x

Z ∞ 0

1

t2e−t2−(x/t)2dt

10 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

Let u = x/t, so t = x/u and

e−t2dt =

√π

2 = c,

in which we used a standard integral that arises often in statistics Then

I(x) =

√π

2 e

−2x.Finally, put x = 3 for the particular integral of interest:

I(3) =

Z ∞ 0

e−t2−(9/t)2dt =

√π

1

P (a − bP )dP = dtand the variables are separated To make the integration easier, write thisequation as

 1a

1

P +

ba

in which k = ac is still constant Then

P (t) < ap0

bp0eateat= a

b.This means that this population function is bounded above Further, bymultiplying the numerator and denominator of P (t) by e(−at, we have

If we attempt an exponential model Q(t) = Aekt, then take A = Q(0) =

3, 929, 214, the population in 1790 To find k, use the fact that

Q(10) = 5308483 = 3929214e10kand we can solve for k to get

k = 1

10ln

 53084833929214



≈ 0.03008667012

12 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS

year population P (t) percent error Q(t) percent error

Table 1.1: Census data for Problem 23

The exponential model, using these two data points (1790 and 1800 ulations), is

pop-Q(t) = 3929214e0.03008667012t.Table 1.1 uses Q(t) and P (t) to predict later populations from these twoinitial figures The logistic model remains quite accurate until about 1960,

at which time it loses accuracy quickly The exponential model becomesquite inaccurate by 1870, after which the error becomes so large that it

is not worth computing further Exponential models do not work wellover time with complex populations, such as fish in the ocean or countriesthroughout the world

1.2 The Linear First-Order Equation

1 With p(x) = −3/x, and integrating factor is

eR (−3/x) dx= e−3 ln(x)= x−3for x > 0 Multiply the differential equation by x−3 to get

x−3y0− 3x−4= 2x−1

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.

y = 1

1 − cx.

y = and y = are singular solutions because these satisfy the... DIFFERENTIAL EQUATIONS

with c constant This implicitly defines a general solution We can go astep further by writing this equation as

x +

+ c

and take the exponential... equation to get the separated differential form

(cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx

Integrate to obtain the implicitly defined general solution

cos(y) + sin(y) = cos(x)