Engineering Applied Math

Notes and Exercises inEngineering Applied

Mathematics

Constantinos V. ChrysikopoulosDepartment of Civil and Environmental Engineering

University of California, Irvine, CA 92697–2175

July, 2002

Corrected and Revised Edition

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS –––––––––––––––––––––––––––––– ii

Preface

The objective of this review course in Engineering Applied Mathematics isto provide the fundamental mathematical background required for problemsolving in the various engineering fields. Emphasis is given on elegant math-ematical tools usefull for analytical derivations, because the best methodof solution is always the analytical one. The lectures are designed for be-ginning graduate or advanced undergraduate students in engineering andapplied sciences. It is assumed that the student is familiar with the deriva-tion of ordinary and partial differential equations from other courses inengineering and applied science. As a result, the material is mainly focusedon the solution of these equations. It is also assumed that the studenthas a good background in basic calculus and ordinary differential equationsand has been introduced to linear algebra. Some familiarity with elemen-tary aspects of partial differential equations is desirable. Numerous usefulexercises together with complete solutions are presented in each topic.

c.v.c.

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C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS ––––––––––––––––––––––––––––– iv

Contents

Chapter 1

Introduction . . . . . . . . . . . . . . . . . . . . . 11.1 Creation and Solution of Engineering Models . . . . . . . . . . 11.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Differential Equations . . . . . . . . . . . . . . . . . . . 41.2.2 Ordinary Differential Equations . . . . . . . . . . . . . . 41.2.3 Partial Differential Equations . . . . . . . . . . . . . . . 51.2.4 Order . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.5 Linearity and Nonlinearity . . . . . . . . . . . . . . . . 61.2.6 Homogeneous Functions . . . . . . . . . . . . . . . . . . 61.2.7 Boundary and Initial Conditions . . . . . . . . . . . . . . 71.2.8 Well–Posedness . . . . . . . . . . . . . . . . . . . . . 7

Chapter 2

Differential Equations of the First Order . . . . . . . . . 92.1 Separable Differential Equations . . . . . . . . . . . . . . . . 92.2 Reduction to Separable Form (Using Change of Variables) . . . 102.3 Exact Differential Equations . . . . . . . . . . . . . . . . 112.4 Integrating Factors (I.F.) . . . . . . . . . . . . . . . . . . 142.5 The General Solution of a First–Order Ordinary

Differential Equation . . . . . . . . . . . . . . . . . . . . 162.6 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . 182.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 20

Chapter 3

Fundamental and Important Properties of LinearOrdinary Differential Equations . . . . . . . . . . . . 23

3.1 Linear Differential Equation of Order n . . . . . . . . . . . . 233.2 Superposition Principle . . . . . . . . . . . . . . . . . . . 243.3 Existence and Uniqueness Theorem . . . . . . . . . . . . . 243.4 Linear Independence . . . . . . . . . . . . . . . . . . . . 253.5 The Wronskian . . . . . . . . . . . . . . . . . . . . . . 253.6 Differential Operators . . . . . . . . . . . . . . . . . . . 263.7 Fundamental Laws of Operation . . . . . . . . . . . . . . . 283.8 General Solution of a Nonhomogeneous Linear Equation

with Constant Coefficients . . . . . . . . . . . . . . . . . 29

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3.9 Homogeneous Solution . . . . . . . . . . . . . . . . . . . 293.9.1 Case 1: Roots All Real and Distinct . . . . . . . . . . . 293.9.2 Case 2: Roots are Complex . . . . . . . . . . . . . . . 303.9.3 Case 3: Repeated Roots . . . . . . . . . . . . . . . . . 30

3.10 Particular Solution . . . . . . . . . . . . . . . . . . . . 333.10.1 Method of Undetermined Coefficients . . . . . . . . . . 333.10.2 Lagrange’s Idea (Introduction to Variation of Parameters) . 363.10.3 Variation of Parameters Method . . . . . . . . . . . . 37

3.11 Particular Solution by Inspection . . . . . . . . . . . . . . 423.12 Equations with Variable Coefficients . . . . . . . . . . . . 43

3.12.1 Euler–Cauchy Equation . . . . . . . . . . . . . . . . 433.12.2 Another Approach . . . . . . . . . . . . . . . . . . 45

3.13 Case where One Solution is Known (Reduction of Order) . . . 473.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 48

Chapter 4

Series Solutions of Differential Equations . . . . . . . . . 494.1 Properties of Power Series . . . . . . . . . . . . . . . . . 494.2 Convergence of Power Series . . . . . . . . . . . . . . . . 504.3 Some Series Expansions . . . . . . . . . . . . . . . . . . 514.4 Operations on Power Series . . . . . . . . . . . . . . . . . 52

4.4.1 Termwise Differentiation or Integration . . . . . . . . . . 524.4.2 Termwise Addition or Subtraction . . . . . . . . . . . . 524.4.3 Termwise Multiplication . . . . . . . . . . . . . . . . 524.4.4 Termwise Division . . . . . . . . . . . . . . . . . . . 53

4.5 Illustrative Examples of the Power Series Method . . . . . . . 534.6 Regular Points of Linear 2nd–Order Differential Equations . . . 574.7 Method of Frobenius . . . . . . . . . . . . . . . . . . . . 584.8 Important 2nd–Order Equations . . . . . . . . . . . . . . . 624.9 Legendre’s Equation . . . . . . . . . . . . . . . . . . . . 634.10 Associated Legendre’s Functions . . . . . . . . . . . . . . 684.11 Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . 684.12 Gamma Function . . . . . . . . . . . . . . . . . . . . . 724.13 Bessel Functions of the Second Kind . . . . . . . . . . . . 734.14 Modified Bessel Functions . . . . . . . . . . . . . . . . . 764.15 Useful Formulas . . . . . . . . . . . . . . . . . . . . . 77

4.15.1 Recurrence Relationships Involving Jν(x) . . . . . . . . 774.15.2 Recurrence Relationships Involving Iν(x) . . . . . . . . . 784.15.3 Recurrence Relationships Involving Kν(x) . . . . . . . . 794.15.4 Integral Representations of Jn(x), In(x) and Kn(x) . . . . 804.15.5 Indefinite Integrals of Bessel Functions . . . . . . . . . . 814.15.6 Definite Integrals Involving Bessel Functions . . . . . . . 82

4.16 Summary of Bessel’s Equations . . . . . . . . . . . . . . . 834.17 Orthogonal Sets of Functions . . . . . . . . . . . . . . . . 88

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4.18 Sturm–Liuville Problem . . . . . . . . . . . . . . . . . . 904.19 Important Orthogonal Functions and Polynomials . . . . . . 93

4.19.1 Bessel Functions . . . . . . . . . . . . . . . . . . . 934.19.2 Legendre Polynomials . . . . . . . . . . . . . . . . . 94

4.20 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 95

Chapter 5

Laplace Transform . . . . . . . . . . . . . . . . . . 995.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 995.2 Characteristics of Problems Suited for Laplace Transform . . . . 995.3 Definition of the Laplace Transform . . . . . . . . . . . . . 995.4 References to Tables of Laplace Transforms . . . . . . . . . . 1025.5 Existence of Laplace Transforms . . . . . . . . . . . . . . . 103

5.5.1 Piecewise Continuity . . . . . . . . . . . . . . . . . . 1035.5.2 Functions of Exponential Order . . . . . . . . . . . . . 103

5.6 Some Important Properties of Laplace Transforms . . . . . . . 1045.6.1 Linearity Property . . . . . . . . . . . . . . . . . . . 1045.6.2 Laplace Transform of Derivatives . . . . . . . . . . . . . 1045.6.3 Laplace Transform of Integrals . . . . . . . . . . . . . . 1055.6.4 First Translation or Shifting Property (Shifting on s–axis) . 1055.6.5 Second Translation or Shifting Property (Shifting on t–axis) 1065.6.6 Change of Scale Property . . . . . . . . . . . . . . . . 1075.6.7 Division by t . . . . . . . . . . . . . . . . . . . . . 1075.6.8 Multiplication by tn . . . . . . . . . . . . . . . . . . 1085.6.9 Periodic Functions . . . . . . . . . . . . . . . . . . . 1085.6.10 Behavior of F (s) as s→∞ . . . . . . . . . . . . . . . 109

5.7 Methods of Finding Laplace Transforms . . . . . . . . . . . 1105.8 The Inverse Laplace Transform . . . . . . . . . . . . . . . 1105.9 Some Important Properties of the Inverse Laplace Transform . . 111

5.9.1 Linearity . . . . . . . . . . . . . . . . . . . . . . . 1115.9.2 First Translation or Shifting Property . . . . . . . . . . . 1115.9.3 Second Translation or Shifting Property . . . . . . . . . . 1125.9.4 Change of Scale Property . . . . . . . . . . . . . . . . 1125.9.5 Inverse Laplace Transform of Derivatives . . . . . . . . . 1125.9.6 Inverse Laplace Transform of Integrals . . . . . . . . . . 1135.9.7 Multiplication by sn . . . . . . . . . . . . . . . . . . 1135.9.8 Division by s . . . . . . . . . . . . . . . . . . . . . 1145.9.9 Convolution Theorem . . . . . . . . . . . . . . . . . . 114

5.10 Some Methods of Finding Inverse Laplace Transforms . . . . . 1155.10.1 Partial Fractions Method . . . . . . . . . . . . . . . . 1165.10.2 Series Method . . . . . . . . . . . . . . . . . . . . 1185.10.3 The Complex Inversion Formula . . . . . . . . . . . . 119

5.11 Some Special Functions . . . . . . . . . . . . . . . . . . 1195.11.1 The Error Function . . . . . . . . . . . . . . . . . . 1195.11.2 The Dirac Delta Function . . . . . . . . . . . . . . . 122

5.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . 123

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS ––––––––––––––––––––––––––– vii

Chapter 6

Application of Laplace Transforms in the Solution ofOrdinary Differential Equations . . . . . . . . . . 125

6.1 Methodology . . . . . . . . . . . . . . . . . . . . . . . 1256.2 Ordinary Differential Equations with Constant Coefficients . . . 1256.3 Ordinary Differential Equations with Variable Coefficients . . . 1286.4 Simultaneous Solution of Ordinary Differential Equations . . . . 1306.5 Application to Beams . . . . . . . . . . . . . . . . . . . 1336.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 136

Chapter 7

Linear Algebra . . . . . . . . . . . . . . . . . . . 1397.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 139

7.1.1 Laws of Vector Algebra . . . . . . . . . . . . . . . . . 1397.1.2 Unit Vectors . . . . . . . . . . . . . . . . . . . . . . 1397.1.3 Components of a Vector . . . . . . . . . . . . . . . . . 1397.1.4 Dot or Scalar Product . . . . . . . . . . . . . . . . . 1407.1.5 General Vector Space . . . . . . . . . . . . . . . . . . 1407.1.6 Gradient and Divergence . . . . . . . . . . . . . . . . 141

7.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . 1417.2.1 Definition of a Matrix . . . . . . . . . . . . . . . . . 1417.2.2 Some Special Definitions and Operations Involving Matrices 1427.2.3 Gaussian Elimination (Triangular Factorization) . . . . . . 1467.2.4 Rank of a Matrix . . . . . . . . . . . . . . . . . . . 1477.2.5 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . 1487.2.6 Gauss–Jordan Elimination . . . . . . . . . . . . . . . . 149

7.3 Determinants . . . . . . . . . . . . . . . . . . . . . . . 1507.3.1 Theorems on Determinants . . . . . . . . . . . . . . . 1517.3.2 Cramer’s Rule (Revisit) . . . . . . . . . . . . . . . . . 1517.3.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . 1527.3.4 Application . . . . . . . . . . . . . . . . . . . . . . 1527.3.5 Multiple Eigenvalues . . . . . . . . . . . . . . . . . . 1567.3.6 Complex Eigenvalues . . . . . . . . . . . . . . . . . . 1577.3.7 Reduction of Matrix to Diagonal Form . . . . . . . . . . 1577.3.8 Properties of Eigenvalues and Eigenvectors . . . . . . . . 158

7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 159

Chapter 8

Fourier Series, Fourier Integrals, Fourier Transforms . . . 1618.1 Periodic Functions . . . . . . . . . . . . . . . . . . . . . 1618.2 Properties of Periodic Functions . . . . . . . . . . . . . . . 1628.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . 163

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8.3.1 Orthogonality Relations for Sine and Cosine Functions . . . 1638.3.2 Evaluation of Fourier Coefficients . . . . . . . . . . . . 1648.3.3 Functions of any Period p = 2L . . . . . . . . . . . . . 1678.3.4 Even and Odd Functions . . . . . . . . . . . . . . . . 1688.3.5 Fourier Series of Even and Odd Functions of Period 2L . . . 1688.3.6 Hidden Symmetry . . . . . . . . . . . . . . . . . . . 1698.3.7 Half–Range Expansions . . . . . . . . . . . . . . . . . 170

8.4 Fourier Integrals . . . . . . . . . . . . . . . . . . . . . . 1728.4.1 Fourier Cosine and Sine Integrals . . . . . . . . . . . . . 1748.4.2 Fourier Transformations . . . . . . . . . . . . . . . . . 1768.4.3 Fourier Transformation (Complex Form) . . . . . . . . . 179

8.5 Properties of Fourier Transformations . . . . . . . . . . . . 1818.5.1 Linearity . . . . . . . . . . . . . . . . . . . . . . . 1818.5.2 Time Shifting . . . . . . . . . . . . . . . . . . . . . 1818.5.3 Frequency Shifting . . . . . . . . . . . . . . . . . . . 1818.5.4 Scaling . . . . . . . . . . . . . . . . . . . . . . . . 1818.5.5 Time–Reversal . . . . . . . . . . . . . . . . . . . . . 1828.5.6 Modulation Theorem . . . . . . . . . . . . . . . . . . 1828.5.7 Additional Properties when f(t) is Real . . . . . . . . . . 1838.5.8 Differentiation and Integration Theorems . . . . . . . . . 1858.5.9 Multidimensional Fourier Transforms . . . . . . . . . . . 1868.5.10 Convolution . . . . . . . . . . . . . . . . . . . . . 1868.5.11 Frequency Convolution . . . . . . . . . . . . . . . . . 187

8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 187

Chapter 9

Partial Differential Equations . . . . . . . . . . . . . 1899.1 Classification of 2nd–Order Partial Differential Equations . . . . 1899.2 Method of Separating Variables (Product Method) . . . . . . . 190

9.2.1 Vibration of a String . . . . . . . . . . . . . . . . . . 1919.2.2 One–Dimensional Acoustic Vibrations . . . . . . . . . . 1979.2.3 Nuclear Reactor Criticality . . . . . . . . . . . . . . . 200

9.3 Laplace Transformation Applied to the Solution ofPartial Differential Equations . . . . . . . . . . . . . . . . . 203

9.4 Fourier Transformations Applied to the Solution ofPartial Differential Equations . . . . . . . . . . . . . . . . . 209

9.5 Self–Similar Solutions . . . . . . . . . . . . . . . . . . . 2129.5.1 Characteristic Scales and Scale–Similar Problems . . . . . . 2129.5.2 Self–Similarity . . . . . . . . . . . . . . . . . . . . . 213

9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 219

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Chapter 10

Solutions to Exercises . . . . . . . . . . . . . . . . 22110.1 Differential Equations of the First Order (Chapter 2) . . . . . 22110.2 Fundamental and Important Properties of Linear Ordinary

Differential Equations (Chapter 3) . . . . . . . . . . . . . . 23110.3 Series Solutions of Differential Equations (Chapter 4) . . . . . 23810.4 Laplace Transform (Chapter 5) . . . . . . . . . . . . . . . 25210.5 Application of Laplace Transform in the Solution of

Ordinary Differential Equations (Chapter 6) . . . . . . . . . . 25910.6 Linear Algebra (Chapter 7) . . . . . . . . . . . . . . . . 26610.7 Fourier Series, Integrals and Transforms (Chapter 8) . . . . . 27410.8 Partial Differential Equations (Chapter 9) . . . . . . . . . . 282

References . . . . . . . . . . . . . . . . . . . . . 289

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS ––––––––––––––––––––––––––––– x

Reality

ODE

PDE

Mathematical Models & Solutions

Power SeriesLaplace Transforms

Numerical Methods

Fourier Series

Separation of Variables

Numerical Methods

Prediction of Reality via Mathematical Model Solutions

Chapter 1

Introduction

1.1 Creation and Solution of Engineering Models

The important steps associated with the creation and solution of an engi-neering mathematical model are:(1) Begin with a physical problem.(2) Define the physics, chemistry and/or biology of the problem in mathe-

matical terms.(3) Collect mathematical relations into a well–posed problem.(4) Solve the mathematical problem.(5) Use the solution to predict, etc.

The mathematical development of an engineering model is also schemati-cally illustrated in Figure 1.1.

Figure 1.1: Pattern for the creation and solution of a physical model. Theabbreviations ODE and PDE represent ordinary and partialdifferential equations, respectively.

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS –––––––––––––––––––––––––––––– 1

Bedr ock

Wate r TableFlow

TCE

It should be noted that an integral transformation produces from givenfunctions new functions which depend on a different variable and appearin the form of an integral to be evaluated. These transformations are ofinterest mainly as tools in solving ordinary differential equations, partialdifferential equations and integral equations, and they often also help inhandling and applying special functions. The Laplace transformation is ofthis kind and is by far the most important integral transformation in engi-neering. From the viewpoint of applications the next in order of importanceare perhaps the Fourier transformations, although these are somewhat moredifficult to handle than the Laplace transformation.

Example

Consider the case of modeling the transient contaminant transport fromnon–aqueous phase liquid pool denser than water in a two–dimensional ho-mogeneous porous medium under steady–state uniform flow conditions. Anon–aqueous phase liquid released into the subsurface environment infil-trates through the vadose zone leaving behind blobs or ganglia which areno longer connected to the main body of the organic liquid. Upon reachingthe water table, non–aqueous phase liquids heavier than water continue tomigrate downward until they encounter an impermeable layer where a flatpool starts to form (see Figure 1.2). As groundwater flows past a non–aqueous phase liquid pool, a plume of dissolved hydrocarbons is created.For mathematical simplicity the physical problem illustrated in Figure 1.2is represented by a conceptual model presented in Figure 1.3.

Figure 1.2: Schematic illustration of the migration in the subsurface ofslightly soluble in water organic liquids which are heavier thanwater, such as trichloroethylene (TCE).


x

z

xlx

lxo lxo+lx

C(t,x ,z)

Cs

Figure 1.3: Profile view of the conceptual model showing the unidirec-tional groundwater velocity Ux, the location of a denser thanwater non–aqueous phase liquid pool with aqueous saturationconcentration Cs and length �x, and the dissolved concentra-tion in the plume C(t, x, z).

Assuming that the organic solvent is sorbing under local equilibrium con-ditions, the governing partial differential equation is given by

R∂C

∂t= Dx

∂2C

∂x2+ Dz

∂2C

∂z2− Ux

∂C

∂x,

where C(t, x, z) is the liquid phase solute concentration; Ux is the averageinterstitial fluid velocity; x, z are the spatial coordinates in the longitudinaland vertical directions, respectively; R is the dimensionless retardation fac-tor for linear, reversible, instantaneous sorption; Dx and Dz are the longi-tudinal and vertical hydrodynamic dispersion coefficients, respectively; andt is time. Assuming that the thickness of the pool is insignificant relative tothe thickness of the aquifer, the non–aqueous phase liquid dissolution at thepool–water interface is described by the following mass transfer relationship

−De∂C(t, x, 0)

∂z= k(t, x) [Cs − C(t, x,∞)] ,

where De = D/τ∗ is the effective molecular diffusion coefficient, D is themolecular diffusion coefficient, τ∗ is the tortuosity coefficient (τ∗ ≥ 1),k(t, x) is the local mass transfer coefficient dependent on time and locationon the pool–water interface, Cs is the aqueous concentration at the inter-face and for a pure organic liquid equals the liquid’s aqueous saturation(solubility) concentration, and C(t, x,∞) � 0 corresponds to the contami-nant concentration outside the boundary layer, the appropriate initial andboundary conditions for this system are:

C(0, x, z) = 0,


y

y=0

C(t,±∞, z) = 0,

De∂C(t, x, 0)

∂z=

{ 0 0 ≤ x ≤ �o,−Csk(t, x) �o < x < �o + �,0 x ≥ �o + �,

C(t, x,∞) = 0.

The solution to the governing partial differential equation subject to theprescribed initial and boundary conditions can be obtined by Laplace trans-form techniques.

1.2 Definitions

1.2.1 Differential Equations

A differential equation contains derivatives of an unknown function, whichwe call y(x) and which we want to determine from the equation.

1.2.2 Ordinary Differential Equations

An ordinary differential equation has only one independent variable andinvolves one or several derivatives of y(x). The equation may also involvey(x) itself, given functions of x, and constants.

Example

The acceleration of a falling stone, as illustrated in Figure 1.4, is equal tothe acceleration of gravity, g, and is described by the following second–orderordinary differential equation

d2y

dt2= g = constant,

where the dependent variable is the vertical distance, y, and the indepen-dent variable is time, t.

Figure 1.4: A falling stone.


1.2.3 Partial Differential Equations

An equation that involves partial derivatives of y of two or more indepen-dent variables is called partial differential equation.

Examples

(a) One–dimensional wave equation (hyperbolic)

∂2u

∂t2= c2 ∂2u

∂x2,

where u(x, t) is the deflection of a string, x is the spatial coordinate and tis time.(b) One–dimensional heat equation (parabolic)

∂u

∂t= c2 ∂2u

∂x2,

where u(x, t) is the temperature and c2 = κ/σρ (κ is the thermal conduc-tivity, σ is the specific heat and ρ is the density).

(c) Two–dimensional Laplace equation (elliptic)

∂2u

∂x2+

∂2u

∂y2= 0.

1.2.4 OrderThe order of the highest derivative of the differential equation is called theorder of the equation.

Examples

∂y

∂t= ky ←− 1st order ODE

∂2u

∂x2+

∂2u

∂y2= F (x, y) ←− 2nd order PDE (Poisson equation)

d2y

dx2+ 2b

(dy

dx

)3

+ y = 0 ←− 2nd order ODE


1.2.5 Linearity and Nonlinearity

A linear equation is one in which the dependent variable and its derivativesappear only to the first power, and never in products. Thus, the generallinear equation of order n may be written as follows

b0(x)dny

dxn+ b1(x)

dn−1y

dxn−1+ · · ·+ bn−1(x)

dy

dx+ bn(x)y = R(x).

If the right–hand side, R(x), is zero for all x, the equation is said to behomogeneous; otherwise, it is said to be nonhomogeneous:

R(x) = 0 ←− homogeneous,

R(x) = 0 ←− nonhomogeneous.

Examples

d2y

dx2+

dy

dx− 6y = 0 ←− 2nd order linear,

dy

dx+ k√

y = 0 ←− 1st order nonlinear.

1.2.6 Homogeneous Functions

The function f(x, y) is said to be homogeneous of degree k in x and y ifand only if

f(λx, λy) = λkf(x, y),

where λ and k are constants.

Examples

4x2 − 3xy + y2 ←− homogeneous, k = 2,

ex ←− nonhomogeneous,

exp(

x

y

)←− homogeneous, k = 0,

y2 tan(

x

y

)←− homogeneous, k = 2,

x3 − xy + y3 ←− nonhomogeneous.


1.2.7 Boundary and Initial Conditions

In addition to the governing differential equation one must have an appro-priate set of boundary and/or initial conditions specified by the physicalsystem examined.

1.2.8 Well–PosednessIf a portion of the needed boundary or initial condition information is miss-ing, the problem is said to be incompletely posed. If too much information,or the wrong type of information, is prescribed, the problem is said to beill–posed. It is important to learn to identify when a problem is not well–posed. A good way to do this is to use physical intuition. There are certainmathematical rules, for simple types of problems, but in dealing with theproblems that typically arise in engineering or applied science one has nochoice but to rely on experience and intuition.


Chapter 2

Differential Equations of the First Order

2.1 Separable Differential Equations

Many first–order differential equations can be reduced to the form

g(y)dy

dx= f(x)

=⇒ g(y) dy = f(x) dx,

that is an equation with separable variables, by integrating on both sideswe obtain ∫

g(y) dy =∫

f(x) dx + c.

Example 1

Solve the following first–order ordinary differential equation

dy

dt= −ky.

dy

y= −k dt, y �= 0

dy

y= d(lny)

=⇒∫

d(ln y) = −k

∫dt + C

=⇒ ln |y| = −kt + C

=⇒y = exp[−kt + C]

=⇒y = eCe−kt

=⇒y = C1e−kt

At t = 0 : y = C1 = initial value �


Example 2


dy

dx=

2y

x, ∀ x > 0, y > 0.

dy

y=

2x

dx

=⇒ ln |y| = 2 ln |x|+ c

=⇒y = exp[2 lnx + c]

=⇒y = ece2 ln x

=⇒y = C1eln x2

=⇒y = C1x2 �

2.2 Reduction to Separable Form (Using Change of Variables)

For equations of the formdy

dx= g

(y

x

)(1)

that is homogeneous functions of degree 0, use

y = ux

(or x = uy; it is sometimes easier to substitute for the variable whosedifferential has the simpler coefficient)

=⇒dy = u dx + x du (2)

(1) & (2) =⇒u + xdu

dx= g

(y

x

)= g(u)

=⇒ du

g(u)− u=

dx

x

the variables are separable.

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS –––––––––––––––––––––––––––– 10

Example


(x2 − xy + y2) dx− xy dy = 0.

=⇒ dx

dy=

xy

x2 − xy + y2

Because the coefficients are both homogeneous and of the same degree(k=2) let

y = ux, dy = u dx + x du

(x2 − x2u + x2u2) dx− x2u(u dx + x du) = 0 divide by x2

=⇒(1− u + u2) dx− u(u dx + x du) = 0=⇒(1− u) dx− ux du = 0 ∀ x > 0, u �= 1

=⇒dx

x+

u du

u− 1= 0

=⇒dx

x+

(1 +

1u− 1

)du = 0

=⇒ ln |x|+ u + ln |u− 1| = constant = ln |C|=⇒ exp [ln|x|+ u + ln|u− 1|] = C

=⇒x(u− 1)eu = C.

In terms of original variables

x(y

x− 1

)exp

[y

x

]= C

=⇒ (y − x) exp[y

x

]= C (Family of solutions) �

2.3 Exact Differential Equations

A first–order differential equation of the form

M(x, y)dx + N(x, y)dy = 0


is exact if a function F exists such that

dF = M dx + N dy,

dF = 0 =⇒ F = C.

But, from calculus the total or exact differential for F (x, y) is given by

dF =∂F

∂xdx +

∂F

∂ydy.

So, M and N must be

M =∂F

∂x, N =

∂F

∂y

=⇒ ∂M

∂y=

∂2F

∂y∂x,

∂N

∂x=

∂2F

∂x∂y

=⇒ ∂M

∂y=

∂N

∂x�

Restrictions: M , N ,∂M

∂y,

∂N

∂xare continuous functions of x and y.

Example 1


3x(xy − 2)︸︷︷︸M

dx + (x3 + 2y)︸︷︷︸N

dy = 0.

∂M

∂y= 3x2,

∂N

∂x= 3x2

Therefore, the governing equation is exact and its solution is F = C.

∂F

∂x= M = 3x2y − 6x, (1)

∂F

∂y= N = x3 + 2y. (2)


Integrating both sides of (1) with respect to x holding y constant, yields

F = x3y − 3x2 + T (y), (3)

where the usual arbitrary constant resulting from the indefinite integrationis now necessarily a function T (y). To determine T (y), we use the fact thatthe function F of equation (3) must also satisfy equation (2). Hence,

x3 +dT (y)

dy= x3 + 2y

=⇒ dT (y)dy

= 2y.

It should be noted that no arbitrary constant is needed because one is beingintroduced on the right–hand side of the solution F = C.∫

dT (y) =∫

2y dy

=⇒ T (y) = y2. (4)

From (3) and (4) =⇒ F = x3y − 3x2 + y2

=⇒ x3y − 3x2 + y2 = C (A set of solutions) �

Example 2


(2x3 − xy2 − 2y + 3)︸︷︷︸M

dx− (x2y + 2x)︸︷︷︸N

dy = 0.

Because∂M

∂y= −2xy − 2 =

∂N

∂x,

a set of solutions is F = C, where

∂F

∂x= 2x3 − xy2 − 2y + 3 = M, (1)

∂F

∂y= −x2y − 2x = N. (2)


Integrating (2) yields

F = −x2y2

2− 2xy + Q(x). (3)

Substituting (3) into (1) yields

−xy2 − 2y +dQ(x)

dx= 2x3 − xy2 − 2y + 3

=⇒ dQ(x)dx

= 2x3 + 3

=⇒ Q(x) =x4

2+ 3x. (4)

Substituting (4) into (3) leads to

F = −x2y2

2− 2xy +

x4

2+ 3x = C

=⇒ x4 − x2y2 − 4xy + 6x = 2C = C∗ �

2.4 Integrating Factors (I.F.)

An inexact ordinary differential equation can be made exact by multiplyingby some nonzero function u(x) or u(y), known as the integrating factor.Consider the inexact equation

P (x, y)dx + Q(x, y)dy = 0, (∗)

assuming that there exists a positive integrating factor u(x). Then, thefollowing equation is exact

u(x)P (x, y)︸︷︷︸M

dx + u(x)Q(x, y)︸︷︷︸N

dy = 0.

=⇒ ∂M

∂y=

∂N

∂x


=⇒ ∂

∂y[u(x)P (x, y)] =

∂

∂x[u(x)Q(x, y)]

=⇒ u(x)∂P (x, y)

∂y=

∂u(x)∂x

Q(x, y) + u(x)∂Q(x, y)

∂x,

dividing by u(x)Q(x, y) yields

=⇒ 1u(x)

du(x)dx

=1

Q(x, y)

[∂P (x, y)

∂y− ∂Q(x, y)

∂x

].

Note that the left–hand side of the preceding expression is only a function ofx. If the right–hand side is also a function of x, then (∗) has an integratingfactor u(x), which is obtained by solving the preceding equation:

=⇒ u(x) = exp{∫

1Q(x, y)

[∂P (x, y)

∂y− ∂Q(x, y)

∂x

]dx

}.

Similarly, if1

P (x, y)

[∂P (x, y)

∂y− ∂Q(x, y)

∂x

]= f(y),

=⇒ u(y) = exp{−

∫1

P (x, y)

[∂P (x, y)

∂y− ∂Q(x, y)

∂x

]dy

}

is an integrating factor. Note the negative exponential argument.

Example 1

The following ordinary differential equation

2y(x2 − y + x)︸︷︷︸P

dx + (x2 − 2y)︸︷︷︸Q

dy = 0,

may be reduced to exact by an integrating factor.

∂P

∂y− ∂Q

∂x= (2x2 − 4y + 2x)− 2x = 2x2 − 4y,

=⇒ 1Q

[∂P

∂y− ∂Q

∂x

]=

2x2 − 4y

x2 − 2y= 2 ←− f(x)


An integrating factor is:

u(x) = exp{∫

1Q

[∂P

∂y− ∂Q

∂x

]dx

}

= exp{∫

2 dx

}= e2x �

Example 2

Obtain an integrating factor for the following ordinary differential equation

y2︸︷︷︸P

dx + (3xy + y2 − 1)︸︷︷︸Q

dy = 0.

∂P

∂y− ∂Q

∂x= 2y − 3y = −y,

1P

[∂P

∂y− ∂Q

∂x

]=−y

y2=−1y

.

An integrating factor is:

u(y) = exp{−

∫1P

[∂P

∂y− ∂Q

∂x

]dy

}

= exp{∫

dy

y

}= exp {ln y} = y �

2.5 The General Solution of a Linear First–Order OrdinaryDifferential Equation

The general form of a first–order ordinary differential equation is given by

dy

dx+ P (x)y = R(x).


For the moment, suppose that there exists for the preceding equation apositive integrating factor, u(x) > 0, a function of x alone. Then, thefollowing equation must be exact

u(x)[

dy

dx+ P (x)y

]= u(x)R(x).

=⇒ [u(x)P (x)y − u(x)R(x)]︸︷︷︸M

dx + u(x)︸︷︷︸N

dy = 0 (1)

∂M

∂y=

∂N

∂x

=⇒u(x)P (x) =du(x)

dx

=⇒P (x) dx =du(x)u(x)

=⇒ lnu(x) =∫

P (x) dx

=⇒u(x) = exp[∫

P (x) dx

]. (2)

Substituting (2) into (1) yields

exp[ · · · ] dy + y exp[ · · · ]P (x) dx = exp[ · · · ]R(x) dx

=⇒ d

dx{y exp[ · · · ]} = exp[ · · · ]R(x)

=⇒y exp[ · · · ] =∫

exp[ · · · ]R(x) dx + c

=⇒y = e−∫

P (x) dx

{∫e∫

P (x) dxR(x) dx + c

}�

Example

Find the solution to the following ordinary differential equation

(x4 + 2y) dx− x dy = 0.

=⇒ x4 + 2y − xdy

dx= 0, divide by − x


=⇒ −x3 − 2y

x+

dy

dx= 0

=⇒ dy

dx+

(− 2

x

)︸︷︷︸

p(x)

y = x3︸︷︷︸R(x)

.

The preceding equation is a linear nonhomogeneous first–order ordinarydifferential equation with a solution directly given as

y = e−∫

− 2x dx

{∫e∫

− 2x dxx3 dx + c

}

=⇒e2 ln x

{∫e−2 ln xx3 dx + c

}

=⇒y = x2

{∫x dx + c

}

=⇒y = x2

{x2

2+ c

}�

2.6 Bernoulli’s Equation

An equation that can be transformed into a type that we know how to solveis the following

dy

dx+ P (x)y = R(x)yn.

If n = 1, the variables are separable, so we concentrate on the case n �= 1.

y−ndy + P (x)y−n+1dx = R(x) dx,

let z = y−n+1, dz = (1− n)y−n dy

=⇒ dz

1− n+ P (x)z dx = R(x) dx

=⇒ dz

dx+ (1− n)P (x)z = (1− n)R(x) �

This transformed equation is a linear equation in the standard form thatcan easily be solved by standard techniques.


Example

Find the solution to the following ordinary differential equation

xy dx +(x2 − 3y

)dy = 0.

=⇒xydx

dy+ x2 − 3y = 0

=⇒dx

dy+

(1y

)x = 3x−1 ← Bernoulli′s eq. in x

P (y) =1y, R(y) = 3, n = −1,

let z = x1−n = x1−(−1) = x2,

dz = 2x dx −→ dx =dz

2x,

=⇒dz

dy+ 2

(1y

)z = (2)(3)

=⇒dz

dy+

(2y

)z = 6 ← 1st order linear

=⇒z = e−∫

2y dy

{∫e∫

2y dy6 dy + c

}

=⇒z = e−2 ln y

{∫6e2 ln y dy + c

}

=⇒z =1y2

{2y3 + c

}=⇒zy2 = 2y3 + c.

Substituting back the original variables leads to

=⇒ x2y2 = 2y3 + c �


2.7 Exercises

(1) Obtain the general solution of (Separable O.D.E.)

(a) xy3dx + ex2dy = 0

(b)dy

dx= xy2

(c) x2ydy

dx= ey

(2) Obtain a family of solutions for the following equations (Reduction toseparable form)

(a) xy dx−(x2 + 3y2

)dy = 0 Ans. x2 = 6y2 ln

y

C

(b) x2 dy

dx= 4x2 + 7xy + 2y2 Ans. x2(y + 2x) = C(y + x)

(c) (x− y)(4x + y)dx + x(5x− y)dy = 0 Ans. x(x + y)2 = C(y− 2x)

(3) Test each of the following equations for exactness and solve the equation

(a) (x + 2y)dx + (2x + y)dy = 0

(b) v(2uv2 − 3

)du +

(3u2v2 − 3u + 4v

)dv = 0

(c)(cos 2y − 3x2y2

)dx +

(cos 2y − 2x sin 2y − 2x3y

)dy = 0


(4) Solve each of the following equations (I.F.):

(a)(x2 + y2 + 1

)dx + x(x− 2y)dy = 0

(b) y(2x− y + 1)dx + x(3x− 4y + 3)dy = 0

(c) y(4x + y)dx− 2(x2 − y

)dy = 0

(5) Find the general solution of the following equations:

(a) y2dx +(xy + y2 − 1

)dy = 0; when x = −1, y = 1

(b)dy

dx= y tanx + cos x

(6) Solve the following equations (Bernoulli’s):

(a) 2x3 dy

dx= y

(y2 + 3x2

)

(b)(2y3 − x3

)dx + 3xy2dy = 0; when x = 1, y = 1


Chapter 3

Fundamental and Important Properties of LinearOrdinary Differential Equations

3.1 Linear Differential Equation of Order n

The general linear differential equation of order n has the form:

b0(x)dny

dxn+ b1(x)

dn−1y

dxn−1+ · · ·+ bn−1(x)

dy

dx+ bn(x)y = R(x). (1)

If R(x) = 0, the preceding equation is said to be linear and homogeneous.If y1 and y2 are solutions of the general homogeneous equation and if

c1 and c2 are constants, then

y = c1y1 + c2y2

is a solution of the general homogeneous equation.The statement that y1 and y2 are solutions of (1) means that

b0yn1 + b1y

n−11 + · · ·+ bn−1y

′1 + bny1 = 0 (2)

b0yn2 + b1y

n−12 + · · ·+ bn−1y

′2 + bny2 = 0 (3)

multiply each member of (2) by c1 and each member of (3) by c2, and addthe results to yield

b0

(c1y

(n)1 + c2y

(n)2

)+b1

(c1y

(n−1)1 + c2y

(n−1)2

)+ · · ·

+bn−1 (c1y′1 + c2y′2) + bn(c1y1 + c2y2) = 0. (4)

Becausec1y

′1 + c2y

′2 = (c1y1 + c2y2)′,

· · · · · ·

c1y(n)1 + c2y

(n)2 = (c1y1 + c2y2)(n),

equation (4) is neither more nor less than the statement that c1y1 + c2y2 isa solution of equation (1).


f(x)

xε

+ε

−ε

Special Case

If c2 = 0, for a homogeneous linear equation any constant times a solutionis also a solution. Similarly: if yi(i = 1, 2, · · · , k) are solutions of equation(1), and if ci(i = 1, 2, · · · , k) are constants, then any linear combination ofthe functions y1, y2, · · · , yk, for example

y = c1y1 + c2y2 + · · ·+ ckyk,

is a solution of equation (1).

3.2 Superposition Principle

Theorem: Any linear combination of solutions of a linear homogeneousdifferential equation is also a solution.

3.3 Existence and Uniqueness Theorem

If b0(x), b1(x), · · · , bn(x) and R(x) are continuous in the interval a < x <b, then there exists one and only one solution to (1) which satisfies thefollowing initial conditions:

y(x0) = y0

y′(x0) = y′0

· · · · · ·

y(n−1)(x0) = y(n−1)0 .

A function f(x) is said to be continuous at ε if limx→ε

f(x) = f(ε). A discon-

tinuous function is shown in Figure 3.1.

Figure 3.1: A function with a discontinuity at x = ε.


3.4 Linear Independence

Given the functions f1, · · · , fn, and

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0, a ≤ x ≤ b

then, f1, · · · , fn are linearly independent when c1 = c2 = · · · = cn = 0.If the functions of a set are linearly dependent, at least one of them

is a linear combination of the others; if they are linearly independent, thennone of them is a linear combination of the others.

Example 1

c1︸︷︷︸−5

f1︷︸︸︷(2e3x

)+ c2︸︷︷︸

2

f2︷︸︸︷(5e3x

)+ c3︸︷︷︸

0

f3︷︸︸︷(e−4x

)= 0

f1, f2 and f3 are linearly dependent since c1 �= c2 �= c3 �= 0.

Example 2

The functions f1 = ex and f2 = xex are linearly independent because

c1ex + c2xe

x = 0

only if c1 = c2 = 0.

3.5 The Wronskian

Theorem: If, on the interval a < x < b, b0, b1, · · · , bn are continuous, andy1, y2, · · · , yn are solutions of the following equation

b0y(n) + b1y

(n−1) + · · ·+ bn−1y′ + bny = 0,

then a necessary and sufficient condition that y1, · · · yn be linearly inde-pendent is the nonvanishing of the Wronskian of y1, · · · , yn on the intervala < x < b. The Wronskian is defined as

W (y1, · · · yn) =

∣∣∣∣∣∣∣∣y1(x) y2(x) · · · yn(x)y′1(x) y′2(x) · · · y′n(x)

......

. . ....

y(n−1)1 (x) y

(n−1)2 (x) · · · y

(n−1)n (x)

∣∣∣∣∣∣∣∣�= 0.


A nonzero Wronskian implies independence. However, it is important tonote that a Wronskian of zero does not imply dependence of vectors unlessthose vectors are known all to be solutions of the same homogeneous linearvector differential equation.

Example

The functions y1(x) = 1, y2(x) = e3x, and y3(x) = e−3x are solutions of

d3y

dx3− 9

dy

dx= 0,

show that they are linearly independent∣∣∣∣∣∣1 e3x e−3x

0 3e3x −3e−3x

0 9e3x 9e−3x

∣∣∣∣∣∣ =∣∣∣∣ 3e3x −3e−3x

9e3x 9e−3x

∣∣∣∣ = 27− (−27) = 54 �= 0,

therefore the functions are independent �

Recall:

D =

∣∣∣∣∣∣α11 α12 α13

α21 α22 α23

α31 α32 α33

∣∣∣∣∣∣ = α11

∣∣∣∣α22 α23

α32 α33

∣∣∣∣−α21

∣∣∣∣α12 α13

α32 α33

∣∣∣∣+α31

∣∣∣∣α12 α13

α22 α23

∣∣∣∣

3.6 Differential Operators

It is sometimes convenient to adopt the following notation

Dy =dy

dx,

D2y =d2y

dx2,

· · · · · ·

Dky =dky

dxk.

The expression

A = b0Dn + b1D

n−1 + · · ·+ bn−1D + bn,


is called a differential operator of order n, which, when applied to anyfunction y, yields

Ay = b0dny

dxn+ b1

dn−1y

dxn−1+ · · ·+ bn−1

dy

dx+ bny.

Example

The equation

xd2y

dx2+ 3

dy

dx− 2xy = sinx

can be written as (xD2 + 3D − 2x

)y = sinx.

• Two operators A and B are said to be equal if, and only if

Ay = By.

• The product AB of two operators A and B is that operator which

ABy = A(By).

(Note that the product of two differential operators always exists and isa differential operator).

• The sum of two differential operators is obtained by expressing each inthe form

b0Dn + b1D

n−1 + · · ·+ bn−1D + bn

and adding corresponding coefficients.

Example

A = 3D2 −D + x− 2,

B = x2D2 + 4D + 7,

=⇒ A+B =(3 + x2

)D2 + 3D + x+ 5 �


3.7 Fundamental Laws of Operation

Let A, B, and C be any differential operators. Then, the differential oper-ators satisfy the following:

A+B = B +A Commutative law for addition(A+B) + C = A+ (B + C) Associative law for addition(AB)C = A(BC) Associative law for multiplicationA(B + C) = AB +AC Distributive law of multiplication

with respect to additionAB = BA Commutative law of multiplication (only

for operators with constant coefficients)DmDn = Dm+n If m and n are positive integers

Example 1

A = D + 2,B = 3D − 1,

AB = (D + 2)(3D − 1) = 3D2 −D + 6D − 2 = 3D2 + 5D − 2,

BA = (3D − 1)(D + 2) = 3D2 + 5D − 2,=⇒AB = BA �

The commutative law of multiplication holds.

Example 2

G = xD + 2,H = D − 1,GH = (xD + 2)(D − 1)

= x(DD)− xD + 2D − 2

= xD2 + (2− x)D − 2,HG = (D − 1)(xD + 2) = D(xD) + 2D − xD − 2

= D + xD2 + (2− x)D − 2 = xD2 + (3− x)D − 2,{D(xD) =

d

dx

(xd

dx

)=

d

dx+ x

d2

dx2

}GH �= HG �

The commutative law of multiplication does not hold because the coeffi-cients are variable.


3.8 General Solution of a Nonhomogeneous Linear Equationwith Constant Coefficients

In view of the superposition principle the general solution of a nonhomoge-neous linear differential equation can be written as:

y(x) = yh(x) + yp(x),

whereyh = c1y1(x) + · · ·+ cnyn(x)

is the solution of the homogeneous equation, it is also known as comple-mentary solution; and

yp(x)

is a solution of the nonhomogeneous equation, it contains no arbitraryconstants.

3.9 Homogeneous Solution

Assume a trial solution of the form

y = eλx, λ = constant,

find roots to satisfy the differential equation(b0D

n + b1Dn−1 + · · ·+ bn

)y = 0

=⇒(b0λ

n + b1λn−1 + · · ·+ bn

)eλx = 0

=⇒b0λn + b1λ

n−1 + · · ·+ bn = 0.

The preceding expression is the characteristic equation or auxiliary equationwhich has roots λ1, λ2, · · · , λn.

3.9.1 Case 1: Roots All Real and Distinct

Then, eλ1x, eλ2x, · · · eλnx are n linearly independent solutions and the re-quired solution is

y = c1eλ1x + c2e

λ2x + · · ·+ cneλnx.


Example 1

Solve the following equation

d2x

dt2− 4x = 0.

The corresponding characteristic equation is

λ2 − 4 = 0,=⇒(λ+ 2)(λ− 2) = 0,

roots : λ1 = −2 and λ2 = 2,=⇒x(t) = c1e

−2t + c2e2t �

Example 2


d2y

dx2− 7

dy

dx+ 12y = 0.


λ2 − 7λ+ 12 = 0,=⇒(λ− 3)(λ− 4) = 0,

roots : λ1 = 3 and λ2 = 4,=⇒y(x) = c1e

3x + c2e4x �

3.9.2 Case 2: Roots are complex, i.e., λ1 = α+ iβ and λ2 = α− iβ

y = c1e(α+iβ)x + c2e

(α−iβ)x,

=⇒y = c1eαxeiβx + c2e

αxe−iβx,{eiβx = cosβx+ i sinβx

e−iβx = cosβx− i sinβx

}

=⇒y = c1eαx(cosβx+ i sinβx) + c2e

αx(cosβx− i sinβx),=⇒y = (c1 + c2)eαx cosβx+ i(c1 − c2)eαx sinβx,

let A = c1 + c2,

B = i(c1 − c2),=⇒y = Aeαx cosβx+Beαx sinβx.


Example 1


d2y

dx2+ 4y = 0.


λ2 + 4 = 0,roots : λ1 = 2i and λ2 = −2i,

=⇒y(x) = Ae0 cos 2x+Be0 sin 2x,=⇒y(x) = A cos 2x+B sin 2x �

Example 2


d3y

dx3− 3

d2y

dx2+ 9

dy

dx+ 13y = 0.


λ3 − 3λ2 + 9λ+ 13 = 0,

=⇒(λ+ 1)(λ2 − 4λ+ 13

)= 0,

roots : λ1 = −1, λ2 = 2 + 3i, and λ3 = 2− 3i

=⇒y(x) = c1e−x + c2e

2x cos 3x+ c3e2x sin 3x �

Illustration of Synthetic Division

λ3 − λ2 − 4λ− 2 = (λ+ 1)(λ2 − 2λ− 2

)�

λ2 − 2λ− 2λ+ 1 | λ3 − λ2 − 4λ− 2

λ3 + λ2

− 2λ2 − 4λ−2λ2 − 2λ

− 2λ− 2−2λ− 2

0


3.9.3 Case 3: Repeated Roots

If λ1 is a root of multiplicity n (λ1 = λ2 = · · · = λn = b) we wish to find nlinearly independent y’s for which

(D − b)ny = 0.

Then,yn = xnebx, n = 0, 1, 2, · · · , (n− 1)

are linearly independent, and the required solution is

y = c1ebx + c2xe

bx + · · ·+ cnxn−1ebx �

Example 1

Solve the following homogeneous differential equation

(D − 1)4y = 0.


(λ− 1)4 = 0,roots : λ1 = λ2 = λ3 = λ4 = 1

=⇒y(x) = c1ex + c2xe

x + c3x2ex + c4x

3ex �

Example 2


(D2 − 1)(D − 1)2(D + 2)2y = 0.


(λ2 − 1)(λ− 1)2(λ+ 2)2 = 0,

=⇒(λ+ 1)(λ− 1)3(λ+ 2)2 = 0,roots : λ1 = −1, λ2 = λ3 = λ4 = 1, λ5 = λ6 = −2

=⇒y(x) = c1e−x + c2e

x + c3xex

+ c4x2ex + c5e

−2x + c6xe−2x �


Example 3


(D4 − 7D3 + 18D2 − 20D + 8

)y = 0.


λ4 − 7λ3 + 18λ2 − 20λ+ 8 = 0,

=⇒(λ− 1)(λ− 2)3 = 0,roots : λ1 = 1, λ2 = λ3 = λ4 = 2

=⇒y(x) = c1ex +

(c2 + c3x+ c4x

2)e2x �

Example 4


d4y

dx4+ 2

d3y

dx3+d2y

dx2= 0.


λ4 + 2λ3 + λ2 = 0,

=⇒λ2(λ2 + 2λ+ 1

)= 0,

=⇒λ2(λ+ 1)2 = 0,roots : λ1 = λ2 = 0, λ3 = λ4 = −1

=⇒y(x) = c1 + c2x+ c3e−x + c4xe

−x �

3.10 Particular Solution

3.10.1 Method of Undetermined CoefficientsThis method is applicable only for ordinary differential equations, for whichR(x) itself is a solution of a homogeneous linear equation with constantcoefficients. In this method we assume a trial solution containing unknownconstants which are to be determined by substitution in the given equation.The assumed solution depends on the special form of R(x) as shown in Table3.1.


Table 3.1: Trial Particular Solutions for Various R(x) Terms.

Term in R(x) Choice for yp

keγx ceγx

kxn cnxn + cn−1x

n−1 + · · ·+ c1x+ c0k cosωx A cosωx+B sinωxk sinωx A cosωx+B sinωxkeax cosωx eax(A cosωx+B sinωx)keax sinωx eax(A cosωx+B sinωx)

This method holds in case no term in the assumed trial solution appearsin the homogeneous solution, yh. If any term of the assumed trial solutiondoes appear in yh, we must multiply this trial solution by the smallestpositive integer power of x which is large enough so that none of the termswhich are then present appear in yh.

Example 1

Solve the following nonhomogeneous ordinary differential equation

d2y

dx2− y = 8xex.

The solution consists of the sum

y = yh + yp.

Homogeneous Solution:

d2y

dx2− y = 0,

characteristic equation : λ2 − 1 = 0,=⇒(λ+ 1)(λ− 1) = 0,

roots : λ1 = 1, λ2 = −1,

=⇒yh = c1ex + c2e

−x.

Particular Solution: assume a trial solution

yp = x (Aex +Bxex) = Axex +Bx2ex


Note that we multiply by x so that no terms in yp are identical to terms ofthe homogeneous solution, yh.

dyp

dx= Aex +Axex + 2Bxex +Bx2ex

= Aex + (A+ 2B)xex +Bx2ex

d2yp

dx2= Aex + (A+ 2B)ex + (A+ 2B)xex + 2Bxex +Bx2ex

= (2A+ 2B)ex + (A+ 4B)xex +Bx2ex

=⇒d2yp

dx2− yp = 8xex

=⇒ (2A+ 2B)ex + (A+ 4B)xex +Bx2ex︸︷︷︸d2yp/dx2

−Axex −Bx2ex︸︷︷︸−yp

= 8xex

=⇒(2A+ 2B)ex + 4Bxex = 8xex

Equating corresponding coefficient on both sides yields2A+ 2B = 0, 4B = 8,

=⇒A = −2, B = 2.

=⇒ yp = −2xex + 2x2ex.

=⇒ y(x) = c1ex + c2e

−x − 2xex + 2x2ex �

Example 2


d2y

dx2+ 2

dy

dx+ 4y = 8x2 + 12e−x.

The solution consists of the sum

y = yh + yp.


d2y

dx2+ 2

dy

dx+ 4y = 0,

characteristic equation : λ2 + 2λ+ 4 = 0,

roots : λ1 = −1 + i√

3, λ2 = −1− i√

3

=⇒yh = c1e−x cos

√3x+ c2e

−x sin√

3x.



y∗p = Ax2 +Bx+ C︸︷︷︸8x2

+De−x︸︷︷︸12e−x

.

Substituting y∗p into the governing differential equation yields

(2A+Dex) + 2(2Ax+B −De−x

)+ 4

(Ax2 +Bx+ C +De−x

)= 8x2 + 12e−x

=⇒4Ax2 + (4A+ 4B)x+ (2A+ 2B + 4C) + 3De−x = 8x2 + 12e−x.

Equating corresponding coefficient on both sides yields

4A = 8, 4A+ 4B = 0, 2A+ 2B + 4C = 0, 3D = 12,=⇒A = 2, B = −2, C = 0, D = 4.

=⇒ yp = 2x2 − 2x+ 4e−x.

=⇒ y(x) =(c1 cos

√3x+ c2 sin

√3x

)e−x + 2x2 − 2x+ 4e−x �

3.10.2 Lagrange’s Idea (Introduction to Variation of Parameters)

The following nonhomogeneous ordinary differential equation

y′′ + p(x)y′ + q(x)y = R(x), (1)

has a general solution y = yh + yp, where

yh = c1y1 + c2y2,

yp = u(x)y1 + ν(x)y2. (2)

Differentiating (2) yields

y′p = u′y1 + uy′1 + ν′y2 + νy′2.

Imposing the arbitrary condition

u′y1 + ν′y2 = 0

leads toy′p = uy′1 + νy′2 (3)


=⇒ y′′p = u′y′1 + uy′′1 + ν′y′2 + νy′′2 . (4)

Substituting (2), (3), and (4) into (1) yields

u (y′′1 + py′1 + qy1)︸︷︷︸0

+ν (y′′2 + py′2 + qy2)︸︷︷︸0

+u′y′1 + ν′y′2 = R.

Recall that y1 and y2 are solutions to the homogeneous equation (1),

=⇒ u′y′1 + ν′y′2 = R and u′y1 + ν′y2 = 0.

Solving the system of algebraic equations for u′ and v′, integrating andbacksubstituting into (2) leads to the following expression

yp = −y1

∫y2R

Wdx+ y2

∫y1R

Wdx,

where W = y1y′2 − y′1y2 is the Wronskian of y1 and y2.

3.10.3 Method of Variation of ParametersLet yh be the homogeneous solution of

b0dny

dxn+ · · ·+ bn−1

dy

dx+ bny = R(x), (1)

yh = c1y1(x) + c2y2(x) + · · ·+ cnyn(x).

Replace the arbitrary constants c1, c2, · · · , cn by functions u1(x), u2(x), · · ·,un(x) and determine these so that

yp = u1(x)y1(x) + u2(x)y2(x) + · · ·+ un(x)yn(x)

is a particular solution of (1). To determine these n functions we mustimpose n restrictions on them:

u′1y1 + u′2y2 + · · · + u′nyn = 0

u′1y′1 + u′2y

′2 + · · · + u′ny

′n = 0

......

. . ....

...

u′1yn−11 + u′2y

n−12 + · · ·+ u′ny

n−1n = R(x)


The last condition indicates that the given differential equation is satisfied.Obtain u′1, u

′2, · · · , u′n by Cramer’s rule and integrate to get u1, u2, · · · , un.

Cramer’s Rule

2α1 − α2 + 2α3 = 2,

α1 + 10α2 − 3α3 = 5,

− α1 + α2 + α3 = −3,

=⇒ D =

∣∣∣∣∣∣2 −1 21 10 −3−1 1 1

∣∣∣∣∣∣ = 46,

α1 =1D

∣∣∣∣∣∣2 −1 25 10 −3−3 1 1

∣∣∣∣∣∣ ,

α2 =1D

∣∣∣∣∣∣2 2 21 5 −3−1 −3 1

∣∣∣∣∣∣ ,

α3 =1D

∣∣∣∣∣∣2 −1 21 10 5−1 1 −3

∣∣∣∣∣∣ .

Example 1


d2y

dx2+ y = secx.

Recall : tanx =sinxcosx

, secx =1

cosx,

cotx =cosxsinx

, cscx =1

sinx.


d2y

dx2+ y = 0,

characteristic equation : λ2 + 1 = 0,roots : λ1 = i1, λ2 = −i1

=⇒yh = c1 cosx+ c2 sinx.



yp = u1(x) cosx+ u2(x) sinx.

Restrictions:

u′1 cosx+ u′2 sinx = 0

−u′1 sinx+ u′2 cosx = secx(y′′p + yp = secx

)

=⇒W =∣∣∣∣ cosx sinx− sinx cosx

∣∣∣∣ = cos2 x+ sin2 x = 1

u′1 =1W

∣∣∣∣ 0 sinxsecx cosx

∣∣∣∣ = − sinx secx = − sinxcosx

= − tanx,

u′2 =1W

∣∣∣∣ cosx 0− sinx secx

∣∣∣∣ = cosx secx = 1,

u1 = −∫

tanxdx = −(− ln cosx) + c∗ = ln cosx+ c∗,

u2 =∫

dx = x+ c∗∗.

=⇒ yp = cosx(ln cosx) + x sinx,

=⇒ y = yh + yp = c1 cosx+ c2 sinx+ cosx(ln cosx) + x sinx,

where c∗ and c∗∗ are incorporated in c1 and c2 �

Example 2 (Same as Example 1 of section 3.10.1)


d2y

dx2− y = 8xex.


yh = c1ex + c2e

−x.

Particular Solution: assume that

yp = u1(x)ex + u2(x)e−x.


Restrictions:u′1e

x + u′2e−x = 0

u′1ex − u′2e

−x = 8xex(y′′p − yp = 8xex

)=⇒W =

∣∣∣∣ ex e−x

ex −e−x

∣∣∣∣ = (−1)− 1 = −2,

u′1 =1−2

∣∣∣∣ 0 e−x

8xex −e−x

∣∣∣∣ =−8x−2

= 4x =⇒ u1 = 2x2 + c∗,

u′2 =1−2

∣∣∣∣ ex 0ex 8xex

∣∣∣∣ =8x(ex)2

−2= −4xe2x =⇒ u2 = −2xe2x + e2x + c.

∫xe2xdx =

xe2x

2−

∫e2x

2dx =

xe2x

2− e2x

4,

Integration by Parts :∫

u dv = uv −∫

v du,

u = x, du = dx, dv = e2xdx, v =e2x

2.

=⇒ yp = 2x2ex + ex − 2xex,

=⇒ y = yh + yp

= c1ex + c2e

−x + 2x2ex + ex − 2xex

= c∗1ex + c2e

−x − 2xex + 2x2ex �

Example 3


(D3 + 4D)y = 4 cot 2x.{

Recall : cotx =cosxsinx

, secx =1

cosx.

}

Homogeneous Solution:(D3 + 4D

)y = 0,

characteristic equation : λ3 + 4λ = 0 =⇒ λ(λ2 + 4

)= 0,

roots : λ1 = 0, λ2 = 2i, λ3 = −2i=⇒yh = c1 + c2 cos 2x+ c3 sin 2x.


Particular Solution: assume thatyp = u1(x) + u2(x) cos 2x+ u3(x) sin 2x.

Restrictions:u′1 + u′2 cos 2x+ u′3 sin 2x = 0

0− 2u′2 sin 2x+ 2u′3 cos 2x = 0

0− 4u′2 cos 2x− 4u′3 sin 2x = 4 cot 2x

=⇒W =

∣∣∣∣∣∣1 cos 2x sin 2x0 −2 sin 2x 2 cos 2x0 −4 cos 2x −4 sin 2x

∣∣∣∣∣∣=

∣∣∣∣−2 sin 2x 2 cos 2x−4 cos 2x −4 sin 2x

∣∣∣∣ = 8(sin2 2x+ cos2 2x

)= 8,

u′1 =1W

∣∣∣∣∣∣0 cos 2x sin 2x0 −2 sin 2x 2 cos 2x

4 cot 2x −4 cos 2x −4 sin 2x

∣∣∣∣∣∣ =4 cot 2x

8

∣∣∣∣ cos 2x sin 2x−2 sin 2x 2 cos 2x

∣∣∣∣= cot 2x,

u1 =∫

cot 2x dx =12

ln sin 2x+ c,

u′2 =1W

∣∣∣∣∣∣1 0 sin 2x0 0 2 cos 2x0 4 cot 2x −4 sin 2x

∣∣∣∣∣∣ = − cos 2x cot 2x,

u2 = −∫

cos 2x cot 2x dx = −12

ln(csc 2x− cot 2x)− 12

cos 2x+ c,

∫cos 2x cot 2x dx =

∫cos2 2xsin 2x

dx

=cos 2x

2+

∫dx

sin 2x

=cos 2x

2+

∫csc 2xdx

=cos 2x

2+

12

ln(csc 2x− cot 2x)

u′3 =1W

∣∣∣∣∣∣1 cos 2x 00 −2 sin 2x 00 −4 cos 2x 4 cot 2x

∣∣∣∣∣∣ = − sin 2x cot 2x = − cos 2x,

u3 = −∫

cos 2x dx = − sin 2x2

+ c.


=⇒ yp =12

ln sin 2x− 12

ln(csc 2x− cot 2x) cos 2x− cos2 2x2

− sin2 2x2︸︷︷︸

−1/2

=⇒ y = yh + yp = c∗1 + c2 cos 2x+ c3 sin 2x+ln sin 2x

2

− cos 2x2

ln(csc 2x− cot 2x),

where c∗1 = c1 − 1/2 �

3.11 Particular Solution by Inspection

(b0D

n + b1Dn−1 + · · ·+ bn−1D + bn

)y = R(x)

• If R(x) = Ro = constant and bn �= 0, then

yp =Ro

bn.

• If R(x) = Ro, bn = 0 and Dky is the lowest–order derivative in theordinary differential equation, then

yp =Rox

k

k! bn−k.

Note that bn−k is always the coefficient ofDky (the lowest–order derivative).

Example 1


(D2 − 3D + 2

)y = (D − 1)(D − 2)y = 16.

yh = c1ex + c2e

2x,

yp =Ro

bn=

162

= 8,

=⇒ y = c1ex + c2e

2x + 8 �


Example 2


d5y

dx5+ 4

d3y

dx3= 7.


d5y

dx5+ 4

d3y

dx3= 0,

characteristic equation : λ5 + 4λ3 = 0 =⇒ λ3(λ2 + 4

)= 0,

roots : λ1 = λ2 = λ3 = 0, λ4 = 2i, λ5 = −2i=⇒yh = c1 + c2x+ c3x

2 + c4 cos 2x+ c5 sin 2x.

Particular Solution: by inspection

yp =Rox

k

k!bn−k=

7x3

3!4=

7x3

24.

=⇒ y = yh + yp = c1 + c2x+ c3x2 + c4 cos 2x+ c5 sin 2x+

7x3

24�

3.12 Equations with Variable Coefficients

3.12.1 Euler–Cauchy Equation

The Euler–Cauchy equation has the form(b0x

nDn + b1xn−1Dn−1 + · · ·+ bn−1xD + bn

)y = R(x).

Apply the following transformation

x = et −→ dx = etdt −→ dx

dt= et,

and use the results

xD = Dt =d

dt

x2D2 = Dt(Dt − 1)

x3D3 = Dt(Dt − 1)(Dt − 2)

· · · · · ·


to reduce the equation to one with constant coefficients.

Derivation of xD and x2D2

Dy =dy

dx=

dy

dt

dt

dx︸︷︷︸chain rule

=dy

dt

(dx

dt

)−1

=dy

dt

(et

)−1 = e−tDty

=⇒ x(Dy) = x(e−tDty

), x = et

=⇒ xDy = Dty

=⇒ xD = Dt �

D2y =d2y

dx2=

d

dx

(e−tDty

)=

d

dt

(e−tDty

) dt

dx

=(−e−tDty + e−tD2

t y) dt

dx︸︷︷︸e−t

= e−2t(D2

t −Dt

)y

=⇒ x2D2y = e2t[e−2t

(D2

t −Dt

)y]

=(D2

t −Dt

)y = Dt(Dt − 1)y

=⇒ x2D2 = Dt(Dt − 1) �

Example

Solve the following nonhomogeneous ordinary differential equation(x2D2 + xD − 4

)y = x3.

Applying the transformationx = et

xD = Dt

x2D2 = Dt(Dt − 1)

yields the following nonhomogeneous ordinary differential equation withconstant coefficients

[Dt(Dt − 1) +Dt − 4] y =[et

]3 =⇒(D2

t − 4)y = e3t.

Homogeneous Solution:(D2

t − 4)y = 0,

characteristic equation : λ2 − 4 = 0,roots : λ1 = 2, λ2 = −2=⇒ yh = c1e

2t + c2e−2t.



yp = Ae3t,dyp

dt= 3Ae3t,

d2yp

dt2= 9Ae3t

=⇒(D2

t − 4)yp = e3t

=⇒9Ae3t − 4Ae3t = e3t =⇒ 5A = 1 =⇒ A =15

=⇒yp =e3t

5.

y = yh + yp = c1e2t + c2e

−2t +e3t

5x = et

=⇒ y = c1x2 + c2x

−2 +x3

5�

3.12.2 Another Approach

Ordinary differential equations with variable coefficients in the form of theEuler–Cauchy equation can also be solved by employing the following trialsolution

y = xp

and then solving for p. Revisiting the previous example:(x2D2 + xD − 4

)y = x3

Dy = pxp−1, D2y = p(p− 1)xp−2

=⇒ x2[p(p− 1)xp−2

]+ xpxp−1 − 4xp = x3.


p(p− 1)xp + pxp − 4xp = 0,characteristic equation : p(p− 1) + p− 4 = 0

=⇒ p2 − 4 = 0roots : p = ±2 (yh = xp)

=⇒ yh = c1x2 + c2x

−2.

Particular Solution: yp is obtained as shown in the previous section

yp =e3t

5.


• Case of Complex Roots

p = µ± iν

yh = c1x(µ+iν) + c2x

(µ−iν)

= xµ(c1x

iν + c2x−iν

)xiν = eiν ln x = cos(ν lnx) + i sin(ν lnx)

x−iν = e−iν ln x = cos(ν lnx)− i sin(ν lnx)

=⇒ yh = xµ [(c1 + c2) cos(ν lnx) + i(c1 − c2) sin(ν lnx)]= xµ [A cos(ν lnx) +B sin(ν lnx)] ,

where A = c1 + c2, B = i(c1 − c2).

• Case of Double Root

y1 = xp,

y2 = (lnx)xp.

Example

Consider the following ordinary differential equation

(x2D2 − 3xD + 4

)y = 0.

Apply the transformationy = xp

=⇒ x2[p(p− 1)xp−2

]− 3xpxp−1 + 4xp = 0

=⇒ p(p− 1)xp − 3pxp + 4xp = 0

=⇒ p2 − 4p+ 4 = 0

=⇒ (p− 2)2 = 0roots : p1 = p2 = 2 (double root)

solutions : y1 = x2, y2 = (lnx)x2

=⇒ y = c1x2 + c2(lnx)x2 �


3.13 Case where One Solution is Known (Reduction of Order)

If one solution y1 = f(x) of g(D)y = R(x) is known, then the substitution

y = u(x)f(x)

transforms the differential equation into one of order n− 1 in u′(x).

Example

Solve the following differential equation, given that y1 = x

(1− x2

) d2y

dx2− 2x

dy

dx+ 2y = 0.

Assume that

y2 = u(x)x,dy2

dx= u+ u′x,

d2y2

dx2= 2u′ + u′′x

=⇒(1− x2

)(2u′ + u′′x)− 2x (u+ u′x) + 2ux = 0

=⇒(x− x3

)u′′ +

(2− 4x2

)u′ = 0 (first order in u′)

=⇒ du′

u′+

2− 4x2

x− x3dx = 0{

2− 4x2

x− x3=

2(1− x2

)− 2x2

x (1− x2)

}

=⇒∫

du′

u′+

∫ (2x− 2x

1− x2

)dx = c

=⇒ lnu′ + 2 lnx+ ln(1− x2

)= c

=⇒ u′x2(1− x2

)= ec = c1

=⇒ u′ =c1

x2 (1− x2)= c1

(1x2

+1

1− x2

){

11− x2

=12

(1

x+ 1+

11− x

)}

=⇒ u = c1

[− 1x

+12

ln(

1 + x

1− x

)]+ c2

=⇒ y2 = ux = c1

[x

2ln

(1 + x

1− x

)− 1

]+ c2x �


3.14 Exercises

(1) Obtain the Wronskian of the functions 1, x, x2, . . ., xn−1 for n > 1.(2) Show that the operators D − 2 and xD + 1 are not commutative with

respect to multiplication.(3) Show that the functions cos 2x, sin2 x, cos2 x are linearly dependent.(4) Let y1(x), y2(x) be solutions of

y′′ + p(x)y′ + q(x)y = 0.

Prove that the Wronskian is

W = y1y′2 − y2y

′1 = C exp

[−

∫pdx

].

(5) Solve the following equations:

(a)(D3 −D2 − 4D + 4

)y = 0

(b)(6D4 + 23D3 + 28D2 + 13D + 2

)y = 0

(c)(D4 − 5D2 − 6D − 2

)y = 0

(d)(D4 + 2D3 + 10D2

)y = 0

(e)(D6 + 9D4 + 24D2 + 16

)y = 0

(6) Use the method of undetermined coefficients to obtain the general so-lution of (

D2 +D)y = sinx.

(7) Use the method of variation of parameters to obtain the general solutionof (

D2 − 4D + 4)y = exp[x].

(8) Find the general solution of the following equations:

(a)(x2D2 − 2

)y = 3x2

(b)(x2D2 + xD − 1

)y = 4


Chapter 4

Series Solutions of Differential Equations

4.1 Properties of Power Series

A large class of ordinary differential equations possesses solutions express-ible, over a certain interval, in terms of power series and related series.Before investigating methods of obtaining such solutions, we review with-out proof certain useful properties of power series. An expression of theform

a0 + a1(x− x0) + · · ·+ an(x− x0)n + · · · =∞∑

n=0

an(x− x0)n (1)

is called power series, where a0, a1, · · ·, an, · · · are the coefficients of theseries, and x0 is a constant known as the center of the series. The expression

Sn(x) = a0 + a1(x− x0) + · · ·+ an(x− x0)n

is the nth partial sum, and

Rn(x) = an+1(x− x0)n+1 + an+2(x− x0)n+2 + · · ·

is the remainder. A power series is defined as the limit

limN→∞

N∑n=0

an(x− x0)n or limN→∞

Sn(x)

for those values of x for which the limit exists. For such values of x theseries is said to converge.

To determine for what values of x the series (1) converges, we maymake use of the ratio test:

ρ = limn→∞

∣∣∣αn+1

αn

∣∣∣|x− x0| = L|x− x0|,

whereL = lim

n→∞

∣∣∣αn+1

αn

∣∣∣.C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS –––––––––––––––––––––––––––– 49

It follows that (1) converges when

|x− x0| <1L,

diverges when

|x− x0| >1L,

and the interval of converge is

(x0 −

1L, x0 +

1L

).

The distance R = 1L is the radius of convergence.

4.2 Convergence of Power Series

Theorem: If∑∞

n=0 αnxn is a power series, then either

(a) The series converges absolutely for all x; or(b) The series converges only when x = 0; or(c) There exists a number R > 0 such that the series converges absolutely

if |x| < R and diverges if |x| > R.

ExampleFind the interval of convergence of

∞∑n=1

xn

n · 3n.

L = limn→∞

∣∣∣αn+1

αn

∣∣∣ = limn→∞

∣∣∣∣1/([n+ 1] · 3n+1)1/(n · 3n)

∣∣∣∣= lim

n→∞n

3(n+ 1)=

13.

Therefore, the interval of convergence is [-3, 3). If |x| < 3 the series con-verges; however, if |x| > 3 the series diverges. Note that when x = 3 wehave the divergent harmonic series

∑1n , and when x = −3, the convergent

alternating series∑ (−1)n

n �


4.3 Some Series Expansions

ex = 1 + x +x2

2!+

x3

3!+

x4

4!+ · · · + xn

n!+ · · · =

!!

i=0

xi

i!

sin x = x ! x3

3!+

x5

5!! x7

7!+

x9

9!! · · · + (!1)i+1 x2i+1

(2i + 1)!+ · · ·

=!!

i=0

(!1)i+1 x2i+1

(2i + 1)!

cos x = 1 ! x2

2!+

x4

4!! x6

6!+

x8

8!! + · · · + (!1)i x2i

(2i)!+ · · ·

=!!

i=0

(!1)i x2i

(2i)!

sinhx = x +x3

3!+

x5

5!+

x7

7!+

x9

9!+ · · · + x2i+1

(2i + 1)!+ · · ·

=!!

i=0

x2i+1

(2i + 1)!

cosh x = 1 +x2

2!+

x4

4!+

x6

6!+

x8

8!+ · · · + x2i

(2i)!+ · · · =

!!

i=0

x2i

(2i)!

e#x = 1 ! x +x2

2!! x3

3!+

x4

4!! x5

5!+ · · · + (!1)n xn

n!+ · · ·

=!!

i=0

(!1)i xi

i!

ex2= 1 + x2 +

x4

2!+

x6

3!+

x8

4!+

x10

5!+ · · · + x2n

n!+ · · · =

!!

i=0

x2i

i!

ln(1 + x) = x ! x2

2+

x3

3! · · · =

!!

n=1

(!1)n+1 xn

nfor |x| < 1

11 ! x

= 1 + x + x2 + · · · =!!

n=0

xn for |x| < 1

(1 + x)p = 1 + px +p(p ! 1)x2

2+

p(p ! 1)(p ! 2)x3

3!+ · · · for |x| < 1


4.4 Operations on Power Series

Convergent power series can be treated, for many purposes, in the sameway as polynomials.

4.4.1 Termwise Di!erentiation or Integration

Inside its interval of convergence, a power series represents a continuousfunction of x with continuous derivatives of all orders. Inside this interval,a power series can be integrated or di!erentiated term by term, and theresultant series will converge in the same interval. For example considerthe di!erentiation of y(x):

y(x) =!!

m=0

am(x ! x0)m !# y$(x) =!!

m=1

mam(x ! x0)m#1.

4.4.2 Termwise Addition or Subtraction

Consider the following two power series

f(x) =!!

m=0

am(x ! x0)m Rf ,

g(x) =!!

m=0

bm(x ! x0)m Rg,

f(x) + g(x) =!!

m=0

(am + bm)(x ! x0)m.

Note that the radius of convergence of f(x) + g(x) is at least equal to thesmaller of Rf and Rg.

4.4.3 Termwise Multiplication

f(x) · g(x) =!!

m=0

am(x ! x0)m!!

n=0

bn(x ! x0)n

=!!

k=0

ck(x ! x0)k,


where

ck = a0bk + a1bk#1 + · · · + akb0 =k!

m=0

ambk#m,

and the resulting series converges at least for |x| < min(Rf , Rg).

4.4.4 Termwise Division

f(x)g(x)

=

!!

m=0

am(x ! x0)m

!!

n=0

bn(x ! x0)n

=!!

k=0

dk(x ! x0)k,

where the d’s are to be determined by the following expressions

am =m!

k=0

bm#kdk (m = 0, 1, 2, · · ·),

a0 = b0d0,

a1 = b1d0 + b0d1,

a2 = b2d0 + b1d1 + b0d2.

4.5 Illustrative Examples of the Power Series Method

Example 1 (Hildebrand, 1976, p. 122)Consider the ordinary di!erential equation

d2y

dx2! y = 0.

Assume a solution in the form

y = a0 + a1x + a2x2 + a3x

3 + a4x4 + a5x

5 + · · ·

and that the series converges in an interval including x = 0. Di!erentiatingtwice term by term the proposed solution y yields

d2y

dx2= 2a2 + 6a3x + 12a4x

2 + 20a5x3 + · · ·


Substituting back into the original ordinary differential equation yields

d2y

dx2−y = (2a2−a0)+(6a3−a1)x+(12a4−a2)x2 +(20a5−a3)x3 + · · · = 0.

In order for this equation to be valid over an interval, it is necessary thatthe coefficients of all powers of x vanish independently. Therefore,

2a2 = a0, ⇒ a2 =a02,

6a3 = a1, ⇒ a3 =a16,

12a4 = a2, ⇒ a4 =a212

=a024,

20a5 = a3, ⇒ a5 =a320

=a1120.

=⇒ y = a0

(1 +

x2

2+x4

24+ · · ·

)︸︷︷︸

cosh x

+a1

(x+

x3

6+x5

120+ · · ·

)︸︷︷︸

sinh x

=⇒ y = a0 coshx+ a1 sinhx �

Note that the expected solution is of the form

y = a0ex + a1e−x,

however, both forms are identical because

sinh =ex − e−x

2, cosh =

ex + e−x

2.

A More Compact Procedure

y =∞∑

k=0

akxk,

d2y

dx2=

∞∑k=0

k(k − 1)akxk−2,

=⇒ d2y

dx2− y =

∞∑k=0

k(k − 1)akxk−2 −

∞∑k=0

akxk = 0.


It should be noted that∞∑

k=0

akxk =

∞∑k−2=0

ak−2xk−2 =

∞∑k=2

ak−2xk−2,

=⇒∞∑

k=0

k(k − 1)akxk−2 −

∞∑k=2

ak−2xk−2 = 0.

The first two terms (k = 0, 1) of the first summation in the precedingequation are zero, thus it may be written as

∞∑k=2

[k(k − 1)ak − ak−2]xk−2 = 0.

By equating to zero the coefficients of all powers yields the following ex-pression known as the Recurrence Formula for ak

k(k − 1)ak = ak−2 (k = 2, 3, · · ·).

Example 2Consider the following ordinary differential equation

Ly = x2 d2y

dx2+

(x2 + x

) dydx− y = 0.

Assuming a solution of the form

y =∞∑

k=0

akxk,

dy

dx=

∞∑k=0

kakxk−1,

d2y

dx2=

∞∑k=0

k(k − 1)akxk−2.

=⇒ Ly =∞∑

k=0

k(k − 1)akxk +

∞∑k=0

kakxk+1 +

∞∑k=0

kakxk −

∞∑k=0

akxk = 0.

Combining the 1st, 3rd and 4th summations yields

∞∑k=0

[k(k − 1) + k − 1]akxk +

∞∑k=0

kakxk+1 = 0,


replacing k by k − 1 in the second summation in the preceding expressionyields

∞∑k=0

(k2 − 1)akxk +

∞∑k=1

(k − 1)ak−1xk = 0

=⇒ −a0 +∞∑

k=1

[(k2 − 1)ak + (k − 1)ak−1]xk = 0.

The constant term, as well as the coefficients of the successive powers of xmust vanish independently. Therefore

a0 = 0,

(k−1) [(k + 1)ak + ak−1] = 0 (k = 1, 2, 3, · · ·) Recurrence Formula.

The temptation to cancel the common factor k−1 before setting k = 1 in therecurrence formula must be resisted. The recurrence formula is identicallysatisfied when k = 1.

For k ≥ 2, ak = − ak−1

k + 1(k = 2, 3, 4, · · ·)

a2 = −a13,

a3 = −a24

=a1

3 · 4 ,

a4 =a35

= − a13 · 4 · 5 .

Note that for this example a0 = 0 and a1 = arbitrary. The solution isexpressed as follows:

y = a1

(x− x

2

3+x3

3 · 4 −x4

3 · 4 · 5 + · · ·)

=2a1x

(x2

2!− x

3

3!+x4

4!− x

5

5!+ · · ·

)

=2a1x

[−1 + x+

(1− x

1!+x2

2!− x

3

3!+x4

4!− x

5

5!+ · · ·

)]

=⇒ y = c[e−x − 1 + x

x

], �

where c = 2a1. It should be noted that for this case only one solution isobtained. This fact indicates that any linear independent solution cannot


be expanded in power series near x = 0, that is, it is not regular at x = 0.A second solution can be obtained by the method of reduction of order.

Example 3Using the power series method obtain the general solution to the following2nd–order ordinary differential equation

Ly = x3 d2y

dx2+ y = 0.

Assume a solution of the form

y =∞∑

k=0

akxk,

=⇒ Ly =∞∑

k=0

k(k − 1)akxk+1 +

∞∑k=0

akxk = 0.

Replacing k by k − 1 in the first summation yields

∞∑k=1

(k − 1)(k − 2)ak−1xk +

∞∑k=0

akxk = 0

=⇒ a0 +∞∑

k=1

[ak + (k − 1)(k − 2)ak−1]xk = 0

a0 = 0,

ak = −(k − 1)(k − 2)ak−1 (k = 1, 2, 3, · · ·) Recurrence Formula

a1 = a2 = a3 = · · · = 0

=⇒ y = 0 �Thus, it follows that the equation possesses no nontrivial solutions whichare regular at x = 0.

4.6 Regular Points of Linear 2nd–Order Differential Equations

The standard form of a 2nd–order ordinary differential equation is:

d2y

dx2+ p(x)

dy

dx+ q(x)y = 0.


The behavior of solutions of the equation near a point x = x0 depends uponthe behavior of the coefficients p(x) and q(x) near x = x0.

Regular (or ordinary) point: if p(x) and q(x) are regular or analytic atx = x0, that is, if p(x) and q(x) can be expanded in power series in aninterval including x = x0.

Singular point: otherwise.

Regular singular point: if the products (x− x0)p(x) and (x− x0)2q(x) areboth regular at x = x0.

Irregular singular point: otherwise.

If x = x0 is a regular point, then there are two linearly independent solu-tions which are regular at x = x0, and hence are both expressible in theform

y =∞∑

k=0

ak(x− x0)k.

If x = x0 is a regular singular point, the equation doesn’t necessarily possessany nontrivial solution which is regular near x = x0, but that at least onesolution exists of the form

y = (x− x0)r∞∑

k=0

ak(x− x0)k.

Such solution is regular at x = x0 only if r is zero or a positive integer. Ifx = x0 is an irregular singular point, a nontrivial solution may or may notexist and the equation cannot have two independent solutions of this type.

4.7 Method of Frobenius

Consider the following 2nd–order differential equation

x2y′′ + xb(x)y′ + c(x)y = 0.

Reducing the equation to standard form yields

y′′ +b(x)xy′ +

c(x)x2y = 0.


Note that x = 0 is a singular point, and b(x), c(x) are analytic. Therefore,one solution of the form

y(x) = xr∞∑

m=0

amxm

= xr(a0 + a1x+ a2x2 + · · ·

)a0 �= 0,

y′(x) =∞∑

m=0

(m+ r)amxm+r−1,

y′′(x) =∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−2.

Note that r may be any real or complex number. Then, assuming that wecan expand b(x) and c(x) as follows

b(x) = b0 + b1x+ b2x2 + · · · ,

c(x) = c0 + c1x+ c2x2 + · · · ,the original ordinary differential equation can be written as

∞∑m=0

(m+ r)(m+ r − 1)amxm+r + (b0 + b1x+ · · ·)

∞∑m=0

(m+ r)amxm+r+

+ (c0 + c1x+ · · ·)∞∑

m=0

amxm+r = 0.

This expression must vanish identically, in the sense that the coefficients ofall powers of x must vanish identically.

For xr:[r(r − 1) + b0r + c0]a0 = 0,

a0 �= 0,

r2 + (b0 − 1)r + c0 = 0, Indicial Equation.

Case1: Distinct Roots (r1, r2)

y1(x) = xr1

∞∑m=0

amxm, y2(x) = xr2

∞∑m=0

Amxm.


Case 2: Double Root

y1(x) = xr∞∑

m=0

amxm, y2(x) = y1(x) lnx+ xr

∞∑m=1

Amxm.

Case 3: Roots Differing by an Integer k

y1(x) = xr1

∞∑m=0

amxm, y2(x) = ky1(x) lnx+ xr2

∞∑m=0

Amxm.

ExampleSolve the following differential equation by the method of Frobenius

x(1− x)y′′ +(

32− 7x

2

)y′ − 3y

2= 0.


y = xr∞∑

m=0

amxm,

y′ =∞∑

m=0

(m+ r)amxm+r−1,

y′′ =∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−2.

The original equation can be written as follows∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−1 −

∞∑m=0

(m+ r)(m+ r − 1)amxm+r

+32

∞∑m=0

(m+ r)amxm+r−1 − 7

2

∞∑m=0

(m+ r)amxm+r − 3

2

∞∑m=0

amxm+r = 0.

Combining the 1st and 3rd summations, as well as the 2nd, 4th and 5thsummations yields

∞∑m=0

[(m+ r)(m+ r − 1) +

32(m+ r)

]amx

m+r−1−

−∞∑

m=0

[(m+ r)(m+ r − 1) +

72(m+ r) +

32

]amx

m+r = 0.


Shifting the 1st summation leads to

∞∑m+1=0

[(m+ 1 + r)(m+ r) +

32(m+ 1 + r)

]am+1x

m+r−

−∞∑

m=0

[(m+ r)(m+ r − 1) +

72(m+ r) +

32

]amx

m+r = 0

=⇒[r(r − 1) +

3r2

]a0x

r−1 +∞∑

m=0

{(m+ 1 + r)

(m+ r +

32

)am+1−

−[(m+ r)

(m+ r +

52

)+

32

]am

}xm+r = 0.

xr−1 terms:

r(r − 1) +3r2

= 0 =⇒ r2 − r +3r2

= 0 =⇒ r2 +r

2= 0 =⇒ r

(r +

12

)= 0

=⇒ Two distinct roots : r1 = −12, r2 = 0.

xm+r terms:

am+1 =(m+ r)

(m+ r + 5

2

)+ 3

2

(m+ 1 + r)(m+ r + 3

2

) am

(a) r1 = − 12

am+1 =

(m− 1

2

)(m+ 2) + 3

2(m+ 1

2

)(m+ 1)

am =m2 + 3

2m+ 12

m2 + 32m+ 1

2

am = am = a0.

=⇒ y1 = x−1/2

∞∑m=0

a0xm

=⇒ y1 =a0x

−1/2

1− x �

for |x| < 1,∞∑

m=0

xm =1

1− x


(b) r2 = 0

Am+1 =m

(m+ 5

2

)+ 3

2

(m+ 1)(m+ 3

2

) Am =m2 + 5

2m+ 32

m2 + 52m+ 3

2

Am = Am = A0

=⇒ y2 =∞∑

m=0

A0xm = A0

11− x �

The desired solution is

y =a0

1− xx−1/2 +

A0

1− x �

4.8 Important 2nd–Order Equations

Bessel’s equation:

x2 d2y

dx2+ x

dy

dx+

(x2 − ν2

)y = 0.

Legendre’s equation:

(1− x2

) d2ydx2− 2x

dy

dx+ n(n+ 1)y = 0.

Gauss’s equation:

x(1− x)d2y

dx2+ [γ − (α+ β + 1)x]

dy

dx− αβy = 0.

Other notable equations:

(1) xd2y

dx2+ (c− x)dy

dx+ ay = 0,

satisfied by the confluent hypergeometric function of Kummer

y =M(a, c, x).


If c = 1 and a = −n (where n is a positive integer or zero) one solution isthe nth Laguerre polynomial

y = Ln(x).

(2)d2y

dx2− 2x

dy

dx+ 2ny = 0,

satisfied by the nth Hermite polynomial

y = Hn(x).

(3)(1− x2

) d2ydx2− xdy

dx+ n2y = 0,

satisfied by the nth Chebyshev polynomial

y = Tn(x).

(4) x(1− x)d2y

dx2+ [a− (1 + b)x]

dy

dx+ n(b+ n)y = 0,

satisfied by the nth Jacobi polynomial

y = Jn(a, b, x).

4.9 Legendre’s Equation

Solutions of the following differential equation

(1− x2

) d2ydx2− 2x

dy

dx+ n(n+ 1)y = 0,

or equivalentlyd

dx

[(1− x2

) dydx

]+ n(n+ 1)y = 0,


are known as Legendre functions of order n, where n is assumed here to bereal and nonnegative. They are of particular use in the solution of potentialproblems involving spherical boundaries.

Since the coefficients of this equation are analytic at x = 0, we mayapply the power series method. Substituting

y =∞∑

m=0

amxm,

y′ =∞∑

m=0

mamxm−1, y′′ =

∞∑m=0

m(m− 1)amxm−2,

into the governing differential equation and letting k = n(n+ 1) yields

∞∑m=0

m(m− 1)amxm−2−

∞∑m=0

m(m− 1)amxm

−2∞∑

m=0

mamxm + k

∞∑m=0

amxm = 0

=⇒∞∑

m+2=0

(m+ 2)(m+ 1)am+2xm −

∞∑m=0

[m(m− 1) + 2m− k]amxm = 0

=⇒∞∑

m=0

{(m+ 2)(m+ 1)am+2 − [m(m− 1) + 2m− k]am}xm = 0.

Coefficients of x0:2a2 + ka0 = 0.

Coefficients of x1:6a3 − (2− k)a1 = 0.

Coefficients of xm (m ≥ 2):

am+2 =m2 +m− k

(m+ 2)(m+ 1)am, k = n(n+ 1)


a2 =−n(n+ 1)

2!a0,

a3 =2− n(n+ 1)

6a1 = − (n− 1)(n+ 2)

3!a1,

a4 =6− n(n+ 1)

4 · 3 a2 =(n− 2)n(n+ 1)(n+ 3)

4!a0,

a5 = − (n− 3)(n+ 4)5 · 4 a3 =

(n− 3)(n− 1)(n+ 2)(n+ 4)5!

a1.

Therefore, the desired solution is of the form

y = a0Un(x) + a1Vn(x),

where

Un(x) = 1− n(n+ 1)2!

x2 +(n− 2)n(n+ 1)(n+ 3)

4!x4 −+ · · ·

Vn(x) = x− (n− 1)(n+ 2)3!

x3 +(n− 3)(n− 1)(n+ 2)(n+ 4)

5!x5 −+ · · ·

Note that these series converge for |x| < 1. Since Un contains even powersof x only, while Vn contains odd powers of x only, the ratio Un/Vn isnot a constant so that Un and Vn are not proportional and thus linearlyindependent solutions.

• If n is even, then Un is a polynomial of degree n, and Vn is an infinite series.

• If n is odd, then Un is an infinite series, and Vn is a polynomial of degree n.

Conventional Terminology

The multiple of the polynomial of degree n which has the value unity whenx = 1 is called the nth Legendre polynomial and is denoted by Pn(x).

Pn(x) =Un(x)Un(1)

n is even

Pn(x) =Vn(x)Vn(1)

n is odd

The first six Legendre polynomials are:


-1

-0.5

0

0.5

1

Pn(x)

-1 -0.5 0 0.5 1

x

P0

P1

P2

P3 P4

P5

Figure 4.1: Plot of the first six Legendre polynomials as a function of x.

P0(x) = 1,

P1(x) = x,

P2(x) =12

(3x2 − 1

),

P3(x) =12

(5x3 − 3x

),

P4(x) =18

(35x4 − 30x2 + 3

),

P5(x) =18

(63x5 − 70x3 + 15x

).

These Legendre polynomials are graphically presented in Figure 4.1, whereit can be seen that:• All the zeros of Pn(x) are real and unrepeated,• All the zeros lie in the interval −1 < x < 1,• In the interval −1 ≤ x ≤ 1 the magnitude of each Pn(x) is maximum at

the end points, so that |Pn(x)| ≤ 1 when |x| ≤ 1,• Outside the interval (-1,1), Pn(x) increases or decreases steadily without

maxima or minima or turning points.


Suitable multiples of Un(x) and Vn(x) are called Legendre Functions of theSecond Kind and are denoted by Qn(x). It is conventional to take

Qn(x) =

−Vn(1)Un(x) n odd,

Un(1)Vn(x) n even.

Thus the general solution is written in the form

y = c1Pn(x) + c2Qn(x).

Rodrigue’s formula: The Legendre polynomials can also be expressed by

Pn(x) =1

2nn!dn

dxn

(x2 − 1

)n.

Note that:

U0(1) = 1, Un(1) = (−1)n2

2 · 4 · 6 · · ·n1 · 3 · 5 · · · (n− 1)

(n = 2, 4, 6, · · ·),

V1(1) = 1, Vn(1) = (−1)n−1

22 · 4 · 6 · · · (n− 1)

1 · 3 · 5 · · ·n (n = 3, 5, 7, · · ·).

Recurrence formulas:

Pn+1(x) =2n+ 1n+ 1

xPn(x)− n

n+ 1Pn−1(x),

P ′n+1(x)− P ′

n−1(x) = (2n+ 1)Pn(x).

Generating function:

1√1− 2xt+ t2

=∞∑

n=0

Pn(x)tn.

Examples

Given that P0(x) = 1, P1(x) = x. Find (a) P2(x) and (b) P3(x).(a) using the recurrence formula with n = 1

P2(x) =32xP1(x)−

12P0(x)

=32x2 − 1

2

=12

(3x2 − 1

)�


(b) using the recurrence formula with n = 2

P3(x) =53xP2(x)−

23P1(x)

=53x

(3x2 − 1

2

)− 2

3x

=5x3

2− 5x

6− 2x

3

=12

(5x3 − 3x

)�

4.10 Associated Legendre Functions

The following differential equation

(1− x2

) d2ydx2− 2x

dy

dx+

[n(n+ 1)− m2

1− x2

]y = 0,

differs from Legendre’s equation only in the presence of the term involvingm. This equation has a solution of the form

y = c1Pmn (x) + c2Qm

n (x),

where

Pmn (x) =

(1− x2

)m2 d

mPn(x)dxm

,

Qmn (x) =

(1− x2

)m2 d

mQn(x)dxm

.

4.11 Bessel’s Equation

One of the most important differential equations in applied mathematics is

x2 d2y

dx2+ x

dy

dx+

(x2 − ν2

)y = 0,


where ν is a given real, nonnegative number. Bessel’s equation has a regularsingular point at x = 0, hence it has solutions of the form

y(x) = xr∞∑

m=0

amxm,

y′(x) =∞∑

m=0

(m+ r)amxm+r−1,

y′′(x) =∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−2.

Substituting the trial solution into Bessel’s equation yields

∞∑m=0

(m+ r)(m+ r − 1)amxm+r +

∞∑m=0

(m+ r)amxm+r+

+∞∑

m=0

amxm+r+2 − ν2

∞∑m=0

amxm+r = 0

⇒∞∑

m=0

[(m+ r)(m+ r − 1) + (m+ r)− ν2

]amx

m+r+∞∑

m=0

amxm+r+2 = 0,

shifting the first summation by 2 leads to

∞∑m+2=0

[(m+ r + 2)(m+ r + 1) + (m+ r + 2)− ν2

]am+2x

m+r+2+

+∞∑

m=0

amxm+r+2 = 0,

expanding the first two terms (m = −2 and m = −1) and combining thesummations yields

∞∑m=0

{[(m+ r + 2)(m+ r + 1) + (m+ r + 2)− ν2

]am+2 + am

}xm+r+2+

+[r(r − 1) + r − ν2

]a0x

r +[(r + 1)r + (r + 1)− ν2

]a1x

r+1 = 0.


xr terms:

[r(r − 1) + r − ν2

]a0 = 0 =⇒ (r + ν)(r − ν) = 0,

roots : r1 = ν,

r2 = −ν.

xr+1 terms: [(r + 1)2 − ν2

]a1 = 0. (∗)

xm+r+2 terms:

[(m+ r + 2)2 − ν2

]am+2 + am = 0 (m = 0, 1, 2, 3, · · ·). (∗∗)

For r = ν:

(∗) −→[(ν + 1)2 − ν2

]a1 = 0 =⇒ [1 + 2ν]a1 = 0,

recall that ν is real nonnegative number (ν ≥ 0), thus

a1 = 0,

(∗∗) −→[(m+ ν + 2)2 − ν2

]am+2 + am = 0

given that a1 = 0 and ν ≥ 0

a3 = a5 = · · · = 0,

=⇒ am+2 =−am

(m+ 2ν + 2)(m+ 2)(m = 0, 2, 4, · · ·).

The coefficient a0 is arbitrary and it is customary to let

a0 =1

2ν Γ(ν + 1),

where Γ(ν + 1) is the Gamma function.

a2 =−a0

22 (ν + 1)=

−122+ν (ν + 1) Γ(ν + 1)

=−1

22+ν 1! Γ(ν + 2)


a4 =−a2

22 2 (ν + 2)=

124+ν 2! (ν + 2) Γ(ν + 2)

=1

24+ν 2! Γ(ν + 3)

=⇒ a2m =(−1)m

22m+ν m! Γ(ν +m+ 1)(m = 1, 2, · · ·).

Given these coefficients, the first solution corresponding to r = ν can beexpressed as

Jν(x) = xν∞∑

m=0

(−1)m x2m

22m+ν m! Γ(ν +m+ 1), ν ≥ 0.

The preceding expression is known as Bessel Function of the First Kindand Order ν. The series converges for all x. If ν = n = integer,

Jn(x) = xn∞∑

m=0

(−1)m x2m

22m+n m! (n+m)!, n = integer.

Note that the following relationship was employed for the derivation of thepreceding expression

Γ(n+m+ 1) = (n+m)! n+m = integer.

For r2 = −ν the solution can be obtained by replacing ν by −ν in Jν(x)

J−ν(x) = x−ν∞∑

m=0

(−1)m x2m

22m−ν m! Γ(m− ν + 1), ν ≥ 0.

Therefore, the solution to Bessel’s equation can also be written as

y(x) = C1Jν(x) + C2J−ν(x),

where ν is real and nonnegative, and C1, C2 are constants. However,

J−n(x) = (−1)n Jn(x).

Hence, Jn(x) and J−n(x) are linearly dependent, so they do not form abasis of solutions.


Proof:

Replace ν with n in the expression for J−ν(x)

J−n(x) = x−n∞∑

m=0

(−1)m x2m

22m−n m! Γ(m− n+ 1).

Note that 1/Γ = 0 for m ≤ n (the gamma function becomes infinite for thefirst n terms). Consequently, the summation begins with m = n

J−n(x) =∞∑

m=n

(−1)m x2m−n

22m−n m! (m− n)!

=∞∑

s=0

(−1)n+s x2s+n

22s+n (n+ s)! s!

= (−1)nxn∞∑

s=0

(−1)s x2s

22s+n s! (n+ s)

= (−1)nJn(x).

4.12 Gamma Function

The gamma function is defined as follows:

Γ(a) =∫ ∞

0

e−t ta−1 dt.

The gamma function is graphically illustrated in Figure 2. Note that 1/Γ(x)is a continuous function, which makes it easier to work with. Some impor-tant properties of the gamma function are:

Γ(a+ 1) = aΓ(a)

Γ(1) = 1,Γ(2) = Γ(1) = 1!, =⇒ Γ(k + 1) = k! (k = 0, 1, · · ·),Γ(3) = 2Γ(2) = 2!,

1Γ(0)

= 0,

Γ(1/2) =√π,

Γ(x) Γ(1− x) =π

sinπx.


-5

-4

-3

-2

-1

0

1

2

3

4

5

-4 -3 -2 -1 0 1 2 3 4

Γ(x )

Γ(x)−1Γ(

x),

Γ(

x)−1

x

Figure 4.2: Plot of Γ(x) and Γ(x)−1 as a function of x.

4.13 Bessel Functions of the Second Kind

The second solution to Bessel’s equation is defined differently, depending onwhether the order ν is an integer. To provide uniformity of formalism andnumerical tabulation, it is desirable to adopt a form of the second solutionthat is valid for all values of the order.

When n = 0, Bessel’s equation can be written as

xd2y

dx2+dy

dx+ xy = 0.

Following the procedure outlined for the solution of Bessel’s equation, theindicial equation resulting from the xr terms is given by

r2a0 = 0,

=⇒ r2 = 0 −→ r1 = r2 = 0, double root.Therefore, according to the Frobenius method a solution is of the form

y2(x) = J0(x) lnx+∞∑

m=1

Amxm,


y′2 = J ′0 lnx+

J0

x+

∞∑m=1

mAmxm−1,

y′′2 = J ′′0 lnx+

2J ′0

x− J0

x2+

∞∑m=1

m(m− 1)Amxm−2.

Substituting into the governing differential equation yields

xJ ′′0 lnx+ 2J ′

0 −J0

x+

∞∑m=1

m(m− 1)Amxm−1 + J ′

0 lnx+J0

x+

+∞∑

m=1

mAmxm−1 + xJ0 lnx+

∞∑m=1

Amxm+1 = 0.

It should be noted that J0 is a solution to Bessel’s equation. Consequently,

[xJ ′′0 + J ′

0 + xJ0]︸︷︷︸0

lnx = 0.

=⇒ 2J ′0 +

∞∑m=1

m(m− 1)Amxm−1 +

∞∑m=1

mAmxm−1 +

∞∑m=1

Amxm+1 = 0

=⇒ 2J ′0 +

∞∑m=1

m2Amxm−1 +

∞∑m=1

Amxm+1 = 0.

Recall that

J0(x) =∞∑

m=0

(−1)m x2m

22m (m!)2, J ′

0(x) =∞∑

m=1

(−1)m x2m−1

22m−1 m! (m− 1)!.

=⇒∞∑

m=1

(−1)m x2m−1

22m−2 m! (m− 1)!+

∞∑m=1

m2Amxm−1 +

∞∑m=1

Amxm+1 = 0.

x0 terms: (2nd sum with m = 1)

A1x0 = 0 −→ A1 = 0.


x2 terms: (2nd sum with m = 3 + 3rd sum with m = 1)[9A3 +A1]x2 = 0,

=⇒ A3 = 0.A1 = 0,Similarly, we can show that

A5 = A7 = · · · = 0.

x terms: (1st sum with m = 1 + 2nd sum with m = 2)

[−1 + 4A2]x = 0 =⇒ A2 =14.

x3 terms: (1st sum with m = 2 + 2nd with m = 4 + 3rd with m = 2)[18

+ 16A4 +A2

]x3 = 0,

A2 =14, =⇒ A4 = − 3

128.

In general

A2m =(−1)m−1

22m (m!)2

(1 +

12

+13

+ · · ·+ 1m

)︸︷︷︸

hm

, (m = 1, 2, 3, · · ·).

=⇒ y2(x) = J0(x) lnx+∞∑

m=1

(−1)m−1 hm x2m

22m (m!)2.

Note that y1 = J0 and y2 are linearly independent, consequently

y1 = J0, and y∗2 = a(y2 + bJ0) a �= 0,

are also linearly independent. Customarily we choose

a =2π, b = γ − ln 2,

whereγ = 0.577215 Euler constant.

We define the Bessel Function of the Second Kind and Order Zero as follows

Y0(x) =2π

[J0(x)

(lnx

2+ γ

)+

∞∑m=1

(−1)m−1 hm x2m

22m (m!)2

].


Similarly, we can obtain a second solution for ν = n = 1, 2, · · ·

Yn(x) =2π

[ln

(x2

)+ γ

]Jn(x) +

xn

π

∞∑m=0

(−1)m−1 [hm + hm+n] x2m

22m+n m! (m+ n)!−

− x−n

π

n−1∑m=0

(n−m− 1)! x2m

22m−n m!.

Therefore, the general complete solution for the case where ν = n is givenby

y(x) = C1Jn(x) + C2Yn(x).

A couple of useful relationships of Bessel functions of the second kind are:

Yn(x) = limν→n

Yν(x),

Yν(x) =Jν(x) cos νπ − J−ν(x)

sin νπ.

4.14 Modified Bessel Functions

The following second–order differential equation is known as the modifiedBessel’s equation

x2 d2y

dx2+ x

dy

dx−

(x2 + ν2

)y = 0.

The general solution of this differential equation is

y = c1Jν(ix) + c2J−ν(ix), ν ≥ 0, real,

y = c1Jn(ix) + c2Yn(ix) n = 0, 1, 2, · · ·Given the following relationships

Iν(x) = i−νJν(ix) =∞∑

m=0

(x2

)2m+ν

m! (m+ n)!,

I−n(x) = In(x),


it is clear that the two solutions are dependent. Therefore, we introducethe following relationships

Kν(x) =π

2I−ν(x)− Iν(x)

2 sin νπ,

Kn(x) =π

2in+1[Jn(ix) + iYn(ix)].

The solution to modified Bessel’s equation can be expressed as follows:

y = c1Iν(x) + c2I−ν(x),

y = c1In(x) + c2Kn(x),

where Iν is the modified Bessel function of the first kind of order ν;In is the modified Bessel function of the first kind of order n;Kn is the modified Bessel function of the second kind of order n.

4.15 Useful Formulas

4.15.1 Recurrence Relationships Involving Jν(x)

Valid for all values of ν as well as when replacing Jν(x) for Yν(x).

(1) J2(x) =2xJ1(x)− J0(x)

(2) J ′0(x) = −J1(x)

(3) Jν+1(x) =2νxJν(x)− Jν−1(x)

(4) J ′ν(x) =

12[Jν−1(x)− Jν+1(x)]

(5) xJ ′ν(x) = νJν(x)− xJν+1(x)

(6) xJ ′ν(x) = xJν−1(x)− νJν(x)


(7)d

dx[xνJν(x)] = xνJν−1(x)

(8)d

dx

[x−νJν(x)

]= −x−νJν+1(x)

(9)(

1x

d

dx

)m

[xνJν(x)] = xν−mJν−m(x)

(10)(

1x

d

dx

)m [x−νJν(x)

]= (−1)mx−ν−mJν+m(x)

(11) J−n(x) = (−1)nJn(x)

(12)d

dx[xnJn(x)] = xnJn−1(x)

4.15.2 Recurrence Relationships Involving Iν(x)

(1) xIν−1(x)− xIν+1(x) = 2νIν(x)

(2) Iν−1(x) + Iν+1(x) = 2I ′ν(x)

(3) xI ′ν(x) + νIν(x) = xIν−1(x)

(4) xI ′ν(x)− νIν(x) = xIν+1(x)

(5)(

1x

d

dx

)m

[xνIν(x)] = xν−mIν−m(x)

(6)(

1x

d

dx

)m [x−νIν(x)

]= x−ν−mIν+m(x)

(7) I−n(x) = In(x)

(8) I2(x) = − 2xI1(x) + I0(x)

(9) I ′0(x) = I1(x)


4.15.3 Recurrence Relationships Involving Kν(x)

(1) xKn−1(x)− xKn+1(x) = −2nKn(x)

(2) Kn−1(x) +Kn+1(x) = −2K ′n(x)

(3) xK ′n(x) + nKn(x) = −xKn−1(x)

(4) xK ′ν(x)− nKn(x) = −xKn+1(x)

(5)(

1x

d

dx

)m

[xnKn(x)] = (−1)mxn−mKn−m(x)

(6)(

1x

d

dx

)m [x−nKn(x)

]= (−1)mx−n−mKn+m(x)

(7) K−ν(x) = Kν(x)

(8) K2(x) =2xK1(x) +K0(x)

(9) K ′0(x) = −K1(x)

(10) K−n(x) = Kn(x)


4.15.4 Integral Representations of Jn(x), In(x) and Kn(x)

(1) J0(x) =2π

∫ π/2

0

cos(x cos θ) dθ

(2) J1(x) =2π

∫ π/2

0

sin(x cos θ) cos θ dθ

(3) Jn(x) =1π

∫ π

0

cos(x sin θ − nθ) dθ

(4) Jn(x) =1π

∫ π

0

cos(x sin θ) cosnθ dθ (n = 0, 2, 4, . . .)

(5) Jn(x) =1π

∫ π

0

sin(x sin θ) sinnθ dθ (n = 1, 3, 5, . . .)

(6) Jn(x) =2√

πΓ(n+ 12 )

(x2

)n∫ π/2

0

cos(x sin θ) (cos θ)2n dθ

(7) Jn(x) =2√

πΓ(n+ 12 )

(x2

)n∫ π/2

0

cos(x sin θ) (sin θ)2n dθ

(8) In(x) =2√

πΓ(n+ 12 )

(x2

)n∫ π/2

0

cosh(x sin θ) (cos θ)2n dθ

(9) In(x) =2√

πΓ(n+ 12 )

(x2

)n∫ π/2

0

cosh(x sin θ) (sin θ)2n dθ

(10) Kn(x) = e−x( π

2x

) 12 1

Γ(n+ 12 )

∫ ∞

0

e−uun− 12

(1 +

u

2x

)n− 12du


4.15.5 Indefinite Integrals of Bessel Functions

Also applicable when replacing Jν(x) for Yν(x).

(1)∫xJ0(ax) dx =

x

aJ1(ax)

(2)∫x2J0(ax) dx =

x2

aJ1(ax) +

x

a2J0(ax)−

1a2

∫J0(ax) dx

(3)∫J0(ax)x2

dx = aJ1(ax)−J0(ax)x

− a2∫J0(ax) dx

(4)∫J1(ax) dx = −1

aJ0(ax)

(5)∫xJ1(ax) dx = −x

aJ0(ax) +

1a

∫J0(ax) dx

(6)∫J1(ax)x

dx = −J1(ax) + a∫J0(ax) dx

(7)∫xJn(ax)Jn(bx) dx =

x[aJn(bx)J ′n(ax)− bJn(ax)J ′

n(bx)]b2 − a2 (a �= b)

(8)∫xJ2

n(ax) dx =x2

2[J ′

n(ax)]2 +x2

2

(1− n2

a2x2

)[Jn(ax)]2

(9)∫xnJn−1(ax) dx =

xn

aJn(ax)

(10)∫x−nJn+1(ax) dx = −x

−n

aJn(ax)

(11)∫xnIn−1(ax) dx =

xn

aIn(ax)

(12)∫x−nIn+1(ax) dx =

x−n

aIn(ax)

(13)∫xnKn−1(ax) dx = −x

n

aKn(ax)

(14)∫x−nKn+1(ax) dx = −x

−n

aKn(ax)


4.15.6 Definite Integrals Involving Bessel Functions

(1)∫ a

0

J1(x) dx = 1− J0(a) (a > 0)

(2)∫ ∞

a

J1(x) dx = J0(a) (a > 0)

(3)∫ ∞

0

Jn(ax) dx =1a

(n > −1, a > 0)

(4)∫ ∞

0

Jn(ax)x

dx =1n

(n = 1, 2, . . .)

(5)∫ 1

0

xJn(ax)Jn(bx) dx =

0, a �= b,

a = b,12 [Jn+1(a)]2, Jn(a) = 0,

Jn(b) = 0.

(n > −1)

(6)∫ ∞

0

e−axJ0(bx) dx =1√

a2 + b2

(7)∫ ∞

0

e−axJn(bx) dx =1√

a2 + b2

[√(a2 + b2)− a

b

]n

(a > 0, n ≥ 0)

(8)∫ ∞

0

Jn(ax) sin(bx) dx =

sin[n sin−1(b/a)]√a2−b2

, 0 < b < a,

an cos(nπ/2)√b2−a2(b+

√b2−a2)n , 0 < a < b,

(n > −2)

(9)∫ ∞

0

Jn(ax) cos(bx) dx =

cos[n cos−1(b/a)]√a2−b2

, 0 < b < a,

−an sin(nπ/2)√b2−a2(b+

√b2−a2)n , 0 < a < b,

(n > −2)

(10)∫ ∞

0

Jm(x)Jn(x)x

dx =

2π(m2−n2) sin (m−n)π

2 , m �= n,

12m , m = n,

(m+n > 0)

(11)∫ ∞

0

J0(ax)J1(bx) dx =

1/b, b2 > a2,

0, b2 < a2,

(12)∫ ∞

0

J0(ax)J1(ax) dx =12a

(a > 0)


-0.5

0

0.5

1

Jn(x)

0 2.5 5 7.5 10

x

J0

J1

( a )

-0.5

-0.3

-0.1

0.1

0.3

0.5

Yn(x)

0 2.5 5 7.5 10

x

Y0

Y1

( b )

0

20

40

60

80

100

0 1 2 3 4 5

x

I 0

100K0

( c )

Kn

(x),

I n

(x)

4.16 Summary of Bessel’s Equations

Bessel Function Modified Bessel Function

x2y′′ + xy′ +(x2 − ν2

)y = 0 x2y′′ + xy′ −

(x2 + ν2

)y = 0

y = C1Jν(x) + C2J−ν(x) y = C1Iν(x) + C2I−ν(x) Real

y = C1Jn(x) + C2Yn(x) y = C1In(x) + C2Kn(x) Integer

Figure 4.3: Plot of (a) J0, J1, (b) Y0, Y1, and (c) K0, I0 as a function of x.


Example 1There are various differential equations which can be reduced to Bessel’sequation. For example consider the following differential equation

xd2y

dx2+dy

dx+

14y = 0.

Make the substitution

√x = z −→ x = z2,

dz

dx=

12x−

12 ,

dy

dx=dy

dz

dz

dx=dy

dz

(12x−

12

),

d2y

dx2=d

dx

(dy

dx

)=d

dx

[dy

dz

(12x−

12

)]

=dy

dz

(−1

4x−

32

)+

(12x−

12

)d

dz

(dy

dz

) (dz

dx

)

=dy

dz

(−1

4x−

32

)+d2y

dz2

(12x−

12

)2

=dy

dz

(−1

4x−

32

)+d2y

dz2

(14x

).

Substituting into the original equation yields

z2[dy

dz

(−1

4x−

32

)+d2y

dz2

(14x

)]+dy

dz

(12x−

12

)+y

4= 0

=⇒14d2y

dz2− 1

4zdy

dz+

12zdy

dz+y

4= 0

=⇒d2y

dz2+

1z

dy

dz+ y = 0

=⇒z2 d2y

dz2+ zdy

dz+

(z2 − 0

)y = 0, Bessel′s Equation.

Therefore, the solution to the transformed differential equation is

y(z) = AJ0(z) +BY0(z).

Backsubstitution of the original variables z =√x yields

=⇒ y(x) = AJ0

(√x)

+BY0

(√x)

�


Example 2Prove that

y =√xJ 1

4

(kx2

2

)and y =

√xY 1

4

(kx2

2

)are solutions of the following second–order ordinary differential equation

d2y

dx2+ k2x2y = 0. (∗)

Let

z =kx2

2=⇒ dz

dx= kx,

dy

dx=dy

dz

dz

dx= kx

dy

dz,

d2y

dx2=d

dx

dy

dx=d

dx

(kxdy

dz

)

= kdy

dz+ kx

d

dz

(dy

dz

)dz

dx

= kdy

dz+ (kx)2

d2y

dz2.

Substituting into the differential equation (*) yields

kdy

dz+ (kx)2

d2y

dz2+ 2kzy = 0

=⇒kx2 d2y

dz2+dy

dz+ 2zy = 0

=⇒2zd2y

dz2+dy

dz+ 2zy = 0

=⇒z2 d2y

dz2+z

2dy

dz+ z2y = 0. (∗∗)

The preceding is a differential equation with a new independent variable.At this stage let

y = uz14 ,

y′ = u′z14 +

14uz−

34 ,

y′′ = u′′z14 +

14u′z−

34 +

14u′z−

34 − 3

16uz−

74

= u′′z14 +

12u′z−

34 − 3

16uz−

74 .


Substituting into into (**) yields

z2[d2u

dz2z

14 +

12du

dzz−

34 − 3

16uz−

74

]+z

2

[du

dzz

14 +

14uz−

34

]+ z2

(uz

14

)= 0

=⇒ z 94d2u

dz2+ z

54du

dz+

(z

94 − z

14

16

)u = 0, divide by z

14

=⇒ z2 d2u

dz2+ zdu

dz+

(z2 − 1

16

)u = 0, Bessel′s equation with ν =

14.

The solution to the preceding Bessel’s equation can be presented either interms of J 1

4and J− 1

4or more generally in terms of J 1

4and Y 1

4

u(z) = AJ 14(z) +BY 1

4(z), ←− u =

y

z14

=⇒ y(z) = Az14 J 1

4(z) +Bz

14Y 1

4(z), ←− z =

kx2

2

=⇒ y(x) = A(kx2

2

) 14

J 14

[kx2

2

]+B

(kx2

2

) 14

Y 14

[kx2

2

]

= A∗√x J 14

[kx2

2

]+B∗√x Y 1

4

[kx2

2

], �

where

A∗ = A(k

2

) 14

, B∗ = B(k

2

) 14

.

It should be noted that because 1/4 is not an integer we can also have asolution of the form:

y(x) = A∗√x J 14

[kx2

2

]+B∗√x J− 1

4

[kx2

2

]�


Example 3Obtain the general solution of the following second–order ordinary differ-ential equation

d2y

dx2+

1x

dy

dx− y − 4y

x2= 0.

Rearranging yields

d2y

dx2+

1x

dy

dx−

(1 +

4x2

)y = 0

=⇒ x2 d2y

dx2+ x

dy

dx−

(x2 + 4

)y = 0.

This is the modified Bessel’s equation with ν = 2 with a general solutionof the form

y(x) = A I2[x] +B K2[x] �

Examples: Gamma FunctionsEvaluate the following:

(a)∫ ∞

0

e−x√x dx.

Recall the definition of the gamma function

Γ(a) =∫ ∞

0

e−tta−1 dt,

and let a− 1 = 12 −→ a = 3

2 in order to obtain the following relationship

=⇒∫ ∞

0

e−xx32−1 dx = Γ

(32

)←− Γ(n+ 1) = nΓ(n)

=⇒ Γ(

32

)= Γ

(12

+ 1)

=12Γ

(12

)←− Γ

(12

)=√π

=⇒∫ ∞

0

e−x√x dx =

√π

2�

(b) Γ(

52

)= Γ

(32

+ 1)

=32Γ

(32

)=

32

(√π

2

)=

3√π

4�


(c) Γ(−2.5) ←− Γ(n+ 1) = nΓ(n) =⇒ Γ(n) =Γ(n+ 1)n

=⇒ Γ(−2.5) =Γ(−1.5)(−2.5)

=Γ(−0.5)

(−2.5)(−1.5)

=Γ(0.5)

(−2.5)(−1.5)(−0.5)

=−√π1.875

�

4.17 Orthogonal Sets of Functions

Two functions gm(x) and gn(x) are said to be orthogonal on the interval(a, b) if the integral of the product gm(x)gn(x) over that interval vanishes

(gm, gn) =∫ b

a

gm(x)gn(x) dx = 0, Orthogonal.

The nonnegative square root of (gm, gm) is called the norm of gm

||gm|| =√

(gm, gm) =

√∫ b

a

g2m(x) dx.

Two functions gm(x) and gn(x) are said to be orthonormal if

(gm, gn) =∫ b

a

gm(x)gn(x) dx =

{ 0 m �= n,

1 m = n,

or equivalently we may write

(gm, gn) = δmn, Kronecker Delta,


where

δmn =

{ 0 m �= n,

1 m = n.From an orthogonal set we may obtain an orthonormal set by dividing eachfunction by its norm.

Orthogonality with Respect to a Weight Function

Two functions gm(x) and gn(x) are said to be orthogonal with respect to aweighting function p(x) on the interval (a, b) if

∫ b

a

p(x)gm(x)gn(x) dx = 0, n �= m.

The norm of gm is now defined as

||gm|| =

√∫ b

a

p(x) g2m(x) dx.

If the norm of each function gm is 1, the set is said to be orthonormal onthat interval with respect to p(x).

ExampleShow that the following set is orthogonal on the interval 0 ≤ x ≤ 2π anddetermine the corresponding orthonormal set

1, cosx, cos 2x, cos 3x, · · · , 0 ≤ x ≤ 2π.

The general term is cosnx (n = 0, 1, 2, 3, · · ·)

(gm, gn) =∫ 2π

0

cosmx cosnx dx = 0,

and recall cosx cos y =12[cos(x+ y) + cos(x− y)],

=⇒ (gm, gn) =12

∫ 2π

0

cos(m+ n)x dx+12

∫ 2π

0

cos(m− n)x dx

=sin(m+ n)x2(m+ n)

∣∣∣∣2π

0

+sin(m− n)x2(m− n)

∣∣∣∣2π

0

= 0.


Therefore, the set

cosnx, n = 0, 1, 2, 3, · · · is orthogonal �

To obtain the norm of each function evaluate

||g0||2 = (g0, g0) =∫ 2π

0

(1 · 1) dx = x∣∣2π

0= 2π,

=⇒ ||g0|| =√

2π,

||gm|| =√

(gm, gm), m = 1, 2, 3, · · ·

(gm, gm) =∫ 2π

0

cos2mx dx

=12

∫ 2π

0

(1 + cos 2mx) dx

=12

(x+

sin 2mx2m

) ∣∣∣∣2π

0

= π,

=⇒ ||gm|| =√π.

The corresponding orthonormal set is:

1√2π,

cosx√π,

cos 2x√π,

cos 3x√π, · · · �

4.18 Sturm–Liouville Problem

A boundary–value problem having the form

d

dx

[r(x)

dy

dx

]+ [q(x) + λp(x)]y = 0, a ≤ x ≤ b,

replace if r(a) = r(b)

k1y(a) + k2y′(a) = 0, ← remove if r(a) = 0 ← y(a) = y(b)

51y(b) + 52y′(b) = 0, ← remove if r(b) = 0 ← y′(a) = y′(b)


where k1, k2, 51, 52 are given constants; r(x), q(x), p(x) are given functionswhich we shall assume to be differentiable; and λ is an unspecified param-eter independent of x, is called a Sturm–Liouville boundary–value problemor Sturm–Liouville system.

A non–trivial solution of this system (one which is not identically zero,y = 0) exists in general only for a particular set of values of the parameterλ. These values are called the characteristic values, or more often eigen-values of the system. The corresponding solutions are called characteristicfunctions or eigenfunctions of the system.

Orthogonality of Eigenfunctions

If ym(x) and yn(x) are eigenfunctions of a Sturm–Liouville problem, whichcorrespond to different eigenvalues λm and λn, then ym, yn are orthogonalon that interval with respect to the weight factor p(x); consequently,

∫ b

a

p(x) ymyn dx = 0.

If p(x) is positive in the interval a ≤ x ≤ b, the λn’s are real.

Example 1Find the eigenvalues and eigenfunctions of the following Sturm–Liouvilleproblem

d2y

dx2+ λy = 0, y(0) = 0, y(π) = 0.

The characteristic equation of the second–order ordinary differential equa-tion is

r2 + λ = 0 =⇒ r = ±i√λ

=⇒ y(x) = A cos√λx+B sin

√λx.

bc1 : y(0) = A+ 0 = 0 =⇒ A = 0,

y(π) = 0 = A cos√λπ +B sin

√λπbc2 :

=⇒ B sin√λπ = 0, B �= 0 because A = 0

=⇒ sin√λπ = 0

=⇒√λπ = nπ, (n = 0, 1, 2, · · ·)


=⇒ λn = n2, (n = 1, 2, 3, · · ·) Eigenvalues �Note that the eigenvalue λ = 0 is excluded because for λ = 0 the solutionis y(x) = B sin(0) = 0. Consequently, the eigenfunctions are given by

y(x) = B sinnx, (n = 1, 2, 3, · · ·) Eigenfunctions �

Also, we may let B = 1.

Example 2

Find the eigenvalues and eigenfunctions of the following Sturm–Liouvileproblem

d2y

dx2+ λy = 0, y(0) = 0,

dy(L)dx

= 0.

The characteristic equation of the second–order ordinary differential equa-tion is

r2 + λ = 0 =⇒ r = ±i√λ

=⇒ y(x) = A cos√λx+B sin

√λx

bc1 : y(0) = A+ 0 = 0 =⇒ A = 0,

bc2 :dy

dx= −√λA sin

√λx+

√λB cos

√λx

=⇒ dy(L)dx

= 0 =√λB cos

√λL

=⇒ cos√λL = 0

=⇒√λL =

(2n+ 1

2

)π, (n = 0, 1, 2, · · ·)

=⇒ λn =[(2n+ 1)π

2L

]2

, (n = 0, 1, 2, · · ·) Eigenvalues �

=⇒ yn(x) = B sin[(2n+ 1)πx

2L

]Eigenfunctions �

Also, we may let B = 1.


4.19 Important Orthogonal Functions and Polynomials

4.19.1 Bessel Functions

Recall Bessel’s equation and the corresponding general solution

x2 d2y

dx2+ x

dy

dx+

(x2 − ν2

)y = 0, (0 ≤ x ≤ R),

y = AJν(x) +BYν(x).

The following differential equation

x2 d2

dx2Jν [x] + x

d

dxJν [x] +

(x2 − ν2

)Jν [x] = 0 (∗)

is satisfied because Jν is a solution to Bessel’s equation. Furthermore, let

x = λs =⇒ ds

dx=

1λ,

dJ [x]dx

=dJ [x]ds

ds

dx=

1λ

dJ [λs]ds

,

d2J [x]dx2

=d

dx

(dJ [x]dx

)=d

dx

(1λ

dJ [λs]ds

)

=d

ds

(1λ

dJ [λs]ds

)ds

dx=

1λ2

d2J [λs]ds2

.

Substituting into (*) yields

=⇒ s2 d2

ds2Jν [λs] + s

d

dsJν [λs] +

(λ2s2 − ν2

)Jν [λs] = 0,

divide by s

=⇒ d

ds

[sdJν [λs]ds

]+

(−ν

2

s+ λ2s

)Jν [λs] = 0, (0 ≥ s ≥ R∗). (∗∗)

This is a Sturm–Liouville problem with λ2 instead of λ, and

r(s) = s, q(s) = −ν2

s, p(s) = s.


Note that r(0) = 0, then the first boundary condition of the Sturm–Liouvilleproblem is dropped and we only have to satisfy the second condition,namely

51Jν [λR∗] + 52J ′ν [λR∗] = 0,

or just the relationshipJν [λR∗] = 0. (∗ ∗ ∗)

LetλR∗ = amn =⇒ λ = λmn =

amn

R∗ .

Then, the solutions of (**) which satisfy (***) form an orthogonal set onthat interval with respect to the weight function p(s) = s; consequently,

∫ R∗

0

s Jν [λmns] Jν [λkns] ds = 0, k �= m.

4.19.2 Legendre Polynomials

Recall Legendre’s equation

(1− x2

) d2ydx2− 2x

dy

dx+ n(n+ 1)y = 0, (−1 ≤ x ≤ 1),

and let λ = n(n+ 1) to obtain the following ordinary differential equation

d

dx

[(1− x2

) dydx

]+ λy = 0.

This is a Sturm–Liouville problem with

r(x) =(1− x2

), q(x) = 0, p(x) = 1.

Note that r(a) = r(−1) = 0 and r(b) = r(1) = 0; therefore, no bound-ary conditions are needed to form a Sturm–Liouville problem. We knowthat, for n = 0, 1, · · · , the Legendre polynomials Pn(x) are solutions of theproblem. Hence, these are eigenfunctions and they are orthogonal on thatinterval with respect to p(x) = 1; consequently,

∫ 1

−1

Pm(x) Pn(x) dx = 0, m �= n.


4.20 Exercises

(1) Determine the interval of convergence for each of the following series,including consideration of the behavior of the series at each end point

(a)∞∑

n=1

(−1)n (x− 2)n

(4n)√n,

(b)∞∑

n=0

2nxn

n!.

(2) For each of the following differential equations, obtain the most generalsolution which is representable by a Maclaurin series (Apply the powerseries method: y =

∑∞k=0 akx

k)

(a) x2 d2y

dx2− dydx

+ y = 0,

(b) x2 d2y

dx2− dydx

= 0,

(c)dy

dx+ x

dy

dx= y.

(3) Show the relation between m, p, and s

∞∑m=2

m(m−1)amxm−2 =

∞∑p=1

p(p+1)ap+1xp−1 =

∞∑s=0

(s+2)(s+1)as+2xs.

(4) Use the method of Frobenius to obtain the general solution of the fol-lowing differential equation, valid near x = 0:

xd2y

dx2+ 2dy

dx+ xy = 0.


(5) Find P6(x).

(6) Obtain the associated Legendre functions

(a) P 12 (x),

(b) P 23 (x),

(c) P 32 (x).

(7) By making an appropriate change of variables, obtain the general solu-tion of the differential equation

(Ax+B)d2y

dx2+A

dy

dx+A2(Ax+B)y = 0.

(8) Using the series definition

Jn[x] =∞∑

m=0

(−1)m(

x2

)2m+n

m! (n+m)!

prove thatd

dx

(x−nJn[ax]

)= −ax−nJn+1[ax].

(9) Prove that

y1 =√x J1/4

[x2

2

]and y2 =

√x Y1/4

[x2

2

]

are solutions ofd2y

dx2+ x2y = 0.

Hint: let z = 12x

2 and then y = uz14 .


(10) Find the eigenvalues and the eigenfunctions of the following Sturm–Liouville problems:

(a)d

dx

[xdy

dx

]+λ

xy = 0, y(1) = 0,

dy(e)dx

= 0,(let x = et

).

(b)d2y

dx2+ λy = 0, y(0) = y(π),

dy(0)dx

=dy(π)dx

.

(11) Show that the following set is orthogonal on the interval −1 ≤ x ≤1, and determine the corresponding orthonormal set; sinπx, sin 2πx,sin 3πx, . . .

(12) Prove that

1∫−1

Pm(x) Pn(x) dx = 0, if m �= n.


Chapter 5

Laplace Transform

5.1 Introduction

The Laplace transform is a method for solving differential equations andcorresponding initial value and boundary problems.

• Laplace transform reduces an ordinary differential equation to an alge-braic equation.

• It solves problems directly

• Initial value problems are solved without first determining a general so-lution.• Nonhomogeneous equations are solved without first solving the corre-

sponding homogeneous equation.

• PDE =⇒ ODE =⇒ Algebraic

5.2 Characteristics of Problems Suited for Laplace Tranform

(a) The differential equation is linear (necessary).(b) The equation has constant coefficients (highly desirable).(c) At least one independent variable has the range 0 to ∞ (highly desirable).(d) There are appropriate initial conditions involving the independent vari-

able in (c) above (desirable).

5.3 Definition of the Laplace Transform

Let f(t) be a function of t specified for t > 0. Then, the Laplace transformof f(t), denoted by L{f(t)}, is defined by

L{f(t)} = F (s) =∫ ∞

0

e−stf(t) dt,

where we assume at present that the parameter s is real and large enough(sometimes it is useful to consider s complex).


Notation

Original functions: lowercase letters (i.e., f(t), g(t), y(t)). Their transforms:same letters in capitals (i.e., F (s), G(s), Y (s)).

t −→ s,

(real space) −→ (Laplace space).In other cases, a tilde can be used to denote the Laplace transform. Thus,for example, the Laplace transform of u(t) is u(s). It should be noted,however, that some authors use opposite notation: F (t) −→ f(s).

Examples

L{1} =∫ ∞

0

e−st(1) dt = −e−st

s

∣∣∣∣∞

0

=1s

(a)

L{a} =a

s, a is a constant

L{t} =∫ ∞

0

e−stt dt = −e−stt

s

∣∣∣∣∞

0

+1s

∫ ∞

0

e−st dt(b)

= 0− e−st

s2

∣∣∣∣∞

0

=1s2


u dv = uv −∫

v du,

u = t, du = dt, dv = e−stdt, v = −e−st

s.

L{at} =a

s2

L{tn} =∫ ∞

0

e−sttn dt (n = 0, 1, 2, 3, · · ·)(c)

=[−tn

(e−st

s

)]∞

0

+∫ ∞

0

n

se−sttn−1dt


u dv = uv −∫

v du,

u = tn, du = ntn−1dt, dv = e−stdt, v = −e−st

s.

C.V. CHRYSIKOPOULOS: ENGINEERING APPLIED MATHEMATICS ––––––––––––––––––––––––––– 100

y

t

y = tn

t

y y = e-st

y = tn e-sty

t

Figure 5.1: An illustration of a function for which a Laplace transformexists.

We assume that s is large enough that e−st diminishes more quickly thantn can increase, so that as t→∞, tne−st → 0 (see Figure 5.1).

=⇒ L{tn} =n

s

∫ ∞

0

e−sttn−1 dt.

Note that An−1 =∫ ∞

0

e−sttn−1 dt is the same as

An =∫ ∞

0

e−sttn dt except that n→ n− 1

An =n

sAn−1 =

n

s

(n− 1s

)An−2

=n

s

(n− 1s

)(n− 2s

)An−3

=n

s

(n− 1s

)(n− 2s

)· · · n− (n− 1)

sA0

=n(n− 1)(n− 2) · · · 3 · 2 · 1

sn+1=

n!sn+1

,

because A0 =∫ ∞

0

e−st(1) dt =1s,


=⇒ L{tn} =n!

sn+1

This is not the easiest approach but shows some aspects of s which lead tothe existence of Laplace transforms.

Another Way of Obtaining the Previous Laplace Transform:

L{ta} =∫ ∞

0

e−stta dt, let x = st → dt =dx

s

=⇒L{ta} =∫ ∞

0

e−x(x

s

)a dx

s=

1sa+1

∫ ∞

0

e−xxa dx

L{ta} =Γ(a + 1)sa+1

, a = real

L{tn} =n!

sn+1a = n = integer

Γ(β) =∫ ∞

0

e−zzβ−1 dz, Gamma function,

Γ(n + 1) = n!

So we can form a table of Laplace transforms.

5.4 References to Tables of Laplace Transforms

Day, W. D., Tables of Laplace Transforms, London Iliffe Books LTD, Eng-land, 1966.

Erdelyi, A. (Editor), Tables of Integral Transforms, Vol. I, McGraw–Hill,1954.

Nixon, F. E., Handbook of Laplace Transformation, Prentice Hall, 1960.Oberhettinger, F. and L. Badii, Tables of Laplace Transforms, Springer–

Verlag, New York, 428pp., 1973.Roberts, G.E. and H. Kaufman, Table of Laplace transforms, W. B. Saun-

ders Co., Philadelphia, 367pp., 1966.Sneddon, I. N., The Use of Integral Transform, McGraw–Hill, 1972.

Widder, D. V., An Introduction to Transform Theory, Academic Press,1971.


f(t)

a t1 t2 t3 b

t

5.5 Existence of Laplace Transforms

Theorem: If f(t) is sectional or piecewise continuous in every finite interval0 ≤ t ≤ N and of exponential order γ for t > N , then its Laplace transformF (s) exists for all s > γ.

5.5.1 Piecewise Continuity

A function is called piecewise continuous in an interval α ≤ t ≤ β if theinterval can be subdivided into a finite number of intervals in each of whichthe function is continuous and has finite right and left hand limits.

Figure 5.2: Illustration of a piecewise continuous function having discon-tinuities at t1, t2 and t3.

Note that in Figure 5.2 the right and left hand limits at t2, for example,are represented by

limε→0

f(t2 + ε) = f(t2 + 0) = f(t2+) (right hand limit),

limε→0

f(t2 − ε) = f(t2 − 0) = f(t2−) (left hand limit).

5.5.2 Functions of Exponential Order

If real constants M > 0 and γ exist such that for all t > N ,∣∣∣e−γtf(t)∣∣∣ < M or |f(t)| < Meγt,


we say that f(t) is a function of exponential order γ as t → ∞ or, briefly,is of exponential order.

Example 1

f(t) = t2 is of exponential order (i.e., γ = 3), because |t2| = t2 < e3t for allt > 0 (let M = 1).

Example 2

f(t) = et3 is not of exponential order because∣∣e−γtet3

∣∣ =∣∣et3−γt

∣∣ < M canbe made larger than any given constant by increasing t.

5.6 Some Important Properties of Laplace Transforms

5.6.1 Linearity Property

The Laplace transformation is a linear operator

L{af(t) + bg(t)} = aL{f(t)}+ bL{g(t)} = aF (s) + bG(s)

Example

L{4t2 − 3 cos 2t + 5e−t

}= 4L

{t2

}− 3L{cos 2t}+ 5L

{e−t

}= 4

(2!s3

)− 3

(s

s2 + 22

)+ 5

(1

s− (−1)

)

=8s3− 3s

s2 + 4+

5s + 1

Where the appropriate Laplace transforms were obtained from Table 6.1 ofKreyszig, 1993, p.265.

5.6.2 Laplace Transform of Derivatives

Assuming that the function f(t) is continuous and of exponential order,and f (n) is sectionally continuous, then

L{f ′(t)} = sF (s)− f(0)

L{f ′′(t)} = s2F (s)− sf(0)− f ′(0)

L{f (n)(t)} = snF (s)− sn−1f(0)− sn−2f ′(0)− · · · − f (n−1)(0)


Example 1

L{f ′(t)} = ? for f(t) = cos 3t

L{cos 3t} = F (s) =s

s2 + 32=

s

s2 + 9(see Table 6.1 [Kreyszig, 1993])

L{f ′(t)} = sF (s)− f(0) = s

(s

s2 + 9

)− cos(0)

=s2

s2 + 9− 1 =

−9s2 + 9

Example 2Another way of obtaining the transforms of given functions.

Let f(t) = eat f ′(t) = aeat, f(0) = 1

L{f ′(t)} = sF (s)− f(0) =⇒ L{aeat

}= sL

{eat

}− 1

=⇒ aL{eat

}= sL

{eat

}− 1

=⇒ 1 = (s− a)L{eat

}=⇒ L

{eat

}=

1s− a

5.6.3 Laplace Transform of Integrals

L{ ∫ t

0

f(τ) dτ

}=

1sL{f(t)} =

F (s)s

Example

L{ ∫ t

0

sin 2τ dτ

}=

1sL{f(t)} =

1s

(2

s2 + 22

)=

2s(s2 + 4)

5.6.4 First Translation or Shifting Property (Shifting on s–axis)

L{eatf(t)} = F (s− a) =∫ ∞

0

e−(s−a)tf(t) dt


a

F(s) F(s-a)

s

1

a

u(t-a)

1

u(t)

(t) (t)

Figure 5.3: Illustration of shifting on the s axis (here, a > 0).

Example 1

L{e−t cos 2t

}= F (s + 1) =

s + 1(s + 1)2 + 4

=s + 1

s2 + 2s + 5{

Since F (s) = L{cos 2t} =s

s2 + 4

}

Example 2

L{e−4tt

}= F (s− a) = F (s + 4) =

1(s + 4)2

{Since F (s) = L{t} =

1s2

}

5.6.5 Second Translation or Shifting Property (Shifting on t–axis)

L{f(t− a)u(t− a)} = e−asF (s),

where u(t− a) ={ 0 t < a,

1 t > a.is the Heaviside function.

Figure 5.4: Schematic representation of u(t− a).


Example 1

L{(t− 2)3u(t− 2)

}= e−2s

(6s4

)

Since F (s) = L

{t3

}=

3!s4

=6s4

,

a = 2.

Example 2

L{u(t− 5)} = e−5s

(1s

)

f(t) = f(t− 5) = 1,

F (s) = L{1} =1s,

a = 5.

5.6.6 Change of Scale Property

L{f(at)} =1aF

( s

a

)

Example

L{sin nt} =1n

(1

( sn )2 + 1

)=

n

s2 + n2

F (s) = L{sin t} =

1s2 + 1

,

a = n.

5.6.7 Division by t

Provided that the limit limt→0

f(t)t

exists

L{f(t)t

}=

∫ ∞

s

F (s) ds


Example

L{

sin t

t

}=

∫ ∞

s

ds

s2 + 1= tan−1 s

∣∣∣∣∞

s

=π

2− tan−1 s

=π

2−

[π

2− tan−1

(1s

) ]= tan−1

(1s

)

F (s) = L{sin t} =1

s2 + 1,

limt→0

sin t

t= 1,

tan−1 x =π

2− tan−1

(1x

)from CRC p.171.

5.6.8 Multiplication by tn

L{tnf(t)

}= (−1)n dn

dsnF (s) = (−1)nF (n)(s)

Example

L{e2t} =1

s− 2,

L{te2t} = − d

ds

(1

s− 2

)=

1(s− 2)2

L{t2e2t} =d2

ds2

(1

s− 2

)=

d

ds

[ −1(s− 2)2

]=

2(s− 2)3

5.6.9 Periodic Functions

Let f(t) have a period p > 0 so that f(t + p) = f(t) then

L{f(t)} =1

1− e−ps

∫ p

0

e−stf(t) dt.


f(t)

t

k

-k

a 2a 3a

Figure 5.5: Illustration of a periodic function with period p = 2a.

Example Obtain the Laplace transform of the function shown in Figure 5.5.

L{f(t)} =1

1− e−2as

{∫ a

0

e−st(k) dt +∫ 2a

a

e−st(−k) dt

}

=1

1− e−2as

{− k

se−st

∣∣∣∣a

0

+k

se−st

∣∣∣∣2a

a

}

=k/s

(1− e−as)(1 + e−as)

{e−2as − 2e−as + 1︸︷︷︸

(1−e−as)2

}

=k

s

[1− e−as

1 + e−as

]

5.6.10 Behavior of F (s) as s→∞

If L{f(t)} = F (s),

then lims→∞

F (s) = 0.

Final value : lims→0

sF (s) = limt→∞

f(t)

Initial value : lims→∞

sF (s) = limt→0

f(t)


Example

f(t) = at

L{at} = F (s) =a

s2

=⇒ lims→∞

F (s) = 0

5.7 Methods of Finding Laplace Transforms

(a) Direct method (use of definition)(b) Series method: If f(t) has a power series expansion given by

f(t) = a0 + a1t + a2t2 + · · · =

∞∑n=0

antn,

its Laplace transform can be obtained by taking the sum of the Laplacetransforms of each term in the series. Thus,

L{f(t)} =a0

s+

a1

s2+

2!a2

s3+ · · · =

∞∑n=0

n!an

sn+1.

(c) Use of tables(d) Miscellaneous

5.8 The Inverse Laplace Transform

If the Laplace transform of f(t) is F (s) or

L{f(t)} = F (s),

then f(t) is the inverse Laplace transform of F (s) or

f(t) = L−1{F (s)},

where L−1 is called the inverse Laplace transformation operator.


Uniqueness of Inverse Laplace Transforms

Theorem: If we restrict ourselves to functions f(t) which are sectionallycontinuous in every finite interval 0 ≤ t ≤ N and of exponential order, thenthe inverse Laplace transform of F (s), i.e., L−1{F (s)} = f(t) is unique.

5.9 Some Important Properties of Inverse Laplace Transforms

5.9.1 Linearity

L−1{c1F1(s) + c2F2(s)} = c1L−1{F1(s)}+ c2L−1{F2(s)}= c1f1(t) + c2f2(t)

Example

L−1{ 4s− 2

− 3ss2 + 16

+5

s2 + 4

}=

= 4L−1{ 1s− 2

}− 3L−1

{ s

s2 + 16

}+ 5L−1

{ 1s2 + 4

}= 4e2t − 3 cos 4t +

52

sin 2t {Since L{sin at} =

a

s2 + a2

}

5.9.2 First Translation or Shifting Property

If L−1{F (s)} = f(t), then L−1{F (s− a)} = eatf(t).

Example

L−1

{1

s2 − 2s + 5

}= L−1

{1

(s− 1)2 + 4

}=

et

2sin 2t {

Since L−1{ 1s2 + 4

}=

12

sin 2t}


5.9.3 Second Translation or Shifting Property

L−1{e−asF (s)} = f(t− a) u(t− a),

where u(t− a) ={ 0 t < a,

1 t > a.

Example

L−1

{e

−πs3

s2 + 1

}= sin

(t− π

3

)u(t− π

3

)=

sin (t− π3 ) t > π

3 ,

0 t < π3 .

{Since L−1

{ 1s2 + 1

}= sin t

}

5.9.4 Change of Scale Property

L−1{F (ks)} =1kf

(t

k

)

Example

L−1

{2s

4s2 + 16

}= L−1

{2s

(2s)2 + 16

}=

12

cos(

4t2

)=

cos 2t2

{Since L−1

{ s

s2 + 16

}= cos 4t

}

5.9.5 Inverse Laplace Transform of Derivatives

L−1{F (n)(s)} = L−1

{dn

dsnF (s)

}= (−1)ntnf(t)

Example

L−1

{ −2s(s2 + 1)2

}= L−1

{d

ds

(1

s2 + 1

)}= (−1)t f(t) = −t sin t


Since L−1{ 1s2 + 1

}= sin t = f(t),

andd

ds

( 1s2 + 1

)=

−2s(s2 + 1)2

.

5.9.6 Inverse Laplace Transform of Integrals

L−1

{ ∫ ∞

s

F (s) ds

}=

f(t)t

Example

L−1

{∫ ∞

s

(1s− 1

s + 1

)ds

}=

f(t)t

=1− e−t

t

{Since L−1

{1s− 1

s + 1

}= 1− e−t

}

5.9.7 Multiplication by s

If L−1{F (s)} = f(t), and f(0) = 0, then

L−1{sF (s)} = f ′(t).

If f(0) = 0, then

L−1{sF (s)− f(0)} = f ′(t).

Thus multiplication by s has the effect of differentiating f(t).

Example

L−1

{s

s2 + 1

}=

d

dtsin t︸︷︷︸

f ′(t)

= cos t

Since L−1

{ 1s2 + 1

}= sin t,

and sin 0 = 0.


5.9.8 Division by s

L−1

{F (s)s

}=

∫ t

0

f(τ) dτ

Thus division by s (or multiplication by 1/s) has the effect of intergratingf(t) from 0 to t.

Example

L−1

{1

s(s2 + 4)

}=

∫ t

0

(sin 2τ

2

)︸︷︷︸

f(τ)

dτ = −cos 2τ4

∣∣∣∣t

0

=1− cos 2t

4

{Since L−1

{1

s2 + 4

}=

sin 2t2

}

5.9.9 Convolution Theorem

The convolution of f(t) and g(t) is written as

f(t) ∗ g(t) =∫ t

0

f(τ)g(t− τ) dτ

Properties:

f(t) ∗ g(t) = g(t) ∗ f(t)f(t) ∗ [g1(t) + g2(t)] = f(t) ∗ g1(t) + f(t) ∗ g2(t)[f(t) ∗ g(t)] ∗ u(t) = f(t) ∗ [g(t) ∗ u(t)]f(t) ∗ 0 = 0

L−1{F (s)G(s)} = f(t) ∗ g(t) =∫ t

0

f(τ)g(t− τ) dτ

Example 1

L−1

{1

(s− 1)(s− 2)

}= et ∗ e2t =

∫ t

0

eτe2(t−τ) dτ =∫ t

0

e2te−τ dτ

= −e2t−τ

∣∣∣∣t

0

= −et + e2t


Since L−1{ 1s− 1

}= et,

and L−1{ 1s− 2

}= e2t.

Example 2

L−1

{1

(s2 + a2)2

}=

(sin at

a

)∗

(sin at

a

)

=∫ t

0

(sin aτ

a

)(sin a(t− τ)

a

)dτ

=1a2

∫ t

0

sin aτ sin a(t− τ) dτ

=1

2a2

∫ t

0

[cos a(2τ − t)− cos at

]dτ

=1

2a2

[sin a(2τ − t)

2a− τ cos at

]t

0

=1

2a2

[sin at

2a− sin(−at)

2a− t cos at + 0

]

=1

2a2

[sin at

a− t cos at

]

=1

2a3

[sin at− at cos at

]

{Since L−1

{1

s2 + a2

}=

sin at

a

}

5.10 Methods of Finding Inverse Laplace Transforms

(a) Partial fractions method(b) Series method(c) Miscellaneous methods using the L−1{} theorems(d) Use of tables(e) The complex inversion formula


5.10.1 Partial Fractions Method

Rules of Partial Fractions

(a) The numerator must be of lower degree than the denominator. If it isnot, first divide out.

(b) Factorize the denominator into its prime factors. These factors deter-mine the shapes of the partial fractions.

(c) A repeated factor (s + a)2 leads to the following partial fraction

A

s + a+

B

(s + a)2,

and a factor (s + a)3 leads to

A

s + a+

B

(s + a)2+

C

(s + a)3.

(d) A quadratic factor (s2 + ps + q) leads to the following partial fraction

As + B

s2 + ps + q,

and the factor (s2 + ps + q)2 leads to

As + B

s2 + ps + q+

Cs + D

(s2 + ps + q)2.

Example 1

L−1

{2s− 8

s2 − 8s + 15

}= L−1

{2s− 8

(s− 3)(s− 5)

}

= L−1

{1

s− 3

}+ L−1

{1

s− 5

}= e3t + e5t

2s− 8(s− 3)(s− 5)

=A

s− 3+

B

s− 5, [multiply by (s− 3)(s− 5)]

=⇒ 2s− 8 = A(s− 5) + B(s− 3)

= s(A + B)− 5A− 3B

=⇒ 2 = A + B and 8 = 5A + 3B

=⇒ A = 1 and B = 1


Example 2

L−1

{s2 − 2s + 3

(s− 2)3

}= L−1

{1

s− 2+

2(s− 2)2

+3

(s− 2)3

}

= e2t

[1 + 2t +

3t2

2

]

s2 − 2s + 3(s− 2)3

=A

s− 2+

B

(s− 2)2+

C

(s− 2)3,

[multiply by (s− 2)3

]=⇒ s2 − 2s + 3 = A(s− 2)2 + B(s− 2) + C

= As2 + s(B − 4A) + 4A− 2B + C

=⇒ A = 1, B − 4A = −2, and 4A− 2B + C = 3

=⇒ A = 1, B = 2, and C = 3

L−1

{1

s− 2

}= e2t,

L−1

{2

(s− 2)2

}= 2e2tt,

L−1

{3

(s− 2)3

}= 3e2t

(t2

2

),

' Translation or shifting : L−1{F (s− a)

}= eatf(t)

L−1

{1s2

}= t,

L−1

{1s3

}=

t2

2

Example 3

L−1

{s2 + 2s + 3

(s2 + 2s + 2)(s2 + 2s + 5)

}= L−1

{1/3

s2 + 2s + 2+

2/3s2 + 2s + 5

}

=13L−1

{1

(s + 1)2 + 1

}+

23L−1

{1

(s + 1)2 + 4

}

=13e−t sin t +

(23

)(12

)e−t sin 2t

=e−t

3(sin t + sin 2t

)


s2 + 2s + 3(s2 + 2s + 2)(s2 + 2s + 5)

=As + B

s2 + 2s + 2+

Cs + D

s2 + 2s + 5

=⇒ s2 + 2s + 3 = (As + B)(s2 + 2s + 5) + (Cs + D)(s2 + 2s + 2)

= (A + C)s3 + (2A + B + 2C + d)s2+

+ (5A + 2B + 2C + 2D)s + 5B + 2

=⇒ A + C = 0, 2A + B + 2C + D = 1,

5A + 2B + 2C + 2D = 2, and 5B + 2D = 3

=⇒ A = 0, B =13, C = 0, and D =

23

' First Translation : L−1{F (s− a)} = eatf(t),

L−1

{1

(s + 1)2 + 1

}= e−tL−1

{1

s2 + 1

}= e−t sin t,

L−1

{1

(s + 1)2 + 4

}= e−tL−1

{1

s2 + 22

}= e−t sin 2t

2

5.10.2 Series Method

If F (s) has a series expansion in inverse powers of s given by

F (s) =a0

s+

a1

s2+

a2

s3+

a3

s4+ · · ·

then

f(t) = a0 + a1t +a2t

2

2!+

a3t3

3!+ · · · .

Example

L−1

{e−1/s

s

}= 1− t +

t2

(2!)2− t3

(3!)2+ · · ·

= 1−(2t1/2

)2

22(1!)2+

(2t1/2

)4

24(2!)2−

(2t1/2

)6

26(3!)2+ · · ·

= J0

[2√t]


B

A

J

K

L

R

γ

γ+iτ

γ−iτ

0x

y

1se−1/s =

1s

{1− 1

s+

12!s2

− 13!s3

+ · · ·}

=1s− 1

s2+

12!s3

− 13!s4

+ · · · ,

L{tn

}=

n!sn+1

,

J0[x] = 1− x2

22(1!)2+

x4

24(2!)2−+.

5.10.3 The Complex Inversion Formula

If F (s) = L{f(t)}, then f(t) =1

2πi

∮c

estF (s) ds

Figure 5.6: The Bromwich contour composed by a line AB and the arcBJKLA of a circle of radius R with center at the origin 0.

5.11 Some Special Functions

5.11.1 The Error Function

The error function as well as the complimentary error function appearsfrequently in solutions to heat and mass transfer problems. The functionsare graphically presented in Figure 5.7 and their numerical values for a widerange of the argument x are tabulated in Table 5.1.


-1

0

1

2

-2 -1 0 1 2 3

er f (x)

er fc(x)

x

erf

(x)

&

erf

c(x)

Figure 5.7: Illustration of erf(x) and erfc(x) as a function of x.

erf(x) =2√π

∫ x

0

e−u2du,

erf(−x) = −erf(x),erf(0) = 0,

erf(∞) =2√π

∫ ∞

0

e−u2du =

2√π

√π

2= 1,

Table 5.1: Values of erf(x) and erfc(x) for various values of x.

x erf(x) erfc(x) x erf(x) erfc(x)

0.1 0.11246 0.88754 1.6 0.97635 0.023650.2 0.22270 0.77730 1.7 0.98379 0.016210.3 0.32863 0.67137 1.8 0.98909 0.010910.4 0.42839 0.57161 1.9 0.99279 0.007210.5 0.52050 0.47950 2 0 0.99532 0.004680.6 0.60386 0.39614 2.1 0.99702 0.002980.7 0.67780 0.32220 2.2 0.99814 0.001860.8 0.74210 0.25790 2.3 0.99886 0.001140.9 0.79691 0.20309 2.4 0.99931 0.000691 0 0.84270 0.15730 2.5 0.99959 0.000411.1 0.88021 0.11979 2.6 0.99976 0.000241.2 0.91031 0.08969 2.7 0.99987 0.000131.3 0.93401 0.06599 2.8 0.99992 0.000081.4 0.95229 0.04771 2.9 0.99996 0.000041.5 0.96611 0.03389 3 0 0.99998 0.00002


erf(ix) =2i√π

∫ x

0

eu2du,

erfc(x) = 1− erf(x) =2√π

∫ ∞

x

e−u2du,

L{erf(t)

}=

es2/4

serfc

(s

2

).

Power Series Representation

erf(x) =2√π

{x− x3

3+

x5

10− x7

42+

x9

216− · · ·

}=

2√π

∞∑k=0

(−1)kx2k+1

k!(2k + 1).

Asymptotic Expansion (x >> 0)

erf(x) � 1− e−x2

x√π

{1− 1

2x2+

1 · 322x4

− 1 · 3 · 523x6

+ · · ·+ (−1)n(2n)!n!22nx2n

}+ Rn,

where the remainder (error) term Rn is less than the magnitude of the firstterm to be neglected in the asymptotic expansion and of the same sign, forx >> 0.Integrals Expressible in Terms of erf(x)∫ ∞

0

e−t2 sin 2txt

dt =π

2erf(x),

∫ ∞

0

e−t2 sin 2tx dx =√πe−x2

2erf(x).

Derivatives of erf(x)

d

dx[erf(x)] =

2√πe−x2

,

d2

dx2[erf(x)] = − 4√

πxe−x2

,


d3

dx3[erf(x)] = − 1√

π

(8x2 − 4

)e−x2

,

d4

dx4[erf(x)] = − 8√

π

(3− 2x2

)xe−x2

.

Integrals of erfc(x)

inerfc(x) =∫ ∞

x

in−1erfc(t) dt (n = 0, 1, 2, . . .),

i−1erfc(x) =2√πe−x2

and i0erfc(x) = erfc(x),

i erfc(x) = − 1πe−x2 − x erfc(x),

i2erfc(x) = −14

[erfc(x)− 2xi erfc(x)] .

Integral and Power Series Representation of ierfc(x)

inerfc(x) =2√π

∫ ∞

x

(t− x)2

n!e−t2dt,

inerfc(x) =∞∑

k=0

(−1)kxk

2n−kk!Γ(

1 +n− k

2

) .

5.11.2 The Dirac Delta Function

The Dirac delta function shown in Figure 5.8 is a mathematical tool andcan only be defined at the limit as k → 0. The Dirac delta function isdenoted by

δ(t− a) =

{∞ t = a,

0 otherwise.

Some very important properties of the Dirac delta function are:

∫ ∞

0

δ(t− a) dt = 1


1/k

Area = 1

a a+k

∫ ∞

0

δ(t− a)f(t) dt = f(a)

L{δ(t− a)} =∫ ∞

0

e−stδ(t− a) dt = e−as

L{δ(t)} =∫ ∞

0

e−stδ(t) dt = 1

Figure 5.8: Schematic illustration of Dirac delta function, defined at thelimit k → 0.

5.12 Exercises

(1) Find the following Laplace transform

L{

cos√t√

t

} (Hint : L

{sin√t}

=√π

2s3/2exp

[− 1

4s

]).

(2) Find L{f(t)

}where

f(t) =

{ sin t 0 < t < π,

0 π < t < 2π.

(3) Evaluate each of the following

(a) L{e−3t cos 2t

}

(b) L{e−tt2

}


(c) L{

4e2t sin t}

(d) L{J0(t)

}

(e) L{J0(at)

}

(4) Evaluate each of the following by use of inverse Laplace transform the-orems

(a) L−1

{4s + 12

s2 + 8s + 16

}

(b) L−1

{3s + 7

s2 − 2s− 3

}

(c) L−1

{se−4πs/5

s2 + 25

}

(d) L−1

{s + 1

s2 + s + 1

}

(e) L−1

{4s− 2

(s2 + 1)2

}

(f) L−1

{1

(s2 + 1)3

}

(g) L−1

{3s + 1

(s− 1)(s2 + 1)

}


O.D.E. t-space

Algebraic eq. s-space

Solution s-space

Semianalytical solution

Solution t-space

L

L-1

Solve

Numerical inversion

Chapter 6

Application of Laplace Transforms in the Solution ofOrdinary Differential Equations

6.1 Methodology

The procedure for the solution of an ordinary differential equation byLaplace transform techniques is illustrated in Figure 6.1.

Figure 6.1: Outline of the Laplace transform procedure for the solutionof ordinary differential equations.

6.2 Ordinary Differential Equations with Constant Coefficients

Example 1

d2y

dt2− 3

dy

dt+ 2y = 2e−t, y(0) = 2,

dy(0)dt

= −1.


Take Laplace transforms of each term of the ODE with respect to timevariable t

[s2y(s)− sy(0)− y′(0)

]− 3[sy(s)− y(0)] + 2y(s) = 2

(1

s + 1

),

L{f ′(t)} = sF (s)− f(0),

L{f ′′(t)} = s2F (s)− sf(0)− f ′(0),

L{e−t

}=

1s + 1

.

employ the boundary conditions

[s2y(s)− 2s + 1]− 3[sy(s)− 2] + 2y(s) =2

s + 1

=⇒ y(s)[s2 − 3s + 2

]− 2s + 7 =

2s + 1

=⇒ y(s)[s2 − 3s + 2

]=

2s2 − 5s− 5s + 1

=⇒ y(s) =2s2 − 5s− 5

(s + 1)(s− 1)(s− 2)

y(t) = L−1{y(s)} = L−1

{1/3s + 1

+4

s− 1+−7/3s− 2

}

=13e−t + 4et − 7

3e2t �

2s2 − 5s− 5(s + 1)(s− 1)(s− 2)

=A

s + 1+

B

s− 1+

C

s− 2,

=⇒ 2s2 − 5s− 5 = A(s− 1)(s− 2) + B(s + 1)(s− 2)

+ C(s + 1)(s− 1)

= s2(A + B + C)− s(3A + B) + 2A− 2B − C,

=⇒ A + B + C = 2, 3A + B = 5, 2A− 2B − C = −5,

=⇒ A =13, B = 4, C = −7

3.


Example 2

d3y

dt3− 3

d2y

dt2+ 3

dy

dt− y = t2et, y(0) = 1, y′(0) = 0, y′′(0) = −2.

Take Laplace transforms of each term of the ODE with respect to timevariable t[

s3y(s)− s2y(0)− sy′(0)− y′′(0)]− 3

[s2y(s)− sy(0)− y′(0)

]+

+ 3[sy(s)− y(0)]− y(s) =2

(s− 1)3

L

{fn(t)

}= snF (s)− sn−1f(0)− sn−2f ′(0)− · · · − f (n−1)(0),

L{eatf(t)} = F (s− a) (First translation theorem),

L{t2} =2s3

.

=⇒ y(s)[s3 − 3s2 + 3s− 1︸︷︷︸

(s−1)3

]− s2 + 3s− 1 =

2(s− 1)3

=⇒ y(s) =s2 − 3s + 1

(s− 1)3+

2(s− 1)6

=s2 − 2s + 1− s

(s− 1)3+

2(s− 1)6

=(s− 1)2 − (s− 1)− 1

(s− 1)3+

2(s− 1)6

=1

s− 1− 1

(s− 1)2− 1

(s− 1)3+

2(s− 1)6

=⇒ y(t) = L−1

{1

s− 1− 1

(s− 1)2− 1

(s− 1)3+

2(s− 1)6

}

= et − tet − t2et

2+ 2

t5et

5!

= et

[1− t− t2

2+

t5

60

]�


6.3 Ordinary Differential Equations with Variable Coefficients

Example 1

td2y

dt2+

dy

dt+ 4ty = 0, y(0) = 3,

dy(0)dt

= 0.

Take Laplace transforms of each term of the ODE with respect to timevariable t

Recall the multiplication by tn theorem :

L{tnf(t)

}= (−1)n dn

dsnF (s).

=⇒ − d

ds

[s2y(s)− sy(0)− dy(0)

dt

]+

[sy(s)− y(0)

]− 4

d

ds

[y(s)

]= 0

=⇒ −2sy(s)− s2 dy(s)ds

+ 3 + sy(s)− 3− 4dy(s)ds

= 0

=⇒ (s2 + 4)dy(s)ds

+ sy(s) = 0

=⇒ dy(s)y(s)

+s ds

s2 + 4= 0 integration yields

=⇒ lny(s) +12ln(s2 + 4) = C

=⇒ y(s) =ec∗

√s2 + 4

=⇒ y(t) = L−1

{C∗

√s2 + 4

}= C∗J0(2t)

Laplace transform relationship No. 31, Kreyszig (1993, p.319 )

L{J0(at)

}=

1√s2 + a2

To determine C∗ note that:

Jo(0) = 1,y(0) = C∗J0(0) = 3 =⇒ C∗ = 3

=⇒ y(t) = 3J0(2t) �


tan

(x

)

−π π 2π

x0

0

5

- 5

Example 2

td2y

dt2+ 2

dy

dt+ ty = 0, y(0) = 1,

dy(0)dt

= C.

Take Laplace transforms of each term of the ODE with respect to timevariable t, and make use of the multiplication by tn theorem.

=⇒ − d

ds

[s2y(s)− sy(0)− dy(0)

dt

]+ 2

[sy(s)− y(0)

]− d

ds

[y(s)

]= 0

=⇒ −2sy(s)− s2 dy(s)ds

+ 1 + 0 + 2sy(s)− 2− dy(s)ds

= 0

=⇒ (s2 + 1)dy(s)ds

= −1

=⇒ dy(s)ds

= − 1s2 + 1

=⇒ y(s) = −tan−1s + A

From Figure 6.2 it is evident that as s→∞,

tan−1(s)∣∣∣s→∞

→ π

2.

Furthermore, recall that the behavior of F (s) as s→∞ is as follows,

lims→∞

F (s) = 0.

Figure 6.2: Plot of tan x as a function of x.


Therefore,A =

π

2.

=⇒ y(s) =π

2− tan−1 s = tan−1

(1s

)

=⇒ y(t) = L−1{y(s)

}= L−1

{tan−1

(1s

)}

=⇒ y(t) =sin t

t�

tan−1

(1x

)=

π

2− tan−1 x (CRC, p. 171),

From example presented in the Division by t section

recall the relationship : L{ sin t

t

}= tan−1

(1s

).

6.4 Simultaneous Solution of Ordinary Differential Equations

Example 1Solve the following system of ordinary differential equations:

dx

dt= 2x− 3y,

dy

dt= y − 2x,

subject to the conditions x(0) = 8, y(0) = 3.Taking Laplace transform with respect to time variable t yields

[sx(s)− x(0)] = 2x(s)− 3y(s)

[sy(s)− y(0)] = y(s)− 2x(s)

Apply the initial conditions{sx(s)− 8 = 2x(s)− 3y(s)

sy(s)− 3 = y(s)− 2x(s)

}=⇒

{(s− 2)x(s) + 3y(s) = 8

2x(s) + (s− 1)y(s) = 3

}


Employ Cramer’s rule to solve for x(s) and y(s)

x(s) =

∣∣∣∣ 8 33 (s− 1)

∣∣∣∣∣∣∣∣ (s− 2) 32 (s− 1)

∣∣∣∣=

8(s− 1)− 9(s− 2)(s− 1)− 6

=8s− 17

s2 − 3s− 4=

8s− 17(s + 1)(s− 4)

=5

s + 1+

3s− 4

y(s) =

∣∣∣∣ (s− 2) 82 3

∣∣∣∣(s + 1)(s− 4)

=3s− 22

(s + 1)(s− 4)=

5s + 1

− 2s− 4

Then,

x(t) = L−1

{5

s + 1+

3s− 4

}= 5e−t + 3e4t �

y(t) = L−1

{5

s + 1− 2

s− 4

}= 5e−t − 2e4t �

Example 2 (From Kreyszig [1993, p. 307])Solve the following system of ordinary differential equations governing thedisplacements of two masses on springs (see Figure 6.3):

d2y1

dt2= k(y2 − 2y1)

d2y2

dt2= k(y1 − 2y2)

subject toy1(0) = 1, y2(0) = 1,

dy1(0)dt

=√

3k,dy2(0)

dt= −√

3k,

where k is the spring modulus and y1, y2 are the displacements of themasses from their positions of static equilibrum.Taking Laplace transform with respect to time variable t yields

[s2y1(s)− sy1(0)− dy1(0)dt

] = ky2(s)− 2ky1(s)

[s2y2(s)− sy2(0)− dy2(0)dt

] = ky1(s)− 2ky2(s)


1

1= 1

2= 1

2

Figure 6.3: Schematic illustration of a mechanical system consisting oftwo masses on three springs.

Apply the initial conditions

s2y1(s)− s−√

3k = ky2(s)− 2ky1(s)

s2y2(s)− s +√

3k = ky1(s)− 2ky2(s)

=⇒(s2 + 2k)y1(s)− ky2(s) = s +

√3k

−ky1(s) + (s2 + 2k)y2(s) = s−√

3k

Employ Cramer’s rule to solve for y1(s) and y2(s)

y1(s) =

∣∣∣∣ s +√

3k −ks−√

3k (s2 + 2k)

∣∣∣∣∣∣∣∣ (s2 + 2k) −k−k (s2 + 2k)

∣∣∣∣=

s3 + 3sk + s2√

3k + k√

3k(s2 + 2k)2 − k2

=s(s2 + 3k) +

√3k(s2 + k)

(s2 + 3k)(s2 + k)=

s

s2 + k+

√3k

s2 + 3k

y2(s) =

∣∣∣∣ (s2 + 2k) s +√

3k−k s−

√3k

∣∣∣∣(s2 + 3k)(s2 + k)

=s3 + 3sk − s2

√3k − k

√3k

(s2 + 3k)2(s2 + k)

=s

s2 + k−√

3ks2 + 3k


o

Then,

y1(t) = L−1{ s

s2 + k

}+ L−1

{ √3k

s2 + 3k

}= cos

√kt + sin

√3kt �

y2(t) = L−1{y2(s)} = cos√kt− sin

√3kt �

6.5 Applications to Beams

Example 1The beam shown in Figure 6.4 is hinged at its ends x = 0, x = � and carriesa uniform load W0 per unit length. Find the deflection at any point.

Figure 6.4: Schematic illustration of a beam hinged at its ends carryinga uniform load per unit length.

The appropriate differential equation and boundary conditions are:

d4y

dx4=

W0

EI0 < x < �

y(0) = 0, y(�) = 0, y′′(0) = 0, y′′(�) = 0,

where E is the modulus of elasticity of material of the beam; I is the secondmoment of the cross–sectional area of the beam. Taking Laplace transformswith respect to x yields

s4y(s)− s3y(0)− s2y′(0)− sy′′(0)− y′′′(0) =W0

EI

(1s

)


Using the boundary conditions, and replacing the unknown conditions bythe following constants y′(0) = c1, y′′′(0) = c2 yields

y(s) =c1s2

+c2s4

+W0

EI

(1s5

)

y(x) = L−1{ c1s2

}+ L−1

{ c2s4

}+

W0

EIL−1

{1s5

}

= c1x +c23!

x3 +W0

EI

x4

4!

= c1x + c2x3

6+

W0

EI

x4

24{Note that : L{tn} =

n!sn+1

}

y(�) = 0 = c1� +c2�

3

6+

W0

EI

�4

24

y′′(�) = 0 = c2� +W0

EI

�2

2

=⇒c1 =

W0�3

24EI,

c2 = −W0�

2EI.

=⇒ y(x) =W0

24EI

[�3x− 2�x3 + x4

]�

Example 2The cantilever beam shown in Figure 6.5 is clamped at the end x = 0 andis free at the end x = �. Find the deflection for the case where the beamcarries a load per unit length given by

W (x) ={

W0 0 < x < �/2,0 �/2 < x < �.

The appropriate differential equation and boundary conditions are:

d4y

dx4=

W (x)EI

0 < x < �

y(0) = 0, y′(0) = 0, y′′(�) = 0, y′′′(�) = 0.

Using the definition of the Heaviside function

u(t− a) ={ 0 t < a

1 t > a,


o

/2

Figure 6.5: Schematic illustration of a cantilever beam with uniform loadcovering only the first half of the beam.

the load distribution can be expressed as

W (x) = W0

[u(x)− u

(x− �

2

)]

Taking Laplace transforms of each term of the governing ordinary differen-tial equation with respect to time variable t yields

s4y(s)− s3y(0)− s2y′(0)− sy′′(0)− y′′′(0) =W0

EI

[1− e−s�/2

s

]

L{f(t− a)u(t− a)} = e−asF (s),

L{

W0

EIu(x)

}=

W0

EI

1s,

L{

W0

EIu

(x− �

2

)}=

W0

EI

e−s�/2

s.

Using the boundary conditions, and replacing the unknown conditions bythe following constants y′′(0) = c1, y′′′(0) = c2 yields

y(s) =c1s3

+c2s4

+W0

EIs5

[1− e−s�/2

]

=⇒ y(x) = L−1{ c1s3

}+ L−1

{ c2s4

}+

W0

EIL−1

{1s5

}− W0

EIL−1

{e−s�/2

s5

}


=c12!

x2 +c23!

x3 +W0

4!EIx4 − W0

4!EI

(x− �

2

)4

u

(x− �

2

)

=⇒ y(x) =

c1x2

2+

c2x3

6+

W0x4

24EI0 < x <

l

2

c1x2

2+

c2x3

6+

W0x4

24EI− W0

24EI

(x− �

2

)4

x >�

2

Now use the conditions at x = � to obtain the constants c1 and c2

y′′(l) = 0 = c1 + c2� +W0�

2

2EI− W0

2EI

(�− �

2

)2

= c1 + c2� +3W0�

2

8EI

y′′′(l) = 0 = c2 +W0�

EI− W0

EI

(�− �

2

)= c2 +

W0�

2EI

=⇒ c1 =W0�

2

8EI, c2 = −W0�

2EI,

y(x) =

W0�2

16EIx2 − W0�

12EIx3 +

W0

24EIx4 0 < x <

�

2

W0�2

16EIx2 − W0�

12EIx3 +

W0

24EIx4 − W0

24EI

(x− �

2

)4�

2< x < �

6.6 Exercises

Use Laplace transforms to solve the following:

(1) y′′(t) + y(t) = 1 given y(0) = 1 and y′(0) = 0.(2) 2x′(t) + 3x(t) = e4t given x(0) = 5.(3) y′′(t) + y(t) = t given y(0) = 1 and y′(0) = −2.(4) x′(t)− 3x(t) = te2t given x(0) = 0.(5) y′′(t)− 3y′(t) + 2y(t) = 4e2t given y(0) = −3 and y′(0) = 5.(6) yIV (t) + 2y′′(t) + y(t) = sin t given y(0) = 1, y′(0) = −2, y′′(0) = 3,

and y′′′(0) = 0.(7) A uniform light horizontal beam PQ, of length 2c end supported at P ,

carries a load which decreases uniformly from w (newtons/m) at Pto zero at x = c. A point load F (newtons) occurs at a distance b


W

P Q

F

c

b

2c

from P . Find an expression for the displacement at any point in thebeam in terms of x.

d4y

dx4=

1EI

W (x)

where

W (x) =[w − w

cx]−

[w − w

cx]u(x− c) + Fδ(x− b)

u(x− c) =

{ 1 x > c,

0 x < c.

given

y(0) = 0, y′(0) = 0, y′′(2c) = 0, and y′′′(2c) = 0.

Figure 6.6: Schematic illustration of the loaded beam for problem 6.7.


y

z

i

jk

0,0,1)

(0,1,0)

( 1,0,0)

Chapter 7

Linear Algebra

7.1 Vectors

A vector quantity is distinquished from a scalar quantity by the fact thata scalar quantity possesses only magnitude, whereas, a vector quantitypossesses both magnitude and direction.

7.1.1 Laws of Vector Algebra (A, B and C are vectors, and m, nscalars)

A + B = B + A Commutative law for additionA + (B + C) = (A + B) + C Associative law for additionm(nA) = (mn)A = n(mA) Associative law for multiplication(m+ n)A = mA + nA Distributive lawm(A + B) = mA +mB Distributive law

7.1.2 Unit VectorsUnit vectors are vectors having unit length. The rectangular unit vectorsi, j and k, shown in Figure 7.1, are unit vectors having the direction of thepositive x, y and z axes of a rectangular (Cartesian) coordinate system.

Figure 7.1: Unit vectors for a Cartesian coordinate system.

7.1.3 Components of a Vector

Any vector in R3 can be represented by component vectors (i.e., A =a1i + a2j + a3k). The magnitude of A is |A| = (a21 + a22 + a23)

1/2.


A

B

P = 0

A

B

P > 0

θ

B

P < 0

θ

7.1.4 Dot or Scalar ProductThe dot product of two vectors A and B denoted A ·B (read A dot B) isdefined as the product of the magnitudes of A and B and the cosine of theangle (θ) between them:

A ·B = |A||B|cosθ, 0 ≤ θ ≤ π.

The projection, P , of A in the direction of B is given by

P = |A| cos θ =A ·B|B| .

Figure 7.2: Projection of A onto B for three different angles.

Note that A ·B is a scalar and the following laws are applicable:(a) A ·B = B ·A Commutative law for the dot product(b) A · (B + C) = A ·B + A ·C Distributive law(c) m(A ·B) = (mA) ·B = A · (mB) = (A ·B)m(d) i · i = j · j = k · k = 1, i · j = j · k = k · i = 0(e) If A = a1i + a2j + a3k and B = b1i + b2j + b3k, then

A ·B = a1b1 + a2b2 + a3b3A ·A = |A|2 = a21 + a22 + a23B ·B = |B|2 = b21 + b22 + b23

(f) If A ·B =0 and A and B are not null vectors, then A and B areperpendicular

7.1.5 General Vector Space

A set of vectors is called a real or linear vector space if the laws of vectoralgebra are defined and contains a zero or null vector,

A + 0 = A,

and for every vector A there is a vector −A such that

A + (−A) = 0.


ajk

row column

7.1.6 Gradient and Divergence

Consider the vector operator ∇ (del) defined by

∇ ≡ i∂

∂x+ j

∂

∂y+ k

∂

∂z.

Then, if φ = f(x, y, z) and A have continuous first partial derivatives in aregion, the following can be defined.Gradient: The gradient of φ is defined by

grad φ = ∇φ =(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)φ =

∂φ

∂xi +

∂φ

∂yj +

∂φ

∂zk.

Interpretation: if φ is the equation of a surface, then ∇φ is a normal to thissurface.Divergence: The divergence of A is defined by

divA = ∇·A =(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)·(a1i+a2j+a3k) =

∂a1∂x

+∂a2∂y

+∂a3∂z.

7.2 Matrices

7.2.1 Definition of a MatrixA matrix of order m× n is a rectangular array of numbers having m rowsand n columns. It can be written in the form

A =

a11 a12 · · · a1n

a21 a22 · · · a2n

· · · · · ·am1 am2 · · · amn

Each number ajk is called an element. The subscripts j and k indicaterespectively the row and column of the matrix in which the element appearsas shown in Figure 7.3.

Figure 7.3: The jth row and kth column element of a matrix.


• A matrix is often denoted by A or by the symbol (ajk) which shows arepresentative element.

• A matrix having one row is called a row matrix or row vector:

a = [ a1 a2 · · · an ] ,

while a matrix having only one column is called a column matrix orcolumn vector:

b =

b1b2...bm

.

• If the number of rows m and columns n are equal the matrix is called asquare matrix of order n× n or briefly n.

• A matrix is said to be real matrix or complex matrix according to whetherits elements are real or complex numbers, respectively.

7.2.2 Some Special Definitions and Operations Involving Matrices

Equality of Matrices

Two matrices A = (ajk) and B = (bjk) of the same order (i.e., equalnumber of rows and columns) are equal if and only if ajk = bjk.

Addition of Matrices

If A = (ajk) and B = (bjk) have the same order we define the sum of Aand B as A + B = (ajk + bjk).

Example

If A =[

2 1 4−3 0 2

], B =

[3 −5 12 1 3

], then

A + B =[

2 + 3 1− 5 4 + 1−3 + 2 0 + 1 2 + 3

]=

[5 −4 5−1 1 5

].

Subtraction of Matrices

If A = (ajk) and B = (bjk) have the same order, we define the differenceof A and B as A−B = (ajk − bjk).


n

m=A mC

n

p

B

p

Example

A−B =[

2 1 4−3 0 2

]−

[3 −5 12 1 3

]

=[

2− 3 1 + 5 4− 1−3− 2 0− 1 2− 3

]

=[−1 6 3−5 −1 −1

]

Multiplication of a Matrix by a Number

If A = (ajk) and λ is any number (or scalar) we define the product of Aby λ as λA = Aλ = (λajk).

Example

λA = 4[

2 1 4−3 0 2

]=

[8 4 16−12 0 8

]

Multiplication of Matrices

If A = (ajk) is an m × n matrix while B = (bjk) is an n × p matrix, thenwe define the product AB of A and B as the matrix C = (cjk) where

cjk =n∑

l=1

ajkblk

and C is of order m× p. Note that matrix multiplication is defined if andonly if the number of columns of A is the same as the number of rows of B,as shown in Figure 7.4. Such matrices are sometimes called conformable.

Figure 7.4: Schematic representation of matrix multiplication: AB=C.


Example

let A =[

2 1 4−3 0 2

], D =

3 5

2 −14 2

then,

AD =[

2 · 3 + 1 · 2 + 4 · 4 2 · 5 + 1 · (−1) + 4 · 2−3 · 3 + 0 · 2 + 2 · 4 −3 · 5 + 0 · (−1) + 2 · 2

]

=[

24 17−1 −11

].

Properties of Matrix Multiplication

(a) AB = BA in general(b) A(BC) = (AB)C(c) A(B + C) = AB + AC(d) (B + C)A = BA + CA(e) AB = 0 does not necessarily imply that A = 0 or B = 0(f) A matrix can be multiplied by itself if and only if it is a square matrix

A A = A2

A A2 = A3

A A3 = A4

(g) Idepotent matrixA2 = A

Transpose of a Matrix

If we interchange rows and columns of a matrix A, the resulting matrix iscalled the transpose of A and is denoted by AT . In symbols, if A = (ajk)then AT = (akj).

Example

The transpose of A =[

2 1 4−3 0 2

]is AT =

2 −3

1 04 2

.

Useful Relationships of the Transpose

(A + B)T = AT + BT

(AB)T = BT AT

(AT )T = A


Symmmetric and Skew–Symmetric Matrices

A square matrix A is called symmetric if AT = A and skew–symmetric ifAT = −A

Example

E =[

2 −4−4 3

]symmetric

F =[

0 −22 0

]skew−symmetric

Complex Conjugate of a Matrix

If all elements ajk of a matrix A are replaced by their complex conjugatesajk, the matrix obtained is called the complex conjugate of A and is denotedby A. Therefore,

A→ A

Hermitian and Skew–Hermitian Matrices

A = AT Hermitian −→ A =[

5 1− 8i1 + 8i 14

]

A = −AT skew−Hermitian −→ A =[

8i 4 + i−4 + i −i

]

Principal Diagonal and Trace of a Matrix

• If A = (ajk) is a square matrix, then the diagonal which contains allelements for which j = k is called the principal or main diagonal and thesum of all such elements is called the trace of A.• A matrix for which ajk = 0 when j = k is called a diagonal matrix

Examples 5 2 0

3 1 −2−1 4 2

trace = 5 + 1 + 2 = 8

5 0 0

0 1 00 0 2

diagonal matrix


Unit Matrix (or Identity Matrix)

A square matrix in which all elements of the principal diagonal are equal to1 while all other elements are zero is called the unit matrix and is denotedby

I =

1 0 0

0 1 00 0 1

.

• A couple important properties of I are the following

AI = IA = A,

In = I (n = 1, 2, 3, · · ·).

• The unit matrix plays a role in matrix algebra similar to that played bythe number one in ordinary algebra.

Zero or Null Matrix

A matrix whose elements are all equal to zero is called the zero or nullmatrix, and is denoted by 0. Given that A and 0 are of the same orderthen

A + 0 = 0 + A = A.

7.2.3 Gaussian Elimination (Triangular Factorization)

The Gauss elimination is a standard method for solving systems of linearequations

Ax = b.

ExampleConsider the following system:

2 1 1

4 1 0−2 2 1

x1

x2

x3

=

1−2

7

Elimination steps: (i) Substract 2 times the first equation from the second,(ii) substract (-1) times the first equation from the third,(iii) substract (-3) times the second equation from the third.


2 1 1

0 −1 −20 0 −4

︸︷︷︸upper triangular U

x1

x2

x3

=

1−4−4

︸︷︷︸c

−→ Ux = c.

Now obtain the vector x = [x1, x2, x3]T by back–substitution. The matricesthat accomplish steps (i), (ii) and (iii) are:

E21 =

1 0 0−2 1 0

0 0 1

, E31 =

1 0 0

0 1 01 0 1

, E32 =

1 0 0

0 1 00 3 1

E32E31E21 =

1 0 0−2 1 0−5 3 1

= L−1 (lower triangular)

E32E31E21A = U

E32E31E21b = c

L = E−121 E−1

31 E−132 =

1 0 0

2 1 0−1 −3 1

−→ A = LU.

7.2.4 Rank of a Matrix

The maximum number of linearly independent row vectors of a matrix Ais called the rank of A. The procedure for obtaining the rank of a matrix isto use Gaussian elimination to reduce the matrix A into an echelon form:

U =

∗ ∗ ∗ ∗ ∗ ∗0 ∗ ∗ ∗ ∗ ∗0 0 ∗ ∗ ∗ ∗0 0 0 0 0 ∗0 0 0 0 0 0

• Rank = Number of pivots

• A non zero diagonal entry of the matrix is called a pivot

• Row rank = Column rankIt should be noted that the previous matrix is a singular matrix becauseits rank is smaller than the number of rows. Furthermore, if the number


of pivots is less than the number of rows, then we have infinitely manysolutions, since some elements of x (Ax = b) remain arbitrary.

A system of linear equations is:Overdetermined: More equations than unknowns,Determined: The number of equations is equal to the unknowns,Underdetermined: Fewer equations than unknowns.Overdetermined and determined systems may or maynot have a solution,whereas underdetermined systems always have solutions.

7.2.5 Inverse of a Matrix

If for a given matrix A there exists a matrix B such that AB = I, then Bis called an inverse of A and is denoted by A−1.Existence: An inverse exists only when the rank is as large as possible.(Non singular or det A = 0).

Useful Formulas

(a) A−1 =adj Adet A

=[Aij ]T

det A,

adj A = adjugate A = [Aij ]T ,

Aij = (−1)i+jdet Mij ,

where Mij is the minor formed by deleting row i and column j of A, andAij is the cofactor.

Example

A =[a11 a12a21 a22

]

adj A =[A11 A21

A12 A22

]=

[a22 −a12−a21 a11

][a11 a21a12 a22

] [A11 A21

A12 A22

]︸︷︷︸

adjugate

=[

det A 00 det A

]

(b) The inverse of a diagonal matrix is easily obtained as follows:

A =

a11 0 · · · 00 a22 · · · 0...

.... . .

...0 0 · · · ann

, A−1 =

1/a11 0 · · · 00 1/a22 · · · 0...

.... . .

...0 0 · · · 1/ann


Properties of the Inverse

(AB)−1 = B−1A−1

(A−1)−1 = A

7.2.6 Gauss–Jordan Elimination

A =

2 1 1

4 1 0−2 2 1

[A|I] =

2 1 1 | 1 0 0

4 1 0 | 0 1 0−2 2 1 | 0 0 1

=

2 1 1 | 1 0 0

0 −1 −2 | −2 1 00 3 2 | 1 0 1

r2 − 2r1r3 + r1

=[U

∣∣L−1]

=

2 1 1 | 1 0 0

0 −1 −2 | −2 1 00 0 −4 | −5 3 1

r3 + 3r2

←− A = LU

=

1 1/2 1/2 | 1/2 0 0

0 1 2 | 2 −1 00 0 1 | 5/4 −3/4 −1/4

r1/2−r2−r3/4

=

1 0 −1/2 | −1/2 1/2 0

0 1 0 | −1/2 1/2 1/20 0 1 | 5/4 −3/4 −1/4

r1 − r2/2r2 − 2r3

=

1 0 0 | 1/8 1/8 −1/8

0 1 0 | −1/2 1/2 1/20 0 1 | 5/4 −3/4 −1/4

r1 + r3/2

=⇒ A−1 =

1/8 1/8 −1/8−1/2 1/2 1/2

5/4 −3/4 −1/4

�

AA−1 = A−1A = I (Check)

Inverse by LU Decomposition

A = LU =⇒ A−1 = (LU)−1 = U−1L−1


7.3 Determinants

A determinant of nth order is a scalar associated with an n× n matrix.

A =

2 −1 1 3−3 2 5 0

1 0 −2 24 −2 3 1

det A =n∑

k=1

ajkAjk

Cofactor (Ajk): If we multiply the determinant of the minor Mjk (of ajk)by (−1)j+k, the result is called the cofactor of ajk and is denoted by Ajk

Ajk = (−1)j+kdet Mjk.

Minor (Mjk): Given any element ajk of A we associate a new determinantof order (n− 1) obtained by removing all elements of the jth row and kthcolumn called the minor of ajk. For instance,

M23 =

2 −1 3

1 0 24 −2 1

.

ExampleEvaluate the following determinant

∣∣∣∣∣∣3 −2 21 2 −34 1 2

∣∣∣∣∣∣by (a) using elements of the first row (b) using elements of the second row.

(a) 3∣∣∣∣ 2 −31 2

∣∣∣∣− (−2)∣∣∣∣ 1 −34 2

∣∣∣∣ + 2∣∣∣∣ 1 24 1

∣∣∣∣ =

= (3)(7)− (−2)(14) + 2(−7) = 35 �


0,0)

(x 2,y2)

(x 1,y1)

(b) (−1)∣∣∣∣−2 2

1 2

∣∣∣∣ + 2∣∣∣∣ 3 24 2

∣∣∣∣− (−3)∣∣∣∣ 3 −24 1

∣∣∣∣ =

= (−1)(−6) + 2(−2)− (−3)(11) = 35 �

7.3.1 Theorems on Determinants

(a) det A = det AT

(b) An interchange of any two rows (or columns) changes the sign of thedeterminant.

(c) If all elements in any row (or column) are multiplied by a number, thedeterminant is also multiplied by this number.Caution: det(kA) = kndet A

(d) If A and B are square matrices of the same order, thendet(AB) = det A det B

(e) If all the entries in a row (or a column) of a determinant are zero, thevalue of the determinant is zero.

7.3.2 Cramer’s Rule (Revisit)

Determinants are also used for the solution of systems of n equations in nunknowns. For instance, assume the following system

x1 + 3x2 = 0,

2x1 + 4x2 = 6,

x1 =

∣∣∣∣ 0 36 4

∣∣∣∣∣∣∣∣ 1 32 4

∣∣∣∣=−18−2

= 9, x2 =

∣∣∣∣ 1 02 6

∣∣∣∣∣∣∣∣ 1 32 4

∣∣∣∣=

6−2

= −3.

The physical interpretation of det E, where E =[x1 y1x2 y2

], is the area of

the corresponding parallelogram as shown in Figure 7.5.

Figure 7.5: Schematic representation of det E.


7.3.3 Eigenvalues and Eigenvectors

Let A = (ajk) be an n × n matrix and x a column vector. The equationAx = λx where λ is a constant can be written as

a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...an1 an2 · · · ann

x1

x2...xn

= λ

x1

x2...xn

or =⇒

x1(a11 − λ) + x2a12 + · · ·+ xna1n = 0x1a21 + x2(a22 − λ) + · · ·+ xna2n = 0

......

. . ....

...x1an1 + x2an2 + · · ·+ xn(ann − λ) = 0

Clearly, x = 0 is a trivial solution. For non–trivial solutions

∣∣∣∣∣∣∣a11 − λ a12 · · · a1n

a21 a22 − λ · · · a2n

· · · · · ·an1 an2 · · · ann − λ

∣∣∣∣∣∣∣ = 0

=⇒ det(A− λI) = 0 (Characteristic equation)

which is a polynomial equation of degree n in λ. The roots of this poly-nomial are called eigenvalues of A. For each eigenvalue there will be asolution x = 0. The vector x is called eigenvector.

7.3.4 Application

Consider the case of vibrating masses on springs (see Figure 7.6)

y′′1 = a1y1 + a2y2y′′2 = b1y1 − b2y2

=⇒ y′′ = Ay =⇒[y′′1y′′2

]=

[a1 a2b1 −b2

] [y1y2

]

Assuming that y = xewt is a trial solution, then

=⇒ w2xewt = Axewt


0

y1

k1

k2

m1

m20

y2

Figure 7.6: Schematic illustration of two vibrating masses on strings.

=⇒ Ax = λx,

where λ = w2.

Therefore, the system examined reduces to an eigenvalue problem.

Example

Find the eigenvalues of the matrix

A =

5 7 −5

0 4 −12 8 −3

If x = [x1, x2, x3]T , we must consider the system of equations Ax = λx,i.e.,

5 7 −50 4 −12 8 −3

x1

x2

x3

= λ

x1

x2

x3

,

=⇒

5x1 +7x2 −5x3

4x2 −x3

2x1 +8x2 −3x3

=

λx1

λx2

λx3

.

Equating corresponding elements of these matrices, we find

(5− λ)x1 + 7x2 − 5x3 = 0(4− λ)x2 − x3 = 0

2x1 + 8x2 − (3 + λ)x3 = 0(∗)


This system will have non–trivial solutions if

∣∣∣∣∣∣5− λ 7 −5

0 4− λ −12 8 −3− λ

∣∣∣∣∣∣ = 0.

Expansion of this determinant yields

(5− λ)∣∣∣∣ 4− λ −1

8 −3− λ

∣∣∣∣ + 2∣∣∣∣ 7 −54− λ −1

∣∣∣∣ =

=(5− λ)[−(4− λ)(3 + λ) + 8] + 2[−7 + 5(4− λ)] =

=6− 11λ+ 6λ2 − λ3 =

=(λ− 1)(λ− 2)(λ− 3) = 0.

=⇒ λ = 1, 2, 3 (eigenvalues).

• Now determine the eigenvectors:

(a) Corresponding to λ = 1, equation (∗) becomes

4x1 +7x2 −5x3 = 03x2 − x3 = 0

2x1 +8x2 −4x3 = 0∣∣∣∣∣∣4 7 −50 3 −12 8 −4

∣∣∣∣∣∣ −→∣∣∣∣∣∣4 7 −50 3 −10 4.5 −1.5

∣∣∣∣∣∣ −→∣∣∣∣∣∣4 7 −50 3 −10 0 0

∣∣∣∣∣∣ .Solving for x1 and x3 in terms of x2 yields

x3 = 3x2, x1 = 2x2.

Then, one of the corresponding eigenvectors is

x1

x2

x3

=

2x2

x2

3x2

= x2

2

13

or simply

2

13

, �

because any scalar (constant) multiple of this eigenvector is also an eigen-vector.(b) Similarly, for λ = 2 equation (∗) becomes


3x1 +7x2 −5x3 = 02x2 − x3 = 0

2x1 +8x2 −5x3 = 0

Solving for x1 and x3 in terms of x2 yields

x1 = x2, x3 = 2x2.

Therefore, one of the corresponding eigenvectors is

x1

x2

x3

=

x2

x2

2x2

= x2

1

12

or simply

1

12

�

(c) Finally, for λ = 3, equation (∗) becomes

2x1 +7x2 −5x3 = 0x2 − x3 = 0

2x1 +8x2 −6x3 = 0


x3 = x2, x1 = −x2.

Thus, one of the corresponding eigenvectors is

x1

x2

x3

=

−x2

x2

x2

= x2

−1

11

or simply

−1

11

�

• Now determine a set of unit vectors.The unit eigenvectors have the property that they have length 1, i.e., thesum of the squares of their components=1. To obtain such eigenvectorswe divide each vector by the square root of the sum of the squares of thecomponents. Thus, the corresponding unit eigenvectors are as follows:

2/√

141/√

143/√

14

,

1/√

61/√

62/√

6

,

−1/

√3

1/√

31/√

3

.


7.3.5 Multiple Eigenvalues

Consider the matrix A which leads to multiple eigenvalues (Kreyszig, 1993,p.389)

A =

−2 2 −3

2 1 −6−1 −2 0

,

=⇒ det(A− λI) = (λ− 5)(λ+ 3)2 = 0,

=⇒ λ1 = 5, λ2 = λ3 = −3.

• Now determine the eigenvectors:

(a) Corresponding to λ1 = 5 the following set of equations is obtained

−7x1 +2x2 −3x3 = 02x1 −4x2 −6x3 = 0−x1 −2x2 −5x3 = 0


x1 = −x3, x2 = −2x3.

The corresponding eigenvector is−1−2

1

, �

(b) Corresponding to λ2 = λ3 = −3 the following set of equations is ob-tained

x1 +2x2 −3x3 = 02x1 +4x2 −6x3 = 0−x1 −2x2 +3x3 = 0∣∣∣∣∣∣

1 2 −32 4 −6−1 −2 3

∣∣∣∣∣∣ −→∣∣∣∣∣∣1 2 −30 0 00 0 0

∣∣∣∣∣∣ .A basis of solutions consists of two linearly independent vectors (Numberof vectors = Number of equations - Rank). It should be noted that a basisor fundamental system of solutions refers to linearly independent solutions.

Let x2 = 0, then x1 = 3x3 −→

3

01

, �


subsequently let x3 = 0, then x1 = −2x2 −→

−2

10

. �

7.3.6 Complex Eigenvalues

For the case of real matrices with complex eigenvalues and eigenvectors theprocedure is as follows (Strang, 1980, p. 193):

det(A− λI) =

∣∣∣∣∣∣−λ −1 0

1 −λ 00 0 1− λ

∣∣∣∣∣∣ = (1− λ)(λ2 + 1) = 0,

=⇒ λ1 = 1, λ2 = i, λ3 = −i.

λ1 = 1 :

∣∣∣∣∣∣−1 −1 0

1 −1 00 0 0

∣∣∣∣∣∣ , x1 = x2 = 0 −→

0

01

,

λ2 = i :

∣∣∣∣∣∣−i −1 01 −i 00 0 1− i

∣∣∣∣∣∣ , x2 = −ix1, x3 = 0 −→

1−i0

,

λ3 = −i :

∣∣∣∣∣∣i −1 01 i 00 0 1 + i

∣∣∣∣∣∣ , x2 = ix1, x3 = 0 −→

1i0

.

7.3.7 Reduction of Matrix to Diagonal Form

If a non–singular matrix A has distinct eigenvalues λ1, λ2, λ3, · · ·, with cor-responding eigenvectors written as columns in matrix B,

B =

b11 b12 · · ·b21 b22 · · ·...

.... . .

, then,

B−1AB =

λ1 0 0 · · ·0 λ2 0 · · ·0 0 λ3 · · ·...

......

. . .


i.e., B−1AB, called the transform of A by B, is a diagonal matrix contain-ing the eigenvalues of A in the main diagonal and zeros elsewhere.

Example

If the columns of B are the eigenvectors of A,

A =

5 7 −5

0 4 −12 8 −3

, where B =

2 1 −1

1 1 13 2 1

,

and the inverse of B is

B−1 =

−1 −3 2

2 5 −3−1 −1 1

,

then

B−1AB =

−1 −3 2

2 5 −3−1 −1 1

5 7 −5

0 4 −12 8 −3

2 1 −1

1 1 13 2 1

=

1 0 0

0 2 00 0 3

.

The diagonal elements are the eigenvalues of matrix A. These eigenvalueshave also been derived in the example presented in section 7.3.4.

7.3.8 Properties of Eigenvalues and Eigenvectors

(a) The eigenvalues of a matrix are unique but not the eigenvectors.(b) If A has n distinct eigenvalues, then it has n independent eigenvectors.

The converse is not always true.(c) The trace of A is equal to the sum of the diagonal elements and also

equal to the sum of the eigenvalues of A.(d) The determinant of A is equal to the product of its eigenvalues.


7.4 Exercises

(1) Consider a vector represented by an arrow running from a point P toa point Q (P −→ Q). The straight line through P and Q is calledthe line of action of the vector, the point P is called the origin of thevector and the point Q is called the terminus of the vector. (a) If aand b are vectors with a common origin O and terminuses A and B,in terms of a and b, find the vector OC where C is the middle point ofAB. (b) If a, b, c and d have a common origin and terminuses A, B,C, and D, and if b− a = c−d show that ABCD is a parallelogram.(c) Prove that the line joining the middle points of any two sides ofa triangle is parallel to the third side, and is equal in length to onehalf of the length of the third side. Note: You are not allowed to usea coordinate system to solve the above problems.

(2) Show that (A + B)2 = A2 + AB + BA + B2 given that

A =

2 1 −1

1 −2 3−2 1 2

, B =

1 −1 2−2 1 3

2 −1 1

.

(3) Use Gauss–Jordan elimination to find the inverse of

A =

1 0 2

2 −1 34 1 8

.

(4) Evaluate the following determinant∣∣∣∣∣∣∣

2 1 −1 4−2 3 2 −5

1 −2 −3 2−4 −3 2 −2

∣∣∣∣∣∣∣ .

(5) Prove that if A is a non–singular matrix, then

det A−1 =1

det A

(6) Use LU decomposition to find the inverse of

A =

2 1 1

4 −1 3−2 4 0

.


(7) Find the eigenvalues and eigenvectors of

A =

5 1 0

0 4 10 1 4

.

(8) Find the eigenvalues and eigenvectors of

A =[

cos θ − sin θsin θ cos θ

].

(9) Use properties of the transpose to show that

A =[


]

is an orthogonal matrix.


Chapter 8

Fourier Series, Fourier Integrals, Fourier Transforms

8.1 Periodic Functions

A periodic function, as illustrated in Figure 8.1, is any function for which

f(t) = f(t + p), ∀ t.

The smallest constant p that satisfies this function is called the period ofthe function. Similarly, we can state that

f(t) = f(t + np), n = 0,±1,±2, . . . .

Figure 8.1: Plot of the periodic function f(t) = 2.5 + sin t + cos 2t, withperiod 2π.

Example

Find the period of the following function

f(t) = cos(t/3) + cos(t/4).

Employing the definition of a periodic function and the trigonometric iden-tity

cos(x + y) = cosx cos y − sinx sin y

=⇒ cos(θ + 2πm) = cos θ (m = integer)


we can write

cos(t + p

3

)+ cos

(t + p

4

)= cos

t

3+ cos

t

4,

=⇒ p

3= 2πm, and

p

4= 2πn (m,n integers)

=⇒ p = 6πm = 8πn

Therefore, as also shown in Figure 8.2, the smallest value of p is obtainedwhen m = 4 and n = 3. Thus,

p = 24π �

Figure 8.2: Plot of the periodic function f(t) = cos (t/3) + cos (t/4),with period 24π.

8.2 Properties of Periodic Functions

For a periodic function f(t) = f(t + p), the following properties are appli-cable: ∫ β

α

f(t) dt =∫ β+p

α+p

f(t) dt,

∫ p

0

f(t) dt =∫ α+p

α

f(t) dt.


8.3 Fourier Series

Let the function f(t) be periodic with period 2π. This function can thenbe represented by the trigonometric series:

f(t) = a0 + a1 cos t + b1 sin t + a2 cos 2t + b2 sin 2t + · · ·

=⇒ f(t) = a0 +∞∑

n=1

(an cos nt + bn sin nt) ←− Fourier series

where a0, an and bn are the Fourier coefficients.

8.3.1 Orthogonality Relations for Sine and Cosine Functions

Assuming that ωo = 2π/p the following orthogonality properties can bederived: ∫ p

2

− p2

cos(mωot) dt = 0 for m �= 0(1)

∫ p2

− p2

sin(mωot) dt = 0 ∀ m(2)

∫ p2

− p2

cos(mωot) cos(nωot) dt =

0 m �= n

p/2 m = n �= 0(3)

∫ p2

− p2

sin(mωot) sin(nωot) dt =

0 m �= n

p/2 m = n �= 0(4)

∫ p2

− p2

sin(mωot) cos(nωot) dt = 0 ∀ m, n(5)

Therefore, 1, cos(nωot), sin(nωot) form an orthogonal set of functions onthe interval −p/2 < t < p/2.

Example

Verify the orthogonality relation (3). In view of the trigonometric identity

cos a cos b =12

[cos(a + b) + cos(a− b)] , and

ωot∣∣t=± p

2=

2πp

(±p

2

)= ±π, where ωo =

2πp

we have


∫ p2

− p2

cos(mωot) cos(nωot) dt =

=12

∫ p2

− p2

{cos[(m + n)ωot] + cos[(m− n)ωot]

}dt

=sin[(m + n)ωot]

2(m + n)ωo

∣∣∣∣∣p2

− p2

+sin[(m− n)ωot]

2(m− n)ωo

∣∣∣∣∣p2

− p2

=sin[(m + n)π] + sin[(m + n)π]

2(m + n)ωo+

sin[(m− n)π] + sin[(m− n)π]2(m− n)ωo

=0 if m �= n (because sin kπ = 0 ∀ k). �For m = n �= 0, by using the following trigonometric identity,

cos2 a =12(1 + cos 2a), we have

∫ p2

− p2

cos(mωot) cos(nωot) dt =∫ p

2

− p2

cos2(mωot) dt

=12

∫ p2

− p2

[1 + cos(2mωot)] dt

=t

2

∣∣∣∣p2

− p2

+sin(2mωot)

4mωo

∣∣∣∣p2

− p2

=p

2+

sin(2mπ)2mωo

=p

2�

8.3.2 Evaluation of Fourier CoefficientsUsing the sine and (or) cosine orthogonality relationship, we can evaluatethe coefficients of the Fourier series

(∗) f(t) = a0 +∞∑

n=1

(an cosnt + bn sinnt)

with period 2π, to be

a0 =12π

∫ π

−π

f(t) dt,(a)


an =1π

∫ π

−π

f(t) cosnt dt (n = 1, 2, · · ·),(b)

bn =1π

∫ π

−π

f(t) sinnt dt (n = 1, 2, · · ·).(c)

Proof

(a) If we integrate the Fourier series (∗) over (−π, π) and use the ortogo-nality relationships (1) and (2), we obtain

∫ π

−π

f(t) dt =∫ π

−π

a0 dt +∫ π

−π

[ ∞∑n=1


]dt

= 2a0π +∞∑

n=1

∫ π

−π

an cosnt dt︸︷︷︸0 (orth.#1)

+∞∑

n=1

∫ π

−π

bn sinnt dt︸︷︷︸0 (orth.#2)

= 2a0π

=⇒ a0 =12π

∫ π

−π

f(t) dt �

(b) Multiplying both sides of (∗) by cosmt and integrating over (−π, π),we get

∫ π

−π

f(t) cosmtdt = a0

∫ π

−π

cosmt dt︸︷︷︸0 (#1)

+∞∑

n=1

an

∫ π

−π

cosnt cosmt dt

︸︷︷︸anπ for m=n (#3)

+∞∑

n=1

bn

∫ π

−π

sinnt cosmt dt︸︷︷︸0 (#5)

= anπ

=⇒ an =1π

∫ π

−π

f(t) cosmt dt �

(c) Multiplying both sides of (∗) by sinmt and intergrating over (−π, π),


f(t)

t

1

0 π−2π − π 3π2π

we get

∫ π

−π

f(t) sinmt dt = a0

∫ π

−π

sinmt dt︸︷︷︸0 (#2)

+∞∑

n=1

an

∫ π

−π

cosnt sinmt dt︸︷︷︸0 (#5)

+∞∑

n=1

bn

∫ π

−π

sinnt sinmt dt

︸︷︷︸bnπ (#4)

= bnπ

=⇒ bn =1π

∫ π

−π

f(t) sinmt dt �

Example

Find the Fourier series for the following periodic function (f(t+2π) = f(t))

f(t) =

0 − π < t < 0

t

π0 < t < π

Figure 8.3: Plot of the periodic function f(t).

As shown in Figure 8.3 the period of f(t) is: p = 2π. Then, the Fouriercoefficients are evaluated as follows

a0 =12π

∫ π

−π

f(t) dt =12π

∫ π

0

t

πdt =

12π

[t2

2π

]π

0

=14


an =1π

∫ π

−π

f(t) cos nt dt =1π

∫ π

0

t

πcos nt dt

=1π2

[t

nsin nt

∣∣∣π0︸︷︷︸

0

− 1n

∫ π

0

sin nt dt

]=

cos nt

n2π2

∣∣∣∣π

0

=cos nπ − 1

n2π2=

(−1)n − 1n2π2

=

0, n even

−2π2n2

, n odd

bn =1π

∫ π

−π

f(t) sin nt dt =1π

∫ π

0

t

πsin nt dt

=1π2

[− t

ncos nt

∣∣∣π0

+1n

∫ π

0

cos nt︸︷︷︸0

dt

]

=1π2

[−π

ncos nπ + 0

]= − (−1)n

nπ

f(t) = a0 +∞∑

n=1


=14− 2

π2

∞∑n=odd

cosntn2

− 1π

∞∑n=1

(−1)n sin t

n

=⇒ f(t) =14− 2

π2

(cos t +

19

cos 3t +125

cos 5t + · · ·)

− 1π

(− sin t +

sin 2t2− sin 3t

3+ · · ·

)�

8.3.3 Functions of any Period p = 2L

Periodic functions rarely have period 2π, but often have some other periodp = 2L. If such a function f(t) can be expressed as a Fourier series, it canbe represented as:

f(t) = a0 +∞∑

n=1

[an cos

(nπt

L

)+ bn sin

(nπt

L

)]

where the Fourier coefficients are given by

a0 =1

2L

∫ L

−L

f(t) dt,(a)


0t

g(t)

0t

f(t)

(a) (b)

an =1L

∫ L

−L

f(t) cos(nπt

L

)dt,(b)

bn =1L

∫ L

−L

f(t) sin(nπt

L

)dt.(c)

8.3.4 Even and Odd FunctionsUnnecessary work in determining Fourier coefficients of a function can beavoided if the function is odd or even. A function is even when it satisfiesthe condition f(−t) = f(t), and it is odd if g(−t) = −g(t)

Figure 8.4: Illustration of (a) an even function, and (b) an odd function.

Even function: symmetric about the vertical axis at the origin∫ L

−L

f(t)dt = 2∫ L

0

f(t) dt.

Odd function: antisymmetric about the vertical axis at the origin∫ L

−L

g(t) dt = 0.

8.3.5 Fourier Series of Even and Odd Functions of Period 2L

Even Function: Fourier Cosine Series

f(t) = a0 +∞∑

n=1

an cos(nπt

L

),


f(t)

-T 0 T 2T

A

f(t)-A/2=g(t)

t-2T

(b)

A/2

t

-A/2

(a)

2TT0-T-2T

with coefficients

a0 =1L

∫ L

0

f(t) dt,

an =2L

∫ L

0

f(t) cos(nπt

L

)dt, (n = 1, 2, · · ·).

Odd Function: Fourier Sine Series

f(t) =∞∑

n=1

bn sin(nπt

L

),

with coefficients

bn =2L

∫ L

0

f(t) sin(nπt

L

)dt.

8.3.6 Hidden Symmetry

Often the symmetry of a periodic function is obscured by a constant term,as shown in Figure 8.5, and the symmetry can be revealed by shifting thefunction accordingly.

Figure 8.5: A periodic function with (a) hidden symmetry, and (b) thecorresponding function with odd symmetry.

Example

Find the Fourier series for the following periodic square wave function

f(t) =

{ 0 − 5 < t < 0,

3 0 < t < 5.


t

f(t)

-5

3

5 10 15t

f(t)-3/2=g(t)

-5

3/2

5 10 1500-3/2

(a) (b)

Figure 8.6: (a) The periodic function f(t), and (b) the correspondingfunction g(t) = f(t)− 3/2 with odd symmetry.

Clearly, the period of the function f(t) is p = 10, therefore, p = 2L =⇒L = 5. Shifting f(t) by -3/2 yields a periodic function with odd symmetry(g(−t) = −g(t)). Therefore, Fourier sine series can be employed

g(t) =∞∑

n=1

bn sin(nπt

L

)�

bn =2L

∫ L

0

g(t) sin(nπt

L

)dt =

25

∫ 5

0

32

sin(nπt

5

)dt

=35

[− 5

nπcos

(nπt

5

)]5

0

=3(1− cos nπ)

nπ=

3[1− (−1)n]nπ

=⇒ g(t) =6π

[sin

πt

5+ 0 +

13

sin3πt5

+ 0 +15

sin5πt5

+ 0 + · · ·]

also f(t) =32

+ g(t)

=⇒ f(t) =32

+6π

[sin

πt

5+

13

sin3πt5

+15

sin5πt5

+ · · ·]

�

8.3.7 Half–Range Expansions

In various physical and engineering problems there is practical need to useFourier series in connection with functions f(t) that are given merely onsome finite interval. Therefore, a nonperiodic function f(t) defined over acertain finite interval (0, τ) can be expanded into a Fourier series defined


τ

f(t)

(a) given

2ττ−τ−2τ

f(t)

(b) even symmetry cosine terms

2ττ−τ−2τ

f(t)

(c) odd symmetry sine terms

ttt

1

π/2 π

f(t)

t

(a) f(t)

t−π ππ/2−π/2 0

1

(b)

Figure 8.7: Schematic illustration of the half–range expansion procedure.

only in the interval (0, τ), as shown in Figure 8.7. It is important to letp = 2τ .

Example

Given the function

f(t) =

0 0 < t < π2 ,

1 π2 < t < π,

expand f(t) in a Fourier cosine series and draw the corresponding periodicextension of f(t).

Figure 8.8: (a) Plot of the function f(t) and (b) the corresponding evensymmetry half–range expansion.

The Fourier cosine series for f(t) is:

f(t) = a0 +∞∑

n=1

an cos(nπt

L

), L = π


a0 =1L

∫ L

0

f(t) dt =1π

∫ π/2

0

(0) dt +1π

∫ π

π/2

(1) dt =1π

[π − π

2

]=

12

an =2L

∫ L

0

f(t) cos(nπt

L

)dt =

2π

∫ π

π/2

cos nt dt

=2nπ

sinnt∣∣∣ππ/2

= − 2nπ

sin(nπ

2

)

=⇒ f(t) =12− 2

π

[cos t− 1

3cos 3t +

15

cos 5t− · · ·]

�

8.4 Fourier Integrals

Fourier integrals are extensively used in solving differential equations. Be-cause many practical problems do not involve periodic functions, it is de-sirable to generalize the method of Fourier series to include non periodicfunctions.

Outline of the Procedure

(1) consider a periodic function f(t) with period 2L.(2) let L→∞ to yield a nonperiodic function assuming that

f∗(t) = limL→∞

f(t)

is absolutely integrable on the t–axis, that is the following integralexists: ∫ ∞

−∞|f∗(t)| dt.

Example

Consider the following periodic function with period 2L

f(t) = e−|t|, −L < t < L,

f(t) = f(t + 2L),

sketched in Figure 8.9(a). By setting L→∞ yields a nonperiodic function,as shown in Figure 8.9(b).


-L 0 L

2L

f(t)

t

f*(t)

0

t

(b)(a)

Figure 8.9: (a) A periodic function and (b) the corresponding nonperiodicfunction resulting by taking the limit L→∞.

Derivation of the Fourier Integral

Recall the definition of Fourier series,

f(t) = a0 +∞∑

n=1

(an cosωnt + bn sinωnt), where ωn =nπ

L, p = 2L,

a0 =1

2L

∫ L

−L

f(t) dt,

an =1L

∫ L

−L

f(t) cosωnt dt,

bn =1L

∫ L

−L

f(t) sinωnt dt.

Substitute the expressions for the Fourier coefficients a0, an and bn into theFourier series relationship and let ν be a dummy integration variable. Theresulting expression is:

f(t) =1

2L

∫ L

−L

f(ν) dν +1L

∞∑n=1

[cosωnt

∫ L

−L

f(ν) cosωnν dν

+ sinωnt

∫ L

−L

f(ν) sinωnν dν

].

Now introduce

∆ω = ωn+1 − ωn =(n + 1)π

L− nπ

L=

π

L=⇒ 1

L=

∆ω

π,


=⇒ f(t) =1

2L

∫ L

−L

f(ν) dν︸︷︷︸0 at L→∞

+1L

∞∑n=1

[cosωnt ∆ω

∫ L

−L

f(ν) cosωnν dν

+ sinωnt ∆ω

∫ L

−L

f(ν) sinωnν dν

].

Now let

L→∞ =⇒ 1L→ 0 =⇒ ∆ω =

π

L→ 0,

∞∑n=1

g(ωn) ∆ω →∫ ∞

0

g(ω) dω. �

Therefore, the infinite series becomes an integral from 0 to ∞,

f∗(t) =∫ ∞

0

[A(ω) cosωt + B(ω) sinωt] dω ← Fourier integral �

A(ω) =1π

∫ ∞

−∞f∗(ν) cosων dν,

B(ω) =1π

∫ ∞

−∞f∗(ν) sinων dν,

where the superscript ∗ indicates nonperiodic function (L→∞).

Conditions for the Validity of f∗(t)

The function f∗(t) must be piecewise continuous in every finite interval andhas right–hand and left–hand derivatives at every point. At a point wheref(t) is discontinuous the value of the Fourier integral equals the average ofthe left– and right–hand limits of f(t) at that point.

8.4.1 Fourier Cosine and Sine Integrals

• If f(t) is an even function then B(ω) = 0, because

∫ L

−L

f(even) sin︸︷︷︸odd

= 0, and


1

-1 1

f(x)

x

f(t) =∫ ∞

0

A(ω) cosωt dω, ←− Fourier cosine integral

A(ω) =2π

∫ ∞

0

f(ν) cosων dν.

• If f(t) is an odd function then A(ω) = 0, because∫ L

−L

f(odd) cos︸︷︷︸odd

= 0, and

f(t) =∫ ∞

0

B(ω) sinωt dω, ←− Fourier sine integral

B(ω) =2π

∫ ∞

0

f(ν) sinων dν.

Example

Find the Fourier integral of the following function, which is also illustratedin Figure 8.10,

f(x) =

1 |x| < 1,

0 |x| > 1.

Figure 8.10: Graphical representation of the function f(t).

Given that f(x) is an even function, that is [f(−x) = f(x)], then B(ω) = 0and

A(ω) =2π

∫ ∞

0

f(ν) cosων dν =2π

∫ 1

0

(1) cosων dν

=2 sinων

ωπ

∣∣∣∣1

0

=2 sinω

ωπ,


f(x) =∫ ∞

0

A(ω) cosωx dω =2π

∫ ∞

0

sinω cosωx

ωdω. �

Use in Integral Evaluation

From the previous relationship it follows that

∫ ∞

0

sinω cosωx

ωdω =

π

2f(x) =

π

20 ≤ x < 1,

π

4x = 1,

0 x > 1,

�

recall that f(x) =

1 0 ≤ x < 1,

1 + 02

=12

x = 1, (note the discontinuity)

0 x > 1.

8.4.2 Fourier Transformations

Fourier Cosine Transformations

For an even function f(t) the following expressions are applicable

f(t) =∫ ∞

0

A(ω) cosωt dω, A(ω) =2π

∫ ∞

0

f(ν) cosων dν.

Conventionally, we let

A(ω) =

√2πfc(ω),

where the subscript c indicates cosine, in order to obtain the following pairof equations

fc(ω) =

√2π

∫ ∞

0

f(t) cosωt dt ←− Fourier cosine transform

f(t) =

√2π

∫ ∞

0

fc(ω) cosωt dω ←− Inverse Fourier cosine transform


Fourier Sine Transformations

For an odd function f(t) the following expressions are applicable

f(t) =∫ ∞

0

B(ω) sinωt dω, B(ω) =2π

∫ ∞

0

f(ν) sinων dν.

Conventionally, we let

B(ω) =

√2πfs(ω),

where the subscript s indicates sine, in order to obtain the following pairof equations

fs(ω) =

√2π

∫ ∞

0

f(t) sinωt dt ←− Fourier sine transform

f(t) =

√2π

∫ ∞

0

fs(ω) sinωt dω ←− Inverse Fourier sine transform

Other Notations

Fc[f(t)] = Fc(ω) = fc(ω), F−1c [Fc(ω)] = f(t),

Fs[f(t)] = Fs(ω) = fs(ω), F−1s [Fs(ω)] = f(t).

Example 1 (Kreyszig, 1993, p. 608)

Find the Fourier cosine and sine transforms of the function

f(x) ={k 0 < x < a,0 x > a.

fc(ω) =

√2π

∫ ∞

0

f(x) cosωx dx

=

√2π

∫ a

0

k cosωx dx = k

√2π

sinωx

ω

∣∣∣∣a

0

= k

√2π

sinωa

ω,

fs(ω) =

√2π

∫ ∞

0

f(x) sinωx dx

=

√2π

∫ a

0

k sinωx dx = −k√

2π

cosωx

ω

∣∣∣∣a

0

= −k√

2π

[cosωa− 1

ω

].


It should be noted that for f(x) = k is constant in the range 0 < x < ∞these transforms do not exist because (f(∞) �= 0) .

Linearity of Fourier Sine and Cosine Transforms

Fc {af(t) + bg(t)} = afc(ω) + bgc(ω),

Fs {af(t) + bg(t)} = afs(ω) + bgs(ω).

Sine and Cosine Fourier Transform of Derivatives

Let f(t) be continuous, f ′(t) piecewise continuous, and f(∞)→ 0.

Fc{f ′(t)} =

√2π

∫ ∞

0

f ′(t) cosωt dt

=

√2π

[f(t) cosωt

∣∣∣∣∞

0

+ ω

∫ ∞

0

f(t) sinωt dt

]

= −√

2πf(0) + ωFs{f(t)} �

Integration by Parts :

∫u dv = uv −

∫v du,

u = cosωt, du = −ω sinωt, dv = f ′(t) dt, v = f(t).

Fs{f ′(t)} =

√2π

∫ ∞

0

f ′(t) sinωt dt

=

√2π

[f(t) sinωt

∣∣∣∣∞

0

− ω

∫ ∞

0

f(t) cosωt dt

]

= −ωFc{f(t)} �

Second Derivatives

Following the same procedure as for the first derivatives but with f ′ insteadf yields

Fc{f ′′(t)} = −√

2πf ′(0) + ωFs{f ′(t)}

= −√

2πf ′(0) + ω [−ωFc{f(t)}]︸︷︷︸

Fs{f ′(t)}

= −√

2πf ′(0)− ω2Fc{f(t)} �


Similarly we obtain

Fs{f ′′(t)} =

√2πωf(0)− ω2Fs{f(t)} �

8.4.3 Fourier Transformation (Complex Form)

Recall the definition of the (real) Fourier integral

f(t) =∫ ∞

0

[A(ω) cosωt + B(ω) sinωt] dω,

A(ω) =1π

∫ ∞

−∞f(ν) cosων dν,

B(ω) =1π

∫ ∞

−∞f(ν) sinων dν.

Substituting the expressions for A(ω) and B(ω) into the Fourier integralyields

f(t) =1π

∫ ∞

0

∫ ∞

−∞f(ν) [cosων cosωt + sinων sinωt] dν dω

=1π

∫ ∞

0

∫ ∞

−∞f(ν) cos(ωt− ων) dν dω

=12π

∫ ∞

−∞

∫ ∞

−∞f(ν) cos(ωt− ων) dν dω

+j

2π

∫ ∞

−∞

∫ ∞

−∞f(ν) sin (ωt− ων) dν dω,

where j =√−1, and the following relationships involving the cosine (an

even function) were employed

cos(a− b) = cos a cos b + sin a sin b,

∫ ∞

0

cos (ξ) dξ =12

∫ ∞

−∞cos (ξ) dξ.


It should be noted that the added last term (complex term) in the previousexpression for f(t) does not change the equality, because

∫ ∞−∞ f(ν) sin(ωt−

ων) dν = G(ω) is an odd function of ω, and∫ ∞−∞ G(ω) dω = 0.

=⇒ f(t) =12π

∫ ∞

−∞

∫ ∞

−∞f(ν)[cos(ωt− ων) + j sin(ωt− ων)] dν dω

=12π

∫ ∞

−∞

∫ ∞

−∞f(ν)ej(ωt−ων) dν dω

=1√2π

∫ ∞

−∞

[1√2π

∫ ∞

−∞f(ν)e−jωνdν

]︸︷︷︸

f(ω)

ejωt dω,

where the Euler Formula was employed

ejt = cos t + j sin t.

Therefore, the Fourier Transform Pair is given by

f(ω) =1√2π

∫ ∞

−∞f(t)e−jωt dt ←− Fourier transform,

f(t) =1√2π

∫ ∞

−∞f(ω)ejωt dω ←− Inverse Fourier transform.

Other Commonly Used Notations

f(ω) =∫ ∞

−∞f(t)e−jωt dt,

f(t) =12π

∫ ∞

−∞f(ω)ejωt dω.

Compare with the definition of the Laplace transform

L{f(t)} = F (s) =∫ ∞

0

e−stf(t) dt,

f(t) =1

2πj

∮c

estF (s) ds.


8.5 Properties of Fourier Transformations

8.5.1 Linearity

If F{f1(t)} = f1(ω) and F{f2(t)} = f2(ω), then

F{af1(t) + bf2(t)} = af1(ω) + bf2(ω).

8.5.2 Time Shifting

F{f(t− to)} = f(ω)e−jωto

Proof

F{f(t− to)} =∫ ∞

−∞f(t− to)e−jωt dt, let t− to = x⇒ dt = dx

F{f(x)} =∫ ∞

−∞f(x)e−jw(to+x) dx

= e−jωto

∫ ∞

−∞f(x)e−jωx dx

= e−jωto f(ω) �

8.5.3 Frequency Shifting

F{f(t)ejωot

}= f(ω − ωo)

Proof

F{f(t)ejωot

}=

∫ ∞

−∞[f(t)ejωot]e−jωt dt

=∫ ∞

−∞f(t)e−j(ω−ωo)t dt

= f(ω − ωo) �

8.5.4 Scaling

F{f(at)} =1|a| f

(ω

a

)(a > 0 and real)


Proof

For a > 0:

F{f(at)} =∫ ∞

−∞f(at)e−jωt dt, let at = x, t =

x

a⇒ dt =

dx

a

=1a

∫ ∞

−∞f(x)e−j( ω

a )x dx

=1af

(ω

a

)(∗)

For a < 0:

F{f(−at)} =∫ ∞

−∞f(−at)e−jωt dt, let − at = x,

= −1a

∫ −∞

∞f(x)e−j( w

a )x dx

(Note the sign change of the integral limits due to a < 0)

=1a

∫ ∞

−∞f(x)e−j( w

−a )x dx

=1af

(ω

−a

)(∗∗)

In view of (∗) and (∗∗) we conclude that

F{f(at)} =1|a| f

(ω

a

)�

8.5.5 Time–Reversal

F{f(−t)} = f(−ω)

Proof

F{f(−t)} = F{f(at)} = f(ω

a

)= f(−ω) �

8.5.6 Modulation Theorem

F{f(t) cosωot} =12f(ω − ωo) +

12f(ω + ωo)


Proof

F{f(t) cosωot} = F{

12f(t)ejωot +

12f(t)e−jωot

}

=12F

{f(t)ejωot

}+

12F

{f(t)e−jωot

}=

12f(ω − ωo) +

12f(ω + ωo). �

where the frequency shifting property and the following trigonometric iden-tity were employed

cosωot =12

[ejωot + e−jωot

].

8.5.7 Additional Properties when f(t) is Real

If f(t) is real, letf(t) = fe(t)︸︷︷︸

even

+ fo(t)︸︷︷︸odd

and

f(ω) = R(ω)︸︷︷︸real

+ jX(w)︸︷︷︸imaginary

then

R(ω) =∫ ∞

−∞f(t) cosωt dt,(a)

X(ω) = −∫ ∞

−∞f(t) sinωt dt,(b)

R(ω) = R(−ω),(c)X(ω) = −X(−ω),(d)

f(−ω) = f†(ω) († indicates complex conjugation),(e)F{fe(t)} = R(ω),(f)F{fo(t)} = jX(ω).(g)

Proof

f(ω) =∫ ∞

−∞f(t)e−jωt dt

=∫ ∞

−∞f(t) cosωt dt− j

∫ ∞

−∞f(t) sinωt dt

= R(ω) + jX(ω). � (a&b)


R(−ω) =∫ ∞

−∞f(t) cos(−ωt)dt =

∫ ∞

−∞f(t) cosωt dt = R(ω). � (c)

X(−ω) = −∫ ∞

−∞f(t) sin(−ωt)dt =

∫ ∞

−∞f(t) sinωt dt = −X(ω). � (d)

Given that R(ω) is an even function and X(ω) is an odd function, then

f(−ω) = R(−ω) + jX(−ω) = R(ω)− jX(w) = f†(ω). � (e)

Given that any function can be expressed as the sum of an even and anodd component, then (Hsu, 1984, p.30)

f(t) = fe(t) + fo(t),

wherefe(t) =

12[f(t) + f(−t)],

fo(t) =12[f(t)− f(−t)].

It can be shown that

F{f(t)} = f(ω) = R(ω) + jX(ω),

F{f(−t)} = f(−ω) = R(ω)− jX(ω) = f†(ω),

then

F{fe(t)} =12f(ω) +

12f†(ω)︸︷︷︸f(−ω)

=12[R(ω) + jX(ω)] +

12[R(ω)− jX(ω)]

= R(ω). � (f)

F{fo(t)} =12f(ω)− 1

2f†(ω)

=12[R(ω) + jX(ω)]− 1

2[R(ω)− jX(ω)] = jX(ω). � (g)


8.5.8 Differentiation and Integration Theorems

F {f ′(t)} = jωF{f(t)} ←− Time domain differentiation(a)

F{f (n)(t)

}= (jω)nf(ω) = (jω)nF{f(t)}(b)

F{−jtf(t)} = f ′(ω) ←− Frequency domain differentiation(c)

if∫ ∞

−∞f(t)dt = f(0) = 0, F

{∫ t

−∞f(x) dx

}=

1jω

f(ω)(d)

Proof

F {f ′(t)} =∫ ∞

−∞f ′(t)e−jωt dt

= f(t)e−jωt∣∣∣∞−∞

+ jω

∫ ∞


= jω

∫ ∞

−∞f(t)e−jωt dt [f(±∞)→ 0]

= jωf(ω) � (a)

Recall f(ω) =∫ ∞


=⇒ df(ω)dω

=d

dω

∫ ∞


=∫ ∞

−∞f(t)

d

dω

(e−jωt

)dt

=∫ ∞

−∞[−jtf(t)]e−jωt dt

= F{−jtf(t)} � (c)

Consider the following function

φ(t) =∫ t

−∞f(x) dx, φ′(t) = f(t), F{φ(t)} = φ(ω)

=⇒ φ(ω) = F{∫ t

−∞f(x) dx

}


F{φ′(t)} = F{f(t)}, F {φ′(t)} = jωφ(ω) =⇒ F{f(t)} = jωφ(ω)

=⇒ φ(ω) =1jωF{f(t)} =

1jω

f(ω)

=⇒ F{∫ t

−∞f(x) dx

}=

1jω

f(ω) � (d)

8.5.9 Multidimensional Fourier TransformsThree–Dimenional Fourier transform pair

f(u, v, w) =∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞f(x, y, z)e−j(ux+vy+wz) dx dy dz,

f(x, y, z) =1

(2π)3

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞f(u, v, w)ej(ux+vy+wz) du dv dw.

8.5.10 Convolution

F{f(t) ∗ g(t)} = f(ω)g(ω), or

F−1{f(ω)g(ω)} = f(t) ∗ g(t).It should be noted that for the definition presented by Kreyszig [1993] theconvolution is defined as follows

F{f(t) ∗ g(t)} =√

2πf(ω)g(ω).

Proof

F{f(t) ∗ g(t)} =∫ ∞

−∞

[∫ ∞

−∞f(x)g(t− x) dx

]︸︷︷︸

f(t)∗g(t)

e−jωt dt

=∫ ∞

−∞f(x)

[∫ ∞

−∞g(t− x)e−jωt dt

]dx

{Time shifting : F{f(t− to)} = f(ω)e−jωto

}=

∫ ∞

−∞f(x)g(ω)e−jωx dx

=[∫ ∞

−∞f(x)e−jωx dx

]g(ω)

=[∫ ∞


]g(ω)

= f(ω) g(ω) �


f(t)

1

-T/4-T/2 T/4 T/2

-1

t0

8.5.11 Frequency Convolution

F−1{f(ω) ∗ g(ω)} = 2πf(t)g(t),

F{f(t)g(t)} =12π

f(ω) ∗ g(ω) =12π

∫ ∞

−∞f(y)g(ω − y) dy.

Proof

F−1{f(ω) ∗ g(ω)} = F−1

{∫ ∞

−∞f(y)g(ω − y)dy

}

=12π

∫ ∞

−∞

[∫ ∞

−∞f(y)g(ω − y) dy

]ejωt dω

{let ω − y = x

}=

12π

∫ ∞

−∞f(y)

[∫ ∞

−∞g(x)ej(x+y)t dx

]dy

{interchange order of integration

}= 2π

[ 12π

∫ ∞

−∞f(ω)ejωt dω

] [12π

∫ ∞

−∞g(ω)ejωt dω

]

= 2π[f(t)g(t)] �

8.6 Exercises

(1) Find the period of the function f(t) = (10 cos t)2.(2) Find the Fourier series for the function whose waveform is (Hint: P =

T [−T/2, T/2])

Figure 8.11: Illustration of the function described in Exercise (2).


1

f(t)

-T 0 Tt

1

0 Tt

f(t)

-T

(3) Find the Fourier series for the function (P = T )

Figure 8.12: Schematic illustration of the function for Exercise (3).

(4) Using the result of Exercise 3, find the Fourier series for the functionf(t) shown below

Figure 8.13: Schematic illustration of the desired function f(t).

(5) Given the function f(t) ={

0 0 < t < π2 ,

1 π2 < t < π,

expand f(t) in a Fourier

sine series and draw the corresponding periodic extension of f(t).

(6) Find the Fourier transform of p(t) ={

1 |t| < d2 ,

0 |t| > d2 .

(7) Find the Fourier transform of f(t) = e−a|t|, where a > 0.

(8) If∞∫0

f(x) cos axdx ={

1− a 0 ≤ a ≤ 1,0 a > 1, find f(x).

(9) Show that∞∫0

sin2 uu2 du = π

2 given that

∞∫0

f(x) cos ax dx ={ 1− a 0 ≤ a ≤ 1,

0 a > 1.

(10) If F{f(t)

}= f(ω) and f(ω) can be differentiated everywhere n times,

show that

F{tpf(t)

}=

1(−j)p

dpf(ω)dωp

for every p ≤ n.


Chapter 9

Partial Differential Equations

9.1 Classification of 2nd–Order Partial Differential Equations

The general linear partial differential equation of order two in two indepen-dent variables has the form:

A∂2u

∂x2+ B

∂2u

∂x∂y+ C

∂2u

∂y2+ D

∂u

∂x+ E

∂u

∂y+ Fu = G, (∗)

where A,B,C,D,E, F and G may depend on x and y but not on u.

• A second–order equation with independent variables x and y which doesnot have the form (*) is called nonlinear.• If G = 0, (*) is homogeneous.• If G �= 0, (*) is nonhomogeneous.

Because of the nature of the solutions of (*) the equation is often classifiedas:Elliptic: B2 − 4AC < 0,Parabolic: B2 − 4AC = 0,Hyperbolic: B2 − 4AC > 0.

Examples

(a)∂2φ

∂x2+

∂2φ

∂y2= 0,

A = 1,

B = 0, B2 − 4AC = −4 < 0 =⇒ Elliptic �C = 1,


(b)∂u

∂t= k

∂2u

∂x2,

A = k,

B = 0, B2 − 4AC = 0 =⇒ Parabolic �C = 0,

9.2 Method of Separating Variables (Product Method)

Procedure:(1) Assume that the solution for the dependent variable (or variables) existsas a product of functions, each of which is a function of only one of theindependent variables. For example,

u(x, y, t) = X(x) Y (y) T (t).

It should be noted that it is customary to use capital letters for the func-tions, and small letters for their arguments.

(2) Manipulate the equation to separate terms that depend upon each of thevariables from one another. This leads to a situation where the equationrequires that a function of one independent variable must be equal to afunction of another independent variable, for any arbitrary values of thetwo independent variables. The only way that this can be true is for thefunctions to be constants.(3) Solve the ordinary differential equation for the assumed functions X(x),Y (y), etc., in which the separation constant appears.

(4) Determine the separation constant (or eigenvalue) from the boundaryconditions.

ExampleSolve the following boundary–value problem

∂u

∂x= 4

∂u

∂y, (∗)

u(0, y) = 8e−3y. (∗∗)



u(x, y) = X(x) Y (y)

and substitute into (*) to yield

X ′(x) Y (y) = 4X(x) Y ′(y)

=⇒ X ′

4X=

Y ′

Y.

Since the left–hand side is a function of x alone, and the right–hand sidedepends upon y only, and x and y are independent variables that may haveany values, the preceding equation can only be true if each side is (thesame) constant.

=⇒ X ′

4X=

Y ′

Y= C,

X ′ = 4CX =⇒ X = Ae4Cx,

Y ′ = CY =⇒ Y = BeCy,

=⇒ u(x, y) =(Ae4Cx

) (BeCy

)= αeC(4x+y),

where α = AB. Using the boundary condition

u(0, y) = αeCy = 8e−3y −→ α = 8, C = −3,

=⇒ u(x, y) = 8e−3(4x+y)

=⇒ u(x, y) = 8e−12x−3y �

9.2.1 Vibration of a String (Kreyszig, 1993, p. 631)

The equation describing the small–amplitude motion of a taut string is thewave equation

α2uxx − utt = 0,

where α2 is a physical constant that depends on the string tension andmass per unit length. For the case illustrated in Figure 9.1 the appropriateboundary conditions are:


-1

-0.5

0

0.5

1

0

2.5 5

7.5 10

u ( x , t )

L

Figure 9.1: An illustration of a vibrating string with length L.

u(0, t) = 0,

u(L, t) = 0.


u(x, t) = X(x) T (t),

and substitute into the governing differential equation to obtain

α2X ′′ T −X T ′′ = 0

=⇒ X ′′

X=

T ′′

α2T= k,

X ′′ − kX = 0,k is still arbitrary.

T ′′ − kα2T = 0,

It should be noted that an artistic choice is to set k = −p2.

Satisfying the Boundary Conditions

u(0, t) = X(0) T (t) = 0 −→ X(0) = 0,

u(L, t) = X(L) T (t) = 0 −→ X(L) = 0.

Consider the case of k = 0:


X ′′ = 0 =⇒ X = C1x + C2,

X(0) = 0 =⇒ C2 = 0,

X(L) = C1L = 0 =⇒ C1 = 0,

=⇒ u(x, t) = 0.

Therefore, k is not a good choice since u(x, t) = 0.

Consider the Case of k = p2:

X ′′ − p2X = 0, r = ±p=⇒ X(x) = Aepx + Be−px.

From the boundary conditions we have

X(0) = 0 = A + B,

X(L) = 0 = AepL + Be−pL,

=⇒ A = B = 0

=⇒ X(x) = 0

=⇒ u(x, t) = 0.

Therefore, k = p2 is not a good choice.

Obviously, k = −p2 is the appropriate choice. Note that the problemdescribes traveling waves, so the solution is expected to be in terms of sinesand cosines. Starting with the X(x) differential equation we have

X ′′ + p2X = 0 =⇒ X(x) = A cos px + B sin px.

Employing the X boundary condition yields

X(0) = 0 = A cos(0) + B sin(0) =⇒ A = 0,

X(L) = 0 = B sin pL,

=⇒ sin pL = 0

=⇒ pL = nπ

=⇒ p =nπ

L(n = 1, 2, 3, · · ·)


-1.5

-1

-0.5

0

0.5

1

1.5

0

2.5 5

7.5 10

mode 1

2

3

L

Figure 9.2: The first three modes of a vibrating string with length L.

=⇒ Xn(x) = B sin(nπL

x)

infinitely many solutions.

Now solve the T (t) differential equation

T ′′ − kα2T = 0,

k = −p2 = −(nπL

)2

,

let λn =nαπ

L,

=⇒ T ′′ + λ2nT = 0

=⇒ T = Γn cosλnt + ∆n sinλnt

=⇒ un(x, t) = B sin(nπx

L

)[Γn cosλnt + ∆n sinλnt] .

Also, assume B = 1. It should be noted that un represents a harmonicmotion having the frequency λn

2π = αn2L cycles per unit time. This motion is

called the nth normal mode of the string. The nth normal mode has n− 1nodes, that is points of the string that do not move (see Figure 9.2). Notethat

sinnπx

L= 0 at x =

L

n,2Ln, · · · , n− 1

nL.


-1.5

-1

-0.5

0

0.5

1

1.5

0

2.5 5

7.5 10

0 L

x

Figure 9.3: Half–range expansion of f(x).

For example, for n = 3

sinnπx

L= 0 at x =

L

3,2L3.

Case 1

• Initial Conditions

u(x, 0) = f(x)∂u(x, 0)

∂t= g(x) ←− (interpretation : string velocity)

In general, a single solution un(x, t) will not satisfy the initial conditions.Now, since the fundamental equation is linear and homogeneous, the sumof infinitely many solutions un is also a solution. Therefore,

u(x, t) =∞∑

n=1

un(x, t) =∞∑

n=1

sin(nπx

L

)[Γn cosλnt + ∆n sinλnt] .

Employing the first initial condition

u(x, 0) =∞∑

n=1

Γn sin(nπx

L

)= f(x). (Fourier sine series)


Note that Γn must be chosen so that u(x, 0) becomes a half–range expansionof f(x), namely, the Fourier sine series of f(x);

Γn =2L

∫ L

0

f(x) sinnπx

Ldx, (Fourier coefficients)

∂u(x, 0)∂t

=

[ ∞∑n=1

sin(nπx

L

){−λnΓn sinλnt + ∆nλn cosλnt}

]t=0

=∞∑

n=1

∆nλn sinnπx

L= g(x),

=⇒ ∆nλn =2L

∫ L

0

g(x) sinnπx

Ldx, (half range expansion)

λn =αnπ

L,

=⇒ ∆n =2

αnπ

∫ L

0

g(x) sinnπx

Ldx. (∗)

Case 2

For the case of zero initial string velocity, g(x) = 0, then

∂u(x, 0)∂t

=∞∑

n=1

∆nλn sinnπx

L= 0 =⇒ ∆n = 0,

which can also be obtained from (*) with g(x) = 0.

=⇒ u(x, t) =∞∑

n=1

Γn cosλnt sinnπx

L,

λn =αnπ

L,

sinx cos y =12[sin(x + y) + sin(x− y)]


f(x) f(x-αt)

αt

x

Figure 9.4: Traveling wave.

=⇒ u(x, t) =12

∞∑n=1

Γn

{sin

[nπ(x + αt)

L

]+ sin

[nπ(x− αt)

L

]},

recall that

f(x) =∞∑

n=1

Γn sinnπx

L,

=⇒ u(x, t) =12[f(x + αt) + f(x− αt)] �

For (α > 0) the function f(x + αt) represents a traveling wave to the leftwith increasing time and f(x−αt) represents a traveling wave to the rightas illustrated in Figure 9.4.

9.2.2 One–Dimensional Acoustic Vibrations (Reynolds, 1981)

The derivation of the fluid pressure from ambient during 1–D acoustic vi-brations is also described by the wave equation

c2∂2p

∂x2− ∂2p

∂t2= 0,

where c is the speed of sound in the fluid. The lateral velocity u associatedwith acoustic motions is related to the pressure field by

ρ∂u

∂t= −∂p

∂x,


where ρ is the fluid density. As an example, let’s consider the case of a tubeclosed at x = 0 and open at x = L, and seek the normal modes of acousticvibration for this case

∂p

∂x= 0 at x = 0,

p = 0 at x = L.


p(x, t) = X(x) T (t).

Substituting into the wave equation yields

c2X ′′

X=

T ′′

T= −w2,

where the constant −ω2 was selected for the same reasons as in the previousexample. Consequently, the following two ordinary differential equationsare obtained

T ′′ + w2T = 0,

X ′′ + λ2X = 0,

whereλ =

w

c,

=⇒ X = c1 sin(λx) + c2 cos(λx).

From the first boundary condition

∂p(0, t)∂x

= X ′(0) T (t) = 0 −→ X ′(0) = 0,

X ′(0) = λc1 cos(0)− λc2 sin(0) −→ c1 = 0.

From the second boundary condition we get

X(L) T (t) = 0 −→ X(L) = 0,

X(L) = c2 cos(λL) −→ c2 cos(λL) = 0

−→ λL =π

2,3π2, · · ·

−→ λL = (2n− 1)π

2


−→ λn = (2n− 1)π

2L

=⇒ Xn = c2 cos(λnx).

Solving the differential equation for T yields

T ′′ + w2T = 0 =⇒ Tn = A1 cos(wnt) + A2 sin(wnt).

Therefore, the solution for the pressure field is

=⇒ p(x, t) = cos(λnx)[A∗1 cos(wnt) + A∗

2 sin(wnt)],

where A∗i = Aic2. We can now calculate the velocity field associated with

each eigenmode

ρ∂u

∂t= −∂p

∂x= − ∂

∂x{cos(λnx) [A∗

1 cos(wnt) + A∗2 sin(wnt)]}

= λn sin(λnx) [A∗1 cos(wnt) + A∗

2 sin(wnt)] ,

integrating yields

=⇒ ρu(x, t) =λn

wnsin(λnx) [A∗

1 sin(wnt)−A∗2 cos(wnt)] ,

wn = λnc,

=⇒ u(x, t) =1ρc

sin(λnx) [A∗1 sin(wnt)−A∗

2 cos(wnt)] .

The constant of integration must be zero for the fluid to remain motionlessat the closed end (at u(0, t) = 0, sin(0) = 0 → const. = 0). Also, notethat the nodes of p at points with zero pressure fluctuations are also theantinodes of u (points where the u field has maximum amplitude).

An Application of this Theory

Suppose you are trying to reduce the noise present in a long room in whichthe acoustic motions of this type occur to bother the occupants.

One solution is to damp the motions by providing a fine fibrous ma-terial that will oppose the fluid motion through viscosity. Obviously, thebest place to locate this material is where the fluid velocity is greatest (au antinode). The material would have little effect if placed at a velocitynode. This means that the material to quiet the room should be placed at


z

r

ro

L/2

-L/2

0

a node in the pressure field, i.e., where the sound you are trying to kill cannot be heard.

It should be noted that in actual situations the acoustic field is muchmore complicated. However, it is in general true that placing the acousticdamping material away from solid walls, out in the room where the air canmove through it, is most effective.

9.2.3 Nuclear Reactor Criticality

A very simple but conceptually useful model for the neutron density in anuclear reactor is that the neutron density φ is described by

∇2φ + µ2φ = 0,

where∇2φ = φrr +

1rφr +

1r2

φθθ + φzz,

µ2 depends upon the reactor size and design and the position of the controlrods. For a cylindrical reactor, the boundary conditions are:

φz = 0 at z = 0,(1)−φr = βφ at r = ro,(2)−φz = βφ at z = L/2.(3)

The first condition is a symmetry relationship and the last two conditionsequate the neutron diffusive flux at the outer surface to the diffusive lossthrough shielding.

Figure 9.5: Schematic diagram of a simplified nuclear reactor.


Note that the lowest value of µ (lowest eigenvalue) determines thecritical mass of the reactor. The objective of our analysis is therefore tocalculate µ.

We seek axisymmetric eigensolutions, consequently, we drop θ depen-dence and assume a solution of the form

φ(r, z) = R(r) Z(z).

Substituting into the governing equation and separating the variables yields

R′′ + 1rR

′

R= −

(Z ′′

Z+ µ2

)= −α2.

The constant −α2 was selected because it will lead to positive real α.

Z–Solution

Z ′′ + (µ2 − α2)Z = 0,

γ2 = µ2 − α2,

=⇒ Z ′′ + γ2Z = 0, =⇒ Z = c1 sin(γz) + c2 cos(γz).

From boundary condition (1)

φz(r, 0) = 0 −→ c1 = 0.

From boundary condition (3)

Z ′ + βZ = 0 at z =L

2,

−γ sin(γL

2

)+ β cos

(γL

2

)= 0

=⇒ β

γ= tan

(γL

2

)=

(βL

2

) (2γL

).

The preceding equation is a transcendental equation and defines an eigen-value problem for γ. The parameter γ can be determined graphically fromthe intersection of the curves tan(γL/2) and (βγ/2)(2/γL) as a function ofγL/2.


-1.5

-1

-0.5

0

0.5

1

0 3 6 9

Jo(z)

Yo(z)

z

R–Solution

R′′ +1rR′ + α2R = 0

=⇒ r2 d2R

dr2+ r

dR

dr+ r2α2R = 0.

This ordinary differential equation looks like Bessel’s equation

r2 d2R

dr2+ r

dR

dr+

(r2 − ν2

)R = 0.

However, the solution to the R problem can be obtained directly fromAbramowitz and Stegun (1965, eq. 9.1.52)

w′′ − 2ν − 1z

w′ + λ2w = 0,

w = zνGν(λz),

w = zν{AJν(λz) + BYν(λz)}.Therefore, the R–solution is of the form

R = B1Jo(αr) + B2Yo(αr).

Because Yo(0) = −∞ (see Figure 9.6), then B2 = 0 and

=⇒ R = B1Jo(αr).

Figure 9.6: Graphical representation of the Bessel functions J0(z) and Yo(z).


From the boundary condition (2)

−R′(ro) = βR(ro)

Recall : J ′o(x) = −J1(x)

=⇒ B1[−αJ1(αro) + βJo(αro)] = 0

=⇒ −αJ1(αro) + βJo(αro) = 0

=⇒ −αroJ1(αro) + βroJo(αro) = 0

=⇒ −λJo(x) + xJ1(x) = 0,

whereλ = βro, and x = αro.

Using appropriate tables (i.e. Table 9.7 from Abramowitz and Stegun,1965) determine the roots. For example, for

βro = 0.2 −→ αro = 0.6170

Finally, we can calculate µ since we know

γ −→ graphically

α −→ from tables, and

µ2 = α2 + γ2 �

9.3 Laplace Transformation Applied to the Solution ofPartial Differential Equations

Procedure

(1) L{Partial differential equation} −→ Ordinary differential equation,

(2) Solve the ordinary differential equation in Laplace space,

(3) Invert.


Example 1

Apply the Laplace transformation method to solve the following first–orderpartial differential equation

∂c

∂x+ x

∂c

∂t= 0,

subject to the initial and boundary conditions:

c(x, 0) = 0,

c(0, t) = t.

Take Laplace transform with respect to t

dc(x, s)dx

+ x{sc(x, s)− c(x, 0)} = 0

=⇒ dc

dx+ xsc = 0.

This is a first–order ordinary differential equation with nonconstant coeffi-cients, which has the following solution

∫dc

c+

∫xs dx = A(s) =⇒ ln c +

x2s

2= A(s)

=⇒ c(x, s) = A∗(s) exp[−x

2

2s

],

where A∗(s) = exp[A(s)]. Take Laplace transform of the boundary condi-tion to yield

c(0, s) =1s2.

In view of the last two expressions it is clear that

A∗(s) =1s2.

Consequently,

c =1s2

exp[−x

2

2s

]


=⇒ c(x, t) = L−1

{1s2

exp[−x

2

2s

]},

L−1

{1s2

}= t,

L−1{f(s)e−αs

}=

f(t− α) t > α,

0 t < α,

=⇒ c(x, t) =

t− x2

2t >

x2

2,

0 t <x2

2.

�

Example 2

A semi–infinite solid x > 0 is initially at temperature zero. At time t = 0, aconstant temperature Uo > 0 is applied and maintained at the face x = 0.Find the temperature at any point of the solid at any later time t > 0.

∂U

∂t= k

∂2U

∂x2, x > 0, t > 0

U(x, 0) = 0,I.C. :U(0, t) = U0, |U(x, t)| < M.B.C. :

It should be noted that the restriction of the boundary condition expressesthe requirement that the temperature must be bounded for all x and t.Taking Laplace transform with respect to t yields

sU(x, s)− U(x, 0) = kd2U(x, s)

dx2

=⇒ d2U(x, s)dx2

− s

kU = 0

=⇒ U(x, s) = c1(s)e√

sk x + c2(s)e−

√sk x.


Because U(x, s) is required to be bounded, c1 = 0 so that U(∞, s) is alsobounded. Consequently,

=⇒ U(x, s) = c2e−√

sk x.

Taking Laplace transform of the boundary condition yields

U(0, s) =Uo

s

=⇒ c2 =Uo

s

=⇒ U(x, s) =Uo

se−√

sk x

=⇒ U(x, t) = L−1

{Uo

se−√

sk x

}

L−1

{e−α

√s

s

}= erfc

[α

2√t

]

=⇒ U(x, t) = Uo erfc[

x

2√kt

]�

Example 3

Rework the previous problem if at t = 0 the temperature applied is givenby G(t) for t > 0,

∂U

∂t= k

∂2U

∂x2, x > 0, t > 0

U(x, 0) = 0,I.C. :U(0, t) = G(t), |U(x, t)| < M.B.C. :

Following the procedure outlined in the previous example we can show that

U(x, s) = c2e−√

sk x,


and taking Laplace transform of the boundary condition yields

U(0, s) = G(s)

=⇒ c2 = G(s)

U(x, s) = G(s)e−√

sk x.

Using the convolution theorem yields

U(x, t) = L−1{e−√

sk x

}∗ L−1{G(s)}

L−1{e−a

√s}

=a

2√πt−32 e−

a24t

L−1{e−√

sk x

}=

x

2√πk

t−32 e−

x24kt

=⇒ U(x, t) =∫ t

0

x

2√πk

u−32 e−

x24kuG(t− u) du �

Example 4

Solve by Laplace transforms the following boundary value problem

∂U

∂t= 4

∂2U

∂x2,

U(x, 0) = 10 sin 2πx− 6 sin 4πx,I.C. :U(0, t) = 0,B.C.1 :U(3, t) = 0.B.C.2 :

Taking Laplace transform with respect to t leads to

sU(x, s)− U(x, 0) = 4d2U(x, s)

dx2,

and employing the initial condition yields

4d2U

dx2− sU = −10 sin 2πx + 6 sin 4πx. (∗)


Homogeneous solution:

UH = Ae√

s2 x + Be−

√s

2 x,

where the unknown coefficients A and B are obtained by the method ofundetermined coefficients. Assume

UP = A∗ sin 2πx + B∗ cos 2πx + C∗ sin 4πx + D∗ cos 4πx,

=⇒ U ′′P =− 4π2A∗ sin 2πx− 4π2B∗ cos 2πx

− 16π2C∗ sin 4πx− 16π2D∗ cos 4πx

Substituting into (*) yields

B∗ = D∗ = 0,

A∗ =10

s + 16π2,

C∗ =−6

s + 64π2.

=⇒ U = UH + UP

= Ae√

s2 x + Be−

√s

2 x +10 sin 2πxs + 16π2

− 6 sin 4πxs + 64π2

.

B.C.1 : −→ U(0, s) = 0 = A + B,

B.C.2 : −→ U(3, s) = 0 = Ae32√

s + Be−32√

s,

=⇒ A = B = 0.

=⇒ U(x, s) =10 sin 2πxs + 16π2

− 6 sin 4πxs + 64π2

=⇒ U(x, t) = L−1

{10 sin 2πxs + 16π2

}− L−1

{6 sin 4πxs + 64π2

}

=⇒ U(x, t) = 10 sin(2πx)e−16π2t − 6 sin(4πx)e−64π2t �


9.4 Fourier Transformations Applied to the Solution ofPartial Differential Equations

Procedure

(1) F{Partial differential equation} −→ Ordinary differential equation,

(2) Solve the ordinary differential equation in Fourier space,

(3) Invert.

Example 1

Find the solution to the following boundary value problem describing thetemperature distribution in an infinite homogeneous rod

∂u(x, t)∂t

= c2∂2u(x, t)

∂x2, (Heat equation)

u(x, 0) = f(x),

u(±∞, t) −→ 0,

∂u(±∞, t)∂x

−→ 0.

First recall the following Fourier transform pair and relationship:

f(w) =∫ ∞

−∞f(x)e−jwx dx,

f(x) =12π

∫ ∞

−∞f(w)ejwx dw,

F{f ′′(x)} = −w2f(w).

Taking Fourier transform with respect to x yields

du(w, t)dt

= −c2w2u(w, t) First order O.D.E.

=⇒ u(w, t) = A(w)e−c2w2t,


I.C. −→ u(w, 0) = f(w),

=⇒ A(w) = f(w) =⇒ u(w, t) = f(w)e−c2w2t

u(x, t) = F−1{f(w)e−c2w2t

}=

12π

∫ ∞

−∞f(w)e−c2w2tejwx dw

where

f(w) =∫ ∞

−∞f(ν)e−jwν dν

=⇒ u(x, t) =12π

∫ ∞

−∞f(ν)

[∫ ∞

−∞e−c2w2tej(wx−wν)dw

]dν.

Note that the bracketed integral in the preceding equation can be rewrittenas follows∫ ∞

−∞{e−c2w2t cos(wx− wν) + je−c2w2t sin(wx− wν)︸︷︷︸

Odd Function

} dw

=∫ ∞

−∞e−c2w2t cos(wx− wν) dw

= 2∫ ∞

0

e−c2w2t cos(wx− wν) dw

=⇒ u(x, t) =1π

∫ ∞

−∞f(ν)

[∫ ∞

0

e−c2w2t cos(wx− wν)dw]dν

=1π

∫ ∞

−∞f(ν)

[12c

(πt

) 12e−

(x−ν)2

4c2t

]dν

=1

2c√πt

∫ ∞

−∞f(ν)e−

(x−ν)2

4c2t dν �

∫ ∞

0

e−s2cos 2bs ds =

√π

2e−b2 ,

let s = cw√t, b =

x− ν

2c√t.


Example 2

Solve the following partial differential equation describing the vibration ofan infinite string

∂2u(x, t)∂x2

− 1c2

∂2u(x, t)∂t2

= 0,

u(x, 0) = f(x), −∞ < x <∞

∂u(x, 0)∂t

= 0.

Taking Fourier transform with respect to x yields

−w2u(w, t)− 1c2

d2u(w, t)dt2

= 0

=⇒ d2u

dt2+ w2c2u = 0 ← Second order O.D.E.

=⇒ u(w, t) = A(w) cos(cwt) + B(w) sin(cwt).

From initial conditionsu(w, 0) = f(w)

=⇒ A(w) = f(w)

F{∂u(x, 0)

∂t

}=

du(w, 0)dt

= 0,

du(w, t)dt

= −cwf(w) sin(cwt) + cwB(w) cos(cwt)

=⇒ B(w) = 0

=⇒ u(w, t) = f(w) cos(cwt),

cos(cwt) =12

[ejcwt + e−jcwt

]=⇒ u(w, t) =

12f(w)

[ejcwt + e−jcwt

].

Recall the following Fourier relationships:

F−1{f(w)ejcwt

}= f(x + ct),


F−1{f(w)e−jcwt

}= f(x− ct),

which are based on the time shifting property:

F−1{f(w)e−jwto

}= f(t− to).

Therefore, the solution may be expressed as:

u(x, t) = F−1

{12f(w)

[ejcwt + e−jcwt

]}

=⇒ u(x, t) =12[f(x + ct) + f(x− ct)] �

9.5 Self–Similar Solutions (From Reynolds, 1981)

9.5.1 Characteristic Scales and Scale–Similar Problems

It is often convenient to present the solution to a PDE problem in a non–dimensional form. This makes the results independent of the size of thesystem for which the solution was obtained. Non–dimensionalization isusually accomplished by choosing some length and time scales characteriz-ing the problem, and then defining non–dimensional independent variablesbased on these scales.

For example, the solution for fluid flow in a rotating sphere might beexpressed non–dimensionally in terms of the dimensionless radius

R =r

ro,

where ro is the radius of the sphere. It should be noted that ro is thecharacteristic length scale of the problem. If the fluid is initially at rest,and at time zero it is put into rotation at angular velocity w, then theperiod of rotation is

τ =2πw

,

where τ would be the characteristic time scale. Then, a suitable dimen-sionless time would be

T =t

τ.


Note that one of the characteristic scales for the independent variables (ro)came from the geometry of the system, and the other (τ) from the boundaryconditions.

The dependent variables also can be represented non–dimensionally.For example, in the rotating sphere problem the equatorial velocity is

uo = wro,

and may be used as a characteristic velocity in the dimensionless velocity

U =u

uo.

The problem may also contain some parameters, such as the kinematicviscosity ν. The parameters also can be reduced to non–dimensional form,and in the case of viscosity it is customary to use a reciprocal dimensionlessviscosity called the Reynolds number,

Re =uoro

ν.

The solution for the velocity within the rotating sphere could then be ex-pressed non-dimensionally as

U = U(R, T,Re).

It might also happen that the flow depends upon the polar angular coordi-nates φ and θ, which are additional non–dimensional independent variables.

Problems which have natural characteristic scales for the independentvariables (here ro and τ) are called scale–similar. Scale–similar solutions forsystems of different size will have the same non–dimensional solution, pro-vided that the two problems also have the same values of the dimensionlessparameters and dimensionless boundary and initial conditions.

9.5.2 Self–Similarity

There are a few very interesting and important problems for which no nat-ural characteristic scales for the independent variables exist in the problemformulation. For example, consider the case of heat conduction in a semi–infinite slab initially at uniform temperature, subjected to a step increasein the surface temperature at time zero. The governing partial differentialequation is

∂2T

∂x2=

1α

∂T

∂t, (∗)


where α is the thermal diffusivity of the medium. Also, the appropriateinitial and boundary conditions are:

T (x, 0) = Ti, x > 0

T (0, t) = Ts,

T (∞, t)→ Ti.

Clearly, there are no characteristic scales for either length or time in thisproblem. This fact is the clue that a self–similar solution must exist. Sincethe solution to all physical problems must be expressible in dimensionlessform (nature is unaware of the length of a meter), there must be some wayto non–dimensionalize the solution to this problem.

Looking at the denominators in (*), it is readily apparent that x2 andαt have the same dimensions, and therefore the quantity x2/αt is dimen-sionless. Somehow the solution must be expressible in terms of this quan-tity, in order to have dimensionless form. Solutions made non–dimensionalby combinations of the independent variables, rather than by characteristicscales imposed by the geometry, boundary, or initial conditions, are calledself–similar solutions.

There is a characteristic temperature for this problem, namely thestep increase in temperature Ts − Ti. Therefore, one might guess that thenon–dimensional form of the solution is

T − Ti

Ts − Ti= f

(x2

αt

). (∗∗)

If the preceding equation is indeed a correct solution, then another fullyequivalent form would be

T − Ti

Ts − Ti= g

(x√αt

),

and anotherT − Ti

Ts − Ti=

x√αt

h

(x√αt

).

All of these solutions would really be the same time, but the functions f ,g, and h would be different. In terms of the similarity variable

η =x√αt

,


T - Ti

1

η

Ts - Ti

0

Ts

Ti0 x

T

t

(a) (b)

Figure 9.7: Transient temperature distribution in a semi–infinite slab (a)and application of the self–similar method (b).

the family of temperature profiles existing at different times will collapseto a single curve as illustrated in Figure 9.7. This is the essence of self–similarity. The solution does not scale on the size of the system, instead itscales on itself.

At first glance, it may appear disadvantageous to seek a solution interms of the non–linear combination of variables η = x/

√αt. However, note

that a single function g(η) would be involved, and therefore one would onlyhave to deal with an ordinary differential equation. This is the practicaladvantage of a self–similar problem in two independent variables. Theexistence of self–similarity will always reduce the number of independentvariables by one.

Example 1: Constant Boundary Conditions

Consider the transient heat transfer problem previously discussed. Let’sassume that the similarity variable of the form

η = Ax

tn(general format),

where A and n are constants to be chosen in a manner that reduces thePDE problem to an ODE problem. Now, assume that the dimensionlesssolution has the form

T − Ti

Ts − Ti= f(η)

=⇒ T − Ti = (Ts − Ti)f(η).


The governing partial differential equation can be transformed as follows

∂2T

∂x2=

1α

∂T

∂t

∂

∂x[T − Ti] =

∂

∂x[(Ts − Ti)f(η)] =⇒ ∂T

∂x= (Ts − Ti)

df(η)dη

∂η

∂x

=⇒ ∂T

∂x= (Ts − Ti)f ′ A

tn,{

Recall :∂η

∂x=

A

tn, f ′ =

df

dη

}

∂2T

∂x2= (Ts − Ti)

A

tndf ′

dη

∂η

∂x= (Ts − Ti)

A

tnf ′′ A

tn,

∂

∂t[T − Ti] =

∂

∂t[(Ts − Ti)f(η)] =⇒ ∂T

∂t= (Ts − Ti)

df

dη

∂η

∂t,

∂η

∂t=

∂

∂t

(Ax

tn

)= −

(nAx

tn+1

)

=⇒ ∂T

∂t= (Ts − Ti) f ′

(−Anxtn+1

).

Then, substituting into (*) we obtain

(Ts − Ti)A2

t2nf ′′ = − 1

α(Ts − Ti)

Anx

tn+1f ′

=⇒ A

tnf ′′ = − 1

α

nx

tf ′

=⇒ f ′′ +1α

nxtn−1

Af ′ = 0.

Using the definition

η = Ax

tn=⇒ x =

tnη

A

remove x and t from the equation so that it will be an ordinary differentialequation for f(η) containing f, f ′, f ′′, and η

=⇒ f ′′ +n

αA2t2n−1ηf ′ = 0, let n =

12


=⇒ f ′′ +1

2αA2ηf ′ = 0, let A =

1√2α

=⇒ f ′′ + ηf ′ = 0 (∗ ∗ ∗)

η =Ax

tn, A =

1√2α

, n =12,

=⇒ η =x√2αt

.

Express the boundary and initial conditions in terms of f(η)

T (x, 0) = Ti,require f(η)→ 0 as η →∞,

T (∞, t)→ Ti,

T (0, t) = Ts, requires f(0) = 1.

(∗ ∗ ∗) −→ df ′

dη= −ηf ′ =⇒ df ′

f ′ = −η dη, integrate

=⇒ ln f ′ = −η2

2+ c0

=⇒ f ′ = e−η2

2 ec0

=⇒ f ′ = c1 exp[−η

2

2

], integrating again

=⇒ f = c1

∫ η

∞exp

[−σ

2

2

]dσ + c2.

The lower limit is arbitrary, and ∞ is a good choice because as t → 0η →∞. Since f(∞)→ 0,

f = c1

∫ ∞

∞exp

[−σ

2

2

]dσ︸︷︷︸

0

+c2 = 0 =⇒ c2 = 0,

f(0) = 1 =⇒ 1 = c1

∫ 0

∞exp

[−σ

2

2

]dσ =⇒ c1 =

1∫ 0

∞ exp[−σ2

2

]dσ

,

=⇒ f(η) =

∫ ∞η

exp[−σ2

2

]dσ∫ ∞

0exp

[−σ2

2

]dσ

.


The solution can be expressed in terms of known special functions by letting

z =σ√2−→ dσ =

√2 dz,

=⇒ f(η) =

∫ ∞η√2=z

e−z2√2 dz∫ ∞

0e−z2√

2 dz=

∫ ∞η√2e−z2

dz∫ ∞0

e−z2dz.

Recall :∫ ∞

0

e−z2dz =

√π

2,

∫ ∞

η√2

e−z2dz =

√π

2erfc

(η√2

).

=⇒ f(η) = erfc(

η√2

), η =

x√2αt

=⇒ T − Ti

Ts − Ti= erfc

(x

2√αt

)�

Example 2: Variable Boundary Conditions

The motion of a viscous fluid, initially at rest, over an infinite plate that isset into motion at time zero is described by

ν∂2u

∂y2=

∂u

∂t,

where u is the velocity tangential to the plate, and ν is the constant kine-matic viscosity. For the following initial and boundary conditions

u(y, 0) = 0,

u(0, t) = atb,

u(y, t)→ 0, as y →∞,

where a, b are fixed parameters, assume a solution of the form

u = Af(η), η = By

tn


=⇒ u(0, t) = atb = Af(0).

This is true only for b = 0. Therefore, it is not a good choice. A betterchoice is to assume a solution of the form

u = Atmf(η), η = By

tn

=⇒ u(0, t) = atb = Atmf(0).

We must choose m = b and A any way we like. If we choose A = a, thenmust impose f(0) = 1. Thus, a good choice is

u = atbf(η), η = By

tn.

9.6 Exercises

(1) Classify each of the following equations accordingly as elliptic, hyper-bolic or parabolic

(a)∂2u

∂t2= k2 ∂

2u

∂x2,

(b)∂2u

∂x2+ 3

∂2u

∂x∂y+ 4

∂2u

∂y2+ 5

∂u

∂x− 2

∂u

∂y+ 4u = 2x− 3y,

(c) x∂2u

∂x2+ y

∂2u

∂y2+ 3y2 ∂u

∂x= 0.

(2) Solve the following boundary–value problem by the method of separa-tion of variables

∂u

∂x= 4

∂u

∂y, u(0, y) = 8e−3y + 4e−5y.

(3) Solve by the method of separation of variables

∂u

∂t= 2

∂2u

∂x2, 0 < x < 3, t > 0


given thatu(0, t) = u(3, t) = 0,

u(x, 0) = 5 sin 4πx− 3 sin 8πx + 2 sin 10πx,

|u(x, t)| < M,

where the last condition states that u is bounded for 0 < x < 3, t > 0.(4) Apply Laplace transformations to show that the Laplace “time” solution

of the following boundary value problem

∂u

∂t= k

∂2u

∂x2, u(a, t) = Qδ(t), u(x, 0) = 0, |u(x, t)| < M,

isu(x, s) = Qe−(x−a)

√s/k.

(5) In the theory of neutron slowing down, one encounters the equation

α(ξ)∂φ

∂ξ= D

∂2φ

∂x2(0 ≤ ξ <∞),

where ξ is a logarithmic energy variable, D is the diffusion coefficient,and α(ξ) is a known positive function. Solve this equation by themethod of Fourier transformations for an infinite medium (in x) subjectto the initial condition: φ(0, x) = δ(x).


Chapter 10

Solutions to Exercises

10.1 Differential Equations of the First Order (Chapter 2)

[1] Obtain the general solution of (Separable ODE)

(a) xy3dx+ ex2dy = 0 =⇒ xy3dx = −ex2

dy

=⇒ xe−x2dx = −dy

y3=⇒

∫xe−x2

dx = −∫y−3dy

=⇒ −e−x2

2=y−2

2+C =⇒ e−x2

+ y−2 = −2C =⇒ e−x2+ y−2 = C∗ �

(b)dy

dx= xy2 =⇒ x dx =

dy

y2=⇒

∫x dx =

∫y−2 dy =⇒ x2

2= −y−1+C

x2

2+

1y

= C =⇒ yx2 + 2 = 2Cy (2C = −C∗)

yx2 + C∗y + 2 = 0 =⇒ y(x2 + C∗) + 2 = 0 �

(c) x2ydy

dx= ey =⇒ x2

dx=

ey

y dy=⇒

∫dx

x2=

∫y dy

ey(∫u dv = uv −

∫v du, u = y, du = dy, dv = e−ydy, v = −e−y

)

=⇒ −1x

= −ye−y −∫ (−e−y

)dy =⇒ − 1

x= −ye−y − e−y + C

=⇒ 1x

+ C = ye−y + e−y =⇒ ey + eyxC = yx+ x

=⇒ x(y + 1) = (1 + xC)ey �


[2] Obtain a family of solutions for the following equations

(a) xy dx−(x2 + 3y2

)dy = 0

(y = ux, dy = u dx+ x du

)ux2dx−

(x2 + 3u2x2

)(udx+ xdu) = 0

=⇒ ux2dx− x2u dx− x3 du− 3u3x2dx− 3u2x3du = 0

=⇒ x3du+ 3u3x2dx+ 3u2x3du = 0 =⇒ x du+ 3u3dx+ 3u2x du = 0

=⇒ x+ 3u3 dx

du+ 3u2x = 0 =⇒ x

dx+

3u3

du+

3u2x

dx= 0

=⇒ x

dx

(1 + 3u2

)= −3

u3

du=⇒ dx

x= −

(1 + 3u2

)du

3u3=⇒ dx

x= − du

3u3− duu

integrate =⇒ ln |x| = u−2

6− ln |u|+ ln |C|

(ln[ab] = ln a+ ln b, ln

[ab

]= ln a− ln b

)

=⇒ ln |xu| − ln |C| = 16u2

=⇒ ln{xuC

}=

16u2

, u =y

x

=⇒ ln{ yC

}=

1

6(yx

)2 =⇒ ln{ yC

}=

x2

6y2=⇒ x2 = 6y2 ln

{ yC

}�

(b) x2 dy

dx= 4x2 + 7xy + 2y2 =⇒ x2 dy = 4x2 dx+ 7xy dx+ 2y2dx

(y = ux, dy = u dx+ x du

)

=⇒ (u dx+ x du)x2 = 4x2 dx+ 7x2u dx+ 2u2x2 dx

=⇒ u dx+ x du = 4 dx+ 7u dx+ 2u2dx

=⇒ x du = 4dx+ 6u dx+ 2u2dx =⇒ x du =(4 + 6u+ 2u2

)dx

=⇒ dx

x=

du

4 + 6u+ 2u2=

du

(2u+ 2)(u+ 2)=

du

2u+ 2− 1

2

(du

u+ 2

)


1(2u+ 2)(u+ 2)

=A

2u+ 2+

B

u+ 2=

12u+ 2

− 12

( 1u+ 2

)A(u+ 2) +B(2u+ 2)

Au+ 2A+ 2Bu+ 2B = 2(A+B) + u(A+ 2B)

A+ 2B = 0 =⇒ B = −A2

2(A+B

)= 1 =⇒ 2

(A− A

2

)= 1 =⇒ A = 1, B = −1

2

integrate =⇒ ln |x| = 12

ln |u+ 1| − 12

ln |u+ 2|+

Const.︷︸︸︷12

ln |C|

=⇒ 2 ln |x| = ln∣∣∣∣(u+ 1u+ 2

)C

∣∣∣∣ =⇒ x2 = C

(u+ 1u+ 2

), u =

y

x

=⇒ x2 = C

( yx + 1yx + 2

)= C

(y + x

y + 2x

)=⇒ x2(y + 2x) = C(y + x) �

(c) (x− y)(4x+ y)dx+ x(5x− y)dy = 0

(y = ux, dy = u dx+ x du)

=⇒ (x− ux)(4x+ ux)dx+ x(5x− ux)(u dx+ x du) = 0

4x2dx+ux2dx−4ux2dx−u2x2dx+5x2u dx+5x3du−u2x2dx−ux3du = 0

4 dx+ 2u dx− 2u2dx+ 5x du− ux du = 0(4 + 2u− 2u2

)dx+ (5x− ux)du = 0

dx

x+

(5− u)du4 + 2u− 2u2

= 0 =⇒ dx

x+

(5− u)du(2 + 2u)(2− u) = 0

5− u(2− u)(2 + 2u)

=A

2− u +B

2 + 2u=

12

(1

2− u

)+

11 + u

A(2 + 2u) +B(2− u)2A+ 2Au+ 2B − uB2(A+B) + u(2A−B)

2(A+B) = 5, (2A−B) = −1 −→ A =12, B = 2


=⇒ dx

x= −1

2

(du

2− u

)− du

1 + u=

12

(du

u− 2

)− du

1 + u

=⇒ ln |x| = 12

ln |u−2|− ln |1+u|+ln |C| =⇒ 2 ln |x(1+u)| = ln |C(u−2)|

=⇒ x2(1 + u)2 = C(u− 2), u =y

x

=⇒ x2(1 +

y

x

)2

= C(−2 +

y

x

)=⇒ x2

(1 +

2yx

+y2

x2

)= C

(−2 +

y

x

)

=⇒ x2 + 2xy + y2 = C(−2 +

y

x

)=⇒ (x+ y)2 =

C

x(y − 2x)

=⇒ x(x+ y)2 = C(y − 2x) �

[3] Test each of the following equations for exactness and solve the equation

(a)

M︷︸︸︷(x+ 2y) dx+

N︷︸︸︷(2x+ y) dy = 0

(exact :

∂M

∂y=∂N

∂x= 2, F = C

)

M =∂F

∂x= x+ 2y =⇒ F =

x2

2+ 2xy + φ(y), N =

∂F

∂y= 2x+ y

∂F

∂y= 2x+ φ′(y) = 2x+ y =⇒ φ′(y) = y =⇒ φ(y) =

y2

2

x2

2+ 2xy +

y2

2= C∗ =⇒ x2 + 4xy + y2 = C �

(b) v(2uv2 − 3

)du+

(3u2v2 − 3u+ 4v

)dv = 0

=⇒M︷︸︸︷(

2uv3 − 3v)du+

N︷︸︸︷(3u2v2 − 3u+ 4v) dv = 0(

exact :∂M

∂v=∂N

∂u= 6uv2 − 3

)


M =∂F

∂u= 2uv3 − 3v =⇒ F = u2v3 − 3uv + φ(v)

N =∂F

∂v= 3u2v2 − 3u+ 4v

∂F

∂v= 3u2v2 − 3u+ φ′(v)

φ′(v) = 4v =⇒ φ(v) = 2v2

F = u2v3 − 3uv + 2v2 = C∗ =⇒ v(u2v2 − 3u+ 2v

)= C∗ �

(c)

M︷︸︸︷(cos 2y − 3x2y2

)dx+

N︷︸︸︷(cos 2y − 2x sin 2y − 2x3y

)dy = 0

(exact :

∂M

∂y=∂N

∂x= −2 sin 2y − 6x2y, F = Const.

)

M =∂F

∂x= cos 2y − 3x2y2 [1]

N =∂F

∂y= cos 2y−2x sin 2y−2x3y =⇒ F =

sin 2y2

+x cos 2y−x3y2 +φ(x)

=⇒ ∂F

∂x= cos 2y − 3x2y2 + φ′(x) [2]

[1]&[2] −→ φ′(x) = 0 =⇒ φ(x) = C∗

=⇒ sin 2y2

+ x cos 2y − x3y2 = C �

[4] Solve each of the following equations (I.F.)

(a)

M︷︸︸︷(x2 + y2 + 1

)dx+

N︷︸︸︷x(x− 2y) dy = 0

∂M

∂y− ∂N

∂x= 2y − (2x− 2y) = 4y − 2x

1N

(∂M

∂y− ∂N

∂x

)=

4y − 2xx(x− 2y)

= − 2(2y − x)x(2y − x) = − 2

x


an I.F. is : u(x) = exp[∫− 2xdx

]= exp

[−2 lnx

]= e− ln x2

= x−2

=⇒ uM dx+ uN dy = 0 must be exact

=⇒ x−2(x2 + y2 + 1

)dx+

1x

(x− 2y)dy = 0

=⇒(

1 +y2

x2+

1x2

)dx+

(1− 2y

x

)dy = 0

M =∂F

∂x= 1 +

y2

x2+

1x2, N =

∂F

∂y= 1− 2y

x=⇒ F = y − y2

x+ φ(x)

∂F

∂x=y2

x2+ φ′(x) = 1 +

y2

x2+

1x2

=⇒ φ′(x) = 1 +1x2

=⇒ φ(x) = x− 1x

y − y2

x+ x− 1

x= C =⇒ x2 − y2 + xy − 1 = Cx �

(b)

M︷︸︸︷y(2x− y + 1) dx+

N︷︸︸︷x(3x− 4y + 3) dy = 0

∂M

∂y− ∂N

∂x= 2x− 2y + 1− (6x− 4y + 3) = −4x+ 2y − 2

1M

(∂M

∂y− ∂N

∂x

)=−2(2x− y + 1)y(2x− y + 1)

= −2y

an I.F. is : u(y) = exp[−

∫ (−2y

)dy

]= e2 ln y = y2

=⇒ y2[y(2x− y + 1)dx+ x(3x− 4y + 3)dy

]= 0(

2xy3 − y4 + y3)dx+

(3x2y2 − 4xy3 + 3xy2

)dy = 0

(exact : F = C

)∂F

∂x= 2xy3 − y4 + y3,

∂F

∂y= 3x2y2 − 4xy3 + 3xy2

=⇒ F = x2y3 − xy4 + xy3 + φ(y)

=⇒ ∂F

∂y= 3x2y2 − 4xy3 + 3xy2 + φ′(y) = 3x2y2 − 4xy3 + 3xy2

=⇒ φ′(y) = 0 =⇒ φ(y) = const.


F = C = x2y3 − xy4 + xy3 + const. =⇒ xy3(x− y + 1) = C∗ �

(c)

M︷︸︸︷y(4x+ y) dx

N︷︸︸︷−2(x2 − y) dy = 0

∂M

∂y− ∂N

∂x= 4x+ 2y − (−4x) = 8x+ 2y

1M

(∂M

∂y− ∂N

∂x

)=

8x+ 2yy(4x+ y)

=2y

an I.F. is : u(y) = exp[−

∫2ydy

]= y−2

=⇒ 1y(4x+ y)dx− 2

y2

(x2 − y

)dy = 0

=⇒(

4xy

+ 1)dx−

(2x2

y2− 2y

)dy = 0 (exact)

∂F

∂x=

4xy

+ 1 =⇒ F =2x2

y+ x+ φ(y) =⇒ ∂F

∂y= −2x2

y2+ φ′(y)

∂F

∂y= −2x2

y2+

2y

φ′(y) =2y

=⇒ φ(y) = 2 ln y

=⇒ F = C =2x2

y+ x+ 2 ln y =⇒ 2x2 + xy + 2y ln y = Cy �

[5] Find the general solution of the following equations

(a) y2dx+(xy + y2 − 1

)dy = 0; when x = −1, y = +1

(dx

dy+ P (y)x = Q(y)

)

=⇒ dx

dy+xy + y2 − 1

y2= 0 =⇒ dx

dy+x

y+ 1− 1

y2= 0


=⇒ dx

dy+

P (y)︷︸︸︷(1y

)x =

Q(y)︷︸︸︷1y2− 1←− general linear 1st− order ODE

x = e−∫

1y dy

{∫e∫

1y dy

(1y2− 1

)dy + C

}

=⇒ x = e− ln y

{∫eln y

(1y2− 1

)dy + C

}=

1y

{∫ (1y− y

)dy + C

}

=⇒ x =1y

(ln y − y2

2+ C

)=⇒ 2xy = 2 ln y − y2 + C

At (x, y) = (−1, 1) : −2 = −1 + C =⇒ C = −1

2xy = 2 ln y − y2 − 1 =⇒ y2 + 2xy + 1 = 2 ln y �

(b)dy

dx= y tanx+ cosx =⇒ dy

dx− y

−P (x)︷︸︸︷tanx =

Q(x)︷︸︸︷cosx

y = e−∫

(− tan x)dx

{∫e∫

(− tan x)dx cosx dx+ C

}(∫

tanx dx = − ln | cosx|+ C

)

= e− ln cos x

{∫eln cos x cosx dx+ C

}

=1

cosx

{∫cos2 x dx+ C

}(∫

cos2 x dx =sinx cosx

2+x

2

)

=1

cosx

(sinx cosx

2+x

2+ C

)

=sinx

2+

x

2 cosx+

C

cosx

=⇒ 2y = sinx+x

cosx+

C1

cosx

=⇒ 2y = sinx+x+ C1

cosx

(recall :

1cos θ

= sec θ)

=⇒ 2y = sinx+ secx(x+ C1) �


[6] Solve the following equations (Bernoulli’s)

(a) 2x3 dy

dx= y

(y2 + 3x2

)

dy

dx=

y3

2x3+

3x2y

2x3

=⇒ dy

dx+

(− 3

2x

)y =

(1

2x3

)y3 ←− Bernoulli′s

let z = y1−3 = y−2, dz = −2y−3dy =⇒ dy =dz

−2y−3

dz

dx+ (−2)

(− 3

2x

)z = (−2)

(1

2x3

)

=⇒ dz

dx+

3xz = − 1

x3(1st− order linear in z)

z = e−∫

3xdx

{∫e∫

3xdx

(− 1x3

)dx+ C

}

=⇒ z = e−3 ln x

{∫e3 ln x

(− 1x3

)dx+ C

}= x−3

{∫x3

(− 1x3

)dx+ C

}

=⇒ z = x−3

{∫(−)dx+ C

}= − 1

x2+C

x3

(recall : z = y−2

)

=⇒ 1y2

= − 1x2

+C

x3=⇒ x3 = −y2x+ y2C =⇒ x3 = y2(C − x) �

(b)(2y3 − x3

)dx+ 3xy2dy = 0; when x = 1, y = 1

=⇒ dy

dx= −2y3 − x3

3xy2=⇒ dy

dx= −2y

3x+x2

3y2

=⇒ dy

dx+

(23x

)y =

(x2

3

)y−2 ←− Bernoulli′s

let z = y1−(−2) = y3, dz = 3y2dy =⇒ dy =dz

3y2


substituting −→ dz

dx+ 3

(23x

)z = 3

(x2

3

)

=⇒ dz

dx+

(2x

)z = x2 ←− 1st− order linear O.D.E.

z = e−∫

2xdx

{∫e∫

2xdx(x2)dx+C

}=

1x2

{∫x4dx+ C

}=

1x2

(x5

5+ C

)

=⇒ z =x3

5+C

x2, z = y3

=⇒ y3 =x3

5+C

x2=⇒ 5x2y3 = x5 + C∗

At (x, y) = (1, 1) : 5 = 1 + C∗ =⇒ C∗ = 4

=⇒ 5x2y3 = x5 + 4 �


10.2 Fundamental and Important Properties of Linear OrdinaryDifferential Equations (Chapter 3)

[1] Obtain the Wronskian of the functions 1, x, x2, · · · , xn−1 for n > 1

W(1, x, x2, · · · , xn−1

)=

∣∣∣∣∣∣∣∣∣∣∣∣

1 x x2 · · · xn−1

0 1 2x · · · (n− 1)xn−2

0 0 2 · · · (n− 1)(n− 2)xn−3

0 0 0 · · · (n− 1)(n− 2)(n− 3)xn−4

......

.... . .

...0 0 0 · · · (n− 1)!

∣∣∣∣∣∣∣∣∣∣∣∣= 1 · 1 · 2 · · · (n− 1)!

= 0! · 1! · 2! · · · (n− 1)! �

[2] Show that the operators D − 2 and xD + 1 are not commutative withrespect to multiplication

let : A = D − 2, B = xD + 1

BA = (xD + 1)(D − 2) = xD(D − 2) + (D − 2) = xD2 − 2xD +D − 2

AB = (D − 2)(xD + 1) = D(xD + 1)− 2(xD + 1)

= xD2 +D +D − 2xD − 2 = xD2 − 2xD + 2D − 2

=⇒ AB �= BA

Therefore, the operators D−2 & xD+1 are not commutative with respectto multiplication. �

[3] Show that the functions cos 2x, sin2 x, cos2 x are linearly dependent.

C1 cos 2x+ C2 sin2 x+ C3 cos2 x = 0

Since cos 2x = cos2 x− sin2 x, we can choose C1 = 1, C2 = 1, and C3 = −1.These constants are not equal to zero, thus the functions are dependent.�

[4] Let y1(x), y2(x) be solutions of y′′ + p(x)y′ + q(x)y = 0. Prove that theWronskian is:

W = y1y′2 − y2y′1 = Ce−

∫p dx.


Because y1 and y2 are solutions: y′′1 + py′1 + qy1 = 0, y′′2 + py′2 + qy2 = 0

=⇒ y2 (y′′1 + py′1 + qy1)− y1 (y′′2 + py′2 + qy2) = 0

=⇒ y2y′′1 + py2y

′1 + qy2y1 − y1y′′2 − py1y′2 − qy1y2 = 0

=⇒ y1y′′2 − y2y′′1 + p (y1y′2 − y2y′1) = 0,

W =∣∣∣∣ y1 y2y′1 y′2

∣∣∣∣ = y1y′2 − y2y′1

=⇒ y1y′′2 − y2y′′1 + pW = 0,

dW

dx= y′1y

′2 + y1y

′′2 − y′2y′1 − y2y′′1 = y1y

′′2 − y2y′′1

=⇒ dW

dx+ pW = 0 =⇒ dW

W= −p dx

=⇒ lnW = −∫p dx+ C =⇒W = Ce−

∫p dx

=⇒W = y1y′2 − y2y′1 = Ce−

∫p dx �

[5] Solve the following equations:

(a)(D3 −D2 − 4D + 4

)y = 0

=⇒ λ3 − λ2 − 4λ+ 4 = 0 =⇒ λ2(λ− 1)− 4(λ− 1) = 0

=⇒ (λ− 1)(λ2 − 4

)= 0 =⇒ (λ− 1)(λ− 2)(λ+ 2) = 0

λ1 = 1, λ2 = 2, λ3 = −2

=⇒ y(x) = C1ex + C2e

2x + C3e−2x �

(b)(6D4 + 23D3 + 28D2 + 13D + 2

)y = 0

=⇒ 6λ4 + 23λ3 + 28λ2 + 13λ+ 2 = 0


6λ3 + 17λ2 + 11λ+ 2λ+ 1 | 6λ4 + 23λ3 + 28λ2 + 13λ+ 2

6λ4 + 6λ3

17λ3 + 28λ2

17λ3 + 17λ2

11λ2 + 13λ11λ2 + 11λ

2λ+ 22λ+ 2

0

=⇒ (λ+ 1)(6λ3 + 17λ2 + 11λ+ 2

)= 0

6λ2 + 5λ+ 1λ+ 2 | 6λ3 + 17λ2 + 11λ+ 2

6λ3 + 12λ2

5λ2 + 11λ5λ2 + 10λ

λ+ 2λ

=⇒ (λ+1)(λ+2)(6λ2+5λ+1

)= 0

(6λ2+5λ+1 = (2λ+1)(3λ+1)

)

=⇒ (λ+ 1)(λ+ 2)(2λ+ 1)(3λ+ 1) = 0

λ1 = −1, λ2 = −2, λ3 = −12, λ = −1

3

=⇒ y(x) = C1e−x + C2e

−2x + C3e− x

2 + C4e− x

3 �

(c)(D4 − 5D2 − 6D − 2

)y = 0

=⇒ λ4 − 5λ2 − 6λ− 2 = 0

λ3 − λ2 − 4λ− 2λ+ 1 | λ4 − 5λ2 − 6λ− 2

λ4 + λ3

− λ3 − 5λ2

−λ3 − λ2

− 4λ2 − 6λ−4λ2 − 4λ

− 2λ− 2


=⇒ (λ+ 1)(λ3 − λ2 − 4λ− 2

)= 0

λ2 − 2λ− 2λ+ 1 | λ3 − λ2 − 4λ− 2

λ3 + λ2

− 2λ2 − 4λ−2λ2 − 2λ

− 2λ− 2

=⇒ (λ+1)2(λ2 − 2λ− 2

)= 0 =⇒ (λ+1)2

(λ− 1−

√3) (λ− 1 +

√3)

= 0

λ1 = λ2 = −1, λ3 = 1 +√

3, λ4 = 1−√

3

=⇒ y(x) = C1e−x + C2xe

−x + C3e(1+

√3)x + C4e

(1−√

3)x �

(d)(D4 + 2D3 + 10D2

)y = 0

=⇒ λ4 + 2λ3 + 10λ2 = 0 =⇒ λ2(λ2 + 2λ+ 10

)= 0(

λ2 + 2λ+ 10 = 0, λ =−2±

√4− 40

2=

{−1 + 3i−1− 3i

)

=⇒ λ2(λ+ 1− 3i)(λ+ 1 + 3i) = 0

λ1 = λ2 = 0, λ3 = −1 + 3i, λ4 = −1− 3i

=⇒ y(x) = C1 + C2x+ C3e−x cos 3x+ C4e

−x sin 3x �

(e)(D6 + 9D4 + 24D2 + 16

)y = 0 =⇒ λ6 + 9λ4 + 24λ2 + 16 = 0

λ4 + 8λ2 + 16λ2 + 1 | λ6 + 9λ4 − 24λ2 + 16

λ6 + λ4

8λ4 + 24λ2

8λ4 + 8λ2

16λ2 + 16λ

=⇒(λ2 + 1

) (λ4 + 8λ2 + 16

)= 0 =⇒

(λ2 + 1

) (λ2 + 4

)2= 0

λ1 = i, λ2 = −i, λ3 = λ4 = 2i, λ5 = λ6 = −2i


y(x) = C1 cosx+ C2 sinx+ (C3 + C4x) cos 2x+ (C5 + C6x) sin 2x �

[6] Use the method of undetermined coefficients to obtain the general so-lution of (

D2 +D)y = sinx.

y = yh + yp

yh :(D2 +D

)y = 0 =⇒ λ2 + λ = 0

=⇒ λ(λ+ 1) = 0 −→ λ1 = 0, λ2 = −1

=⇒ yh = C1 + C2e−x

yp : Assume yp = A cosx+B sinx,

dypdx

= −A sinx+B cosx,d2ypdx2

= −A cosx−B sinx

d2ypdx2

+dypdx

= sinx

=⇒ −A cosx−B sinx−A sinx+B cosx = sinx

=⇒ −(A+B) sinx+ (B −A) cosx = sinx

−A−B = 1, B −A = 0 −→ A = B =⇒ A = B = −12

=⇒ yp = −cosx2− sinx

2

y = yh + yp =⇒ y(x) = C1 + C2e−x − cosx

2− sinx

2�

[7] Use the method of variation of parameters to obtain the general solutionof (

D2 − 4D + 4)y = ex.

y = yh + yp

yh :(D2 − 4D + 4

)y = 0 =⇒ λ2 − 4λ+ 4 = 0

=⇒ (λ− 2)(λ− 2) = 0 −→ λ1 = λ2 = 2

yh = C1e2x + C2xe

2x

yp : Assume yp = u1(x)e2x + u2(x)xe2x


u′1e

2x + u′2xe2x = 0

2u′1e2x + u′2

(e2x + 2xe2x

)= ex

W =∣∣∣∣ e2x xe2x

2e2x e2x + 2xe2x

∣∣∣∣ = e4x + 2xe4x − 2xe4x = e4x

u′1 =1e4x

∣∣∣∣ 0 xe2x

ex e2x + 2xe2x

∣∣∣∣ = −xe3x

e4x= −xe−x,

u1 = −∫xe−xdx = xe−x + e−x.

∫u dv = uv −

∫v du

u = x, du = dx

dv = e−xdx, v = −e−x

− xe−x −∫

(−e−x)dx = −xe−x − e−x

u′2 =1e4x

∣∣∣∣ e2x 02e2x ex

∣∣∣∣ =1ex

= e−x, u2 =∫e−xdx = −e−x.

=⇒ yp =(xe−x + e−x

)e2x − e−xxe2x = xex + ex − xex = ex

=⇒ y = yh + yp =⇒ y(x) = C1e2x + C2xe

2x + ex �

[8] Find a general solution of the following equations

(a)(x2D2 − 2

)y = 3x2 Euler− Cauchy type

Let x = et, x2D2 = Dt (Dt − 1)

=⇒[Dt(Dt − 1)− 2

]y = 3

(et

)2 =⇒(D2

t −Dt − 2)y = 3e2t

λ2 − λ− 2 = 0 −→ λ =1±√

1 + 82

={ 2−1

=⇒ yh = C1e2t + C2e

−t

For the particular solution assume

yp = Ate2t


multiply by t to differentiate from yh.

dypdt

= Ae2t + 2Ate2t

d2ypdt2

= 2Ae2t + 2Ae2t + 4Ate2t = 4Ae2t + 4Ate2t

=⇒ 4Ae2t + 4Ate2t −Ae2t − 2Ate2t − 2Ate2t = 3e2t

=⇒ 3Ae2t = 3e2t =⇒ A = 1

yp = te2t

y(t) = C1e2t + C2e

−t + te2t, x = et (lnx = t)

y(x) = C1x2 + C2x

−1 + x2 lnx �

(b)(x2D2 + xD − 1

)y = 4 Euler− Cauchy type

let x = et, xD = Dt, x2D2 = Dt (Dt − 1)[Dt(Dt − 1) +Dt − 1

]y = 4 =⇒

(D2

t − 1)y = 4

=⇒ λ2 − 1 = 0 −→ λ1 = 1, λ2 = −1

yh = C1et + C2e

−t

For the particular solution, assume

yp = A =⇒ D2t yp − yp = 4 =⇒ 0−A = 4 =⇒ A = −4

y(t) = C1et + C2e

−t − 4, x = et

=⇒ y(x) = C1x+ C2x−1 − 4 �


10.3 Series Solutions of Differential Equations (Chapter 4)

[1] Determine the interval of convergence for each of the following series,including consideration of the behavior of the series at each end point

(a)∞∑n=1

(−1)n(x− 2)n

(4n)√n

=∞∑n=1

(−1)n(x− x0)n

(4n)√n

where x0 = 2

L = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣∣ (−1)n+1/[(4n+1)

√n+ 1

](−1)n/ [(4n)

√n]

∣∣∣∣∣= lim

n→∞14

(n

n+ 1

)1/2

=14

The radius of convergence is: R =(

14

)−1

= 4.

The series converges when x0 −R < x < x0 +R −→ −2 < x < 6.The series diverges when x < −2 or x > 6.

For x = 6, reduces to the convergent alternating series∞∑n=1

(−1)n√n

For x = −2, reduces to the divergent series∞∑n=1

1√n

.

=⇒ Interval of convergence is : (−2, 6 ] �

(b)∞∑n=0

2nxn

n!where x0 = 0

L = limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣∣∣2n+1

(n+1)!

2n

n!

∣∣∣∣∣∣ = limn→∞

∣∣∣∣∣2n·2

n!(n+1)

2n

n!

∣∣∣∣∣ = limn→∞

∣∣∣∣ 2n+ 1

∣∣∣∣ = 0

Interval of convergence :(− 1L,1L

)= (−∞,∞)

Therefore, the series converges for all x �


[2] For each of the following differential equations, obtain the most generalsolution which is representable by a Maclaurin series (Apply the powerseries method: y =

∑∞k=0 akx

k)

(a) x2 d2y

dx2− dy

dx+ y = 0

Let y =∞∑k=0

akxk,

dy

dx=

∞∑k=0

kakxk−1,

d2y

dx2=

∞∑k=0

k(k − 1)akxk−2

Ly =∞∑k=0

k(k − 1)akxk −∞∑k=0

kakxk−1 +

∞∑k=0

akxk = 0

=⇒∞∑k=0

[k(k − 1) + 1]akxk −∞∑k=0

kakxk−1 = 0

=⇒∞∑k=1

[(k − 1)(k − 2) + 1]ak−1xk−1 −

∞∑k=0

kakxk−1

︸︷︷︸∑∞k=1

kakxk−1

= 0

=⇒∞∑k=1

[(k2 − 3k + 3

)ak−1 − kak

]xk−1 = 0

recurrence formula : ak =(k2 − 3k + 3

k

)ak−1 (k = 1, 2, 3, · · ·)

a1 = a0, a2 =a1

2=a0

2, a3 = a2 =

a0

2, a4 =

7a3

4=

7a0

8

=⇒ y = a0

(1 + x+

x2

2+x3

2+

7x4

8+ · · ·

)�

(b) x2 d2y

dx2− dy

dx= 0

Let y =∞∑k=0

akxk,

dy

dx=

∞∑k=0

kakxk−1,

d2y

dx2=

∞∑k=0

k(k − 1)akxk−2


Ly =∞∑k=0

k(k − 1)akxk −∞∑k=0

kakxk−1 = 0

=⇒∞∑k=1

[(k − 1)(k − 2)] ak−1xk−1 −

∞∑k=0

kakxk−1 = 0

=⇒∞∑k=1

(k2 − 3k + 2

)ak−1x

k−1 −∞∑k=1

kakxk−1 = 0

=⇒∞∑k=0

[(k2 − 3k + 2)ak−1 − kak

]xk−1 = 0

ak =(k2 − 3k + 2

k

)ak−1 (k = 1, 2, 3, · · ·)

a1 = 0, a2 = 0, a3 =2a2

3= 0, a4 =

6a3

4= 0

=⇒ y = a0 �

(c)dy

dx+ x

dy

dx= y

Let y =∞∑k=0

akxk,

dy

dx=

∞∑k=0

kakxk−1

∞∑k=0

kakxk−1 +

∞∑k=0

kakxk −

∞∑k=0

akxk = 0

=⇒∞∑k=0

kakxk−1 +

∞∑k=0

(k − 1)akxk = 0

=⇒∞∑

k+1=0

(k + 1)ak+1xk +

∞∑k=0

(k − 1)akxk = 0

=⇒∞∑k=0

[(k + 1)ak+1 + (k − 1)ak]xk = 0

(k + 1)ak+1 + (k − 1)ak = 0 =⇒ ak+1 = −(k − 1k + 1

)ak


a1 = a0, a2 = a3 = · · · = 0

y =∞∑k=0

akxk = a0 + a1x =⇒ y = a0(1 + x) �

[3] Show the relation between m, p, and s

α︷︸︸︷∞∑

m=2

m(m− 1)amxm−2 =

β︷︸︸︷∞∑p=1

(p+ 1)pap+1xp−1 =

γ︷︸︸︷∞∑s=0

(s+ 2)(s+ 1)as+2xs

∞∑m=2

m(m− 1)amxm−2 =∞∑

m+1=1

(m+ 1)mam+1xm−1

comparing with β =⇒ m = p+ 1 �∞∑

m=2

m(m− 1)amxm−2 =∞∑

m+2=0

(m+ 2)(m+ 1)am+2xm

comparing to γ =⇒ m = s+ 2 �

[4] Use the method of Frobenius to obtain the general solution of the fol-lowing differential equation, valid near x = 0:

xd2y

dx2+ 2

dy

dx+ xy = 0, at x = 0 regular singular point

let y = xr∞∑

m=0

amxm, y′ =

∞∑m=0

(m+ r)amxm+r−1,

y′′ =∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−2

=⇒∞∑

m=0

(m+ r)(m+ r − 1)amxm+r−1 + 2∞∑

m=0

(m+ r)amxm+r−1

+∞∑

m=0

amxm+r+1 = 0


=⇒∞∑

m=0

[(m+ r)(m+ r − 1) + 2(m+ r)] amxm+r−1 +∞∑

m=0

amxm+r+1 = 0

=⇒∞∑

m+2=0

[(m+ r + 2)(m+ r + 1) + 2(m+ r + 2)] am+2xm+r+1

+∞∑

m=0

amxm+r+1 = 0

=⇒ [r(r − 1) + 2r] a0xr−1 + [(r + 1)r + 2(r + 1)] a1x

r

+∞∑

m=0

{[(m+ r + 2)(m+ r + 1) + 2(m+ r + 2)] am+2 + am}xm+r+1 = 0

xr−1 terms : r(r − 1) + 2r = 0←− indicial equation (a0 �= 0)

=⇒ r2 + r = 0 =⇒ r(r + 1) = 0

=⇒ r1 = 0, r2 = −1

xr terms :[(r + 1)r + 2(r + 1)

]a1 = 0

for r1 = 0 −→ 2a1 = 0 =⇒ a1 = 0

r2 = −1 −→ 0a1 = 0 =⇒ a1 = arbitrary constant

xm+r+1 terms :

am+2 =−am

(m+ r + 2)(m+ r + 1) + 2(m+ r + 2)(m = 0, 1, 2, · · ·)

r1 = 0 −→ a1 = 0

am+2 =−am

(m+ 2)(m+ 1) + 2(m+ 2)=

−am(m+ 2)(m+ 3)

a2 = −a0

6= −a0

3!

a3 = −a1

12= 0 = a5 = a7 · · ·

a4 = −a2

20= − 1

20

(−a0

6

)=

a0

120=a0

5!

a6 = −a4

42= − 1

42

( a0

120

)= − a0

5040= −a0

7!


r2 = −1 am+2 =−am

(m+ 1)m+ 2(m+ 1)=

−am(m+ 1)(m+ 2)

a2 = −a0

2!

a3 = −a1

6= −a1

3!

a4 = −a2

12= − 1

12

(−a0

2

)=a0

24=a0

4!

a5 = −a3

20=

(−120

) (−a1

6

)=

a1

120=a1

5!

=⇒ y(x) = C1y1 + C2y2

= C1

∞∑m=0

amxm + C2

∞∑m=0

amxm−1

= C1

[a0 −

a0

3!x2 +

a0

5!x4 − a0

7!x6 +− · · ·

]+ x−1C2

[a0 + a1x−

a0

2!x2 − a1

3!x3 +

a0

4!x4 +

a1

5!x5 −−+ + · · ·

]

=⇒ y(x) = C1x−1a0

[x− 1

3!x3 +

x5

5!− x7

7!+− · · ·

]

+ C2a0x−1

[1− x2

2!+x4

4!−+ · · ·

]

+ C2a1x−1

[x− x3

3!+x5

5!−+ · · ·

]

=⇒ y(x) = Ax−1

[1− x2

2!+x4

4!−+ · · ·

]

+Bx−1

[x− x3

3!+x5

5!− x7

7!+− · · ·

]

(Let A = C2a0 , B = C1a0 + C2a1)

=⇒ y(x) = x−1 (A cosx+B sinx) �


[5] Find P6(x)

Pn+1(x) =2n+ 1n+ 1

xPn(x)− n

n+ 1Pn−1(x)

P4 =18

(35x4 − 30x2 + 3

)P5 =

18

(63x5 − 70x3 + 15x

)P6 =

11x6

[18

(63x5 − 70x3 + 15x

)]− 5

6

[18

(35x4 − 30x2 + 3

)]

=[69348

x6 − 77048

x4 +16548

x2

]−

[17548

x4 − 15048

x2 +1548

]

=⇒ P6(x) =116

(231x6 − 315x4 + 105x2 − 5

)�

[6] Obtain the associated Legendre functions: P 12 (x), P 2

3 (x), and P 32 (x).

Pmn (x) =

(1− x2

)m/2 dmPn(x)dxm

P2(x) =12

(3x2 − 1

)P3(x) =

12

(5x3 − 3x

)

(a) P 12 (x) =

(1− x2

)1/2 d

dxP2(x) =

(1− x2

)1/2 d

dx

[12

(3x2 − 1

)]

=⇒ P 12 (x) = 3x

(1− x2

)1/2 �

(b) P 23 (x) =

(1− x2

)2/2 d2

dx2P3(x) =

(1− x2

) d2

dx2

[12

(5x3 − 3x

)]

=(1− x2

) d

dx

[15x2 − 3

2

]

=⇒ P 23 (x) = 15

(x− x3

)�

(c) P 32 (x) =

(1− x2

)3/2 d3

dx3P2(x) =

(1− x2

)3/2 d3

dx3

[12

(3x2 − 1

)]

=(1− x2

)3/2 d2

dx2(3x) =

(1− x2

)3/2 d

dx(3)

=⇒ P 32 (x) = 0 �


[7] By making an appropriate change of variables, obtain the general solu-tion of the differential equation:

(Ax+B)d2y

dx2+A

dy

dx+A2(Ax+B)y = 0.

Let N = Ax+B =⇒ dN

dx= A

dy

dx=

dy

dN

dN

dx=

dy

dNA

d2y

dx2=

d

dx

(dy

dx

)=

d

dN

(dy

dx

)dN

dx= A2 d

2y

dN2

=⇒ NA2 d2y

dN2+A

[Ady

dN

]+A2Ny = 0

=⇒ Nd2y

dN2+dy

dN+Ny = 0

=⇒ N2 d2y

dN2+N

dy

dN+N2y = 0

(Bessel′s eq. with ν = 0

)=⇒ y(N) = C1J0(N) + C2Y0(N), N = Ax+B

=⇒ y(Ax+B) = y(x) = C1J0(Ax+B) + C2Y0(Ax+B) �

[8] Using the series definition,

Jn(x) =∞∑

m=0

(−1)m(x2 )2m+n

m!(n+m)!

Prove thatd

dx

[x−nJn(ax)

]= −ax−nJn+1(ax)

(Note :

1m!

=1

Γ(m+ 1),

1Γ(0)

= 0)


d

dx

[x−nJn(ax)

]=

d

dx

[x−n

∞∑m=0

(−1)m(xa2 )2m+n

m!(n+m)!

]

=d

dx

[ ∞∑m=0

(−1)ma2m+nx2m

22m+nm!(n+m)!

]

=∞∑

m=0

(−1)ma2m+n2mx2m−1

22m+nm!(n+m)!

=∞∑

m=0

(−1)ma2m+nx2m−1

22m+n−1(m− 1)!(n+m)!

= ax−n∞∑

m=0

(−1)ma2m+n−1x2m+n−1

22m+n−1(m− 1)!(n+m)! (m− 1)! =

m!m

(m+ n)! = (m+ n)(m+ n− 1) · · · (m+ 1)m(m− 1) · · ·

= ax−n∞∑

m=0

(−1)m(ax2 )2m+n−1

(m!m )[(m+ n)(m+ n− 1) · · · (m+ 1)m(m− 1) · · ·]

(let m = m+ 1)

= −ax−n∞∑

m+1=0

−(−1)m+1(ax2 )2(m+1)+n−1

(m+ 1)![(m+ n+ 1)(m+ n) · · · (m+ 2)︸︷︷︸(m+n+1)!

(m+1)!

m(m− 1) · · ·︸︷︷︸m!

]

= −ax−n∞∑

m=−1

(−1)m(ax2 )2m+n+1

(m+ 1)! (m+n+1)!(m+1)! m!

Note : m! = Γ(m+ 1)1

Γ(m+ 1)= 0 at m = −1

the summation term for m = −1 is zero

= −ax−n∞∑

m=0

(−1)m(ax2 )2m+n+1

m!(m+ n+ 1)!

(comparing with the definition of Jn(x))

= −ax−nJn+1(ax) �


[9] Prove that y1 =√xJ1/4

[x2

2

]and y2 =

√xY1/4

[x2

2

]are the solutions

ofd2y

dx2+ x2y = 0

(Hint : let z =

12x2 and then y = uz

14

)

z =12x2 =⇒ dz

dx= x

dy

dx=dy

dz

dz

dx= x

dy

dz

d2y

dx2=

d

dx

(dy

dx

)=

d

dx

(xdy

dz

)=dy

dz+ x

d

dz

(dy

dz

)dz

dx=dy

dz+ x2 d

2y

dz2

=⇒

x2 d

2y

dz2+dy

dz+ x2y = 0

z =x2

2

=⇒2zd2y

dz2+dy

dz+ 2zy = 0

y = uz14

dy

dz=

(z

14

)dudz

+14uz−

34

d2y

dz2=d2u

dz2

(z

14

)+

12du

dz

(z−

34

)− 3

16uz−

74

=⇒ 2z[z

14d2u

dz2+

12du

dzz−

34 − 3

16uz−

74

]+ z

14du

dz+

14uz−

34 + 2z

54u = 0

=⇒ 2z54d2u

dz2+ 2z

14du

dz− 1

8z−

34u+ 2z

54u = 0

(multiply by

z34

2

)

=⇒ z2 d2u

dz2+ z

du

dz+

(z2 − 1

16

)u = 0

(Bessel′s eq. with ν =

14

)

=⇒ u(z) = AJ1/4[z] +BY1/4[z], u =y

z14

=⇒ y(z) = Az14 J1/4[z] +Bz

14Y1/4[z], z =

12x2

=⇒ y(x) = A

(x2

2

) 14

J1/4

[x2

2

]+B

(x2

2

) 14

Y1/4

[x2

2

],


let A∗ = A

(12

) 14

, B∗ = B

(12

) 14

=⇒ y(x) = A∗√xJ1/4

[x2

2

]+B∗√xY1/4

[x2

2

]�

[10] Find the eigenvalues and the eigenfunctions of the following Sturm-Liouville problems:

(a)d

dx

[xdy

dx

]+λ

xy = 0, y(1) = 0,

dy(e)dx

= 0

let x = et =⇒ dx

dt= et =⇒ dt

dx= e−t,

dy

dx=dy

dt

dt

dx=dy

dte−t

=⇒ d

dx

[etdy

dte−t

]+ λe−ty = 0 =⇒ d

dt

(dy

dt

)dt

dx+ λe−ty = 0

=⇒ e−t

[d2y

dt2+λy

]= 0 =⇒ d2y

dt2+λy = 0 =⇒ y(t) = A cos

√λt+B sin

√λt

BC1 : x = et =⇒ 1 = e0 −→ y(x = 1) = y(t = 0) = 0

BC2 :dy

dx=dy

dte−t, x = et =⇒ e = e1

−→ dy(x = e)dx

= e−t dy(t = 1)dt

= 0

y(t = 0) = 0 =⇒ A = 0

e−t

(dy

dt

)= e−t

(−√λA sin

√λt+

√λB cos

√λt

)

=⇒ e−t dy(t = 1)dt

=1e

(√λB cos

√λ)

= 0 =⇒ B∗√λ cos√λ = 0

=⇒√λ =

(2n+ 1

2

)π (n = 0, 1, · · ·)

eigenvalues : λ =[(2n+ 1)π

2

]2

(n = 0, 1, · · ·) �

=⇒ y(t) = B sin[(2n+ 1)

πt

2

], x = et =⇒ t = lnx


eigenfunctions : y(x) = B sin[(2n+ 1)π

2lnx

](n = 0, 1, · · ·) �

(b)d2y

dx2+ λy = 0, y(0) = y(π),

dy(0)dx

=dy(π)dx

.

(For solution outline see previous problem)

y(x) = A cos√λx+B sin

√λx

dy(x)dx

= −A√λ sin

√λx+B

√λ cos

√λx

BC1 : y(0) = y(π) =⇒ A = A cos√λπ +B sin

√λπ

BC2 :dy(0)dx

=dy(π)dx

=⇒ B√λ = −A

√λ sin

√λπ +B

√λ cos

√λπ

A = A cos

√λπ +B sin

√λπ

B = −A sin√λπ +B cos

√λπ

=⇒

A

(cos√λπ − 1

)+B sin

√λπ = 0

−A sin√λπ +B

(cos√λπ − 1

)= 0

A non-trivial solution of this system (A = B �= 0) can be obtained if:

cos√λπ = 1︸︷︷︸

√λ=2n∗ (n∗=0,1,2,···)

and sin√λπ = 0︸︷︷︸

√λ=n∗∗ (n∗∗=0,1,2,···)

=⇒√λ = 2n

eigenvalues : λn = 4n2 (n = 0, 1, 2, · · ·)eigenfunctions : yn(x) = A cos 2nx+B sin 2nx (n = 0, 1, 2, · · ·) �


[11] Show that the following set is orthogonal on the interval −1 ≤ x ≤ 1,and determine the corresponding orthonormal set;

sinπx, sin 2πx, sin 3πx, . . .

The general term is sinnπx (n = 1, 2, 3, · · ·)

(gm, gn) =

1∫−1

(sinmπx)(sinnπx)dx

(sinx sin y =

12[cos(x− y)− cos(x+ y)

])

=⇒ (gm, gn) =12

1∫−1

cos(m− n)πxdx− 12

1∫−1

cos(m+ n)πxdx

=sin(m− n)πx2(m− n)π

∣∣∣∣1

−1

− sin(m+ n)πx2(m+ n)π

∣∣∣∣1

−1

= 0.

Therefore, it is an orthogonal set. Now obtain the norm of each function.

||gm||2 = (gm, gm) =

1∫−1

sin2mπxdx =12

1∫−1

(1− cos 2mπx)dx

=12

(x− sin 2mπx

2mπ

) ∣∣∣∣1

−1

= 1

=⇒ ||gm|| =√

1 = 1 =⇒ The set is also orthonormal �

[12] Prove that

1∫−1

Pm(x)Pn(x)dx = 0 if m �= n.

Since Pm(x) and Pn(x) satisfy Legendre’s equation

(1− x2

)P ′′m − 2xP ′

m +m(m+ 1)Pm = 0

(1− x2

)P ′′n − 2xP ′

n + n(n+ 1)Pn = 0


Multiply the first by Pn, the second by Pm and subtract

(1− x2

)[PnP ′′

m − PmP ′′n ]− 2x [PnP ′

m − PmP ′n]

= [n(n+ 1)−m(m+ 1)]PmPn

=⇒(1− x2

) d

dx[PnP ′

m − PmP ′n]− 2x

[PnP

′m − PmP ′

n

]=

[n(n+ 1)−m(m+ 1)

]PmPn

=⇒ d

dx

{(1− x2

)[PnP

′m − PmP ′

n

]}=

[n(n+ 1)−m(m+ 1)

]PmPn

Integrating

−→(1− x2

)[PnP

′m − PmP ′

n

]∣∣∣1−1

=[n(n+ 1)−m(m+ 1)

] 1∫−1

PmPndx

=⇒ 0 =[n(n+ 1)−m(m+ 1)

] 1∫−1

PmPndx

Because m �= n,

1∫−1

Pm(x)Pn(x)dx = 0 �


10.4 Laplace Transform (Chapter 5)

[1] Find the following Laplace transform

L{

cos√t√

t

} (Hint : L

{sin√t}

=√π

2s3/2exp

[− 1

4s

]).

let f(t) = sin√t, f(0) = 0

=⇒ f ′(t) =cos√t

2√t

(L{f ′(t)} = sF (s)− f(0))

L{df(t)dt

}=

12L

{cos√t√

t

}= sF (s)− f(0) = s

√π

2s32e−

14s − 0

=⇒ L{

cos√t√

t

}=

(πs

) 12e−

14s �

[2] Find L{f(t)

}where f(t) =

{ sin t 0 < t < π,

0 π < t < 2π.

Periodic functions : L{f(t)} =

p∫0

e−stf(t)dt

1− e−ps, p = 2π∫

eat sin btdt =eat(a sin bt− b cos bt)

a2 + b2

L{f(t)} =1

1− e−2πs

2π∫0

e−stf(t)dt

=1

1− e−2πs

π∫0

e−st sin tdt

=1

1− e−2πs

{e−st(−s sin t− cos t)

s2 + 1

}∣∣∣∣π

0

=1

1− e−2πs

{e−πs + 1s2 + 1

}

=1 + e−πs

(1 + e−πs)(1− e−πs)(s2 + 1)

=1

(1− e−πs)(s2 + 1)�


[3] Evaluate each of the following

(a) L{e−3t cos 2t

}=

s+ 3(s+ 3)2 + 4

=s+ 3

s2 + 6s+ 13�

L

{eatf(t)

}= F (s− a), First translation

L{cos 2t} =s

s2 + 4, (a = −3)

(b) L{e−tt2

}=

2(s+ 1)3

�

{L

{t2

}=

2!s3

=2s3, (a = −1)

}

(c) L{4e2t sin t

}=

4(s− 2)2 + 1

=4

s2 − 4s+ 5�

{L{4 sin t} =

4s2 + 1

, (a = 2)}

(d) L{J0(t)} =?{J0(t) = 1− t2

22(1!)2+

t4

24(2!)2− t6

26(3!)2+− · · ·

}

L{J0(t)} =1s− 1

22

2!s3

+1

2242

4!s5− 1

224262

6!s7

+ · · ·

=1s

{1− 1

2

(1s2

)+

1 · 32 · 4

(1s4

)− 1 · 3 · 5

2 · 4 · 6

(1s6

)+ · · ·

}

=1s

{(1 +

1s2

)− 12}

=1s

(s2

s2 + 1

) 12

=1√s2 + 1

�

(e) L{J0(at)

}=

1a

1√(sa

)2 + 1=

1a

(a2

s2 + a2

) 12

=1√

s2 + a2�


L{f(at)} =

1aF

( sa

), change of scale property

F (s) = L{J0(t)} =1√s2 + 1

.

[4] Evaluate each of the following by use of inverse Laplace transform the-orems

(a) L−1

{4s+ 12

s2 + 8s+ 16

}= L−1

{4s+ 12(s+ 4)2

}= L−1

{4(s+ 4)− 4

(s+ 4)2

}

= 4L−1

{1

s+ 4

}− 4L−1

{1

(s+ 4)2

}= 4e−4t − 4e−4tt = 4e−4t(1− t) �L−1

{F (s− a)

}= eatf(t), first translation

L−1

{1s2

}= t.

(b) L−1

{3s+ 7

s2 − 2s− 3

}= L−1

{3s+ 7

(s− 1)2 − 4

}= L−1

{3(s− 1) + 10(s− 1)2 − 4

}

= 3L−1

{s− 1

(s− 1)2 − 4

}+ 5L−1

{2

(s− 1)2 − 4

}= 3et cosh 2t+ 5et sinh 2t = et [3 cosh 2t+ 5 sinh 2t] �L−1

{s

s2 − a2

}= cosh at, from Kreyszig Table 6.10

L−1

{a

s2 − a2

}= sinh at.

L−1

{se−

4πs5

s2 + 25

}= cos 5

(t− 4π

5

)u

(t− 4π

5

)(c)

=

cos 5(t− 4π

5

)t >

4π5

0 t <4π5

�


L−1{e−asF (s)

}= f(t− a)u(t− a), second shifting property

L−1

{s

s2 + 25

}= cos 5t,

u(t− a) ={ 0 t < a

1 t > a

(d) L−1

{s+ 1

s2 + s+ 1

}= L−1

{s+ 1

(s+ 12 )2 + 3

4

}= L−1

{s+ 1

2 + 12

(s+ 12 )2 + 3

4

}

= L−1

{s+ 1

2

(s+ 12 )2 + 3

4

}+

1√3L−1

{ √3

2

(s+ 12 )2 + 3

4

}

= e−t2 cos

√3t2

+1√3e−

t2 sin

√3t2

=e−

t2

√3

{√

3 cos√

3t2

+ sin√

3t2

}�

L−1 {F (s− a)} = eatf(t), first shifting property

L−1

{s

s2 + a2

}= cos at, L−1

{a

s2 + a2

}= sin at

(e) L−1

{4s− 2

(s2 + 1)2

}=?

L−1

{1

s2 + 1

}= sin t

L−1

{4s− 2s2 + 1

}= 4L−1

{s

s2 + 1

}− 2L−1

{1

s2 + 1

}= 4 cos t− 2 sin t


L−1

{4s− 2

(s2 + 1)2

}= (4 cos t− 2 sin t) ∗ sin t

=

t∫0

(4 cos τ − 2 sin τ) sin(t− τ)dτ

=12

t∫0

{4[sin t+ sin(t− 2τ)]− 2[− cos t+ cos(2τ − t)]} dτ

= 2[τ sin t+

cos(t− 2τ)2

]t0

+[τ cos t− sin(2τ − t)

2

]t0

= 2t sin t+ cos(−t)− cos t+ t cos t− sin t2

+sin(−t)

2= 2t sin t− sin t+ t cos t �

(f) L−1

{1

(s2 + 1)3

}=?

L−1

{1

(s2 + 1)

}= sin t

L−1

{1

(s2 + 1)2

}= sin t ∗ sin t =

t∫0

sin τ sin(t− τ)dτ

=12

t∫0

[cos(2τ − t)− cos t] dτ

(sinx sin y =

12

[− cos(x+ y) + cos(x− y)])

=12

[sin(2τ − t)

2− τ cos t

]t0

=12

[sin t2− sin(−t)

2− t cos t

]

=12

[sin t− t cos t]


L−1

{1

(s2 + 1)3

}=

(sin t− t cos t

2

)∗ sin t

=

t∫0

(sin τ − τ cos τ

2

)sin(t− τ)dτ

=14

t∫0

[− cos t+ cos(2τ − t)− τ sin t− τ sin(t− 2τ)︸︷︷︸by parts

]dτ

cos y sinx =12

[sin(x+ y) + sin(x− y)]

by parts : uv −∫vdu, u = τ, du = dτ

dv = sin(t− 2τ)dτ, v =cos(t− 2τ)

2

=14

[−τ cos t+

sin(2τ − t)2

− τ2 sin t2

− τ cos(t− 2τ)2

− sin(t− 2τ)2 · 2

]t0

=14

[−t cos t+

sin t2− sin(−t)

2− t2 sin t

2− t cos(−t)

2− sin(−t)

4+

sin t4

]

= − t cos t4− t2 sin t

8− t cos t

8+

3 sin t8

=3 sin t

8− 3t cos t

8− t2 sin t

8�

L−1

{3s+ 1

(s− 1)(s2 + 1)

}= L−1

{2

s− 1

}+ L−1

{−2s+ 1s2 + 1

}(g)

= 2L−1

{1

s− 1

}+ L−1

{1

s2 + 1

}− 2L−1

{s

s2 + 1

}= 2et + sin t− 2 cos t �


3s+ 1(s− 1)(s2 + 1)

=A

s− 1+Bs+ C

s2 + 1

=⇒ 3s+ 1 = (s2 + 1)A+ (s− 1)(Bs+ C)

= s2(A+B) + s(C −B) +A− C=⇒ A+B = 0, C −B = 3, A− C = 1

−→ A = 2, B = −2, C = 1


10.5 Application of Laplace Transforms in the Solution ofOrdinary Differential Equations (Chapter 6)

Use Laplace transforms to solve the following:[1] y′′(t) + y(t) = 1 given y(0) = 1 y′(0) = 0. Taking the Laplacetransform of both sides of the differential equation yields:

[s2y(s)− s y(0)︸︷︷︸

1

− y′(0)︸︷︷︸0

]+ y(s) =

1s

=⇒ s2y(s)− s+ y(s) =1s

=⇒ y(s)[s2 + 1

]= s+

1s

=⇒ y(s) =s+ 1

s

s2 + 1=

1s

=⇒ y(t) = L−1{y(s)

}= L−1

{1s

}= 1 �

[2] 2x′(t) + 3x(t) = e4t given x(0) = 5.Taking the Laplace transforms

2[sx(s)− x(0)︸︷︷︸

5

]+ 3x(s) =

1s− 4

=⇒ x(s)[2s+ 3]− 10 =1

s− 4

=⇒ x(s) =[10 +

1s− 4

]1

2s+ 3

=10

2s+ 3+

1(s− 4)(2s+ 3)

=5

s+ 32

−211

2s+ 3+

111

s− 4

=5

s+ 32

−111

s+ 32

+111

s− 4

=(

5411

)1

s+ 32

+(

111

)1

s− 4

=⇒ x(t) = L−1{x(s)

}=

5411L−1

{1

s+ 32

}+

111L−1

{1

s− 4

}

=⇒ x(t) =111

{54e−

32 t + e4t

}�


[3] y′′ + y = t given y(0) = 1, and y′(0) = −2.Taking the Laplace transform of both sides of the ordinary differential equa-tion and using the given conditions yields:

L{y′′ + y

}= L{t} =⇒ L

{y′′

}+ L{y} = L{t}

=⇒[s2y(s)− s y(0)︸︷︷︸

1

− y′(0)︸︷︷︸−2

]+ y(s) =

1s2

=⇒ s2y(s)− s+ 2 + y(s) =1s2

=⇒ y(s)[s2 + 1

]=

1s2

+ s− 2

=⇒ y(s) =1

s2(s2 + 1)+s− 2s2 + 1

=1s2− 1s2 + 1

+s

s2 + 1− 2s2 + 1

=⇒ y(t) = L−1

{1s2

}− 3L−1

{1

s2 + 1

}+ L−1

{s

s2 + 1

}=⇒ y(t) = t− 3 sin t+ cos t �

[4] x′(t)− 3x(t) = te2t given x(0) = 0.

[sx(s)− x(0)︸︷︷︸

0

]− 3x(s) =

1(s− 2)2

=⇒ x(s)[s− 3] =1

(s− 2)2

=⇒ x(s) =1

(s− 3)(s− 2)2=

1s− 3

− 1s− 2

− 1(s− 2)2

=⇒ x(t) = L−1

{1

s− 3

}− L−1

{1

s− 2

}− L−1

{1

(s− 2)2

}

=⇒ x(t) = e3t − e2t − te2t =⇒ x(t) = e3t − (1 + t)e2t �

[5] y′′ − 3y′ + 2y = 4e2t given y(0) = −3, and y′(0) = 5.Taking the Laplace transform of both sides of the differential equationyields:

L{y′′

}− 3L

{y′

}+ 2L{y} = 4L

{e2t

} (L

{eat

}=

1s− a

)


=⇒[s2y(s)− s y(0)︸︷︷︸

−3

− y′(0)︸︷︷︸5

]− 3

[sy(s)− y(0)︸︷︷︸

−3

]+ 2y(s) = 4

(1

s− 2

)

=⇒[s2y(s) + 3s− 5

]− 3 [sy(s) + 3] + 2y(s) =

4s− 2

=⇒ y(s)[s2 − 3s+ 2

]+ 3s− 14 =

4s− 2

=⇒ y(s) =4

(s− 2)(s2 − 3s+ 2)+

14− 3ss2 − 3s+ 2

=20s− 3s2 − 24(s− 2)2(s− 1)

=A

s− 1+

B

s− 2+

C

(s− 2)2

= − 7s− 1

+4

s− 2+

4(s− 2)2

20s− 3s2 − 24 = A(s− 2)2 +B(s− 2)(s− 1) + C(s− 1)

= s2(A+B)− s(4A+ 3B − C) + 4A+ 2B − CA+B = −3, 4A+ 3B − C = −20, 4A+ 2B − C = −24

−→ A = −7, B = 4, C = 4

=⇒ y(t) = L−1

{ −7s− 1

}+ L−1

{4

s− 2

}+ L−1

{4

(s− 2)2

}L−1

{F (s− a)

}= eatf(t) shifting property

L−1

{4s2

}= 4t

=⇒ y(t) = −7et + 4e2t + 4te2t �

[6] yIV (t) + 2y′′(t) + y(t) = sin t given y(0) = 1, y′(0) = −2, y′′(0) =3 and y′′′(0) = 0.(

L{f (n)(t)

}= snF (s)− sn−1f(0)− sn−2f ′(0)− · · · − f (n−1)(0)

)[s4y(s)− s3 y(0)︸︷︷︸

1

− s2 y′(0)︸︷︷︸−2

−s y′′(0)︸︷︷︸3

− y′′′(0)︸︷︷︸0

]

+ 2

[s2y(s)− s y(0)︸︷︷︸

1

− y′(0)︸︷︷︸−2

]+ y(s) =

1s2 + 1


=⇒[s4y(s)− s3 + 2s2 − 3s

]+ 2

[s2y(s)− s+ 2

]+ y(s) =

1s2 + 1

=⇒ y(s)[s4 + 2s2 + 1︸︷︷︸

(s2+1)2

]=

1s2 + 1

+ s3 − 2s2 + 5s− 4

=⇒ y(s) =1

(s2 + 1)3+s3 − 2s2 + 5s− 4

(s2 + 1)2

=1

(s2 + 1)3+

(s3 + s)− 2(s2 + 1) + 4s− 2(s2 + 1)2

=1

(s2 + 1)3+

s

s2 + 1− 2s2 + 1

+4s− 2

(s2 + 1)2

L−1

{1

(s2 + 1)3

}=

38

sin t− 38t cos t− 1

8t2 sin t (HW#5 4f)

L−1

{4s− 2

(s2 + 1)2

}= 2t sin t− sin t+ t cos t (HW#5 4e)

=⇒ y(t) = L−1{y(s)

}= L−1

{1

(s2 + 1)3

}+ L−1

{s

s2 + 1

}

− 2L−1

{1

s2 + 1

}+ L−1

{4s− 2

(s2 + 1)2

}

=[38

sin t− 38t cos t− 1

8t2 sin t

]+ cos t− 2 sin t

+ [2t sin t− sin t+ t cos t]

= cos t+58t cos t− 21

8sin t+ 2t sin t− t2

8sin t

=⇒ y(t) =(

1 +5t8

)cos t−

(218− 2t+

t2

8

)sin t �

[7] A uniform light horizontal beam PQ, of length 2c end supported at P ,carries a load which decreases uniformly from w (newtons/m) at P to zeroat x = c. A point load F (newtons) occurs at a distance b from P . Findan expression for the displacement at any point in the beam in terms of x.

d4y

dx4=

1EI

W (x)


W

P Q

F

c

b

2c

where W (x) =[w − w

cx]−

[w − w

cx]u(x− c) + Fδ(x− b),

u(x− c) =

{ 1 x > c,

0 x < c.

given y(0) = 0, y′(0) = 0, y′′(2c) = 0, and y′′′(2c) = 0.

=⇒ d4y

dx4=

1EI

{w

c

[c− x− (c− x)u(x− c)

]+ Fδ(x− b)

}

=1EI

{w

c

[c− x+ (x− c)u(x− c)

]+ Fδ(x− b)

}

=⇒ EI

s4y(s)− s3 y(0)︸︷︷︸

0

−s2 y′(0)︸︷︷︸0

−s y′′(0)︸︷︷︸A

− y′′′(0)︸︷︷︸B

=w

c

[c

s− 1s2

+e−cs

s2

]+ Fe−bs

{L−1

{e−asF (s)

}= f(t− a)u(t− a)

L{δ(t− a)

}= e−as

}

EI{s4y(s)−As−B

}=w

c

{c

s− 1s2

+e−cs

s2

}+ Fe−bs


=⇒ EIy(s) =A

s3+B

s4+w

c

{c

s5− 1s6

+e−cs

s6

}+Fe−bs

s4

L−1 =⇒ EIy(x) =Ax2

2+Bx3

6+w

c

{cx4

24− x5

120+

(x− c)5120

u(x− c)}

+F (x− b)3

6u(x− b)

Now evaluate A & B

EIy′(x) = Ax+Bx2

2+w

c

{cx3

6− x4

24+

(x− c)424

u(x− c)}

+F (x− b)2

2u(x− b)

EIy′′(x) = A+Bx+w

c

{cx2

2− x3

6+

(x− c)36

u(x− c)}

+ F (x− b)u(x− b)

EIy′′′(x) = B +w

c

{cx− x2

2+

(x− c)22

u(x− c)}

+ Fu(x− b)

At x = 2c, y′′(2c) = 0

=⇒ 0 =A+ 2Bc+w

c

{2c3 − 4c3

3+c3

6

}+ F (2c− b)

At x = 2c, y′′′(2c) = 0

=⇒ 0 =B +w

c

{2c2 − 4c2

2+c2

2

}+ F

=⇒ B = −[wc

2+ F

]

=⇒ A = 2[wc

2+ F

]c− w

c

[56c3

]− F (2c− b)

= wc2 + 2cF − 56wc2 − F (2c− b)

= wc2(

1− 56

)+ Fb

=wc2

6+ Fb


Substituting the value of A and B into the previous result yields:

EIy(x) =[wc2

6+ Fb

]x2

2−

[wc

2+ F

]x3

6

+[cx4

24− x5

120+

(x− c)5120

u(x− c)]w

c

+F (x− b)3

6u(x− b) �


B

A

b

c

a

a+b

D

B C

D

O

a

b

c

d

a

b c

p'1

p'2

p1

p2

c'

10.6 Linear Algebra (Chapter 7)

[1] Consider a vector represented by an arrow running from a point P to apoint Q (P −→ Q). The straight line through P and Q is called the line ofaction of the vector, the point P is called the origin of the vector and thepoint Q is called the terminus of the vector. (a) If a and b are vectors witha common origin O and terminuses A and B, in terms of a and b, find thevector OC where C is the middle point of AB. (b) If a, b, c and d havea common origin and terminuses A, B, C, and D, and if b − a = c − dshow that ABCD is a parallelogram. (c) Prove that the line joining themiddle points of any two sides of a triangle is parallel to the third side, andis equal in length to one half of the length of the third side. NOTE: Youare not allowed to use a coordinate system to solve the above problems.

(1a) The vectors a and b form the twosides of a parallelogram. The two linesdrawn from A to B and from O to Dbisect with each other. Hence, the vec-tor drawn from O to the middle pointof AB is equal to half of vector drawnfrom O to D. That is c = 1

2 (a + b).

(1b) From the given condition that b−a = c−d, we see that the vector drawnfrom A to B and the vector drawn fromD to C are equal: AB = DC. Thecondition rewritten in the form b−c =a−d shows that CB = DA. The oppo-site sides are parallel and equal, henceABCD is a parallelogram.

(1c) Let the position vectors of pointsp1 and p2 be denoted by a and b, re-spectively. The mid–points of the sidesOp1 and Op2 (i.e. Op′1 and Op′2) arethen a/2 and b/2. The vectors drawnfrom p1 to p2 and from p′1 to p′2 are thenc = b−a, c′ = b/2−a/2 = c/2. Obvi-ously, c′ is parallel to c and |c′| = |c|/2


[2] Show that (A + B)2 = A2 + AB + BA + B2 given that

A =

2 1 −1

1 −2 3−2 1 2

, B =

1 −1 2−2 1 3

2 −1 1

.

A + B =

3 0 1−1 −1 6

0 0 3

(A + B)2 = (A + B)(A + B)

=

3 0 1−1 −1 6

0 0 3

3 0 1−1 −1 6

0 0 3

=

9 0 6−2 1 11

0 0 9

A2 =

2 1 −1

1 −2 3−2 1 2

2 1 −1

1 −2 3−2 1 2

=

7 −1 −1−6 8 −1−7 −2 9

AB =

2 1 −1

1 −2 3−2 1 2

1 −1 2−2 1 3

2 −1 1

=

−2 0 6

11 −6 −10 1 1

BA =

1 −1 2−2 1 3

2 −1 1

2 1 −1

1 −2 3−2 1 2

=

−3 5 0−9 −1 11

1 5 −3

B2 =

1 −1 2−2 1 3

2 −1 1

1 −1 2−2 1 3

2 −1 1

=

7 −4 1

2 0 26 −4 2

=⇒ A2 + AB + BA + B2 =

9 0 6−2 1 11

0 0 9

= (A + B)2 �


[3] Use Gauss–Jordan elimination to find the inverse of

A =

1 0 2

2 −1 34 1 8

.

[A|I] =

1 0 2 | 1 0 0

2 −1 3 | 0 1 04 1 8 | 0 0 1

r1r2r3

1 0 2 | 1 0 00 −1 −1 | −2 1 04 1 8 | 0 0 1

r1r2 − 2r1r3

1 0 2 | 1 0 00 −1 −1 | −2 1 00 1 0 | −4 0 1

r1r2 − 2r1r3 − 4r1

1 0 2 | 1 0 00 0 −1 | −6 1 10 1 0 | −4 0 1

r1r2 − 2r1 + (r3 − 4r1)

(−1)× −→

1 0 0 | −11 2 2

0 0 −1 | −6 1 10 1 0 | −4 0 1

1 0 0 | −11 2 2

0 1 0 | −4 0 10 0 1 | 6 −1 −1

=⇒ A−1 =

−11 2 2−4 0 1

6 −1 −1

�

[4] Evaluate

∣∣∣∣∣∣∣2 1 −1 4−2 3 2 −5

1 −2 −3 2−4 −3 2 −2

∣∣∣∣∣∣∣ .If we multiply the elements of any row (or column) by a given number andadd to corresponding elements of any other row (or column), then the valueof the determinant remains the same. Now, multiplying the elements of thefirst row by −3, 2, 3, and adding to the elements of the second, third, andfourth rows, respectively, yields


∣∣∣∣∣∣∣2 1 −1 4−8 0 5 −17

5 0 −5 102 0 −1 10

∣∣∣∣∣∣∣ = (−1)

∣∣∣∣∣∣−8 5 −17

5 −5 102 −1 10

∣∣∣∣∣∣

= (−1)(−8)∣∣∣∣−5 10−1 10

∣∣∣∣− (−5)∣∣∣∣ 5 −17−1 10

∣∣∣∣− (2)∣∣∣∣ 5 −17−5 10

∣∣∣∣ = −85 �

[5] Prove that if A is a non–singular matrix, then

det A−1 =1

det A

AA−1 = I =⇒

det(AA−1

)= det I = 1

det(AA−1

)= det A det A−1

=⇒ det A det A−1 = 1 =⇒ det A−1 =1

det A�

[6] Use LU decomposition to find the inverse of

A =

2 1 1

4 −1 3−2 4 0

.

A =LU =⇒ A−1 = U−1L−1

[A|I] =

2 1 1 | 1 0 0

4 −1 3 | 0 1 0−2 4 0 | 0 0 1

−→

2 1 1 | 1 0 0

0 −3 1 | −2 1 00 5 1 | 1 0 1

−→

2 1 1 | 1 0 0

0 −3 1 | −2 1 00 0 8/3 | −7/3 5/3 1

= [U|L−1]


Now obtain U−1

2 1 1 | 1 0 0

0 −3 1 | 0 1 00 0 8/3 | 0 0 1

−→

2 1 0 | 1 0 −3/8

0 −3 0 | 0 1 −3/80 0 1 | 0 0 3/8

−→

2 0 0 | 1 1/3 −4/8

0 1 0 | 0 −1/3 1/80 0 1 | 0 0 3/8

−→

1 0 0 | 1/2 1/6 −1/4

0 1 0 | 0 −1/3 1/80 0 1 | 0 0 3/8

=⇒ A−1 = U−1L−1 =

1/2 1/6 −1/4

0 −1/3 1/80 0 3/8

1 0 0−2 1 0−7/3 5/3 1

A−1 =

3/4 −1/4 −1/4

3/8 −1/8 1/8−7/8 5/8 3/8

�

[7] Find the eigenvalues and eigenvectors of

A =

5 1 0

0 4 10 1 4

.

∣∣∣∣∣∣5− λ 1 0

0 4− λ 10 1 4− λ

∣∣∣∣∣∣ = 0

=⇒ (5− λ)∣∣∣∣ 4− λ 1

1 4− λ

∣∣∣∣ = (5− λ)[(4− λ)2 − 1

]= 0

=⇒ λ1 = 5, 4− λ = ±1 −→{λ2 = 3

λ3 = 5


=⇒ Eigenvalues : λ1 = 5, λ2 = 3, λ3 = 5 �For λ1 = λ3 = 5

0 1 0

0 −1 10 1 −1

−→

0 1 0

0 −1 10 0 0

−→

0 1 0

0 0 10 0 0

=⇒ x2 = x3 = 0

The corresponding eigenvector is:

x1

00

= x1

1

00

or simply

1

00

For λ2 = 3 2 1 0

0 1 10 1 1

−→

2 1 0

0 1 10 0 0

x2 = −x3, 2x1 = −x2 = x3 =⇒ x1 =x3

2


x3/2−x3

x3

= x3

1/2−1

1

or

1/2−1

1

=⇒ Eigenvectors :

1

00

,

1/2−1

1

�

[8] Find the eigenvalues and eigenvectors of A =[


].

∣∣∣∣ cos θ − λ − sin θsin θ cos θ − λ

∣∣∣∣ = 0 =⇒ (cos θ − λ)2 + sin2 θ = 0

=⇒ cos2 θ − 2λ cos θ + λ2 + sin2 θ = 0 =⇒ 1− 2λ cos θ + λ2 = 0


=⇒ λ =2 cos θ ±

√4 cos2 θ − 42

=2 cos θ ±

√−4 sin2 θ

2= cos θ ± i sin θ

(eiθ = cos θ + i sin θ

)=⇒ λ = e±iθ �

[cos θ − λ − sin θ

sin θ cos θ − λ

] [x1

x2

]= 0 =⇒

(cos θ − λ)x1 − (sin θ)x2 = 0

(sin θ)x1 + (cos θ − λ)x2 = 0

(For λ = eiθ

)=⇒

(cos θ − eiθ

)x1 − (sin θ)x2 = 0

(sin θ)x1 +(cos θ − eiθ

)x2 = 0

=⇒

−i(sin θ)x1 − (sin θ)x2 = 0

(sin θ)x1 − i(sin θ)x2 = 0

=⇒ x2 = −ix1


[x1

x2

]=

[x1

−ix1

]= x1

[1−i

]or simply

[1−i

].

(For λ = e−iθ

)=⇒

(cos θ − e−iθ

)x1 − (sin θ)x2 = 0

(sin θ)x1 +(cos θ − e−iθ

)x2 = 0{

e−iθ = cos θ − i sin θ}

=⇒

i(sin θ)x1 − (sin θ)x2 = 0

(sin θ)x1 + i(sin θ)x2 = 0

=⇒ x2 = ix1


[x1

x2

]=

[x1

ix1

]= x1

[1i

]or simply

[1i

].


[9] Use properties of the transpose to show that A =[


]is

an orthogonal matrix.

ATA =[

cos θ sin θ− sin θ cos θ

] [cos θ − sin θsin θ cos θ

]

=[

cos2 θ + sin2 θ 00 sin2 θ + cos2 θ

]

=[

1 00 1

]= I

(For an orthogonal matrix AT = A−1

)=⇒ A is an orthogonal matrix �


f(t)

1

-T/4-T/2 T/4 T/2

-1

t0

10.7 Fourier Series, Integrals, and Transforms (Chapter 8)

[1] Find the period of the function f(t) = (10 cos t)2.

f(t) = 100 cos2 t, cos2 θ =12(1 + cos 2θ)

=⇒ f(t) =1002

(1 + cos 2t) = 50 + 50 cos 2t

Since a constant is a periodic function of period P for any value of P andthe period of cos 2t is π, we conclude that the period of f(t) is π. �

[2] Find the Fourier series for the function whose waveform is (Hint: P =T [−T/2, T/2])

The function f(t) can be expressed analytically as

f(t) =

1 +4tT

− T

2≤ t < 0

1− 4tT

0 ≤ t < T

2

f(t) = a0 +∞∑n=1

(an cos

2nπTt+ bn sin

2nπTt

)

a0 =1T

T/2∫−T/2

f(t)dt =1T

0∫−T/2

(1 +

4tT

)dt+

1T

T/2∫0

(1− 4t

T

)dt

=1T

[t+

2t2

T

]0

−T/2

+1T

[t− 2t2

T

]T/20

=1T

[T

2− T

2

]+

1T

[T

2− T

2

]= 0


an =2T

T/2∫−T/2

f(t) cos(

2nπTt

)dt

=2T

T/2∫−T/2

cos(

2nπTt

)dt+

2T

0∫−T/2

4tT

cos(

2nπTt

)dt

− 2T

T/2∫0

4tT

cos(

2nπTt

)dt

The first integral on the right–hand side equals zero:

=⇒ an =8T 2

0∫−T/2

t cos(

2nπTt

)dt− 8

T 2

T/2∫0

t cos(

2nπTt

)dt

Letting t = −τ in the first integral, we get:

an =8T 2

0∫+T/2

(−τ) cos(

2nπT

(−τ))

(−dτ)− 8T 2

T/2∫0

t cos(

2nπTt

)dt

=8T 2

0∫T/2

τ cos(

2nπTτ

)dτ − 8

T 2

T/2∫0

t cos(

2nπTt

)dt

= − 8T 2

T/2∫0

τ cos(

2nπTτ

)dτ − 8

T 2

T/2∫0

t cos(

2nπTt

)dt

= − 16T 2

T/2∫0

t cos(

2nπTt

)dt


1

f(t)

-T 0 Tt

1/2

-1/2

0-T Tt

g(t)=f(t)-1/2

= − 16T 2

T

2nπt sin

(2nπTt

)∣∣∣∣T/2

0

− T

2nπ

T/2∫0

sin(

2nπTt

)dt

= − 16T 2

{0 +

(T

2nπ

)2

cos(

2nπTt

)∣∣∣∣T/2

0

}

= − 16T 2

{(T

2nπ

)2[cosnπ − 1

]}

=4

n2π2

[1− cosnπ

]=

4n2π2

[1− (−1)n

]

=⇒ an =

0 n even

8n2π2

n odd

bn =2T

T/2∫−T/2

f(t) sin(

2nπTt

)dt = 0,

because f(−t) = f(t), even function.

=⇒ f(t) =8π2

[cos

(2πTt

)+

132

cos(

6πTt

)+

152

cos(

10πTt

)+ · · ·

]�

[3] Find the Fourier series for the function (P = T )

Let w0 =2πT

. Also, because g(t) has odd symmetry,

g(t) =∞∑n=1

bn sinnw0t


1

0 Tt

f(t)

-T

1

-T 0 Tt

g(t)=1-f(t)

bn =4T

T/2∫0

g(t) sin(nw0t)dt, g(t) =12− t

T(0 < t < T )

=4T

T/2∫0

(12− t

T

)sin(nw0t)dt

=4T

[−

(12− t

T

)cosnw0t

nw0− sinnw0t

T (nw0)2

]T/20

=1nπ

Thus, g(t) =1π

∞∑n=1

1n

sinnw0t

=⇒ f(t) =12

+1π

[sinw0t+

12

sin 2w0t+13

sin 3w0t+ · · ·]

�

[4] Using the result of problem 3, find the Fourier series for the functionf(t) shown below

g(t) = 1− f(t) =12

+1π

∞∑n=1

1n

sin(nw0t), w0 =2πT

=⇒ f(t) = 1− 12− 1π

∞∑n=1

1n

sin(nw0t)

=12− 1π

[sinw0t+

12

sin 2w0t+13

sin 3w0t+ · · ·]

�


−π −π/2 0 π/2 πt

f(t)

1

-1

[5] Given the function f(t) =

0 0 < t < π2 ,

1 π2 < t < π,

expand f(t) in a Fourier

sine series and draw the corresponding periodic extension of f(t).

Fourier sine series : f(t) =∞∑n=1

bn sin(nt)

bn =2π

π∫0

f(t) sin(nt)dt =2π

π∫π/2

sin(nt)dt

= − 2nπ

cos(nt)∣∣∣ππ/2

= − 2nπ

[cos(nπ)− cos

(nπ2

)]

=

2nπ

n = 1, 3, 5, · · ·

− 4nπ

n = 2, 6, 10, · · ·

0 n = 4, 8, 12, · · ·

=⇒ f(t) =2π

[sin t+

13

sin 3t+15

sin 5t+ · · ·]

− 2π

[sin 2t+

13

sin 6t+15

sin 10t+ · · ·]

�


P(t)

-d/2 0 d/2

1

t

f(t)

t0

eat e-at

[6] Find the fourier transform of p(t) =

1 |t| < d2 ,

0 |t| > d2 .

p(w) = F{p(t)

}=

∞∫−∞

p(t)e−jwtdt =

d/2∫−d/2

e−jwtdt

=1−jwe

−jwt

∣∣∣∣d/2

−d/2

=1jw

[ejwd/2 − e−jwd/2

] (eit = cos t+ i sin t

)

=2w

sin(wd

2

)�

[7] Find the Fourier transform of f(t) = e−a|t|, where a > 0.

f(t) = e−a|t| =

{e−at t > 0

eat t < 0


f(w) =

∞∫−∞

f(t)e−jwtdt =

0∫−∞

eate−jwtdt+

∞∫0

e−ate−jwtdt

=

0∫−∞

e−(−a+jw)tdt+

∞∫0

e−(a+jw)tdt

=−1

−a+ jw+

1a+ jw

=2a

a2 + w2�

[8] If

∞∫0

f(x) cos ax dx =

{ 1− a 0 ≤ a ≤ 1,

0 a > 1,find f(x).

Let f(a) =

√2π

∞∫0

f(x) cos ax dx

and choose f(a) =

√2π (1− a) 0 ≤ a ≤ 1

0 a > 1

Then : f(x) =

√2π

∞∫0

f(a) cos ax da =

√2π

1∫0

√2π

(1− a) cos ax da

=2π

1∫0

(1− a) cos ax da =2(1− cosx)

πx2�

[9] Show that

∞∫0

sin2 u

u2du =

π

2given that

∞∫0

f(x) cos ax dx =

{ 1− a 0 ≤ a ≤ 1,

0 a > 1.

Using the result from problem #8

2π

∞∫0

1− cosxx2

cos ax dx =

{ 1− a 0 ≤ a ≤ 1

0 a > 1


Taking the limit as a −→ 0+, we find∞∫0

1− cosxx2

dx =π

2, sin2 x =

12(1− cos 2x)

=⇒∞∫0

2 sin2(x2 )x2

dx =π

2, let x = 2u

=⇒∞∫0

sin2 u

u2du =

π

2�

[10] If F{f(t)

}= f(ω) and f(ω) can be differentiated everywhere n times,

show that

F{tpf(t)

}=

1(−j)p

dpf(ω)dωp

for every p ≤ n.

Since F{− jtf(t)

}= f ′(w) −→ Frequency domain differentiation

F{(−jt)2f(t)

}=d2f(w)dw2

· · · · · · · · · · · ·

F{(−jt)pf(t)

}=dpf(w)dwp

, p ≤ n

f(w) =

∞∫−∞

f(t)e−jwtdt

=⇒ d2f(w)dw2

=d2

dw2

∞∫−∞

f(t)e−jwtdt =

∞∫−∞

f(t)d2

dw2

(e−jwt

)dt

=

∞∫−∞

[−jt]2f(t)e−jwtdt = F{(−jt)2f(t)

}

=⇒ 1(−j)pF

{(−jt)pf(t)

}=

1(−j)p

dpf(w)dwp

=⇒ F{tpf(t)

}=

1(−j)p

dpf(w)dwp

�


10.8 Partial Differential Equations (Chapter 9)

[1] Classify each of the following partial differential equations as elliptic,hyperbolic or parabolic

(a)∂2u

∂t2= k2 ∂

2u

∂x2

(A∂2u

∂x2+B

∂2u

∂x∂y+ C

∂2u

∂y2+D

∂u

∂x+ E

∂u

∂y+ Fu = G

)

A = k2, B = 0, C = −1

=⇒ B2 − 4AC = 4k2 > 0.

The equation is hyperbolic �

(b)∂2u

∂x2+ 3

∂2u

∂x∂y+ 4

∂2u

∂y2+ 5

∂u

∂x− 2

∂u

∂y+ 4u = 2x− 3y,

A = 1, B = 3, C = 4

=⇒ B2 − 4AC = −7 < 0.

The equation is elliptic �

(c) x∂2u

∂x2+ y

∂2u

∂y2+ 3y2 ∂u

∂x= 0.

A = x, B = 0, C = y

=⇒ B2 − 4AC = −4xy.

In the region xy > 0 the equation is elliptic. In the region xy < 0 theequation is hyperbolic. If xy = 0, the equation is parabolic �

[2] Solve the following boundary–value problem by the method of separa-tion of variables

∂u

∂x= 4

∂u

∂y, u(0, y) = 8e−3y + 4e−5y.

Assume u(x, y) = X(x)Y (y).

X ′(x)Y (y) = 4X(x)Y ′(y)


=⇒ X ′

4X=Y ′

Y= c

X ′ = 4cX =⇒ X = Ae4cx & Y ′ = cY =⇒ Y = Becy

=⇒ u(x, y) =(Ae4cx

)(Becy) = aec(4x+y) (a = AB)

By the principle of superposition u(x, y) = aec1(4x+y) + bec2(4x+y) is also asolution.

b.c. −→ u(0, y) = 8e−3y + 4e−5y = aec1y + bec2y.

a = 8, c1 = −3, b = 4, c2 = −5

=⇒ u(x, y) = 8e−3(4x+y) + 4e−5(4x+y)

=⇒ u(x, y) = 8e−12x−3y + 4e−20x−5y �

[3] Solve by the method of separation of variables

∂u

∂t= 2

∂2u

∂x2, 0 < x < 3, t > 0

given that u(0, t) = u(3, t) = 0, u(x, 0) = 5 sin 4πx− 3 sin 8πx+ 2 sin 10πx,|u(x, t)| < M , where the last condition states that u is bounded for 0 <x < 3, t > 0.Assume u(x, t) = X(x)T (t)

XT ′ = 2X ′′T

=⇒ X ′′

X=T ′

2T= −λ2

(Note: if we use +λ2, the resulting solution does not satisfy the boundednesscondition for real values of λ.)

X ′′ + λ2X = 0 =⇒ X = A1 cosλx+B1 sinλx

T ′ + 2λ2T = 0 =⇒ T = c1e−2λ2t

=⇒ u(x, t) = c1e−2λ2t

[A1 cosλx+B1 sinλx

]= e−2λ2t

[A cosλx+B sinλx

]From u(0, t) = Ae−2λ2t = 0 −→ A = 0


=⇒ u(x, t) = Be−2λ2t sinλx

From u(3, t) = Be−2λ2t sin 3λ = 0 (B �= 0)

=⇒ sin 3λ = 0 −→ 3λ = nπ −→ λ =nπ

3(n = 0,±1,±2, · · ·)

=⇒ u(x, t) = Be−2(

n2π29

)t sin

nπx

3Also, by the principle of superposition

u(x, t) = b1e− 2n2

1π2t

9 sinn1πx

3+ b2e

− 2n22π2t

9 sinn2πx

3+ b3e

− 2n23π2t

9 sinn3πx

3

is also a solution. From the last boundary condition:

u(x, 0) = b1 sinn1πx

3+ b2 sin

n2πx

3+ b3 sin

n3πx

3= 5 sin 4πx− 3 sin 8πx+ 2 sin 10πx

This is possible if and only if:

{b1 = 5 n1 = 12b2 = −3 n2 = 24b3 = 2 n3 = 30

=⇒ u(x, t) = 5e−32π2t sin 4πx−3e−128π2t sin 8πx+2e−200π2t sin 10πx �

[4] Apply Laplace transformations to show that the Laplace “time” solutionof the following boundary value problem

∂u

∂t= k

∂2u

∂x2,

u(a, t) = Qδ(t), u(x, 0) = 0, |u(x, t)| < M, is

u(x, s) = Qe−(x−a)√

s/k.

Taking Laplace transforms w.r.t. t yields:

{su(x, s)− u(x, 0)

}= k

d2u(x, s)dx2

, u(x, 0) = 0

=⇒ d2u(x, s)dx2

− s

ku(x, s) = 0


=⇒ u(x, s) = Ae√

s/kx +Be−√

s/kx

From the boundedness condition, we require A = 0

=⇒ u(x, s) = Be−√

s/kx

From the boundary condition

u(a, s) = L{Qδ(t)

}= Q

=⇒ u(a, s) = Be−√

s/ka = Q

=⇒ B = Qe√

s/ka

=⇒ u(x, s) = Qe−(x−a)√

s/k �

[5] In the theory of neutron slowing down, one encounters the equation

α(ξ)∂φ

∂ξ= D

∂2φ

∂x2(0 ≤ ξ <∞),

where ξ is a logarithmic energy variable, D is the diffusion coefficient, andα(ξ) is a known positive function. Solve this equation by the method ofFourier transformations for an infinite medium (in x) subject to the initialcondition: φ(0, x) = δ(x).

Taking Fourier transforms with respect to x yields:

α(ξ)dφ(ξ, w)dξ

= −Dw2φ(ξ, w)

=⇒ dφ(ξ, w)

φ(ξ, w)= −Dw2 dξ

α(ξ)f(w) =

∞∫−∞

f(x)e−jwxdx, & f(x) =12π

∞∫−∞

f(w)ejwxdw

F{f ′′(x)

}= −w2f(w)

Integrate and let∫

dξ

α(ξ)= β(ξ), 0 < ξ <∞


=⇒ ln φ(ξ, w) = −Dw2β(ξ) +A(w)

=⇒ φ(ξ, w) = A∗(w) exp[−Dw2β(ξ)

]From the initial condition

F{φ(0, x)} = F{δ(x)}

=⇒ φ(0, w) =

∞∫−∞

δ(x)e−jwxdx = 1

At ξ = 0 −→ β(0) =

0∫0

dξ

α(ξ)= 0

=⇒ φ(0, w) = 1 = A∗(w)e0 =⇒ A∗(w) = 1

−→ φ(ξ, w) = exp[−Dw2β(ξ)

]

=⇒ φ(ξ, x) = F−1{

exp[−Dw2β(ξ)

]}

=12π

∞∫−∞

e−Dw2β(ξ)ejwxdw

=12π

∞∫−∞

e−[Dw2β(ξ)−jwx

]dw

Dw2β(ξ)− jwx =(√

Dβ(ξ)w − jx

2√Dβ(ξ)

)2

+x2

4Dβ(ξ)

=⇒ φ(ξ, x) =12πe−

x24Dβ(ξ)

∞∫−∞

e−(√

Dβ(ξ)w− jx

2√

Dβ(ξ)

)2

dw

let z =√Dβ(ξ)w − jx

2√Dβ(ξ)

=⇒ dz =√Dβ(ξ)dw =⇒ dw =

dz√Dβ(ξ)


=⇒ φ(ξ, x) =12π

1√Dβ(ξ)

exp[ −x2

4Dβ(ξ)

] ∞∫−∞

e−z2dz

︸︷︷︸√π(∫

e−z2dz =

√π

2erf[z], erf[−∞] = −1, erf[∞] = 1

)

=⇒ φ(ξ, x) =e−

x24Dβ(ξ)√

4πDβ(ξ)where β(ξ) =

∫dξ

α(ξ)�


References

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