Table of Contents
Part I Ordinary Differential Equations
1 Introduction to Differential Equations 1
2 First-Order Differential Equations 22
3 Higher-Order Differential Equations 99
4 The Laplace Transform 198
5 Series Solutions of Linear Differential Equations 252
6 Numerical Solutions of Ordinary Differential Equations 317
Part II Vectors, Matrices, and Vector Calculus
7 Vectors 339
8 Matrices 373
9 Vector Calculus 438
Part III Systems of Differential Equations
10 Systems of Linear Differential Equations 551
11 Systems of Nonlinear Differential Equations 604
Part IV Fourier Series and Partial Differential Equations
12 Orthogonal Functions and Fourier Series 634
13 Boundary-Value Problems in Rectangular Coordinates 680
14 Boundary-Value Problems in Other Coordinate Systems 755
15 Integral Transform Method 793
16 Numerical Solutions of Partial Differential Equations 832
, Part V Complex Analysis
17 Functions of a Complex Variable 854
18 Integration in the Complex Plane 877
19 Series and Residues 896
20 Conformal Mappings 919
Appendices
Appendix II Gamma function 942
Projects
3.7 Road Mirages 944
3.10 The Ballistic Pendulum 946
8.1 Two-Ports in Electrical Circuits 947
8.2 Traffic Flow 948
8.15 Temperature Dependence of Resistivity 949
9.16 Minimal Surfaces 950
14.3 The Hydrogen Atom 952
15.4 The Uncertainity Inequality in Signal Processing 955
15.4 Fraunhofer Diffraction by a Circular Aperture 958
16.2 Instabilities of Numerical Methods 960
, Part I Ordinary Differential Equations
Introduction to
1 Differential Equations
EXERCISES 1.1
Definitions and Terminology
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
!
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of ẋ2
9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 .
However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u.
11. From y = e−x/2 we obtain y ′ = − 12 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.
12. From y = 6
5 − 65 e−20t we obtain dy/dt = 24e−20t , so that
" #
dy 6 6 −20t
+ 20y = 24e−20t + 20 − e = 24.
dt 5 5
13. From y = e3x cos 2x we obtain y ′ = 3e3x cos 2x − 2e3x sin 2x and y ′′ = 5e3x cos 2x − 12e3x sin 2x, so that
y ′′ − 6y ′ + 13y = 0.
14. From y = − cos x ln(sec x + tan x) we obtain y ′ = −1 + sin x ln(sec x + tan x) and
y ′′ = tan x + cos x ln(sec x + tan x). Then y ′′ + y = tan x.
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y ′ = 1 + 2(x + 2)−1/2 we have
(y − x)y ′ = (y − x)[1 + (2(x + 2)−1/2 ]
= y − x + 2(y − x)(x + 2)−1/2
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8.
1
, 1.1 Definitions and Terminology
An interval of definition for the solution of the differential equation is (−2, ∞) because y ′ is not defined at
x = −2.
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
$ $
{x $ 5x ̸= π/2 + nπ} or {x $ x =
̸ π/10 + nπ/5}. From y ′ = 25 sec2 5x we have
y ′ = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is
(π/10, 3π/10), and so on.
$ $
17. The domain of the function is {x $ 4 − x2 ̸= 0} or {x $ x ̸= −2 or x ̸= 2}. From y ′ = 2x/(4 − x2 )2 we have
" #2
1
y = 2x
′
= 2xy.
4 − x2
An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞, −2)
and (2, ∞).
√
18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x ̸= 0 or sin x ̸= 1. Thus, the domain
$
is {x $ x ̸= π/2 + 2nπ}. From y ′ = − 12 (1 − sin x)−3/2 (− cos x) we have
2y ′ = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.
An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2),
and so on.
19. Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain X
2 dX 1 dX
− =1 4
2X − 1 dt X − 1 dt
" #
2 1 dX
− =1 2
2X − 1 X − 1 dt
2X − 2 − 2X + 1 dX t
=1 -4 -2 2 4
(2X − 1)(X − 1) dt
dX -2
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt
Exponentiating both sides of the implicit solution we obtain -4
2X − 1
= et
X −1
2X − 1 = Xet − et
(et − 1) = (et − 2)X
et − 1
X= .
et − 2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the
solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid.
2
Part I Ordinary Differential Equations
1 Introduction to Differential Equations 1
2 First-Order Differential Equations 22
3 Higher-Order Differential Equations 99
4 The Laplace Transform 198
5 Series Solutions of Linear Differential Equations 252
6 Numerical Solutions of Ordinary Differential Equations 317
Part II Vectors, Matrices, and Vector Calculus
7 Vectors 339
8 Matrices 373
9 Vector Calculus 438
Part III Systems of Differential Equations
10 Systems of Linear Differential Equations 551
11 Systems of Nonlinear Differential Equations 604
Part IV Fourier Series and Partial Differential Equations
12 Orthogonal Functions and Fourier Series 634
13 Boundary-Value Problems in Rectangular Coordinates 680
14 Boundary-Value Problems in Other Coordinate Systems 755
15 Integral Transform Method 793
16 Numerical Solutions of Partial Differential Equations 832
, Part V Complex Analysis
17 Functions of a Complex Variable 854
18 Integration in the Complex Plane 877
19 Series and Residues 896
20 Conformal Mappings 919
Appendices
Appendix II Gamma function 942
Projects
3.7 Road Mirages 944
3.10 The Ballistic Pendulum 946
8.1 Two-Ports in Electrical Circuits 947
8.2 Traffic Flow 948
8.15 Temperature Dependence of Resistivity 949
9.16 Minimal Surfaces 950
14.3 The Hydrogen Atom 952
15.4 The Uncertainity Inequality in Signal Processing 955
15.4 Fraunhofer Diffraction by a Circular Aperture 958
16.2 Instabilities of Numerical Methods 960
, Part I Ordinary Differential Equations
Introduction to
1 Differential Equations
EXERCISES 1.1
Definitions and Terminology
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
!
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of ẋ2
9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 .
However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u.
11. From y = e−x/2 we obtain y ′ = − 12 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.
12. From y = 6
5 − 65 e−20t we obtain dy/dt = 24e−20t , so that
" #
dy 6 6 −20t
+ 20y = 24e−20t + 20 − e = 24.
dt 5 5
13. From y = e3x cos 2x we obtain y ′ = 3e3x cos 2x − 2e3x sin 2x and y ′′ = 5e3x cos 2x − 12e3x sin 2x, so that
y ′′ − 6y ′ + 13y = 0.
14. From y = − cos x ln(sec x + tan x) we obtain y ′ = −1 + sin x ln(sec x + tan x) and
y ′′ = tan x + cos x ln(sec x + tan x). Then y ′′ + y = tan x.
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y ′ = 1 + 2(x + 2)−1/2 we have
(y − x)y ′ = (y − x)[1 + (2(x + 2)−1/2 ]
= y − x + 2(y − x)(x + 2)−1/2
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8.
1
, 1.1 Definitions and Terminology
An interval of definition for the solution of the differential equation is (−2, ∞) because y ′ is not defined at
x = −2.
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
$ $
{x $ 5x ̸= π/2 + nπ} or {x $ x =
̸ π/10 + nπ/5}. From y ′ = 25 sec2 5x we have
y ′ = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another interval is
(π/10, 3π/10), and so on.
$ $
17. The domain of the function is {x $ 4 − x2 ̸= 0} or {x $ x ̸= −2 or x ̸= 2}. From y ′ = 2x/(4 − x2 )2 we have
" #2
1
y = 2x
′
= 2xy.
4 − x2
An interval of definition for the solution of the differential equation is (−2, 2). Other intervals are (−∞, −2)
and (2, ∞).
√
18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x ̸= 0 or sin x ̸= 1. Thus, the domain
$
is {x $ x ̸= π/2 + 2nπ}. From y ′ = − 12 (1 − sin x)−3/2 (− cos x) we have
2y ′ = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.
An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2),
and so on.
19. Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain X
2 dX 1 dX
− =1 4
2X − 1 dt X − 1 dt
" #
2 1 dX
− =1 2
2X − 1 X − 1 dt
2X − 2 − 2X + 1 dX t
=1 -4 -2 2 4
(2X − 1)(X − 1) dt
dX -2
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt
Exponentiating both sides of the implicit solution we obtain -4
2X − 1
= et
X −1
2X − 1 = Xet − et
(et − 1) = (et − 2)X
et − 1
X= .
et − 2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞). The graph of the
solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid.
2
