DIFFERENTIAL EQUATIONS FOR ENGINEERS
This book presents a systematic and comprehensive introduction to ordinary differential
equations for engineering students and practitioners. Mathematical concepts and various
techniques are presented in a clear, logical, and concise manner. Various visual features
are used to highlight focus areas. Complete illustrative diagrams are used to facilitate
mathematical modeling of application problems. Readers are motivated by a focus on the
relevance of differential equations through their applications in various engineering
disciplines. Studies of various types of differential equations are determined by
engineering applications. Theory and techniques for solving differential equations are
then applied to solve practical engineering problems. Detailed step-by-step analysis is
presented to model the engineering problems using differential equations from physical
principles and to solve the differential equations using the easiest possible method. Such a
detailed, step-by-step approach, especially when applied to practical engineering
problems, helps the readers to develop problem-solving skills.
This book is suitable for use not only as a textbook on ordinary differential equations
for undergraduate students in an engineering program but also as a guide to self-study. It
can also be used as a reference after students have completed learning the subject.
Wei-Chau Xie is a Professor in the Department of Civil and Environmental Engineering
and the Department of Applied Mathematics at the University of Waterloo. He is the
author of Dynamic Stability of Structures and has published numerous journal articles on
dynamic stability, structural dynamics and random vibration, nonlinear dynamics and
stochastic mechanics, reliability and safety analysis of engineering systems, and seismic
analysis and design of engineering structures. He has been teaching differential equations
to engineering students for almost twenty years. He received the Teaching Excellence
Award in 2001 in recognition of his exemplary record of outstanding teaching, concern
for students, and commitment to the development and enrichment of engineering
education at Waterloo. He is the recipient of the Distinguished Teacher Award in 2007,
which is the highest formal recognition given by the University of Waterloo for a superior
record of continued excellence in teaching.
84
2 first-order and simple higher-order differential equations
2.100
3 y y ′ y ′′ − y ′3 + 1 = 0
2.101
y ′′ − y ′2 − 1 = 0
2.102
x 3 y ′′ − x 2 y ′ = 3 − x 2
2.103
2.104
2.105
3(C1 y + 1)2/3 − 2C1 x = C2 ; y = x
y = − ln cos(x +C1 ) + C2
ANS
ANS
1
+ x + C1 x 2 + C2
x
2 y ′′ = y ′3 sin 2x, y(0) = 1, y ′(0) = 1 ANS y = 1+ ln sec x + tan x ANS
y=
dy
d2 y
=
2
−
NS y = 2x + C1 lnx + C2
A
dx 2
dx
4
√
y ′′ = 3 y, y(0) = 1, y ′(0) = 2
ANS y = ± 12 x + 1
x
1 dy d2 y
dy
2.106 x 2 =
+ x sin
·
dx
dx
x dx
1 x
ANS y = x2 + 2 tan−1 C1 x − C +C2 ;
C1
1
y=
kπ 2
x +C, k = 0, ±1, ±2, . . .
2
y y ′′ = y ′2 (1 − y ′ sin y − y y ′ cos y)
ANS y = C; x = − cos y + C1 ln y + C2
Z
1 2
2.108 y ′′ + x y ′ = x
y
=
x
+
C
e− 2 x dx + C2
NS
A
1
2.107
2.109
2.110
ANS
x y ′′ − y ′3 − y ′ = 0
ANS
x 2 + ( y − C 1 )2 = C 2 ; y = C
y (1 − ln y) y ′′ + (1 + ln y) y ′2 = 0
y = C; (C1 x + C2 )(ln y − 1) + 1 = 0
Review Problems
2.111
x y 2 (x y ′ + y) = 1
2.112
5 y + y ′2 = x (x + y ′ )
2.113
y′ =
2.114
y ′′ (ex + 1) + y ′ = 0
2.115
x y ′ = y − x e y/x
2.116
(1 + y 2 sin 2x)dx − 2 y cos2 x dy = 0
√
(2 x y − y)dx − x dy = 0, x >0, y >0
2.117
2.118
ANS
y −2x
y +2
+ tan
x +1
x +1
y ′′ + y ′2 = 2e− y
ANS
ANS
2x 3 y 3 − 3x 2 = C
ANS
4 y = x 2 ; 5 y = −5x 2 + 5Cx − C 2
ANS
sin
y −2x
= C (x +1)
x +1
y = C1 (x − e−x ) + C2
y = −x ln lnCx ANS
x − y 2 cos2 x = C
√
ANS x y − x = C
ANS
e y + C1 = (x + C2 )2
234
5 applications of linear differential equations
5.4 A mass m is dropped with zero initial velocity from a height of h above a
spring of stiffness k as shown in the following figure. Determine the maximum
compression of the spring and the duration between the time when the mass contacts the spring and the time when the spring reaches maximum compression.
r r
r
mg mg
mg m π
−1 mg
ANS ymax = k 2h+ k + k , T = k 2 + tan
2hk
m
h
k
5.5 A uniform chain of length L with mass density per unit length ρ is laid on a
rough horizontal table with an initial hang of length l, i.e., y = l at t = 0 as shown
in the following figure. The coefficients of static and kinetic friction between the
chain and the surface have the same value µ. The chain is released from rest at time
t = 0 and it starts sliding off the table if (1+µ)l >µL. Show that the time T it
takes for the chain to leave the table is
s
i
h
L
L
−1
.
cosh
T=
(1+µ)g
(1+µ)l −µL
L −y
y, y, y
y(t)
5.6 A uniform chain of length L with mass density per unit length ρ is laid on a
smooth inclined surface with y = 0 at t = 0 as shown in the following figure. The
chain is released from rest at time t = 0. Show that the time T it takes for the chain
to leave the surface is
s
1 L
T=
.
cosh−1
(1− sin θ )g
sin θ
352
7.11
ANS
7.12
ANS
7.13
ANS
7 systems of linear differential equations
( D −4)x + 3 y = sin t,
−2x + ( D +1) y = −2 cos t
x = C1 et + C2 e2t + cos t − 2 sin t, y = C1 et + 23 C2 e2t + 2 cos t − 2 sin t
dx
dy
− y = 0, −x +
= et + e−t
dt
dt
x = C1 et + C2 e−t + 12 tet − 12 te−t
y = C1 + 12 et + −C2 − 21 e−t + 12 tet + 21 te−t
( D +2)x + 5 y = 0, −x + ( D −2) y = sin 2t
x = A cos t + B sin t + 35 sin 2t
y = − 15 (2A + B) cos t + 51 (A − 2B) sin t − 23 (cos 2t + sin 2t)
7.14
ANS
7.15
ANS
7.16
ANS
7.17
ANS
7.18
ANS
7.19
ANS
( D −2)x + 2 D y = −4e2t ,
(2 D −3)x + (3 D −1) y = 0
y = −C1 e−2t + 12 C2 et − e2t
(3 D +2)x + ( D −6) y = 5et , (4 D +2)x + ( D −8) y = 5et + 2t − 3
x = C1 e−2t + C2 et + 5e2t ,
x = A cos 2t + B sin 2t + 2et − 3t + 5, y = B cos 2t − A sin 2t + et − t
−x + ( D −1) y = 5e−t
( D −5)x + 3 y = 2e3t ,
x = C1 e2t + 3C2 e4t − e−t − 4e3t , y = C1 e2t + C2 e4t − 2e−t − 2e3t
( D −2)x + y = 0,
x + ( D −2) y = −5et sin t
x = C1 et +C2 e3t +et (2 cos t − sin t), y = C1 et −C2 e3t +et (3 cos t + sin t)
( D +4)x + 2 y =
2
et −1
,
6x − ( D −3) y =
3
et −1
x = C1 + 2C2 e−t + 2e−t lnet −1, y = −2C1 − 3C2 e−t − 3e−t lnet −1
( D −1)x + y = sec t,
−2x + ( D +1) y = 0
x = C1 cos t + C2 sin t + t (cos t + sin t) + (cos t − sin t) ln cos t y = (C1 −C2 ) cos t + (C1 +C2 ) sin t + 2t sin t + 2 cos t ln cos t The Method of Laplace Transform
Solve the following differential equations using the method of Laplace transform.
7.20
ANS
7.21
ANS
dx
− x − 2 y = 16t et ,
dt
2x −
dy
− 2 y = 0,
dt
x(0) = 4, y(0) = 0
x = −et (12t +13) + e−3t + 16e2t , y = −2et (4t +3) − 2e−3t + 8e2t
dx
dy
− 2x + y = 5et cos t, x +
− 2 y = 10et sin t, x(0) = y(0) = 0
dt
dt
x = 5et (1 − cos t + sin t), y = 5et (1 − cos t)
Problems
7.22
ANS
353
dx
− 4x + 3 y = sin t,
dt
2x −
dy
− y = 2 cos t,
dt
x(0) = x0 , y(0) = y0
x = (−2x0 +3 y0 −4)et + 3(x0 − y0 +1)e2t + cos t − 2 sin t
y = (−2x0 +3 y0 −4)et + 2(x0 − y0 +1)e2t + 2 cos t − 2 sin t
7.23
ANS
dx
− 2x − y = 2et ,
dt
x−
dy
+ 2 y = 3e4t ,
dt
x(0) = x0 , y(0) = y0
x = 21 (x0 + y0 +4)e3t + 12 (x0 − y0 +2t −2)et − e4t
y = 12 (x0 + y0 +4)e3t − 12 (x0 − y0 +2t)et − 2e4t
7.24
ANS
7.25
ANS
dy
d2 x
dx
+
− 2 y = 40e3t ,
+
2
dt
dt
dt
x(0) = 1, y(0) = 3, x ′(0) = 1
dy
dx
+x−
= 36et
dt
dt
x = 22e−2t −33e−t −3et (2t −3)+3e3t , y = 11e−2t −12et (t +1)+4e3t
dx
dy
− 2x − y = 2et ,
− 2 y − 4z = 4e2t ,
dt
dt
x(0) = 9, y(0) = 3, z(0) = 1
x−
dz
−z =0
dt
x = 3t + 2 + 2et − 3e2t + 8e3t , y = −6t − 1 − 4et + 8e3t
z = 3t − 1 + et − e2t + 2e3t
7.26
ANS
d2 x
dy
dx
d2 y
+
2x
−
2
− 8 y = 240et
=
0,
3
+
dt 2
dt
dt
dt 2
x(0) = y(0) = x ′(0) = y ′(0) = 0
x = 12 cos 2t − 24 sin 2t − 10e−2t + 30e2t − 32et
y = −12 cos 2t − 6 sin 2t + 15e−2t + 45e2t − 48et
7.27
ANS
7.28
ANS
dx
− x − 2 y = 0,
dt
dy
= 15 cos t H(t −π), x(0) = x0 , y(0) = y0
dt
x = 32 (x0 + y0 )e2t + 13 (x0 −2 y0 )e−t + 4e2(t−π ) + 5e−(t−π )
+ 9 cos t + 3 sin t H(t −π)
y = 13 (x0 + y0 )e2t − 31 (x0 −2 y0 )e−t + 2e2(t−π ) − 5e−(t−π )
− 3 cos t − 6 sin t H(t −π)
x−
dx
dy
−x +y = 2 sin t 1−H(t −π) , 2x −
−y = 0, x(0) = y(0) = 0
dt
dt
x = (t +1) sin t − t cos t + −(t −π +1) sin t + (t −π) cos t H(t −π)
y = 2(sin t − t cos t) + 2 − sin t + (t −π) cos t H(t −π)
354
7.29
ANS
7 systems of linear differential equations
dx
dy
dy
dx
+x−5
− 4 y = 28et H(t −2), 3
− 2x − 4
+y =0
dt
dt
dt
dt
x(0) = 2, y(0) = 0
x = −e−t + 3et + 5e4−t − (6t −7)et H(t −2)
y = −e−t + et + 5e4−t − (2t +1)et H(t −2)
2
The Matrix Method
Solve the following differential equations using the matrix method, in which
( · )′ = d( · )/dt.
7.30
ANS
7.31
ANS
7.32
ANS
7.33
ANS
7.34
ANS
x1′ = x1 − x2 ,
x2′ = −4x1 + x2
x1 = C1 e−t + C2 e3t ,
x1′ = x1 − 3x2 ,
x2 = 2C1 e−t − 2C2 e3t
x2′ = 3x1 + x2
x1 = et (A cos 3t + B sin 3t),
x2 = et (A sin 3t − B cos 3t)
x1′ = 5x1 + 3x2 , x2′ = −3x1 − x2 , x1 (0) = 1, x2 (0) = −2
x1 = (−3t +1)e2t ,
x1′ = 2x1 − x2 + x3 ,
x1 = C2 e2t + C3 e3t ,
x1′ = 3x1 − x2 + x3 ,
x2 = (3t −2)e2t
x2′ = x1 + 2x2 − x3 ,
x3′ = x1 − x2 + 2x3
x2 = C1 et + C2 e2t ,
x3 = C1 et + C2 e2t + C3 e3t
x2′ = x1 + x2 + x3 ,
x1 = C1 et + C2 e2t + C3 e5t ,
x3′ = 4x1 − x2 + 4x3
x2 = C1 et − 2C2 e2t + C3 e5t
x3 = −C1 et − 3C2 e2t + 3C3 e5t
7.35
ANS
7.36
ANS
7.37
x1′ = 2x1 + x2 ,
x3′ = −x1 + 2x2 + 3x3
x1 = C e2t + e3t (A cos t +B sin t), x2 = e3t (A+B) cos t +(B−A) sin t
x3 = C e2t + e3t (2A−B) cos t + (2B+A) sin t
x1′ = 3x1 − 2x2 − x3 ,
7.38
ANS
x2′ = 3x1 − 4x2 − 3x3 ,
x3′ = 2x1 − 4x2
x1 = (C1 +2C2 )e2t +C3 e−5t , x2 = C2 e2t +3C3 e−5t , x3 = C1 e2t +2C3 e−5t
x1′ = x1 − x2 + x3 ,
x1 (0) = 1,
ANS
x2′ = x1 + 3x2 − x3 ,
x2′ = x1 + x2 − x3 ,
x2 (0) = −2,
x1 = t et + e2t ,
x3′ = −x2 + 2x3
x3 (0) = 0
x2 = (t −2)et ,
x3 = (t −1)et + e2t
x1′ = −x1 + x2 − 2x3 , x2′ = 4x1 + x2 , x3′ = 2x1 + x2 − x3
x1 = C1 +C2 (t −1) e−t , x2 = − 2C1 +C2 (2t −1) e−t + 2C3 et
x3 = −(C1 +C2 t)e−t + C3 et
386
8 applications of systems of linear differential equations
F (t) = F0 sint
Machine
x1
m1
c
Vibration isolator k
x2
m2
Supporting structure
2. For the case when c →0, show that the steady-state responses of both the
machine and the supporting structure, i.e., x1P (t) and x2P (t) are given by
x1P (t) =
(k −m2 2 )F0 sin t
,
m1 m2 4 −k(m1 +m2 )2
x2P (t) =
kF0 sin t
.
m1 m2 4 −k(m1 +m2 )2
8.5 A mass m1 hangs by a spring of stiffness k1 from another mass m2 which in
turn hangs by a spring of stiffness k2 from the support as shown.
k2
x0 = asint
m2
k1
x2
m1
x1
1. Show that the two natural circular frequencies of vibration are given by the
equation
m1 m2 ω4 − m1 (k1 +k2 ) + m2 k1 ω2 + k1 k2 = 0.
2. If the support vibrates with x0 (t) = a sin t, show that the amplitudes of the
forced vibration are
a1 =
a2 =
m1 m2
4
k1 k2 a
,
− m1 (k1 +k2 ) + m2 k1 2 + k1 k2
(k − m1 2 )k2 a
1
.
m1 m2 4 − m1 (k1 +k2 ) + m2 k1 2 + k1 k2
8.6 A mass m, supported by an elastic structure which may be modeled as a spring
with stiffness k, is subjected to a simple harmonic disturbing force of maximum
12.1 closed-form solutions of differential equations
507
12.1.3 The Laplace Transform
Integral transforms, such as the Laplace transform and the Fourier transform, are
available by loading the inttrans package using with(inttrans). In Maple,
the Heaviside step function H(t −a) is Heaviside(t-a), and the Dirac delta
function δ(t −a) is Dirac(t-a).
Load the inttrans package.
>with(inttrans):
Given a function f (t).
>f:=t*cosh(2*t)+t^2*sin(5*t)+t^3+sin(t)*Heaviside(t-Pi);
f := t cosh 2t + t 2 sin 5t + t 3 + sin t Heaviside(t −π)
Evaluate the Laplace transform using laplace.
>F:=laplace(f,t,s);
F :=
1
1
10(3s2 −25)
6
e−π s
+
+
+
+
2(s −2)2 2(s +2)2
(s2 +25)3
s4 s2 +1
Given the Laplace transform of a function G(s).
>G:=(s-3)/(s^2-6*s+25)+2/(s+2)^3+exp(-2*s)*(2+1/(s^2+1));
G :=
1 s −3
2
−2s
2
+
+
e
+
s2 −6s +25 (s +2)3
s2 +1
Evaluate inverse Laplace transform using invlaplace.
>g:=invlaplace(G,s,t);
g := e3t cos 4t + t 2 e−2t + 2 Dirac(t −2) + Heaviside(t −2) sin(t −2)
When evaluating the inverse Laplace transform by hand, one frequently needs
to perform partial fraction decomposition, which can be easily done using Maple.
>F:=8*(s+2)/(s-1)/(s+1)^2/(s^2+1)/(s^2+9);
Define a fraction.
F :=
8(s +2)
(s −1)(s +1)2 (s2 +1)(s2 +9)
Perform partial fraction decomposition using convert with option parfrac,
in which s is the variable.
>convert(F,parfrac,s);
27
3
1
s −3
s −11
−
−
+
−
2
2
10(s −1) 5(s +1)
50(s +1) 4(s +1) 100(s2 +9)
When an ODE is solved using dsolve with the option method=laplace, Maple
forces the equation to be solved by the method of Laplace transform.
a.2 table of derivatives
533
A.2 Table of Derivatives
1.
d n
x = nx n−1
dx
2.
d x
e = ex
dx
3.
d
1
ln x =
dx
x
=⇒
d x
a = ax ln a,
dx
=⇒
1
d
log a x =
,
dx
x ln a
∵ ax = ex ln a
∵ log a x =
ln x
ln a
Trigonometric Functions
Hyperbolic Functions
4.
d
sin x = cos x
dx
d
sinh x = cosh x
dx
5.
d
cos x = − sin x
dx
d
cosh x = sinh x
dx
6.
1
d
tan x =
= sec2 x
dx
cos2 x
d
1
= sech2 x
tanh x =
2
dx
cosh x
7.
1
d
cot x = − 2 = − csc2 x
dx
sin x
d
1
= −csch2 x
coth x = −
dx
sinh2 x
8.
sin x
d
sec x =
= tan x sec x
dx
cos2 x
d
sinh x
sech x = −
dx
cosh2 x
= − tanh x sech x
9.
d
cos x
csc x = − 2
dx
sin x
= − cot x csc x
cosh x
d
csch x = −
dx
sinh2 x
= − coth x csch x
Inverse Trigonometric Functions
10.
1
d
sin−1 x = √
dx
1−x 2
d
1
cos−1 x = − √
dx
1−x 2
11.
1
d
tan−1 x =
dx
1+x 2
1
d
cot−1 x = −
dx
1+x 2
12.
d
1
sec−1 x = √
dx
x x 2 −1
d
1
csc−1 x = − √
dx
x x 2 −1