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BC: 8831 - Probability and Statistical Theory
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PST˙ws
World Scientific
Published by
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British Library Cataloguing-in-Publication Data
A catalogue record for this book is available from the British Library.
LECTURES, PROBLEMS AND SOLUTIONS FOR ORDINARY DIFFERENTIAL
EQUATIONS
Copyright © 2015 by World Scientific Publishing Co. Pte. Ltd.
All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,
electronic or mechanical, including photocopying, recording or any information storage and retrieval
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ISBN 978-981-4632-24-9
ISBN 978-981-4632-25-6 (pbk)
Printed in Singapore
ZhangJi - Lectures, problems and solutions.indd 1
12/8/2014 5:13:32 PM
Preface
This book, Lectures, Problems and Solutions for Ordinary Differential
Equations, results from more than 20 revisions of lectures, exams, and
homework assignments to approximately 5,000 students in the College of
Engineering and Applied Sciences at Stony Brook University over the
past 30 semesters. The book contains notes for twenty-five 80-minute
lectures and approximately 1,000 problems with solutions. Creating
another ODEs book is probably more unnecessary than reinventing the
wheel. Yet, I constructed this manuscript differently by focusing on
gathering and inventing examples from physical and life sciences,
engineering, economics, and other real-world applications. These
examples, partly adapted from well-known textbooks and partly freshly
composed, focus on illustrating the applications of ordinary differential
equations. The examples are much stronger stimuli to students for
learning to prove the theorems while mastering the art of applying them.
While preparing this manuscript, I benefit immensely from many people
including undergraduate students who took the class and graduate
teaching assistants who helped teach it. Changnian Han has triple roles:
one of the undergraduate students, one of the TAs, and one of the editors.
I also thank the World Scientific Publishing and its fantastic senior editor
Ms. Ji Zhang.
Stony Brook, New York
August 1, 2014
v
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PST˙ws
Problems
&
Solutions
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PST˙ws
Contents
PREFACE ..........................................................................................................v
CHAPTER 1
FIRST-ORDER DIFFERENTIAL EQUATIONS ................................. 1
1.1 DEFINITION OF DIFFERENTIAL EQUATIONS ........................................................1
1.2 MATHEMATICAL MODELS .............................................................................9
1.2.1
Newton’s Law of Cooling ..............................................................9
1.2.2
Newton’s Law of Motion.............................................................11
1.2.3
Torricelli’s Law for Draining ........................................................13
1.2.4
Population Models ......................................................................17
1.2.5
A Swimmer’s Problem .................................................................18
1.2.6
Slope Fields & Solution Curves ....................................................21
1.3 SEPARATION OF VARIABLES .........................................................................26
1.4 LINEAR FIRST-ORDER DES ..........................................................................32
1.5 SUBSTITUTION METHODS ...........................................................................40
1.5.1
Polynomial Substitution ..............................................................40
1.5.2
Homogeneous DEs ......................................................................42
1.5.3
Bernoulli DEs ...............................................................................44
1.6 THE EXACT DES ........................................................................................51
1.7 RICCATI DES ............................................................................................72
CHAPTER 2
MATHEMATICAL MODELS .......................................................76
2.1 POPULATION MODEL .................................................................................76
2.1.1
General Population Equation ......................................................76
2.1.2
The Logistic Equation ..................................................................78
2.1.3
Doomsday vs. Extinction .............................................................81
2.2 ACCELERATION-VELOCITY MODEL ................................................................87
2.2.1
Velocity and Acceleration Models...............................................87
2.2.2
Air Resistance Model ..................................................................88
2.2.3
Gravitational Acceleration ..........................................................93
2.3 AN EXAMPLE IN FINANCE ..........................................................................101
ix
Contents
CHAPTER 3
3.1
3.2
3.3
3.4
3.5
3.6
CLASSIFICATION OF DES ........................................................................... 107
LINEAR INDEPENDENCE ............................................................................ 112
CONSTANT COEFFICIENT HOMOGENEOUS DES ............................................. 121
CAUCHY-EULER DES................................................................................ 135
INHOMOGENEOUS HIGHER ORDER DES ...................................................... 139
VARIATION OF PARAMETERS ..................................................................... 151
CHAPTER 4
4.1
4.2
4.3
4.4
4.5
LINEAR DES OF HIGHER ORDER ............................................. 107
SYSTEMS OF LINEAR DES ....................................................... 162
BASICS OF DE SYSTEMS ............................................................................ 162
FIRST-ORDER SYSTEMS AND APPLICATIONS .................................................. 165
SUBSTITUTION METHOD........................................................................... 173
OPERATOR METHOD ............................................................................... 179
EIGEN-ANALYSIS METHOD........................................................................ 184
CHAPTER 5
LAPLACE TRANSFORMS ......................................................... 193
5.1 LAPLACE TRANSFORMS ............................................................................ 193
5.2 PROPERTIES OF LAPLACE TRANSFORMS ....................................................... 195
5.2.1
Laplace Transforms for Polynomials......................................... 196
5.2.2
The Translator Property ............................................................ 199
5.2.3
Shifting Property ....................................................................... 203
5.2.4
The t-multiplication property ................................................... 206
5.2.5
Periodic Functions..................................................................... 209
5.2.6
Differentiation and Integration Property ................................. 210
5.3 INVERSE LAPLACE TRANSFORMS ................................................................. 215
5.4 THE CONVOLUTION OF TWO FUNCTIONS ..................................................... 220
5.5 APPLICATION OF LAPLACE TRANSFORMS ...................................................... 224
APPENDIX A
SOLUTIONS TO SELECTED PROBLEMS ............................... 238
CHAPTER 1 FIRST-ORDER DES ............................................................................ 238
1.1 Definition of DEs ............................................................................... 238
1.2 Mathematical Models....................................................................... 244
1.3 Separation of Variables..................................................................... 254
1.4 Linear First-Order DEs ....................................................................... 262
x
Contents
1.5 Substitution Methods ........................................................................273
1.6 The Exact DEs ....................................................................................296
1.7 Riccati DEs .........................................................................................308
CHAPTER 2 MATHEMATICAL MODELS...................................................................316
2.1 Population Model ..............................................................................316
2.2 Acceleration-Velocity Model .............................................................326
2.3 An example in Finance ......................................................................358
CHAPTER 3 LINEAR DES OF HIGHER ORDER ...........................................................368
3.1 Classification of DEs ..........................................................................368
3.2 Linear Independence .........................................................................370
3.3 Constant Coefficient Homogeneous DEs ...........................................378
3.4 Cauchy-Euler DEs...............................................................................391
3.5 Inhomogeneous Higher Order DEs ....................................................396
3.6 Variation of Parameters ....................................................................415
CHAPTER 4 SYSTEMS OF LINEAR DES ....................................................................426
4.2 First-Order Systems and Applications ...............................................426
4.3 Substitution Method .........................................................................431
4.4 Operator Method ..............................................................................439
4.5 Eigen-Analysis Method......................................................................447
CHAPTER 5 LAPLACE TRANSFORMS ......................................................................455
5.2 Properties of Laplace Transforms......................................................455
5.3 Inverse Laplace Transforms...............................................................463
5.4 The Convolution of Two Functions ....................................................467
5.5 Application of Laplace Transforms ....................................................470
APPENDIX B
LAPLACE TRANSFORMS ....................................................505
SELECTED LAPLACE TRANSFORMS.........................................................................505
SELECTED PROPERTIES OF LAPLACE TRANSFORMS ...................................................506
APPENDIX C
DERIVATIVES & INTEGRALS ..............................................509
APPENDIX D
ABBREVIATIONS ...............................................................511
APPENDIX E
TEACHING PLANS..............................................................513
xi
Contents
REFERENCES ................................................................................................ 515
INDEX .......................................................................................................... 517
xii
Chapter 1
First-Order
Differential Equations
1.1
Definition of Differential Equations
A differential equation (DE) is a mathematical equation that relates some
functions of one or more variables with its derivatives. A DE is used to
describe changing quantities and it plays a major role in quantitative
studies in many disciplines such as all areas of engineering, physical
sciences, life sciences, and economics.
Examples
Are they DEs or not?
+
+ =0
No
1
Chapter 1 First-Order Differential Equations
+
+
′′ =
′+ =0
′+
Yes
′=0
Yes
Yes
Here ′ =
Here ′ =
Here ′ =
and ′ =
To solve a DE is to express the solution of the unknown function (the
dependent variable) in mathematical terms without the derivatives.
Example
+
=0
′=−
=−
is not a solution
is a solution
In general, there are two common ways in solving DEs, analytic and
numerical. Most DEs, difficult to solve by analytical methods, must be
“solved” by numerical methods although many DEs are too stiff to solve
using numerical techniques. Solving DEs by numerical methods is a
different subject requiring basic knowledge of computer programming
and numerical analysis; this book focuses on introducing analytical
methods for solving very small families of DEs that are truly solvable.
Although the DEs are quite simple, the solution methods are not and the
essential solution steps and terminologies involved are fully applicable to
much more complicated DEs.
Classification of DEs
Classification of DEs is itself another subject in studying DEs. We will
introduce classifications and terminologies for flowing the contents of
the book but one may still need to lookup terms undefined here. First, we
introduce the terms of dependent and independent variables or functions.
A dependent variable represents the output or effect while the
independent variable represents the input or the cause. Truly, a DE is an
equation that relates these two variables. A DE may have more than one
variable for each and the DE with one independent variable and one
2
1.1 Definition of Differential Equations
dependent variable is called ordinary differential equation or ODE. The
ODE, and simply referred to as DE, is the object of our study.
Continuing, you understand why this 1-to-1 relationship is called
ordinary. A DE that has one dependent variable and ≥ 2 independent
variables, 1-to-N relationship, is called partial differential equation or
PDE. Studying PDEs, out of the scope of this book, requires solid
understanding of partial derivatives and, more desirably, full knowledge
of multiple variable calculus. By now, you certainly want to see two
other types of DEs. With ≥ 2 dependent variables and 1 independent
variable, you may compose an N-to-1 system of ODEs. Similarly, with
≥ 2 dependent variables and ≥ 2 independent variables, you may
compose an N-to-M system of PDEs.
We have several other classifications to categorize DEs.
Order of DEs
The order of a DE is determined by the highest order derivative of the
dependent variable (and sometimes we interchangeably call it unknown
function). If you love to generalize things, the algebraic equations may
be classified as the 0th order DEs as there are no derivatives for the
unknowns in the algebraic equations.
Examples
Determine the order of the following DEs:
+
+
+
+
+
=0
=0
=0
=0
=0
1st-order DE
2nd-order DE
n-th order DE
3
Chapter 1 First-Order Differential Equations
Definition
ODEs
vs.
PDEs
ODEs contain only one
independent variable.
PDEs contain two or more
independent variables.
Order of DEs
The highest order of the
derivatives of the
independent variables.
Linear
vs.
Nonlinear
Homogenous
vs.
DEs containing nonlinear
term(s) for the dependent
variables are nonlinear
regardless the nature of the
independent variables
A 1st-order DE whose both
sides are homogeneous
functions of the same
degree is a 1st-order
homogeneous and
otherwise it
inhomogeneous.
Inhomogeneous
A linear DE whose terms
contain only independent
variable or its derivatives is
homogeneous. (See text)
4
Examples
ODE:
+
PDE:
=
=
+
1st order:
′= +
2nd order:
′′ = +
Linear:
+
′=
Nonlinear:
′′ = +
1st order
Homogeneous:
=
+
1st order
Inhomogeneous:
′′ =
+
Homogeneous:
+ cos = 0
Inhomogeneous:
+ = cos
1.1 Definition of Differential Equations
Linearity of DEs
DEs can also be classified as linear or nonlinear according to the linearity
of the dependent variables regardless of the nature of the independent
variables. A DE that contains only linear terms for the dependent
variable or its derivative(s) is a linear DE. Otherwise, a DE that contains
at least one nonlinear term for the dependent variable or its derivative(s)
is a nonlinear DE.
Linear
+ =0
+ + =0
+ + =0 $
Nonlinear
+
=0
+
+
=0
% +
+
=0
Solutions
A function that satisfies the DE is a solution to that DE. Seeking such
function is the main content of the book while composing the DEs,
which may excite engineering majors more, is the secondary objective of
this book.
Solutions can also be classified into several categories, for example,
general solutions, particular solutions and singular solutions. A general
solution (G.S.), i.e., complete solution, is the set of all solutions to the
DE and it usually can be expressed in a function with arbitrary
constant(s). A particular solution (P.S.) is a subset of the general
solutions whose arbitrary constant(s) are determined by initial or
boundary conditions or both. A singular solution is a solution that is
singular. In a less convoluting term, a singular solution is one for which
the DE (or the initial value problem or the Cauchy problem) fails to have
a unique solution at some point on the solution. The set on which the
solution is singular can be a single point or the entire real line.
5
Chapter 1 First-Order Differential Equations
Examples
Try to guess the solutions of the following DEs:
(1)
(2)
(3)
(4)
(5)
+
+
+
+
+
=0
=
=0
= sin
= sin
(
where, in general,
(
≠
As briefly mentioned before, there are several methods to find the
solution of DEs. The following two are common:
1. To obtain analytical (closed form) solutions. Only a small percentage
of linear DEs and a few special nonlinear DEs are simple enough to
allow findings of analytical solutions. In such studies, basic concepts
and theorems concerning the properties of the DEs or their solutions
are introduces.
2. To obtain numerical solutions. Most DEs in science, engineering,
and finance are too complicated to allow findings of analytical
solutions and numerical methods are the only approach to finding
approximate numerical solutions. Unfortunately, most of these DEs
are the most important for practical purposes. Indeed, every rose has
its thorn. To solve DEs numerically, one has to acquire a different set
of skills: numerical analysis and computer programming.
Problems
Problem 1.1.1 Verify by substitution that each given function is a solution of
the given DE. Throughout these problems, primes denote derivatives with
respect to (WRT) .
+ = 3 cos 2 ,
= sin − cos 2
( = cos − cos 2 ,
6
1.1 Definition of Differential Equations
Problem 1.1.2 Verify by substitution that each given function is a solution of
the given DE. Throughout these problems, primes denote derivatives WRT .
−
+ 2 = 0,
,
= sin ln
( = cos ln
Problem 1.1.3 A function = is described by some geometric property
of its graph. Write a DE of the form ⁄ =
, having the function - as
its solution (or as one of its solutions). The graph of - is normal to every curve
+ / (/ is a constant) where they meet.
of the form =
+ =
Problem 1.1.4 Determine by inspection at least one solution of DE
3 . That is, use your knowledge of derivatives to make an intelligent guess.
Then test your hypothesis.
+ =
Problem 1.1.5 Determine by inspection at least one solution of DE
0. That is, use your knowledge of derivatives to make an intelligent guess.
Then test your hypothesis.
Problem 1.1.6 Verify by substitution that the given function is a solution of
the given DE. Primes denote derivatives WRT .
1
+2
= 0,
=
1+
Problem 1.1.7 Verify that
value of the constant 1so that
+3
= 0,
satisfies the given DE and then determine a
satisfies the given initial condition (I.C.).
= 1exp − ,
0 =7
Problem 1.1.8 Verify that
value of the constant 1 so that
+ tan = cos ,
satisfies the given DE and then determine a
satisfies the given I.C..
= + 1 cos ,
8 =0
Problem 1.1.9 Verify that
value of the constant 1 so that
=3
+1 ,
satisfies the given DE and then determine a
satisfies the given I.C..
= tan
+1 ,
0 =1
7
Chapter 1 First-Order Differential Equations
Problem 1.1.10 Verify that
value of the constant 1 so that
+3 = 2
9,
satisfies the given DE and then determine a
satisfies the given I.C..
1 9 1
=
+ ,
2 =1
4
Problem 1.1.11 Verify by substitution that the given functions are solutions of
the given DE. Primes denote derivatives WRT .
=9 ,
,
= exp −3
( = exp 3
Problem 1.1.12 Verify by substitution that the given functions are solutions of
the given DE. Primes denote derivatives WRT .
exp
= 1,
= ln + 1 ,
0 =0
Problem 1.1.13 Verify that
satisfies the given DE and then determine a
value of the constant 1 so that
satisfies the given I.C.. In the equation and
the solution = is a given constant parameter.
= 1 exp −
,
0 = 2014
′ + = >( = 0;
8
1.2 Mathematical Models
1.2
Mathematical Models
Given below are a few examples that illustrate the process of translating
scientific laws and principles into DEs. The independent variable in these
examples is the time while the other variables are dependent variables.
1.2.1 Newton’s Law of Cooling
The time rate of change (the rate of change WRT time) of the
temperature @
of a body is proportional to the difference between
@
and the temperature A of the surrounding medium, which is
assumed to be so big that the energy exchange between the medium and
the body does not change the temperature of the medium. This is an
approximation of true heat transfer in which the total energy is
conserved.
@
@
A
Figure 1.1 Place an object in a medium of temperature A.
Consider an object at a temperature @
that is placed in a medium at a
constant temperature A (Figure 1.1). We establish the DE that describes
the temperature change of the object while it exchanges energy (heat)
with the medium.
•
•
A is the temperature of the medium.
@ is the temperature of the object at time ; @
the initial temperature.
= 0 = @B is
9
Chapter 1 First-Order Differential Equations
•
= / A − @ is the rate of temperature variation, or the DE,
is the time.
•
C
and / is a positive constant that characterizes the heat
conductivity of the medium with the body.
Solution
@
=/ A−@
@
= −/
@−A
Integrating both sides of the DE, we get
C
CD
@
=−
@−A
B
/
ln @ − A − ln @B − A = −/
ln E
@
@0
A
@0
@−A
F = −/
@B − A
@0 > A, H < 0
@0 = A
@0 > A, H > 0
@0 < A, H > 0
@0 < A, H < 0
Figure 1.2 Plot of the object’s temperature change @ with time .
10
1.2 Mathematical Models
@
@−A
= exp −/
@B − A
@ − A = @B − A exp −/
=A
+ @B − A exp −/
Finally, we get the solution of the DE,
@
= @B exp −/
+ 1 − exp −/
A
Thus, we have converted a scientific law into a DE and found its
solution. Now, if given the values of A and / , we can predict the
temperature of the object at any time.
Remarks
(1) If A ≈ @B we can get
C
≈ 0 according to the original DE,
meaning that @
does not change with time much,
approximately a constant.
(2) If A > @B , we have
C
(3) If A < @B , we have
C
>0.
The temperature of the object increases while heat is transferred
from the medium to the object. This process will last till the
object reaches the same temperature of the medium.
< 0.
The temperature of the object decreases while heat is transferred
to the medium from the object. This process will last till the
object reaches the same temperature of the medium.
1.2.2 Newton’s Law of Motion
Let the motion of a particle (a mathematical dot) of mass K along the
straight line be described by its position function =
. Then, the
velocity of the particle is
11
Chapter 1 First-Order Differential Equations
and the acceleration is
If a force M
motion, then
=
L
L
=
=
=
=
=
=L
=
acts on a particle and it is directed along its line of
M
=K
Initial position:
Initial speed: L
0
0
LB
B
We can establish an equation of motion for the particle as
SK
Solution
RL
Q
0
0
M
B
0
LB
Let us suppose that we are given the acceleration of the particle and that
it is a constant.
L
So to find L we integrate both sides of the above equation.
L
1(
1(
We know that the initial velocity at time
0 is LB . So after plugging
this information into the above equation we get 1( LB , and thus
12
1.2 Mathematical Models
L=
=
+ LB
Now, integrating the speed, we get the position of the particle as
=
=
Applying the I.C.
=
=0 =
1
2
L
+1
+ LB
+1
+ LB + 1
B,
we get 1 =
1
2
+ LB +
B.
Finally, we get the position of the particle as a function of time (the
independent variable) and other parameters such as the initial position,
speed, and the acceleration
=
B
So, if given the acceleration (through the force), the initial position and
speed, we can determine the position and speed of the particle at any
time.
1.2.3 Torricelli’s Law for Draining
The time rate of change of the volume T of water or other liquid in a
draining sink is proportional to the square root of the depth of water in
the tank.
T
is the volume of water in the sink at time . Although a hemisphere
is used in this example (Figure 1.3), the draining model is applicable to
containers of any conceivable shapes.
13
Chapter 1 First-Order Differential Equations
is the height of the water surface, from the draining point, at time .
The draining model introduced by Torricelli can be written as
V
= / V ∝ −V
where / is the so-called draining constant which depends on
A
Y
Y
X
X
X−
X
1) Liquid property such as viscosity. The / value will be different
for the following liquids: tea, honey, water, or gasoline, etc.
2) Size of the hole.
3) Pressure differential between inside and outside of the container.
Figure 1.3 A hemisphere draining model of radius X.
Let A
be the cross section area of the container at height . We can
write the volume of the container as a function of area and height.
14
1.2 Mathematical Models
V=A
where V is the volume element of the container as a function of area
and height variation from +
to . With the draining equation
V
A
Z
We can now write
= −/V
= −/V
=0 =
(1.1)
B
where B is the initial height of liquid surface in the container. We have
converted the original draining equation to involving independent
variable and the dependent variable , the height.
Solution
Let us solve the above draining problem for a peculiar case: the container
is a hemisphere whose inner radius is X. We can easily write the crosssection area at any height as
A
= 8Y = 8 X − X −
= 8 X − X + 2X −
= 8 2X −
Plugging (1.2) into (1.1), we get
8 2X −
8V 2X −
[
8V 2X −
(1.2)
= −/V
= −/
=
B
−/
15
Chapter 1 First-Order Differential Equations
[
8 \2X
8
8 E2X
[
4
8E X
3
(
(
−
−
−
2
5
F
[
9
= −/
] = −/
F_ = −/
[
4
2
4
2
E X−
F − 8X E X − XF = −/
3
5
3
5
8
4
2
14 9
E X−
F − 8X = −/
3
5
15
Therefore, the resulting solution of the draining equation can be written
as:
8
4
2
14 9
E X−
F=
8X − /
3
5
15
This equation relates the height of the surface of the water in hemisphere
container with time, in an implicit form. The initial state is that the
hemisphere container is filled with water to the rim.
Remarks
(1) For = 0, the only way the solution to make sense is = X.
(2) The time for the tank to be empty is that = 0. Plugging in =
0 to the above equation, we can compute the time needed to
empty the container with
=0⇒/ =
empty
16
=
14 9
8X
15
14 8 9
X
15 /
1.2 Mathematical Models
a) From the above equation we have, empty ∝ X. So the bigger
the tank (the larger its radius), the longer the time needed to
empty it: X ↑ ⇒ empty ↑.
A special case, if the container is tiny, little time is needed to
empty it: X → 0 ⇒ empty → 0.
empty
∝ e . Thus,
(
/ ↑ ⇒ empty ↓.
A special case, if the draining constant is huge (a huge
leaking hole, or a very slippery liquid, or a massive pressure
differential), little time is needed to empty it: / → ∞ ⇒
empty → 0.
The opposite is also true: if the draining constant is tiny (a
tiny leaking hole, or a very stuffy liquid, or no pressure
differential), it takes eternity to empty it: / → 0 ⇒ empty →
∞.
b) From the above equation we also have,
So far, everything appears to make sense!
1.2.4 Population Models
The time rate of change of a population h
with constant birth and
death rates is in many cases proportional to the size of the population.
h
= Population at time
We can establish a simple initial value problem (IVP) to describe the
population as a function of time.
h
∝h⇒
h
Z
h( = 0) = hB
= /h
17
Chapter 1 First-Order Differential Equations
Solution
h
l
For example, if h
lD
= /h
h
=/
k
h
= /
h
B
ln h − ln hB = /
h
= hB exp /
= 0 = 1000 and h
m
hB = 1000
/ = ln 2
= 1 = 2000, we can get
with which we can construct the model of exponential growth of
populations as
h
= 1000 exp ln 2 = 1000 ⋅ 2
In this case, the population doubles every time unit.
1.2.5 A Swimmer’s Problem
The picture (Figure 1.4) shows a northward flowing river of width o =
2 . The lines = ± represent the banks of the river and the -axis is
the center. Suppose that the velocity L[ at which the water flows
increases as one approaches the center of the river and is given in terms
of distance from the center by
L[ = LB \1 −
18
]
1.2 Mathematical Models
where L[
is the water speed at location , LB is the water speed at the
center of the river and Lq is the swimmer speed.
Yr
st
2
Figure 1.4 A swimmer model.
Remarks
(1) The water speed at the center of the river is maximum:
= 0, L[ = LB 1 − 0 = LB
(2) The water speed at either bank of the river is zero:
At left bank where = −
L[ = LB 1 − 1 = 0
At right bank where =
L[ = LB 1 − 1 = 0
The two cases where the speed of water is zero at the riverbanks
are called no-slip conditions, a well-known result in fluid
mechanics.
19
Chapter 1 First-Order Differential Equations
(3) The water speed profile formula can be derived from more basic
laws of physics. The derivation is out of the scope and we use it
without proof.
A Swimmer’s Problem
Suppose that the swimmer starts at − , 0 on the west bank and swims
at the east direction, as shown in Figure 1.4, the swimmer’s velocity
vector relative to the ground had horizontal component Lu and vertical
component L[ . Hence the swimmer’s instantaneous directional angle v is
given by
=
= tan v =
L[ LB
= \1 −
Lu Lu
]
Thus, we have an IVP that describes the trajectory of the swimmer.
=
Z
Solution
LB
\1 − ]
Lu
=− =0
Solving this problem, we have
LB
\1 − ]
Lu
LB
=
\1 −
>w Lu
=
B
=
LB
\ −
Lu
3
]x
]
>w
Finally, we obtain the equation for describing the trajectory of the
swimmer as
20
1.2 Mathematical Models
=
LB
\ −
Lu
3
+
2
]
3
Remarks
We can see at the east bank of the river
have drifted by
drift
LB
\ −
Lu
3
LB 4
=
Lu 3
=
=
= , the swimmer would
+
2
]
3
(1) If ↑ ⇒ drift ↑, meaning that the wider the river, the longer the
drift.
(2) If TB ↑ ⇒ drift ↑ , meaning that the faster the river flow, the
longer the drift.
(2a) LB → 0 ⟹ → 0, a special case: if the river water does
not move (e.g., a big lake), you would not drift.
(2b) LB → ∞ ⟹ → ∞, a special case: if the river flows very
fast (e.g., Niagara River), you would drift very far.
(3) If Tu ↑ ⇒ drift ↓ , meaning that the faster the swimmer, the
shorter the drift.
(3a) Lu → ∞ ⟹ → 0
(3b) Lq → 0 ⟹ → ∞
1.2.6 Slope Fields & Solution Curves
Suppose we have a DE
|
=
=
,
21
Chapter 1 First-Order Differential Equations
(1) Does the solution exist?
(2) How many solutions exist? Is it unique?
(3) How to find it when it exists?
Here we define some concepts that can help us study the properties of the
given DE
Figure 1.5 Examples of slope field and isocline.
Slope Curves: A line in , plane whose slope tan v is
Slope Field: A collection of the slope curves. (Figure 1.5)
,
.
Isocline: A series of lines with the same slope, such as a family of
curves
, = . (Figure 1.5)
Example 1
Find the isoclines
Solution
The Isocline equation
It is a circle
22
=
=
+
=1
,
=
+
=1
1.2 Mathematical Models
Example 2
Find the isoclines
Solution
= sin
−
− = arcsin
= − arcsin
Problems
Problem 1.2.1 A car traveling at 60 mph (88ft/s) skids 176ft after its brakes
are suddenly applied. Under the assumption that the braking system provides
constant deceleration, what is that deceleration? For how long does the skid
continue?
Problem 1.2.2 Find the position function
of a moving particle with the
given acceleration
, initial position 0 = B and initial velocity L 0 =
LB .
= 50 sin 5
LB = −10
•
B =8
Problem 1.2.3 Find the position function
of a moving particle with the
given acceleration
, initial position 0 = B and initial velocity L 0 =
LB .
1
=
+1
, where = ≥ 3
€
LB = 0
B =0
Problem 1.2.4 We drain a fully filled spherical container of inner radius X by
drilling a hole at the bottom. We assume the draining process follows
Torricelli’s law and the draining constant k.
(1) Compute the time needed to empty the container.
(2) If we double the radius of the container which is also fully filled and keep
all other conditions unchanged, compute the draining time again.
23
Chapter 1 First-Order Differential Equations
Problem 1.2.5 A complete spherical container of radius R has two small holes.
One at the top for air to come in, and one at the bottom for liquid to drain out
(think of it as a hollowed bowling ball). At = 0, the container is filled with
liquid. Besides the radius R, you are also given the draining constant /. The
draining process follows Torricelli’s law. Now please do the following,
(1) Compute the time @( needed to drain the upper half of the container (i.e.,
the water level dropping from the top hole to the equator line of the ball).
(2) Compute the time @ needed to drain the lower half of the container (i.e.,
the water level dropping from the equator line of the ball to the bottom hole).
(3) Which relationship is correct @( > @ , @( < @ , or @( = @ ?
Problem 1.2.6 A coffee cup from the “Moonbucks” coffee shop is like a
cylinder with decreasing horizontal cross-section area. The radius of the top
circle is X, and that for the bottom is 10% smaller. The cup height is „. Now, a
nutty kid drills a pinhole at the half height point of the cup, allowing coffee to
leak under the Draining Equation with a draining constant /. Compute the time
for coffee to leak to the pinhole. Compute the time again if we turn the cup
upside down with full cup to start.
Problem 1.2.7 A spherical container of inner radius X is originally filled up
with liquid. We drill one hole at the bottom, and another hole at the top (for air
in-take), to allow liquid to drain from the bottom hole. The draining process
follows the Torricelli’s law, and the draining constant depends on the liquid
properties; the hole size, and the inside-outside pressure differential, etc. While
draining the upper half, we make the draining constant as /. What draining
constant must we set (e.g., by adjusting the bottom hole size) so that the lower
half will take the same amount of time to drain as the upper half did?
Problem 1.2.8 A spherical container of inner radius X is filled up with liquid.
We drill one hole at the bottom (and another at the top to allow air to come in).
The draining process follows the Torricelli’s law with the usual draining
constant /. We start draining the container, when it is full, until the leftover
volume reaches a “magic” value at which the draining time for the top portion
is equal to that for the bottom (left-over) portion. Please calculate the leftover
volume.
24
1.2 Mathematical Models
Problem 1.2.9 During the 2014 Stony Brook University Roth Quad Regatta
boat race, students propelled their boats from one end to the other of a long
pond (not a river) of length …. Since it is not a river, water movement and wind
have little effect on the boats. Suppose your boat’s mass is and each of your
boaters’ mass is K (all boaters are equal in mass). The propelling force from
each boater during the entire race is an equal constant B . Water resistance is
approximated to be proportional to the speed of the boat with a constant †B
regardless of the weight of the boat plus boaters. Your boat is always still at the
start, and the boaters are the only power source.
(1) Establish the DE, for = boats for relating the boat’s speed with time and the
given parameters.
(2) Solve the DE established above to find the speed as a function of time.
(3) Which leads to the shortest travel time with all other conditions fixed? One
boater or as many as possible?
25
Chapter 1 First-Order Differential Equations
1.3
Separation of Variables
All 1st-order IVP can be written as
If M
,
can be written as
=M
=
|
=
-
,
=
Then the 1st-order IVP is said to be separable.
Solution
=
=
M
-
=‡
-
+1
where 1 is a constant determined by the I.C.:
Example 1A
Solve
Solution
|
= −6
= −6
0 =7
ˆ‰Š‹Œ‹•Ž‰ ••
‘’’’’’’’’“
=
= .
−6
= −3 + ln 1
= 1 exp −3
which is the G.S. of the DE. Plugging in I.C.
1 exp −3 , we will get
0 = 1 exp 0 = 7
This gives
1=7
26
ln
0 = 7 into
=
1.3 Separation of Variables
= 7 exp −3
Therefore, the P.S. is
= −6
Example 1B
Is = 0 a solution for equation
?
Left Hand Side LHS = = 0
Right Hand Side RHS = −6 = −6 ⋅ 0 = 0
LHS = RHS
Therefore
=0
is a P.S. for = −6 .
Solution
Example 2
Solve IVP
Z
4−2
3 −5
1 =3
=
Solution
3
=
−5
=
4−2
3 −5
4−2
Integrate with both side, we have
−5 =4 − +1
as the G.S.. Plugging the I.C. 1 = 3 gives
3 − 15 = 4 − 1 + 1
1=9
Therefore, the P.S. for the IVP is
−5 = 4 − +9
Example 3
Solve
Solution
+
+
=0
=0
27
Chapter 1 First-Order Differential Equations
=−
1
1
=−
−
1
1
Therefore,
is the G.S. of the given DE.
=−
=
+
1
1
1
1
+1
=1
Singular Solutions
Suppose we have a DE
=„
‡
To solve this DE, we have to divide both sides by „
That gives
=‡
„
=
„
‡
.
Such division is mathematically allowed if and only if „
Therefore
„
=0
≠ 0.
is a solution for the equation because it indeed satisfies the original DE.
This solution is peculiar so it is called singular solution.
28
1.3 Separation of Variables
Problems
Problem 1.3.1 Find G.S. of the DE.
3
−2
+
+1
=0
Problem 1.3.2 Find G.S. (implicit if necessary, explicit if convenient) of the
DE in
′ = 64
Prime denotes derivatives WRT .
(
Problem 1.3.3 Find G.S. (implicit if necessary, explicit if convenient) of the
DE in
′=1+ + +
Prime denotes derivatives WRT . (Suggestion: Factor the right-hand side.)
Problem 1.3.4 Find a function satisfying the given DE and the prescribed I.C..
Z
=
exp −
0 =1
Problem 1.3.5 Find explicit P.S. of the IVP in
= −2 cos 2
|
0 = 2014
Problem 1.3.6 Find explicit P.S. of the IVP in
2√ ′ = cos
8
•
4 =
4
Problem 1.3.7 Find the explicit P.S. of the IVP in
= 2 +3
exp
|
0 =5
Problem 1.3.8 Find the G.S. (implicit if necessary, explicit if convenient) of
the DE in the following problem. Primes denote derivatives WRT .
29
Chapter 1 First-Order Differential Equations
′=
+ 1 cos
Problem 1.3.9 Find the G.S. of following DE.
+ =5
4
Problem 1.3.10 Find explicit P.S. of the IVP in
+3
′=2
|
1 = −1
Problem 1.3.11 Find the G.S. (implicit if necessary, explicit if convenient) of
the DE in the following problem. Primes denote derivatives WRT .
′ = 2 sec
Problem 1.3.12 Find the explicit P.S. of the IVP in
tan ′ =
8
8
•
š ›=
2
2
Problem 1.3.13 Find the G.S. (implicit if necessary, explicit if convenient) of
the DE in the following problem. Primes denote derivatives WRT .
+ 1 tan
′=
Problem 1.3.14 Find the G.S. (implicit if necessary, explicit if convenient) of
the DE in the following problem.
= −2
+3
Problem 1.3.15 Solve the following DE.
−
=3
|
1 =1
Problem 1.3.16 Solve the following DE.
1+
+ = cos
|
0 =1
Problem 1.3.17 Solve the following DE
30
1.3 Separation of Variables
= œ v − ln
where, obviously, v and œ are parameters independent of and . If ( =
0) = B and ( → ∞) = • , can you express v and/or œ in terms
of B and • ? If so, do it.
31
Chapter 1 First-Order Differential Equations
1.4
Linear First-Order DEs
All linear 1st-order DEs can be written in the following form:
If A
If ž
A
+ž
= 0, then it is not a DE. Now suppose A
= 0,
= 0,
=1
A
It is a separable DE.
If 1
=1
A
+ž
=0
Dividing both side of above equation by
=−
It is a separable DE.
≠ 0.
(1.3)
gives
ž
A
For the general form of linear 1st-order DE, we divide both sides of (1.3)
by A ,
Let
We have
32
+
ž
A
ž
A
=h
=
and
+h
1
A
1
A
=Ÿ
=Ÿ
(1.4)
1.4 Linear First-Order DEs
Let us now derive the method to solve the general linear 1st-order DE in
the form of (1.4). The idea to get rid of h
term is to compose a
whole derivative as a function of
:
+
+h
If we can make the LHS of (1.5)
+
=Ÿ
h
=Ÿ
h
(1.5)
=
(1.6)
then (1.5) can be easily solved as a function of
. We will prove
later that the above relation we introduced is proper. Eq-(1.6) means we
have
Cancelling
+
h
=
+
′ on both sides of the above equation, we have
h
=
h
=
This is a separable DE with unknown function
we have
=h
Therefore, we have
ln
=
=
. To solve this DE,
h
h
= exp E h
F
(1.7)
This is called the integrating factor of DE (1.4). Now we know that
from Eq-(1.7) satisfies the property of (1.6). Plugging (1.6) back to (1.5),
we have
33
Chapter 1 First-Order Differential Equations
which is easy to solve
=
=Ÿ
(1.8)
Ÿ
Thus, the solution to the original DE (1.4) is
1
E Ÿ
=
Plugging back (1.7) generates the solution
= exp E−
h
FE Ÿ
F
exp E
F
h
F
(1.9)
Now we show the proof of how (1.6) satisfies. Since we have the
integrating factor
We can prove
exp E h
Here we have
RHS =
F
′ + exp E h
Eexp E h
Fh
F F
=
=
exp E
h
F + Eexp E h
=
exp E
h
F + exp E h
=
exp E
= LHS
Hence the proof.
34
= exp E h
h
F + exp E h
F
Eexp E
FF
FE h
Fh
h
F F
F
1.4 Linear First-Order DEs
We now outline the steps on how to solve linear DE without memorizing
the formula.
STEP 1: Convert DE to the form + h
STEP 2: Identify h
STEP 3: Compute integrating factor
=Ÿ
= exp E h
F
STEP 4: Multiply both sides of the converted DE with the integrating
factor
F ′ + exp E h
exp E h
Fh
=Ÿ
exp E h
F
STEP 5: Write the LHS as a whole derivative
F ′ + exp E h
exp E h
Therefore
Eexp E h
exp E h
F
=
h
Example 1
Solve
Solution
STEP 1: Converting the DE
STEP 2: Identify
=
F F=Ÿ
Finally, the solution of the DE is
= exp E−
Fh
Ÿ
+
+3
3
+1
exp E
h
exp E
h
exp E h
FE Ÿ
+1
Eexp E h
=
F
F F
F
F
F
=6
6
+1
35
Chapter 1 First-Order Differential Equations
3
+1
STEP 3: Computing the integrating factor.
3
F
= exp E
+1
3
+1
]
= exp \
+1
2
3
+1 F
= exp E ln
2
h
=
=
+1
STEP 4: Multiplying the DE by the integrating factor.
3
6
+1 +
+1 =
+1
+1
+1
+3
+1
(
=6
STEP 5: Completing the solution.
E
=
=
+1
1
+1 F = 6
+1
+1
=
>
3
E2
=2+1
This is the solution to the original DE.
Example 2
Solve
|
6
+1
+1
+1
(
+1
(
+1
+1
>
+1
(
(
+1
+ 1F
+
= sin
1 = B
Solution
Method I: Step-by-step solution with the integrating factor.
sin
+ =
The integrating factor
36
= exp E
1
F = exp ln
=
1.4 Linear First-Order DEs
Multiplying the integrating factor yields
sin
=
⋅
=
1
=
This is the G.S. to the original DE.
Using the I.C. 1 = B
1 =
1=
sin
=
B
sin
1
D
=
B
+
sin
1
1
(
D
(
−
D
1
+
sin
+
sin
Now plugging the 1 back into
, we have
sin
1
1
B
+ −
=
=
B
D
+
1
sin
(
Method II: Direct use the formula (1.9)
= exp E−
h
= exp E
Using the I.C.
Therefore,
=
1
1 =
D
B,
h
1
sin
=
FE Ÿ
1
,
FE
+
we have
1=
=
B
1
sin
−
B
Ÿ
+
(
1
D
1
(
D
1
1
sin
exp E h
=
sin
exp E
1
F
F
F
+ 1F
sin
(
sin
37
Chapter 1 First-Order Differential Equations
Problems
Problem 1.4.1 Find G.S. and the corresponding P.S. of the following DE.
3
+1
+3
= 6 exp E−
F
Z
2
0 =1
Problem 1.4.2 (a) Show that
is a G.S. of + h
(b) Show that
¢
= 1 exp E−
¡
= 0.
= exp E−
h
h
FE Ÿ
=Ÿ .
is a P.S. of + h
(c) Suppose that ¡
is any G.S. of
P.S. of + h
= Ÿ . Show that
h
=Ÿ .
+h
=
F
exp E h
¡
F
F
= 0 and that ¢
is any
+ ¢
is a G.S. of +
Problem 1.4.3 Solve the following IVP.
+2 = 7
|
2 =5
Problem 1.4.4 Find the G.S. of the following DE.
+ cot = cos
Problem 1.4.5 Find the G.S. of the following DE by two different methods.
= 3 +7
Problem 1.4.6 Find the G.S. of the following DE.
= 2 + cos
Problem 1.4.7 Find the G.S. of the following DE.
2 +1
38
+
= 2 +1
1.4 Linear First-Order DEs
Problem 1.4.8 Find the G.S. of the following DE by two different methods.
2 +2
=
+1
Problem 1.4.9 Find the G.S. of the following DE.
1+2
=1+
(Hint: regard
as depending variable and
as independent)
Problem 1.4.10 Find the G.S. of the following DE.
+ = 10√
2
Problem 1.4.11 Find the G.S. of the following DE.
=1
+ exp
(Hint: regarding
as depending variable and
as independent)
Problem 1.4.12 Find the G.S. of the following DE.
2 + +1
=3 +1
Problem 1.4.13 Solve the following IVP.
1+
′ + = sin
|
=0 =1
Problem 1.4.14 Find the G.S. of the following second-order DE.
′′ + 3 ′ = 4
(Hint: introduce substitution L = ′)
39
Chapter 1 First-Order Differential Equations
1.5
Substitution Methods
Substitution methods are those of introducing one or more new variables
to represent one or more variables that will convert the original DE to a
separable or other more easily solvable DE.
to .
For example, one can change one variable
M(
,
£
→¤
= 0 ‘’’’’“ M
One may also consider changing both variables.
‡(
,
¥
¦
,
,
→¤
→§
= 0 ‘’’’’’’’“ ‡
,
,L
Let us see experience power of the substitution methods by solving
several families of DEs.
1.5.1 Polynomial Substitution
Solve
=M
+
where , , and are constants.
STEP 1: Introducing a new variable L.
L=
+
+
+
STEP 2: Transform the DE into a DE of L.
L =
=
40
1
+
L −
′
1.5 Substitution Methods
After the substitution, the original equation now becomes
=M L
Plugging back the equation for ′, we have
1
1
which is now separable.
L −
=M L
L =M L +
STEP 3: Solve the above DE.
1 L
1
=M L +
L
M L +
L
M L +
=
Example
Solve
Solution
Let
=
=
L
M L +
=
+
+3
L = + +3
L =1+
=L −1
After the substitution, the original equation becomes
L −1=L
41
Chapter 1 First-Order Differential Equations
L
=
L +1
L
=
L +1
>(
tan L = + 1
L = tan + 1
Substituting the value of L back into the above equation, we now obtain
+ + 3 = tan + 1
That is
= tan + 1 − − 3
1.5.2 Homogeneous DEs
For the 1st-order homogeneous DEs
we make a different substitution,
Let
= Mš ›
L=
L=
= L
= L +L
Substituting this in the original DE we have
L +L =M L
L =
M L −L
which is a separable DE. Solving this DE, we have
42
1.5 Substitution Methods
L
=
M L −L
L
+ 1 = ln
M L −L
L
+ 1F
M L −L
L
= 1 exp E
F
M L −L
= exp E
Example
Solve
2
Solution
Diving the original DE by 2
This DE is homogeneous.
Let
=4
=
2
+3
+
3
2
L=
=L
= L + L′
Substituting this back to the DE, we have
2 3
L+ L = + L
L 2
1 2 L
L = E + F
L 2
2L L
=
4+L
L +4
=
4+L
ln 4 + L = ln + 1
4+L = 1
Plugging L back, we have
4+š › =1
43
Chapter 1 First-Order Differential Equations
+4
=1
1.5.3 Bernoulli DEs
Given the DE
+h
=Ÿ
(1.10)
We have a few cases depending on the value of =.
CASE 1 (= < 0):
(1.10) is a general nonlinear 1st-order DE.
CASE 2 (= = 0):
(1.10) is the linear 1st-order DE + h
solution method was introduced before.
CASE 3 (= = ):
(
=Ÿ
whose
(1.10) is nonlinear 1st-order DE.
CASE 4 (= = 1):
(1.10) is linear 1st-order DE
is actually separable.
CASE 5 (= = 2):
+ $h
−Ÿ
% = 0 which
(1.10) is nonlinear 1st-order DE.
CASE 6 (= > 2):
(1.10) is nonlinear 1st-order DE.
To sum up, for = ≠ 0, 1, we know that (1.10) is a nonlinear 1st-order DE
and we have not learned any methods to solve the DE for such cases.
Now, let us use a substitution to solve the Bernoulli DEs for = ≠ 0, 1.
44
1.5 Substitution Methods
Solution steps
STEP 1: Select a proper substitution (why this sub?)
L=
STEP 2: Find the relation of L = -
(>
L = 1−=
>
′
STEP 3: Substituting L into the DE by inserting the above formula
Since = ≠ 1, we have
>
1
$ 1−=
1−=
+h
>
(>
%+h
Plugging L and L′ into the DE yields
L
+h
1−=
That is
L + =−1 h
=Ÿ
(>
=Ÿ
L=Ÿ
L = =−1 Ÿ
(1.11)
STEP 4: Solve this DE in terms of L. Since it becomes a linear DE, we
can solve this DE by any of the methods that were introduced earlier.
STEP 5: After solving (1.11), we find L.
STEP 6: Back-substitute variable L with to express the solution in the
original variables.
Example
Solve
+6 =3
Solution
It is a Bernoulli DE. Let us follow the step-by-step procedure.
45
Chapter 1 First-Order Differential Equations
STEP 1:
L=
STEP 2:
(>
=
= L>
L =−
1
3
>
′
>
= −3L L >
= −3L L >
STEP 3:
(
−3L L > + 6L > = 3 L >
−3 L L > + 6L > = 3 L >
L − 2L = −
2L
L −
= −1
STEP 4: Now we solve the above DE as a linear 1st-order DE
= exp E h
h
h
=−
=−
2
2
F
= −2 ln
= exp −2 ln = >
Multiplying the integrating factor to both side of the DE
1
2L
1
L −
=−
E
1
STEP 5: Plug
back
1
L=−
L
L=
>
(
=
=
46
LF = −
=
=
1
1
+1
+1
+1
1
+1
1
1+1
1
1.5 Substitution Methods
Problems
Problem 1.5.1 Find G.S. of the following DE.
=
+
=
+ V4
Problem 1.5.2 Find G.S. of the following DE.
+
Problem 1.5.3 Show that the substitution L = ln
h
=Ÿ
ln into the linear DEy L + h
transforms the DE
=Ÿ L
+
Problem 1.5.4 Find the G.S. of the following DE. Prime denotes derivatives
WRT .
5
=
+2
Problem 1.5.5 Find the G.S. of the following DE.
−4
+ 2 ln = 0
Problem 1.5.6 Find the G.S. of the following DE.
′ − K + 1 ¨ + 2 ln
=0
Problem 1.5.7 Find the G.S. of the following DE.
= 6 + 12
Problem 1.5.8 Find the G.S. of the following 2nd-order DE.
=3
Problem 1.5.9 Find the G.S. of the following DE by two different methods.
−
′=
Problem 1.5.10 Show that the solution curves of the following DE.
2 −
=−
2 −
47
Chapter 1 First-Order Differential Equations
are of the form
+
(Hint: Use substitution
= 31
= ⁄ )
.
Problem 1.5.11 Find the G.S. of the following DE.
=
+
exp š ›
Problem 1.5.12 Find the G.S. of the following DE.
+
=1
Problem 1.5.13 Find the G.S. of the following DE.
= 4 + sin
2 sin cos
Problem 1.5.14 Find the G.S. of the following second-order DE.
= +
(Hint: Introduce substitution L =
)
Problem 1.5.15 Find the G.S. of the following DE.
+ exp
= exp − − 1
Problem 1.5.16 Find the G.S. of the following DE.
+
3
=
(
Problem 1.5.17 Find the G.S. of the following DE.
+1 1+
=
Problem 1.5.18 Find the G.S. of the following DE.
exp
= 2 exp
+ exp 2
Problem 1.5.19 Use your favorite method to find the G.S. of the following 1storder nonlinear DE.
−4
ln + 2
where = ≥ 2 is a constant integer.
48
ln
=0
1.5 Substitution Methods
Problem 1.5.20 Find the G.S. of the following DE.
+
=
Problem 1.5.21 Find the G.S. of the following DE.
6
+2 + 9
+8
′=0
Problem 1.5.22 Given a 1st-order nonlinear DE
′+7
−3 = 3
+
(1) Prove that after substitution,
=
DE to a Bernoulli equation.
(2) Solve the resulting Bernoulli equation.
, one can transform this
Problem 1.5.23 Find the G.S. of the following DE.
+3
=4
(Hint: introduce substitution
= ′)
Problem 1.5.24 Find the G.S. of the following DE.
′ − 1000
+2 =0
Problem 1.5.25 Find the G.S. of the following DE.
=0
+
−
+
Problem 1.5.26 Find the G.S. of the following DE.
2
′+2
+1=0
Problem 1.5.27 Find the G.S. of the following DE by two different methods.
+1 ′−2 −2 = 0
Problem 1.5.28 Solve the following IVP.
=
−
where $ and are two constants and
(1) Sketch the solution for < 1.
(2) Sketch the solution for = 1.
1+
! "
/
=$ =0
in %0, $'. Please also do the following,
49
Chapter 1 First-Order Differential Equations
(3) Sketch the solution for > 1.
(4) In which of the above cases,
= 0 = 0 is possible.
Problem 1.5.29 Find the G.S. of the following DE.
8
4
+
′=3 −
(Hint: introduce substitution
= + )
(
Problem 1.5.30 Find the G.S. of the following DE.
′= H
+ +œ
−œ
When the function H
is given as H
= B(( and œ = 1, find the specific
form of the solution.
Problem 1.5.31 Find the G.S. of the following DE.
− =œ
+
Problem 1.5.32 Find the G.S. of the following DE.
′=
+
where , ≠ 0 are constants and = = 0, −2. You need to consider both values
of =.
50
1.6 The Exact DEs
1.6
The Exact DEs
Consider DE
It is
•
•
•
•
•
6
−
−3
+ 4 +3
=0
1st-order
Nonlinear
Not homogeneous
Not Separable
Not Bernoulli
New methods must be discovered in order to solve DEs of this kind.
The G.S. of all DEs can be written as
M
With this claim we can produce
$M
Therefore we can see
Let
This gives
,
,
=1
%= 1=0
¬M ¬M
+
⋅
¬
¬
¬M
E F
¬
,
=
,
¬M
+E F
¬
¬M
and
¬
+
=0
=0
,
,
=
¬M
¬
=0
(1.12)
51
Chapter 1 First-Order Differential Equations
All 1st-order DEs can be written in the above manner.
The condition for the DEs to be exact is
¬ ¬M
¬ ¬M
E F=
E F
¬ ¬
¬ ¬
That is, if
=
, then (1.12) is exact. In fact, a theorem can be proven
=
that the necessary and sufficient condition for (1.12) to be exact is
.
Example 1
Is DE
6
exact?
Solution
We have
Here
,
Since
−
=6
+ 4 +3
−
=
Example 2
Is DE
52
=6 −3
+3
exact?
Here
=4 +3
¬
¬
6 −
=6 −3
=
¬
¬
¬
¬
4 +3 −3
=6 −3
=
¬
¬
The DE is exact.
Solution
From DE, we get
,
and
=0
−3
,
=
and
=
¬
¬
=0
,
=1
=3
−3
1.6 The Exact DEs
=
That means
¬
3 )=3
¬
≠
The DE is not exact.
Example 3
Is DE
+3
exact?
Solution
We get
Here
, )=
, )=3
and
¬
¬
¬
=
3
¬
=
=
That means
=0
The DE is exact.
=3
)=3
=3
Since the above DE is exact, let us try to find its solution.
There is a new set of methods for solving exact DEs. As mentioned in
the beginning of this section the solution of any DE can be written as
M
And from above example, we have
, )=1
(1.13)
¬M
=
¬
(1.14)
( , )=
( , )=
So from (1.14) we have
¬M
=3
¬
(1.15)
53
Chapter 1 First-Order Differential Equations
M
, )=
1
=
Here we have 1 as a constant WRT
function of : 1 = - . Now we have
M
. Consider the constant be a
, )=
-( )
Plugging (1.16) back into (1.15), we get
¬
$
¬
-( )% = 3
-( )=3
3
So we find that -
Plugging -
-( )=0
-
M
,
Therefore, from (1.13) we have
Thus, the solution is
Steps for solving exact DEs
STEP 1: Write the DE into form
54
1(
is actually a constant.
1( into (1.16), we get
and get the formula
(1.16)
,
1(
1
,
1
1(
0
(1.17)
1.6 The Exact DEs
,
STEP 2: Find
,
M
,
=
=
=
,
¬M
¬
¬M
¬
STEP 3: Plug M , into (1.19) to obtain STEP 4: The solution is
,
Remarks
+-
(1.18)
(1.19)
+.
=1
In the above steps, we start from (1.18), integrate it and then plug it
back to (1.19) to get the solution. Similarly, we can start from (1.19),
integrate it and plug back to (1.18) to get the same solution, different
by a constant.
Now let us clarify a few points:
(1) Not all DEs are exact. Some non-exact DEs can be converted to
exact DEs and some cannot. So, the question is that under what
condition(s), a non-exact DE can be converted to an exact DE.
(2) If a non-exact DE can be converted to an exact DE, is the
conversion unique? In other words, is there more than one
method to convert a non-exact DE to exact?
Now let us try to convert a non-exact DE into an exact DE. Suppose we
have DE
,
If we have integrating factor -
+
,
such that
=0
55
Chapter 1 First-Order Differential Equations
-
be exact, we should have
That means
-
¬
¬
, )
¬
$¬
Here the integrand
,
,
-⋅
-$
-
=0
-( ) ( , )
%
,
-
¬
$¬
¬
¬
⋅%
-⋅
,
%
¬
¬
-
- ⋅
,
F
exp E
must be a function with a single variable .
Exact DE Theorem
For DE
1) If
,
,
0
is a function of one variable , we have the integrating factor
56
1.6 The Exact DEs
= exp E
2) If
−
F
=-
is a function of one variable , we have the integrating factor
= exp E−
Remarks
-
F
(1) For one DE,
and
may both exist. In this case, use the
most convenient form.
(2) It is possible none of the
and
exist.
(3) You must have noticed that
has no negative sign in the
exponential term while
has a negative sign in the
exponential term. So given below is the proof of why we need
the negative sign for
but is not required for
.
Proof of the negative sign in
in
= exp E−
-
F
,
+
,
integrating factor as
= 0 is the given DE. We will choose our
Let
®
,
,
® ,
+
,
=
,
=
,
=0
57
Chapter 1 First-Order Differential Equations
Therefore, we have the new DE as ®
+® ,
,
® = ®
=0
Since we want the above new DE to be exact, we should have
That is
,
¬
$
¬
¬
¬
,
+
⋅
+
where
as noted above.
Solving this DE, we have
¬
¬
=
That is
¬
$
¬
%=
⋅
,
−
= −-
=
= exp E−
,
=
=
⋅
%
¬
¬
−
-
F
We sum up above discussion in the following theorems:
Theorem 1
For a given 1st-order DE
58
,
+
,
= 0, if
,
1.6 The Exact DEs
−
=
is a function of pure , then the DE can be converted to an exact DE by
multiplying the original DE with
= exp E
Theorem 2
For a given 1st-order DE
,
F
+
−
=-
,
= 0, if
is a function of pure , then the DE can be converted to a exact DE by
multiplying the original DE with
= exp E−
-
F
Next, let us work on one example to understand the above theorem and
make several remarks.
Example
Determine whether the DE
+3
=0
is exact or not? If not convert it into an exact DE and solve it.
Solution
From the given DE, we get
,
=
and
,
=3
¬
¬
=1
=
¬
¬
¬
¬
=
3 =3
¬
¬
≠ , we know that the given DE is not exact. For solving this DE,
Since
there are two methods of doing this.
Here, we have
59
Chapter 1 First-Order Differential Equations
Method I:
It is clear that
−
1−3
2
=−
=
3
3
is a function of pure
=
= exp E
2
3
= exp E−
2
= exp E− ln F
3
Multiplying
F
F
= >
on both sides of the original DE, we have
+3 ⋅ >
=0
This is now our new DE. Here we have
¬M
(1.20)
= ® , = >
¬
(
¬M
(1.21)
= ® , =3
¬
Thus, we have
¬
® =
E > F= >
¬
(
¬
® =
E3 F = >
¬
Since ® = ® we know the new DE is now exact.
So we can now use either (1.20) or (1.21) to find M , . Since the
function in (1.21) is easier to integrate, we choose (1.21) to find M , .
We have
(
¬M
= ® , =3
¬
>
M
,
=
3
(
(
+-
=3
Substituting (1.22) back into (1.20), we now have
(
¬
\3
+- ] = > += >
¬
This solves = 0. That means = 1( . Substituting this back to
(1.22), we now have
M
60
,
=3
(
+ 1(
(1.22)
1.6 The Exact DEs
That gives the solution of the DE
M
That is
,
(
=1
=1
Method II:
We now discuss the second method of converting a non-exact DE to an
exact DE.
−
1−3
2
=
=− =3
The integrating factor
= exp E−
= exp E
2
-
F
F
=
on both sides of the original DE, we have
+3
=0
as our new DE. Here we have
¬M
(1.23)
= ® , =
¬
¬M
(1.24)
= ® , =3
¬
and
¬
® =
=3
¬
¬
® =
3
=3
¬
Since ® = ® , the new DE formed is now exact.
We can now use either (1.23) or (1.24) to find M , . Since the function in
(1.23) is easier to integrate, we choose (1.23) to find M , . We now have
Multiplying
M
,
=
+-
=
+Substituting (1.25) back into (1.24), we now have
¬
=3
$
+- %=3
+¬
It gives
=0
which means
(1.25)
61
Chapter 1 First-Order Differential Equations
= 1(
Substituting this back into (1.25), we now have
M , =
+ 1(
Thus, the solution of the DE is
+ 1( = 1
That is
=1
This is the same answer as we found in Method I.
SUMMARY
We have 4 ways of solving a non-exact DE, which is summarized as
follows.
Steps for solving a non-Exact DE by converting it into an Exact DE
( ) = exp E
Exact DEs
( )
−
where ( ) =
No
= exp E−
where -( ) =
-
−
F
F
¬M
= ® ,
¬
R¬M
® ,
Q¬ =
S
¬M
= ®
¬
R¬M
®
Q¬ =
S
Figure 1.6 Steps of converting a non-Exact DE to an Exact DE.
STEP 1: Determine ( , and
, .
STEP 2: Check if
= ? If YES then go to STEP 6:.
STEP 3: Check if
−
is a function of pure
62
or if
= ( )
,
,
1.6 The Exact DEs
−
=-
is a function of pure .
STEP 4: Find the easiest integrating factor to solve
= exp E
F or
= exp E−
STEP 5: Compute the new ® and ® for either
®
,
® ,
=
,
=
STEP 6: Construct the partial DEs
¬M
S = ® ,
¬
R ¬M = ® ,
Q¬
,
=
®
,
+-
,
=
® ,
+-
®
,
M
,
M
F
,
STEP 7: Use either of the above DE to find M
or equivalently,
or
-
STEP 8: Substituting the DE we get from STEP 7: back to the DE of
STEP 6:, we have either
or
¬
\
¬
+-
]= ® ,
63
Chapter 1 First-Order Differential Equations
¬
\
¬
® ,
,
+
+-
]= ®
,
STEP 9: Solving the above DE produces the final solution to the DE
M
,
=C.
,
=
−
−
=0
?
= ( )
Convertible ( )
= -( )
Convertible ( )
Exact Equation
Not an Exact Equation
Figure 1.7 Steps of solving a non-Exact DE.
Example 1
Solve exact DE
Solution
64
,
+
,
=0
1.6 The Exact DEs
We can compose two simple partial DEs with M as the dependent variable
and and as the independent variables
¬M
(1.26)
=
,
¬
¬M
(1.27)
=
,
¬
Solving (1.26), we get
M
,
=
,
+-
Plugging the above into (1.27), we get
¬
E
,
F+¬
¬
=
, −
E
¬
Solving this, we have
-
=
,
−
Thus, we get the solution of the DE as
,
+
Example 2
Is DE exact? Solve it.
6 −
,
−
°
=
¬
E
¬
°
+ 4 +3
¬
E
¬
−3
,
,
,
F
F±
,
F±
=1
=0
Solution
From the given DE, we have
¬M
=
, =6 −
¬
¬M
=
, = 4 +3 −3
¬
Now we evaluate
¬
=
6 −
=6 −3
¬
¬
4 +3 −3
=6 −3
=
¬
Since
=
, we know that the given DE is exact. Now we can find
M ,
65
Chapter 1 First-Order Differential Equations
M
¬M
=
¬
,
Plugging it back to
We have
¬
$3
¬
3
−
=
=3
,
6
¬M
=
¬
+-
=6
−
−
,
−
+-
+-
%=4 +3
=4 +3
+=4
=2
Therefore, the solution of the DE is
3
−
+2 =1
−3
−3
−3
Let us do one more example to show that not all DEs can be solved in
this way.
Example 3
Solve the DE
3
+5
Solution
From the given DE, we have
+ 3
,
,
=3
=3
+2
+5
+2
=0
¬
3 +5
= 6 +5
¬
¬
3 +2
=3 +6
=
¬
Since
≠ , the given DE is not exact. Now let us see if we can convert
it into an exact DE.
−
3 −
=
3 +2
is not a function of pure , and
−
3 −
=
3 +5
=
66
1.6 The Exact DEs
is not a function of pure . Therefore, the given DE cannot be converted
into an exact DE. Therefore it is not possible to solve this DE.
Finally, we show the relation between the exact DE method and the
linear DE we previously studied.
Converting a Linear DE into an Exact DE
Now the question arises whether linear DEs are exact. And if they are
not, is it possible to convert them into such a form? So now let us try to
answer these queries. The most general form of a linear DE is
+h
That is
Now we have
Since in general
$h
≠
=Ÿ
+h
,
−Ÿ
−Ÿ
=h
,
%
+
(1.28)
=0
−Ÿ
=0
(1.29)
=1
=h
=0
, (1.28) is not an exact DE.
As we have shown that the given general form of linear DE is not exact,
we have to now try to convert it into an exact DE.
Compare the following two formulas.
−
−
=
h
=h
h
−Ÿ
67
Chapter 1 First-Order Differential Equations
As we can see, working with first one is much easier than working with
second one. Hence we have our integrating factor.
Multiplying
exp E h
= exp E h
F
on both sides of (1.29), we have
F $h
−Ÿ
,
= exp E h
%
which is now the new DE. Thus, we have
¬M
= ®
¬
® =
¬M
= ®
¬
,
¬
\exp E h
¬
= exp E h
=h
F
F $h
¬
$h
¬
F
¬
exp E h
¬
=h
F $h
= exp E h
exp E h
® =
+ exp E h
exp E h
F
F
−Ÿ
−Ÿ
−Ÿ
F
Since ® = ® , the new DE is exact.
=0
%
(1.30)
(1.31)
%
%]
F
Now we can use either (1.30) or (1.31) to find M , . Since the function
in (1.31) is easier to integrate, we use this to find M ,
M
68
,
&=
exp E h
F
1.6 The Exact DEs
=
exp E h
F+-
Substituting this back to (1.30), we have
¬
\ exp E h
¬
=
h
h
F+-
-
= −Ÿ
=−
Substituting this back to M
,
=
exp E h
exp E h
-
M
] = exp E h
exp E h
F−Ÿ
Ÿ
,
exp E h
That is
= exp E−
F−
h
exp E h
, we now have
Therefore, the solution to the DE is
exp E h
F+-
F
exp E h
F−
Ÿ
FE Ÿ
F $h
Ÿ
−Ÿ
%
F
F
exp E h
exp E h
F
F
exp E h
=1
F
+ 1F
Problems
Problem 1.6.1 Solving the following DE using the exact DE method.
1 + ln
+E F
=0
69
Chapter 1 First-Order Differential Equations
Problem 1.6.2 Given a 1st-order DE
A ³,
+ž ,
=0
which is not an exact equation in general, but the function A , and ž ,
satisfy the following relationship
¬A ,
¬ž ,
\
−
] /ž , = h
¬
¬
where h
is a function of one variable . Prove that, after multiplying the
original DE by an integrating factor
= exp ´ h
, one can
transform the original non-exact DE to an exact DE.
Problem 1.6.3 Given 1st-order linear DE
+œ
v
+µ
=0
where v( ) ≠ 0.
(1) Using exact equation method to solve the equation and express the solution
in terms of v( ), œ , and µ( ).
(2) Using 1st-order linear equation method to solve the equation and express
the solution in terms of v( ), œ , and µ( ).
Problem 1.6.4 The 1st-order linear DE can be expressed alternatively as
(h( ) − Ÿ( ))
=0
where functionsh( ) ≠ 0 and Ÿ( ) ≠ 0 are given. Please
(1) Check if this DE is Exact.
(2) If not, convert it to an Exact DE.
(3) Solve the DE using the Exact Equation method and your solution may be
expressed in terms of the functions h( ) and Ÿ( ).
(
(4) If h( ) = and Ÿ( ) =
¶·¸
, get the specific solution.
Problem 1.6.5 Solving the following DE using the exact DE method.
(2 − )
=0
Problem 1.6.6 Solving the following DE by two different methods.
3 +2
′=−
4
Problem 1.6.7 Solving the following DE by two different methods.
70
1.6 The Exact DEs
=
+3
−3
Problem 1.6.8 Solving the following DE using the exact DE method.
+ 2 +
=0
Problem 1.6.9 Solving the following DE using the exact DE method.
′ 2 +1 =2 −
Problem 1.6.10 Solving the following DE using the exact DE method.
+
′− ′ =0
Problem 1.6.11 Check whether the following equation is exact
cos + ln
+ E + exp
F
=0
If it is an exact equation according to your verification above, solve the
equation using the exact equation solution technique. If it is not, then use some
other technique to solve the equation.
Problem 1.6.12 Solving the following DE by two different methods.
exp
+ cos + exp
+ sin
′=0
71
Chapter 1 First-Order Differential Equations
1.7 Riccati DEs
DEs that can be expressed as follows are called Riccati DEs:
If A
If 1
=A
+ž
+1
= 0, it reduces to the Bernoulli’s DE.
= 0, it reduces to the 1st-order linear DE.
By some method (including guessing), one may obtain one P.S.
With this P.S., one may propose the following G.S.:
1
+
= (
¹
Thus,
¹′
= (−
¹
=
(
+
(
.
1 2 (
+
¹
¹
Plugging in the above trial G.S. and other terms to the original Riccati
DE, we have
(
−
¹
= A+žE
¹
(
1
+ F+1E
¹
(
+
1 2 (
+
F
¹
¹
which can be written as follows, after re-organizing the terms:
(
Because
zero:
Thus,
72
(
−A−ž
(
−1
(
=
¹
1
1 2 (
+žE F+1E +
F
¹
¹
¹
¹
is the P.S. of the original Riccati DE, the LHS above must be
(
−A−ž
(
−1
(
=0
1.7 Riccati DEs
¹
1
1 2 (
+žE F+1E +
F=0
¹
¹
¹
¹
Or, multiplying the above equation by ¹ , we get
¹ + ž+2
(
¹+1 =0
which is a simple 1st-order linear DE that can be easily solved.
Example
Find the G.S. of the following DE
+
=
2
Solution
This is a simple Riccati DE and, by inspection, we may assume a P.S. in the
following form
(
=
º
where A is a constant to be determined. Since it is a
P.S., it must satisfy the original DE, i.e.,
A A
2
− +
=
We can easily find two roots A = −1 and A = 2 to satisfy the above
equation. Thus, we found two P.S.’s:
(
= − or
(
(
= .
Next, we may select any one of the P.S.’s to compose the G.S. as (We
selected P.S.: ( = − 1» :
1 1
=− +
¹
Thus,
1
¹′
=
−
¹
1
2
1
=
−
+
¹ ¹
Plugging in the above two formulas to the original DE:
1 ¹
1
2
1
2
\ − ]+E −
+ F=
¹
¹ ¹
We get
2
¹ + ¹=1
whose solution is
73
Chapter 1 First-Order Differential Equations
¹=
3
1
+
1
=− +
Finally, the G.S. for the DE is
3
+ 1(
Problems
=A
+ž
+1
is called Riccati DE. Suppose that one P.S. ( of this DE is known. Show that
the substitution = ( + 1/L can transform the Riccati DE into the linear
DE L + ž + 2A ( L = −A.
Problem 1.7.1 The DE
+2
is a solution.
Problem 1.7.2 Solve the DE
given that
(
=
Problem 1.7.3 Solve the DE
given that
(
=
is a solution.
′=1+
Problem 1.7.4 Solve the DE
′ − 13
given that ( = is a solution.
Problem 1.7.5 Solve the DE
given that
(
=
=
is a solution.
Problem 1.7.6 Solve the DE
′=−
given that ( = is a solution.
74
=1+
1
4
+
+
+4
+
−
+ 26 = 1
+
+4
−
+
1.7 Riccati DEs
Problem 1.7.7 Solve the DE
given that
(
′=
2 cos
= sin is a solution.
− sin
2 cos
+
Problem 1.7.8 Find the G.S. of the following DE
+2 −
+v
−
′=
and express it (of course) in terms of the given function v
this equation has one solution = .
Problem 1.7.9 Solve the DE
given that
(
=
is a solution.
+
−
. We know that
=2
75
Chapter 2
Mathematical Models
2.1
Population Model
2.1.1 General Population Equation
It is customary to track the growth or decline of a population in terms of
its birth rate and death rate functions, which are defined as follows:
•
•
œ
is the number of births per unit population per unit time at
time
¼
is the number of deaths per unit population per unit time at
time
76
2.1 Population Model
Then the numbers of births and deaths that occur during the time interval
, + Δ is approximately given by births: œ h Δ and deaths:
¼ h Δ . The change Δh in the population during the time
interval , + Δ of the length Δ is
Thus,
∆h = births − deaths
=œ h Δ −¼
= $œ
− ¼ %h
∆h
= $œ
∆
−¼
Taking the limit ∆ → 0, we get the DE
h
h
Δ
Δ
%h
= œ−¼ h
(2.1)
which is the General Population Equation where œ = œ
and ¼ = ¼
where œ and ¼ can be either a constant or functions of or they may,
indirectly, depend on the unknown function h .
Example 1
Given the following information, find the population after 10 years.
h 0 = hB = 100
œ = 0.0005h
¼=0
Solution
Substituting the given information in the above formula we have
h
= œ−¼ h
h
= 0.0005h
h
= 0.0005
h
h
= 0.0005
h
77
Chapter 2 Mathematical Models
1
= 0.0005 + 1
h
Substitution of h 0 = 100 in the above DE given
1
1=−
100
Therefore
2000
h
=
20 −
and
h 10 = 200
Thus, the population after 10 years is 200.
−
2.1.2 The Logistic Equation
Let us consider the case of a fruit-fly population in a container. It is often
observed that the birth rate decreases as the population itself increases
due to limited space and food supply. Suppose that the birth rate œ is a
linear decreasing function of the population size h so that œ = œB − œ( h
where œB and œ( are positive constants. If the death rate ¼ = ¼B remains
constant, then the general population equation takes the form
Let
= œB − ¼B and
h
= œB − œ( h − ¼B h
= œ( , then
h
= h− h
(2.2)
Eq-(2.2) is called the logistic equation for population model. We can
rewrite the logistic equation as
h
= /h
−h
(2.3)
where / = and
= ⁄ are constants.
is sometimes called the
carrying capacity of the environment, that is to say, the maximum
78
2.1 Population Model
population that the environment can support on a long-term basis. This
DE can be solved using the separation of variables as follows.
h
=/
Zh
−h
h = 0 = hB
l
1
l
lD
lD
E
h
l
lD
h
=
−h
h
B
/
h=/
−h
1
1
+ F h=
−h h
ln h − ln
h
Eln E F + ln E
hB
h
hB
−h
l
lD
=
/
− hB
FF =
−h
− hB
= exp
−h
/
/
/
This gives us a solution as an implicit function of h. We can write it
explicitly.
h=
h \1 +
−h
hB exp /
− hB
hB exp /
]=
− hB
h=
h=
hB exp /
− hB
hB exp /
− hB + hB exp
hB +
/
hB
− hB exp − /
(2.4)
79
Chapter 2 Mathematical Models
This is the final formula for the logistic equation.
Remarks
(1) If = 0, h
= 0 = hB
hB , / = 0
=• , />0
→•
0, / < 0
(3) Limited Environment Situation
A certain environment can support at most individuals. It is
then reasonable to expect the growth rate œ − ¼ (the combined
birth and death rates) to be proportional to − h because we
may think of − h as the potential for further expansion. Then
œ−¼ =/
−h
so that
h
= œ − ¼ h = /h
−h
(2) The limit cases: lim h
An example for the limited environment situation is the fruit-fly
population.
(4) Competition Situation
If the birth rate œ is constant but the death rate ¼ is proportional
to h, so that ¼ = vh, then
h
= œ − vh h = /h
−h
Example
Suppose that at time = 0, 10,000 people in a city with population =
100,000 are attacked by a virus. After a week the number h
of those
attacked with the virus has increased to h 1 = 20,000. Assuming that
h
satisfies a logistic equation, when will 80% of the city’s population
have been attacked by that particular virus?
80
2.1 Population Model
Solution
In order to eliminate the unnecessary manipulation of large numbers, we
can consider the population unit of our problem as thousand. Now,
substituting PB = 10 and M = 100 into the logistic equation we get
hB
h
=
hB +
− hB exp − /
1000
h
=
10 + 90 exp −100/
Then substituting h
We solve for
With h
= 1 = 20 into the equation
1000
20 =
1 + 90 exp −100/
= 80, we have
Which we solve for
4
9
1
4
/=−
ln
100 9
exp −100/ =
80 =
1000
10 + 90 exp −100/
exp −100/
=
1
36
ln 36
ln 36
=
≈ 4.42
100/ ln 9/4
It follows that 80% of the population has been attacked by the virus after
4.42 weeks.
Thus,
=
2.1.3 Doomsday vs. Extinction
Consider a population h
of unsophisticated animals in which females
rely solely on chance of encountering males for reproductive purposes.
The rate of which encounters occur is proportional to the product of the
number h⁄2 of males and h⁄2 of females, hence at a rate proportional
to h . We therefore assume the births occur at the rate of /h (per unit
time and / is constant). The birth rate (births per unit time per
81
Chapter 2 Mathematical Models
population) is then given by œ = /h. If the death rate ¼ is constant then
the general population equation gives the following
Let
= ¼ ⁄/ > 0, we have
h
h
Note
= /h − ¼h
= /h h −
(2.5)
The RHS of the above DE is the NEGATIVE of the RHS of the logistic
equation. The constant is now called the threshold population. The
behavior of the future population depends critically on whether the initial
population hB is less than or greater than .
Let us now solve this DE that is very similar to the logistic equation.
h
h h−
1
Plugging the I.C. h
82
h
h h−
=/
h h−
1
E
h−
=
/
h=/
1
− F h=
h
/
= 0 = hB to the left integration, we have
ln h −
hB h −
h hB −
− ln h |llD =
= exp
/
/
2.1 Population Model
CASE 1: hB >
Let
1( =
hB
hB −
We know that 1( > 1. To solve an explicit DE for h, we have
h= h−
h 1( exp − /
h
=
1( exp − /
−1 =
1( exp − /
1( exp − /
1( exp − / − 1
Note that the denominator in the above DE approaches zero as
approaches
That is, lim h
→C
@=
1
hB
ln 1(
=
ln
/
/
hB −
= ∞ → Doomsday.
CASE 2: 0 < hB <
Let
1 =
>0
hB
− hB
We know that 1 > 0. A similar solution for h will give
− h 1 exp − /
h=
h 1 + 1 exp − /
h
=
=
1 exp − /
1 exp − /
1 exp − / + 1
Since 1 > 0, it follows that lim h
→•
= 0 → Extinction.
83
Chapter 2 Mathematical Models
Problems
Problem 2.1.1 Suppose that when a certain lake is stocked with fish, the birth
and death rates œ and ¼ are both inversely proportional to √h.
(1) Show that
1
h
= E / + VhB F
2
where / is constant.
(2) If hB = 100 and after 6 months there are 169 fish in the lake, how many
will there be after 1 year?
Problem 2.1.2 Consider a prolific breed of rabbits whose birth and death rates,
œ and ¼, are each proportional to the rabbit population h = h , with œ > ¼
and h = 0 = hB.
(1) Show that
hB
h
=
1 − /hB
where / is a constant.
(2) Prove that h
→ ∞ as → 1⁄/hB . This is doomsday.
(3) Suppose that hB = 6 and there are 9 rabbits after ten months. When does
doomsday occur?
(4) If œ < ¼, what happens to the rabbit population in the long run?
Problem 2.1.3 Consider a population h
satisfying the logistic equation
h
= h− h
where ž = h is the time rate at which births occur and  = h is the rate at
which deaths occur. If initial population is h 0 = hB and žB births per month
and ÂB deaths per month are occurring at time = 0, show that the limiting
population is
žB hB
=
ÂB
Problem 2.1.4 For the modified logistic population equation h =
/h ln − ln h , h = 0 = hB where h is the population count as a function
of time ; /, hB and are positive constants.
(1) Solve the DE.
84
2.1 Population Model
(2) Sketch the solution for hB >
(3) Sketch the solution for hB <
and show the limit when → ∞.
and show the limit when → ∞.
Problem 2.1.5 A tumor may be regarded as a population of multiplying cells.
It is found empirically that birth rate of the cells in a tumor decreases
exponentially with time, so that
œ
= œB exp −v
where v and œB are positive constants. Hence
h
= œB exp −v h
Z
h 0 = hB
Solve this IVP to see that
Observe that h
as → ∞.
h
= hB exp °
œB
1 − exp −v
v
±
approaches the finite limiting population
œB
hB exp E F
v
and h
Problem 2.1.6 Consider two population functions h(
which satisfy the logistic equation with the limiting population
different values /( and / of the constant / in DE
h
= /h
−h
, both of
, but with
Assume that /( < / . Which population approaches the most rapidly? You
can reason geometrically by examining slope fields (especially if appropriate
software is available), symbolically by analyzing the solution given in DE
hB
h
=
hB +
− hB exp − /
or numerically by substituting successive values of .
Problem 2.1.7 Suppose that the fish population h
in a lake is attacked by a
disease at time = 0, with the result that the fish cease to reproduce (so that
the birth rate is œ = 0) and the death rate ¼ (deaths per week per fish) is
thereafter proportional to 1⁄√h . If there were initially 900 fish in the lake and
441 were left after 6 weeks, how long did it take all the fish in the lake to die?
85
Chapter 2 Mathematical Models
Problem 2.1.8 Consider a rabbit population h
satisfying the logistic
equation h = h − h , where ž = h is the time rate at which births occur
and  = h is the rate at which deaths occur. If the initial population is 120
rabbits and there are 8 births per month and 6 deaths per month occurring at
time = 0, how many months does it take for h
to reach 95% of the
limiting population ? Please sketch h
with given parameters.
Problem 2.1.9 Consider a prolific animal whose birth and death rates, œ and v,
are each proportional to its population with œ > v and hB = h = 0 .
(1) Compute the population as a function of time and parameters and I.C.
given.
(2) Find the time for doomsday.
(3) Suppose that hB = 2011 and that there are 4027 animals after 12 time units
(days or months or years), when is the doomsday?
(4) If œ < v, compute the population limit when time approaches infinity.
Problem 2.1.10 Fish population h
in a lake is attacked by a disease (such as
human beings who eats them) at time = 0, with the result that the fish cease
to reproduce (so that the birth rate is œ = 0) and the death rate ¼ (death per
week per fish) is thereafter proportional to h >(/ . If there were initially 4000
fish in the lake and 2011 were left after 11 weeks, how long did it take all the
fish in the lake to die? Can you change the “2011” to a different number such
that the fish count never changes with time? To what number if so?
86
2.2 Acceleration-Velocity Model
2.2
Acceleration-Velocity Model
2.2.1 Velocity and Acceleration Models
We know that
L=
where
=
L
=
L
=
L=
L
=
= acceleration
= velocity
Ä of the body
= position
Newton’s Second Law of Motion, M = K , implies that to gain an
acceleration for a mass K, one must exert force M.
Example
Suppose that a ball is thrown upwards from the ground ( B = 0) with
initial velocity LB = 49 m⁄s . Neglecting air resistance, calculate the
maximum height it can reach and the time required for it come back to its
original position.
Solution
The body if thrown upwards reaches the maximum height when the
velocity is zero that is max = (L( ) = 0). The time required for the body
to return back to its original is calculated when its position function is zero
that is when = 0. From the Newton’s Law,
L
K
= MÅ
where MÅ = −K- is the (downward-directed) force of gravity, where the
gravitational acceleration is - ≈ 9.8 K⁄³ . Using the above equation and
plugging the I.C., we have
87
Chapter 2 Mathematical Models
L
= −9.8
L( ) = −9.8 + 49
Hence, the ball’s height function ( ) is given by
( )=
B
=
L
(−9.8 + 49)
= −4.9
= 0, we have
+ 49 +
B
( ) = −4.9 + 49
So the ball reaches its maximum height when L = 0
−9.8 + 49 = 0
= 5 sec
Hence its maximum height is
max = (5) = 122.5 K
The time required for the ball to return is
( )=0
−4.9 + 49 = 0
= 10 sec
Since
2.2.2 Air Resistance Model
Now let us add the consideration of the air resistance in problems. In
order to account for air resistance, the force M[ exerted by air, on the
moving mass K must be added. So now we have
K
L
= MÅ + M[
It is found that M[ = /L ¢ where 1 ≤ k ≤ 2 and the value of / depends
on the size and shape of the body as well as the density and viscosity of
the air. Generally speaking k = 1 is for relatively low speeds and k = 2
for high speeds and we consider an intermediate value 1 < k < 2 for
intermediate speeds.
88
2.2 Acceleration-Velocity Model
L
MX
M‡
Figure 2.1 A rocket model with the consideration of air resistance.
The vertical motion of a body with mass K near the surface of the earth
is subject to two kinds of force: a downward gravitational force MÅ and a
force M[ (we have taken k = 1 to simplify situations) of the air resistance
that is proportional to velocity and directed in the opposite direction of
motion of the body. We then have, MÅ = −K- and M[ = −/L where / is
a positive constant and L is the velocity of the body. Note that the
negative sign for M[ makes the force positive (an upward force) when the
body is falling down (L is negative) and negative (a downward force) if
the body is thrown upwards (L is positive). So the total force on the body
is now
Therefore
Let
M = M[ + MÅ = −/L − KK
L
= −/L − K-
89
Chapter 2 Mathematical Models
Then the DE becomes
L
=
/
>0
K
=− L−-
This DE is a separable DE that has a solution
L( ) = ELB + F exp(− ) −
where LB is the initial velocity of the body L(0) = LB . We now have
LÇ = lim L( ) = −
→•
(2.6)
-
Thus the speed of a falling body with air resistance does not increase
indefinitely. In fact it approaches a finite limiting speed or terminal
speed.
|LÇ | =
-
=
K/
Thus the velocity of the body in terms of its terminal speed under the
influence of air resistance is written as
L( ) = (LB − LÇ ) exp(− ) + LÇ
(2.7)
Integration of the above DE will give us the position of the body in terms
of its terminal velocity under the influence of air resistance as
( )=
=
L
$(LB − LÇ ) exp(− ) + LÇ %
1
= − (LB − LÇ ) exp(− ) + LÇ + 1
90
2.2 Acceleration-Velocity Model
The initial height of the body at time = 0 is (0) =
this is the above equation we have
1=
B
+
LB − LÇ
B.
Substituting
Thus we get the position function of the body as
( )=
B
1
+ LÇ + (LB − LÇ )(1 − exp(− ))
(2.8)
Example
We will again consider the ball that is thrown from the ground level with
initial velocityLB = 49 K⁄³. But now we will take the air resistance =
0.04 into account and find out the maximum height and the total time that
the ball spent in the air.
Solution
We first find out the terminal velocity
9.8
= −245
LÇ = − = −
0.004
Substituting the values of B = 0, LB = 49 and LÇ = 245 in the equation of
velocity and position we get
L( ) = (LB − LÇ ) exp(− ) + LÇ
= $49 − (−245)% exp(−0.04 ) + (−245)
( )=
B
= 294 exp E−
F − 245
25
1
+ LÇ + (LB − LÇ )(1 − exp(− ))
= 0 + (−245) +
1
(49 + 245)(1 − exp(−0.04 ))
0.04
= 7350 − 245 − 7350 exp E−
F
25
The ball reaches maximum height when its velocity is zero. That is
L( ) = 0
F − 245 = 0
25
245
= −25 ln
294
≈ 4.558 sec
294 exp E−
¨
91
Chapter 2 Mathematical Models
Here we notice that the time required for the ball to reach maximum
height is ¨ = 4.558 sec with air resistance as compared to 5 sec without
air resistance. So for the maximum height of the ball substitute the value of
¨ into the position function. So we get
max = ( ¨ )
= 7350 − 245
¨
− 7350 exp E−
F
25
¨
≈ 108.3 K
Again here we have the maximum height as 108.3 K with air resistance as
opposed to 122.5 K without air-resistance. Similarly, the time required for
the ball to reach back in its original position is 9.41sec as compared to 10
sec from previous example.
We can also formulate DEs for the velocity and position of the ball
without using their terminal velocity. This follows directly from (2.6)
We can substitute
Remarks
L( ) = −
KK/
+ šLB +
› exp E−
F
/
/
K
back to /⁄K.
(2.9)
(1) The time required for the ball to reach maximum height is when
its velocity is zero. At that point in time the ball reverses its
motion to start falling down. So the time required for the ball to
reverse its direction is rev.
KK/
−
+ šLB +
› exp E−
F=0
/
/
K Çȧ
/
/LB
exp E−
+1
Çȧ F =
K
KK
/LB
ln E
+ 1F
rev =
K/
(1a) When the resistance is huge, it takes no time to reverse
lim Œ‰É = 0
e→•
(1b) When the resistance is tiny, we have to evaluate
care.
92
rev
with
2.2 Acceleration-Velocity Model
K
/LB
ln E
+ 1F
e→• /
KK /LB
= lim
e→• / KLB
=
which is just the value for no air resistance case.
(2) If → ∞,
K= Lterminal
L( ) ⇒ −
/
We see that:
Lterminal has no dependence on LB
Lterminal ∝ K
Lterminal ∝ 1
∞, / → 0
Lterminal ∝ = |
0, / → ∞
/
rev
= lim
2.2.3 Gravitational Acceleration
According to Newton’s Law of Gravitation, the gravitational force of
attraction between two point masses
and K located at a distance Y
apart is given by
M=
‡ K
Y
where ‡ is gravitational constant given as ‡ = 6.67 × 10>((
(K⁄/-) in MKS units.
⋅
Also the initial velocity necessary for a projectile to escape from earth
altogether is called the escape velocity and is found out to be
93
Chapter 2 Mathematical Models
2‡
LB = Ë
X
where X is the radius of the earth.
Problems
Problem 2.2.1 Suppose that a body moves horizontally through a medium
whose resistance is proportional to its velocity L, so that L = −/L.
(1) Show that its velocity and position at time are given by
L
L( ) = LB exp(−/ ) and
= B + š › 1 − exp −/
/
(2) Conclude that the body travels only a finite distance, and find that distance.
L
MX
Figure 2.2 The motion model for Problem 2.2.1 and Problem 2.2.2.
Problem 2.2.2 Consider a body that moves horizontally through a medium
whose resistance is of the L = −/L Í . Show that
4LB
2
2
L
=
and
= B + VLB \1 −
]
/
/
L
V B+2
$/ VLB + 2%
Conclude that under a 3⁄2 -power resistance a body coasts only a finite
distance before coming to a stop.
Ì
Problem 2.2.3 In Jules Verne’s original problem, the projectile launched from
the surface of the earth is attracted by both the earth and the moon, so its
distance Y
form the center of the earth satisfies the IVP
‡ È
‡ ¨
Y
S
=−
+
Î−Y
Y
Y 0 =X
R
Y 0 = LB
Q
94
2.2 Acceleration-Velocity Model
where È and ¨ denote the masses of the earth and the moon, respectively; X
is the radius of the earth and Î = 384,400 /K is the distance between the
centers of the earth and the moon. To reach the moon, the projectile must only
just pass the point between the moon and earth where its net acceleration
vanishes. Thereafter it is “under the control” of the moon, and falls from there
to the lunar surface. Find the minimal launch velocity LB that suffices for the
projectile to make it “From the Earth to the Moon.”
X
Y
MÏ
MK
Î
Figure 2.3 The projectile model for Problem 2.2.3.
Problem 2.2.4 One pushes straight up a toy rocket of fixed mass m at an initial
velocity of LB from the surface of earth. Assume earth’s gravitational constant
to be g), and air frictional force on the rocket −/L where k is the frictional
constant and v is the velocity of the rock. (Figure 2.1) Ignore the rocket
“launcher” height. Please
(1) Compute the max height ℎ¨w the rocket can reach.
(2) Compute the time @¤¢ for the rocket to reach max height.
Problem 2.2.5 Suppose that a motorboat is moving at 40 ft/s when its motor
suddenly quits, and that 10s later the boat has slowed to 20 ft/s.
(1) Assume that the resistance it encounters while coasting is proportional to its
velocity. How far will the boat coast in all?
(2) Assuming that the resistance is proportional to the square of the velocity.
How far does the motorboat coast in the first minute after its motor quits?
40 ft/s
MX
20 ft/s
Figure 2.4 The motorboat model for Problem 2.2.5.
95
Chapter 2 Mathematical Models
Problem 2.2.6 The resistance of an object (mass K ) moving in Earth’s
gravitational field (with gravitational constant - ) is proportional to its
speed L( ).
(1) Write down the equation of motion by Newton’s second law, in terms of
speed of the object, with assumption that the object was initially thrown
upward at speed LB, and air frictional constant /.
(2) Naturally, the object will move upward in the sky until at one point when it
reverses its direction to move downward to the earth. Find the time when the
direction of the object has just reversed. Does this time (at which reversing
occurs) depend on the initial velocity? How?
(3) Eventually, the object will move at a constant speed, the Terminal Speed.
Find this terminal speed. Does this terminal speed depend on the initial
velocity? How?
Problem 2.2.7 On January 15, 2009, the US Airways Flight 1549 “landed”
safely in the Hudson River a couple of minutes after departing from LGA. The
air (straight line) distance between LGA and the landing spot on the river is .
We assume (I) the flight’s Captain Sullenberger flies the plane at a constant
speed LB relative to the wind, after realizing the bird problem immediately after
taking off; (II) the wind blows at a constant speed o and at a direction
perpendicular to the line linking the landing spot and LGA; (III) the plane
maintains its heading directly toward the landing spot. Compute the critical
wind speed o¡ above which the plane would have been blown away. Compute
the lengths of the flight trajectories for o < o¡ and o ≠ 0 and o = 0.
o
M
v
L0
LGA
Figure 2.5 The aircraft model for Problem 2.2.7.
Problem 2.2.8 On a straight stretch of road of length … there were two stop
signs at the two ends. Assume you follow traffic laws and stop at these signs.
Your poor car can accelerate at a constant ( and decelerate at another
constant . Compute the shortest time to travel from one stop sign to another
96
2.2 Acceleration-Velocity Model
if there is no speed limit in between. Do it again if the speed limit between the
signs is L¨ .
L
STOP
1
STOP
2
Figure 2.6 The moving car model for Problem 2.2.8.
Problem 2.2.9 A man with a parachute jumps out of a hovering helicopter at
height „ . During the fall, the man's drag coefficient is / (with closed
parachute) and =/ (with open parachute) and air resistance is taken as
proportional to velocity. The total weight of the man and his parachute is K.
Take the initial velocity when he jumped to be zero. Gravitational constant
is -. Find the best time for the man to open his parachute after he leaves the
helicopter for the quickest fall and yet “soft” landing at touchdown speed ≤ LB .
MX
M‡
Figure 2.7 The jumper model for Problem 2.2.9 and Problem 2.2.14.
Problem 2.2.10 If a projectile leaves Earth’s surface at velocity LB , please
derivate the formula for the later velocity as a function of the distance Y from
the projectile to the center of the earth. The earth’s radius X and gravitational
constant ‡ are given. Also, naturally, the velocity has to be a real number.
Compute the condition at which the velocity will become imaginary.
97
Chapter 2 Mathematical Models
Problem 2.2.11 Suppose that a projectile is fired straight upward from the
surface of the earth with initial velocity LB and assume its height ( ) above
the surface satisfies the IVP
‡
− œ exp(− )
S =−
( + X)
(0) = 0
R
(0) = LB
Q
Compute the maximum altitude the projectile reaches and the time (from
launch) for the projectile to reach the maximum altitude. In the above, ‡, , LB
and œ are all given constants.
Problem 2.2.12 A guy (of body mass K) jumps out of a horizontally flying
plane of speed LB at height H. The horizontal component of the jumper’s initial
velocity is equal to that of the plane’s and the jumper’s initial vertical speed
can be regarded as 0. The air resistance is −vLÑ with the parachute open where
v is a given constant and LÑ is the jumper’s velocity. Please compute the total
distance, from when the guy was thrown off the plane to the landing spot, if the
parachute is opened at the middle height point. For clarification, you compute
the length of the trajectory the poor guy draws while in sky (before hitting
ground). You may express your solution implicitly.
(Hint: For a curve described by M( , ) = 0, one can calculate its length with
by … =
Ô
´w V
+
=
Ô
´w Ó1
+š ›
)
L0
∈[ , ]
MX
M‡
Figure 2.8 The jumper model for Problem 2.2.12 and Problem 2.2.13.
Problem 2.2.13 A guy (of body mass K) jumps out of a horizontally flying
plane of speed LB at height „. The horizontal component of the jumper’s initial
velocity is equal to that of the plane’s and the jumper’s initial vertical speed
can be regarded as 0. The air resistance is v|LÑ| with the parachute open where
98
2.2 Acceleration-Velocity Model
v is a given constant and |LÑ| = $L + L %
is the magnitude of jumper’s
velocity. Please compute the times taken the jumper to reach land if the
parachute is opened at heights „ and „.
(
(
(/
Problem 2.2.14 A man with a parachute jumps out of a “frozen” helicopter at
height „ . During the fall, the man's drag coefficient is / (with closed
parachute) and =/ (with open parachute) and air resistance is taken as
proportional to velocity. The total weight of the man and his parachute is K.
Take the initial velocity when he jumped to be zero. Gravitational constant
is -. The man opens the parachute B time after he jumps off the helicopter.
Please find the total falling time, and the speed he hits the ground. Can you
adjust the B to enable the quickest fall, and lightest hit on the ground?
Problem 2.2.15 A fighter jet taking off from one aircraft carrier 1( wishes to
land on another carrier 1 , currently located precisely on the northeast of 1( .
The distance between the two carriers is …. The carrier 1 moves east at a
constant speed of L just after the jet took off. Wind blows from south to north
at a constant speed of LÕ and its impact to the carriers is negligible while the
impact to the jet must be taken into account fully. Obviously, given the scale of
flight distance (usually a thousand miles), the carriers and jet can be treated as
mathematical points. Compute the shortest trajectory of the jet to land as it
wishes.
Problem 2.2.16 A bullet of mass m is fired off the barrel of a gun at an initial
speed LB to a medium whose resistance follows vL + œL where v and œ are
constants and L is the bullet’s instantaneous speed. Compute the maximal
distance that the bullet can travel in this medium. If we double the initial speed
and keep other conditions unchanged, compute the maximal distance again.
You may ignore gravity.
Problem 2.2.17 Consider shooting a bullet of mass K to three connected (and
fixed) media each of equal length …. The resistance in Medium-1 is /L, in
Medium-2 is /L , and Medium-3 is /L. Compute the bullet’s speed to enter
Medium-1 such that the bullet will pass through Medium-1 and then Medium-2
and then Medium-3, and magically stop precisely at the edge of Medium-3.
You may neglect gravity.
99
Chapter 2 Mathematical Models
Problem 2.2.18 Consider shooting a bullet of mass K to a first medium whose
resistance is proportional to the speed, to a second medium whose resistance is
proportional to L Í , and a third medium whose resistance is proportional to L .
Compute in all three cases the farthest the bullet can travel with a given initial
speed of LB. The medium in all three cases is big (you may imagine the water
of the ocean).
Ì
Problem 2.2.19 Consider shooting, at an initial speed LB , a bullet of mass K to
a medium whose resistance is – v(œ + L ) where L is the instatenous speed, v
is the given resistance constant and œ is another given constant. You may
assume the medium is infinite (for example, the ocean) and you may also
neglect the gravity for this problem. Please compute
(1) the farthest distances the bullet can travel in the medium;
(2) the time the bullet is in motion.
Problem 2.2.20 We set a 2-dimensional Cartesian coordinate system for a
north running river, so that the y-axis lies on the river’s west bank (pointing
from south to north), and the x-axis across the river, pointing from west to east.
A ferryboat wants to cross the river from a point ( , 0) on the east bank to the
west bank, and it hopes to dock at (0,0). The water speed, depending on the
location in the river, is measured by a function where ×( ) = B ( − )/
where both B and are positive constants. Now, we further assume the
ferryboat keeps facing the docking point (0, 0) at all times during the trip, and
it moves at a constant speed LB relative to the water. Please do the following:
(1) Establish the equation for the boat’s trajectory.
(2) Solve the equation.
(3) Sketch the boat’s trajectory if the river flows much slower than the boat.
Problem 2.2.21 An egg is dropped vertically to a fully-filled swimming pool
of depth „ at an initial speed 0 at the surface. Three forces act on the egg:
gravity, water floating force, and friction with water. Assume water density 1,
egg density > 1, egg mass K. Floating force is given as × 1 × - where Ø̈
is the usual gradational constant while the friction force is †L where † is a
friction-related constant and L is egg’s instantaneous speed. Compute the egg
impact speed at the pool bottom. You may express the solution in implicit
form.
100
2.3 An Example in Finance
2.3
An Example in Finance
Consider a situation when one owes a certain amount of money to the
bank.
We consider the following terms:
•
•
•
•
•
Ù( ) is the amount owed to the bank at time .
Y is the interest rate which is a constant
is the time period from the time interval [ , +
o is the payment rate which is also a constant
Ù is the change of debts to bank during time
]
So now let us calculate the decrement of loan due to payments made to
the bank.
The decrement of loan due to payments made
total amount with interest
where
•
•
•
Ù
Ù
Y
o
payments
Ù is the decrement of loan due to payments made
Ù Y is the increase of loan due to interest
o is the payment rate×time=payments made
Therefore
is the loan equation.
Ù
Ù
ÙY
ÙY
o
o
(2.10)
To solve this DE, we simply use the separation of variable method.
101
Chapter 2 Mathematical Models
Ù
o
=Ù−
Y
Y
Ù
o=Y
Ù− Y
Ù
Y
o=
ÛD Ù −
ÜB
Y
o Û
ln šÙ − ›Ý = Y
Y ÛD
Û
o
Ù− Y
ln Þ
oß = Y
ÙB − Y
o
o
Ù − = šÙB − › exp(Y )
Y
Y
Thus the solution to the financial model problem is
Remarks
Ù( ) =
o
o
− š − ÙB › exp(Y )
Y
Y
(2.11)
(1) At = 0, Ù( = 0) = ÙB
(2) If Y = 0, Ù ≠ ÙB (Since o⁄Y will become indeterminate)
(3) If the rate by which the payments are made is greater than the
interest rate
o > ÙB Y
then we have
o
− ÙB > 0
Y
As the payments are higher than the interest rate, the loan will
decrease.
102
2.3 An Example in Finance
If he increases his rate of payments (i.e., he starts paying
biweekly/weekly/daily instead of monthly) his loan will decrease
faster as shown in the figure.
(4) If the rate by which the payments are made is equal to the
interest rate then
o
− ÙB = 0
Y
The time required to complete the entire payment amount would
be when Ù = 0
o
o
− š − ÙB › exp(Y ) = 0
Y
Y
o
o
= š − ÙB › exp(Y )
Y
Y
o
exp(Y ) = o Y
− ÙB
Y
YÙB
Y = − ln š1 −
›
o
YÙB
1
›
= − ln š1 −
Y
o
Since time cannot be negative, we have
YÙB
ln š1 −
›<0
o
YÙB
0 < š1 −
›<1
o
YÙB
<1
0<
o
Hence the time required to complete the loan is
1
YÙB
= − ln š1 −
›
Y
o
(5) If the rate by which the payments are made is smaller to the
interest rate, then
o
š − ÙB › < 0
Y
103
Chapter 2 Mathematical Models
the payment will become indefinite and the loan will never be
paid off.
Problems
Problem 2.3.1 A person borrowed ¹B amount of funds from a bank which
charges a fixed (constant) interest rate Y. The periodical payment the borrower
makes to the bank is ŸB , where ŸB is a constant, and is the time.
(1) Establish the DE governing the time-varying loan balance ¹( ) with other
parameters.
(2) Solve the equation to express ¹( ) explicitly as a function of .
Problem 2.3.2 Mr. and Mrs. Young borrowed ¹ amount of money from a bank
at a daily interest rate Y. The Youngs make the same “micro-payment” of Â
amount to the bank each day.
(1) Set up the DE for amount x owed to the bank at the end of each day after
the loan closing.
(2) Solve the above DE for x as a function of , ¹, and Y.
(3) Obviously, if  is too small (paid too little) to the bank, the Youngs will
never pay off the loan. Find the critical  at which the loan will not change.
(4) The Youngs wish to pay off the loan in days. What do they have to pay
each day? Write it as a function of ¹, , and Y.
(5) The Youngs wish to pay off the loan in /2 days. Find the daily payment
 as a function of ¹, , and Y.
Problem 2.3.3 Mr. A borrowed ¹B from a bank at a fixed interest rate r. Mr. A
pays the bank fixed amount W periodically so that he can pay of the loan at
time T. Please do the following.
(1) Derive a formula for T in terms of ¹B , W, and r. In other words, find the
exact form for function @ = @(¹B , ×, Y).
(2) For given values of ¹B and W, compute
(a) @( = @ š¹B , ×, Y → 0› =?
(Hint: àrK à= 1
(b) @
104
→B
@ š¹B , ×, Y =
Õ
áD
)
› =?
2.3 An Example in Finance
(3) For given values of ¹B and r, how much does Mr. A have to add to his
current payment × to pay off the loan at @⁄2 instead of T?
Problem 2.3.4 Mr. Wyze and Mr. Fulesch each borrows ÙB from a bank at the
same time. Wyze got it at a fixed interest rate Y and will pay it off at time @
with a fixed periodic payment ×B . Fulesch’s rate changes with time according
to Yâ = Y(1 + ) so Fulesch’s rate is only 1⁄5 of Wyze’s at the start. For
(
9
example, if Wyze’s rate is 6.0 %, Fulesch’s starting rate is only 1.2 %. If
Fulesch also pays ×B periodically, his debt drops faster than Wyze’s initially
before it drops slower and then grows. Compute the time when both loaners
own the bank the same amount, again. The first time when they own the same
is when they just got the loans.
Problem 2.3.5 Given that a loan of ¹B with a fixed interest rate r would be paid
off at time @ with a fixed periodic payment ×B . Now, one pays a new periodic
payment (1 + v)×B , compute the new time to pay off the loan. Compute total
interests paid to the bank in both cases.
Problem 2.3.6 One borrows ¹B from a bank at a fixed rate Y. Now, we assume
the time to pay off the loan with fixed periodic payment × is @( . Please do the
following.
(1) If the periodic payment is doubled while all other terms remain unchanged,
derive a formula for the new payoff time.
(2) If the interested rate is doubled while all other terms remain unchanged,
derive a formula for the new periodic payment.
Problem 2.3.7 One borrows ¹B dollars from a bank at a fixed rate Y. Now, we
assume the time to pay off the loan with fixed periodic payment × is @. Please
do the following.
(1) If the periodic payment is =× while all other terms remain unchanged,
derive a formula for the new payoff time.
(2) If the interested rate is Y while all other terms remain unchanged, derive a
formula for the new periodic payment.
Problem 2.3.8 One borrows ¹B from a bank at a fixed interest rate Y . We
assume the time to pay off the loan with a fixed periodical payment ×B is @B .
Now, if the payment is changed to v×B while all other terms remain
105
Chapter 2 Mathematical Models
unchanged, derive a formula for the new payoff time @( in terms of the given
parameters.
(1) If v > 1, which of the following is true?
(a) @( > @B
(b) @( = @B
(c) @( < @B
(2) If 0 < v < 1, which of the following is true?
(a) @( > @B
(b) @( = @B
(c) @( < @B
(3) If v < 0, compute @(.
106
Chapter 3
Linear DEs of Higher Order
3.1
Classification of DEs
Linear DE
Constant Coefficients
Variable Coefficients
+
=0
+ h( ) = 0
Nonlinear DE
+
+ h( )
=0
=0
We will study two types of DEs, namely,
(1) The general form of a variable coefficient, linear, 2nd-order DE.
+ h( ) + Ÿ( ) = X( )
If X( ) = 0, the DE is “reduced” to a homogeneous DE.
(2) The general form of a constant coefficient, linear, 2nd-order DE.
107
Chapter 3 Linear DEs of Higher Order
+
+
=
where , and are constants and if = 0, the DE is “reduced” to a
homogeneous DE.
Theorem 1
For homogeneous DE
+ h( )
+ Ÿ( ) = 0
(3.1)
+ h( ) + Ÿ( ) = X( ), if ( ( ) is a
of a inhomogeneous DE
solution and ( ) is another solution, then the linear combination of (
and
is also a solution. Further, if ( ( ) and ( ) are linearly
independent (L.I.) then their linear combination ( ) = 1( ( ( ) +
1 (x) is the G.S..
Corollaries of Theorem 1
If
(,
,…,
are = solutions for DE
( )
+ h( ( )
( >()
+ ⋯+ h ( ) = 0
then the linear combination of ( , , … , is also a solution to the DE.
are = L.I. solutions of the DE, then their linear
If ( , , … ,
combination is the G.S. to the DE.
Proof of Theorem 1
Let
= 1(
(
+1
be the linear combination of ( and
where 1( and 1 are constants.
Since ( and are the solutions to the DE (3.1), they satisfy
(
108
+ h( )
+ h( )
(
+ Ÿ( )
+ Ÿ( )
(
=0
=0
(3.2)
(3.3)
3.1 Classification of DEs
1( × (3.2 gives
Since
1(
1(
we have
1(
1( h
(
1(
(
h
(
Similarly, 1 × 3.3 gives
1
h
(
(
1(
1( Ÿ
and 1(
Ÿ
(
1
1
(
$1(
0
(
h
1
That is
Therefore
1(
(
1
1(
Í
%
h
Ÿ
h
(
1
1(
1(
1(
1(
Ÿ
Adding the two equations above generates
1(
(
(
(
0
(
(
(
1
Ÿ
1
1
Ÿ
′
0
0
Solution
Check (
(
0
1
Í
%
(3.6)
0
is a solution to the DE.
are solutions of the DE
=0
′′ +
is a constant and find . Given ( = cos
(
(3.5)
$1(
Example 1
Check if ( and
where
(3.4)
= − sin
and
+
(
Thus ( is a solution of the given DE.
(
=−
=0
(
and
= sin
cos
109
Chapter 3 Linear DEs of Higher Order
=
and
= − sin
+
=0
Thus is a solution of the given DE.
As per Theorem 1, we choose
= 1( ( + 1
= 1( cos
+ 1 sin
Check
= − 1( sin
+ 1 cos
and
= − 1( cos
+
=0
Therefore is also a solution of the DE
+
=0
Check
Example 2
Given ( = sin
,
cos
= 5 sin
Find the third solution.
Solution
According to theorem 1
−
1 sin
. Both are solutions of the DE
+
=0
= 1( ( + 1
= 1( sin
+ 51 sin
= (1( + 51 ) sin
= 1 sin
where 1 = 1( + 51
Problems
Problem 3.1.1 Show that = 1/ is a solution of
0 and ≠ 1, then = / is not a solution.
Problem 3.1.2 Show that =
is not a solution.
then =
Problem 3.1.3 Show that
0, but that their sum =
110
=6
= 0, but that if
≠
, but that if ≠ 1,
= 1 and
= √ are solutions of
+ is not a solution.
(
(
is a solution of
+
+( ) =
3.1 Classification of DEs
Problem 3.1.4 Let ¢ be a P.S. of the inhomogeneous DE
+k +å =
( ) and let ¡ be a solution of its associated homogeneous DE
+k +
å = 0. Show that = ¡ + ¢ is a solution of the given inhomogeneous DE.
With ¢ = 1 and ¡ = 1( cos + 1 sin in the above notation, find a solution
+ = 1 satisfying the I.C. (0) = (0) = −1.
of
111
Chapter 3 Linear DEs of Higher Order
3.2
Linear Independence
Definition
For functions ( and , if ( is not a constant multiple of , then ( is
L.I. of . That is, if one can find two constants ( and that are not zero
= 0 , then ( and
are called
simultaneously to make ( ( +
linear dependent (L.D.). O.W., they are L.I..
æ
Í
For the two-function case, if
Example
Determine if
çè
sin
sin 2
2 sin
2 exp
exp 2
exp
(,
)constant, then they are L.I..
çè
çé
tan
2 cos
2
2
exp
exp$
are L.I.
çé
cos
sin
sin
exp
exp
exp
L.I.
Y
Y
N
N
Y
Y( ≠ )
%
Now Let us define a new operator
Â=
Example
and so on. So for the =
112
Â
¦
Â
 =E
=E
F
F
=
order we have
=E
F
=
=
=
=
=
( )
3.2 Linear Independence
So now, our =
¦
order DE
( )
can be written as
Â
That is
Definition
+ h( ( )
+⋯+ h ( ) = 0
(3.7)
+ ⋯ + h ( )% = 0
(3.8)
( >()
+ h( ( )Â
$Â + h( ( )Â
+ ⋯+ h ( ) = 0
>(
>(
A new linear differential operator … as
… = Â + h( ( )Â
>(
+ ⋯+ h ( )
so that (3.8) now can be written as … = 0.
Example 1
Find the linear differential operator … in
+
+ 1]
=0
Solution
\
+
=0
… =Â +1
Example 2
Find the linear differential operator … in
Solution
\
=0
−5
−5
−5
+6 = 0
+ 6]
+6 =0
=0
… = Â − 5Â + 6
113
Chapter 3 Linear DEs of Higher Order
Properties of the Linear Differential Operator
(a) (…( + … ) = …( + …
(b) 1(… ) = …(1 )
From the above two properties, we can conclude that the linear
differential operator … satisfies the linearity property.
Theorem 2
For an = ¦ order linear DE … = 0, if
the DE, then the linear combination
combo
= ê 1ë
ëÜ(
is the G.S. to the DE.
Proof
Since ∀ 1 ≤ r ≤ =,
ë
(,
…
ë
is a solution to … = 0, then
…
=0
ë
Multiplying both sides of (3.9) by 1ë , we have
By the linearity of …, that gives
… 1ë
Summing over r, we have
That is
114
1ë …
ë
ë
=0 ∀1≤r ≤=
ê … 1ë
ëÜ(
=0
ë
=0
are = L.I. solutions to
(3.9)
3.2 Linear Independence
… °ê 1ë ë ± = 0
ëÜ(
By the definition, we know this is
…
combo
L.I. for More than Two Functions
(3.10)
=0
For more than two functions, the definition of linear independence
becomes a lot more complicated. The definition at the beginning of the
section is still valid as the definition for pair wise linear independence.
However, a set of functions that are mutually pair-wise L.I. does not
necessarily indicate they are L.I. as a whole.
Example
Given ( = sin ,
= cos and
= sin + cos , prove that they are
mutually pair wise linear independent.
Solution
It is easy to verify that
(
(
= tan
=1
cot
= tan + 1
These prove that they are mutually linear independent.
Remarks
Despite the fact they are mutually pair wise L.I., intuitively we think they are
= ( + . From the
not L.I. as a whole since we can easily find that
definition we will introduce below, they are indeed L.D..
For a formal definition of the linear independence for more than two
functions, we have to expand the definition for two functions from a new
115
Chapter 3 Linear DEs of Higher Order
direction. For any two L.I. functions ( and
that we cannot find a constant such that
(
, from definition, we know
=
(3.11)
for all where ( and are defined. Now let us write this definition in
a balanced fashion. Consider the equation
1(
(
+1
=0
(3.12)
If we can find a non-zero vector [1( , 1 ]C that satisfies (3.12), we know
we can find a constant 1 that satisfies (3.11). Note that the zero in (3.12)
means the zero function that is 0 everywhere ( and
are defined.
Thus, we know that the linear independence of ( and means that the
only solution for (3.12) is [1( , 1 ]C = [0,0]C . Now we can expand this
definition to = functions.
Definition
For = functions
(,
,…,
1(
said to be L.I. if
(
+1
+ ⋯+ 1
=0
(3.13)
has only one solution which is [1( , 1 , … , 1 ]C = [0,0, … ,0]C . The zero
in the RHS of (3.13) is the zero function defined where ( , … , are
defined.
It is very difficult to verify directly from the above original definition of
L.I. if a set of functions are L.I.. In practice, several other approaches can
be adopted. One of them is that of the Wronskian method.
Let us now introduce the Wronskian of = functions.
Definition
The Wronskian of = functions
116
(,
…
is the following determinant
3.2 Linear Independence
)=í
×( ( , … ,
Theorem 3
(
(
⋮
⋮
( >()
(
( >()
⋯
⋯
⋱
⋯
⋮
( >()
í
(3.14)
×
For = functions ( , , … , , if the Wronskian ×( ( , … , ) ≠ 0, then
are L.D., then the
they are L.I.. Conversely, if ( , , … ,
Wronskian ×( ( , … , ) = 0.
Example 1
Check if ( = exp(Y( )and
Solution
= exp(Y ) (Y( ≠ Y ) are L.I..
Example 2
Check if ( = exp(Y ) and
=
exp(Y( )
exp(Y )
)=_
_
Y( exp(Y( ) Y exp(Y )
= exp(Y( ) ⋅ Y exp(Y ) − exp(Y ) ⋅ Y( exp(Y( )
= (Y − Y( ) exp$(Y − Y() %
Since Y( ≠ Y , we know that ×( ( , ) ≠ 0. Thus, ( and are L.I..
×( ( ,
Solution
Thus,
(
exp(Y )
exp(Y )
)=_
_
Y exp(Y ) exp(Y ) + Y exp(Y )
= (1 + Y ) exp(2Y ) − Y exp(2Y )
= exp(2Y ) ≠ 0
are L.I..
×( ( ,
and
Example 3
Check if ( = sin
Solution
exp(Y ) are L.I..
and
×( ( ,
= cos
are L.I., where
) = Ý sin
cos
= − sin
cos
− sin
− cos
≠ 0 is a constant.
Ý
117
Chapter 3 Linear DEs of Higher Order
Thus,
Remarks
(
and
are L.I..
= − (sin
=− ≠0
+ cos
)
Please note that if Wronskian is non-zero, the functions are L.I.. But,
if the Wronskian is zero, it does not necessarily imply that the
functions are L.D.. For example, for ( =
and = | |, though
they are L.D. on either [0, +∞) or (−∞, 0], we cannot find a set of
non-zero [1( , 1 ]C such that (3.12) holds for the entire real axis. This
means ( and
are L.I. when considered as a function
on (−∞, +∞), but the Wronskian gives for ∈ [0, +∞)
and for
×( ( ,
∈ (−∞, 0), we have
×( ( ,
)=_
2
)=_
2
2
_=0
− _=0
−2
These two mean ×( ( , ) = 0 ∀ . This example indicates the
Wronskian is zero does not give much information about the linear
independence of the functions.
Problems
Problem 3.2.1 Define two operators with given constants v( , v , œ( , œ .
…( ≡  + v(  + œ(
… ≡ +v Â+œ
Applying these two operators to a function ( ), we may get …( … ( ) and
… …( ( ) check if …(… ( ) = … …( ( ). What if we define the two operators
as follows, check it again.
…( ( ) ≡ Â ( ) + ( )
… ( )≡ Â ( )+ ( )
118
3.2 Linear Independence
Problem 3.2.2 Check the linear independence of the following three functions:
exp
, cosh , sinh
Problem 3.2.3 Check the linear independence of the following three functions:
0, sin , exp
Problem 3.2.4 Check the linear independence of the following three functions:
8
2 , 3 , 5
Problem 3.2.5 Determine if
8
and -
are L.I..
cos
sin
be two solutions of A
ž
Problem 3.2.6 Let ( and
1
0 on an open closed interval - where, A, ž and 1 are continuous and
A
is never zero.
(1) Let × × (,
, Show that
×
A (
A
( A
Then substitute for A
(
and A
A
from the original DE to show that
×
ž ×
(2) Solve this 1st-order DE to deduce Abel’s formula
ž
]
×
H exp \
A
Problem 3.2.7 Prove directly that the functions
are L.I..
= ,…, ( ) =
B
1,
((
)= ,
( )
Problem 3.2.8 Show that the Wronskian of the following =ñò order
inhomogeneous DE
( )
h( ( ) ( >() ⋯ h ( ) = ( )
is a function of h( ( ) only, and find the expression of the Wronskian in terms
of this h( ( ).
119
Chapter 3 Linear DEs of Higher Order
Problem 3.2.9 Suppose that the three numbers Y( , Y and Y are distinct. Show
and exp Y
are L.I. by showing
that the three functions exp Y( , exp Y
that their Wronskian
1 1 1
Y ó ) 0, ∀
× exp$ Y( Y
Y % ó Y( Y
Y( Y
Y
Problem 3.2.10 Compute the Vandermonde determinant:
1
1
⋯
1
Y(
Y
⋯
Y
T í
í
⋮
⋮
⋱
⋮
>
>
>(
Y(
Y
⋯ Y
Problem 3.2.11 Given that the Vandermonde determinant:
1
1
⋯
1
Y
⋯
Y
Y(
í
í
⋮
⋮
⋱
⋮
>
>
>(
Y(
Y
⋯ Y
is non-zero if the numbers Y( , Y , … , Y are distinct. Prove that the functions
, 1 Æ r Æ = are L.I..
ë ( ) = exp Yë
120
3.3 Constant Coefficient Homogeneous DEs
3.3
Constant Coefficient Homogeneous DEs
Consider a 2nd-order DE with constant coefficients
+
+
=0
≠ 0,
and
(3.15)
Let us try out a solution. Without proper preparation, we can assume the
solution to DE (3.15) is exp(Y ). So, we can try a trial solution
( ) = exp(Y )
= Y exp(Y ) ,
Plugging these to (3.15), we have
That is
= Y exp(Y )
Y exp(Y ) + Y exp(Y ) + exp(Y ) = 0
( Y + Y + ) exp(Y ) = 0
an algebraic equation. Since exp(Y ) > 0, to enable the above equation,
we must have
Y + Y+ =0
(3.16)
This is called the characteristic equation (C-Eq) of the original DE
(3.15). Solving this C-Eq, we can get two solutions.
− +√ −4
2
− −√ −4
Y =
2
Y( =
So through “back substitution”, we get the solutions of the DE.
(
= exp(Y( ) and
= exp Y
Since ≠ 0, there are three possible cases for Y( and Y depending on the
value of − 4 . They are:
121
Chapter 3 Linear DEs of Higher Order
•
•
•
Case (1)
Case (2)
Case (3)
CASE 1: (
−4
−4
−4
−4
> 0)
>0
=0
<0
The C-Eq (3.16) has two distinct real roots Y( ≠ Y . The two trial
solutions are
(
= exp(Y( ) and
= exp Y
and exp Y
From last section, we know that exp Y(
the G.S. for the DE is
= 1( exp Y(
CHECK
are L.I. and thus
+ 1 exp Y
(3.17)
Since Y( and Y are the solutions of the algebraic equation (3.16), we
have
ô
Y( + Y( + = 0
Y + Y + =0
Plugging the two trial solutions exp Y(
(3.16), we have
exp Y(
+ exp Y(
= Y( exp Y( + Y( exp Y(
= Y( + Y( + exp Y(
and exp Y
(3.18)
into the DE
+ exp Y(
+ exp Y(
From (3.18), we know that the above formula gives 0. Hence exp Y(
a solution of the DE and similarly, we can prove it for exp Y .
Example
Solve the DE.
122
−5
+6 = 0
is
3.3 Constant Coefficient Homogeneous DEs
Solution
STEP 1: Introduce the trial solution = exp(Y )
STEP 2: Plugging the trial solution into the DE, we have
= Y exp(Y ) and
= Y exp(Y )
Y exp(Y ) − 5Y exp(Y ) + 6 exp(Y ) = 0
(Y − 5Y + 6) exp(Y ) = 0
We get the C-Eq
Y − 5Y + 6 = 0
(Y − 2)(Y − 3) = 0
Y( = 2 and Y = 3
STEP 3: Because Y( ≠ Y and both are real, we conclude it is case 1
STEP 4: Compose the G.S.
= 1( exp(2 ) + 1 exp(3 )
STEP 5: Check
= 1( exp(2 ) + 1 exp(3 )
= 21( exp(2 ) + 31 exp(3 )
= 41( exp(2 ) + 91 exp(3 )
−5 +6
= 41( exp(2 ) + 91 exp(3 ) − 5(21( exp(2 ) + 31 exp(3 ))
+6(1( exp(2 ) + 1 exp(3 ))
= 1((4 − 10 + 6) exp(2 ) + 1 (9 − 15 + 6) exp(3 )
=0
Thus = 1( exp(2 ) + 1 exp(3 ) is the G.S..
CASE 2: (
−4
= 0)
The C-Eq (3.16) has two equal real roots
Y( = Y = Y = −
2
It is clear that exp(Y( ) = exp(Y ) and they are not L.I. In order to get
the G.S., let us now try a new trial solution.
We have
=
exp(Y )
= Y exp(Y ) + exp(Y )
= 2Y exp(Y ) + Y
exp(Y )
123
Chapter 3 Linear DEs of Higher Order
Substituting these back to the given DE we will have
exp(Y ) + 2Y exp(Y )) +
(Y exp(Y ) + exp(Y )) + exp(Y )
= ( Y + Y + ) exp(Y ) + (2 Y + ) exp(Y )
(Y
From discussion of case 1, we know that the first term is 0. Noticing
that Y = − /2 , it is clear the second term of the above formula is
also 0. This means that is a solution to the original DE. Hence, we
now have
(
And the G.S. is
= exp(Y )
=
exp(Y )
= 1( exp(Y ) + 1
exp(Y )
Sometimes, we also write it as form
= (1( + 1 ) exp(Y )
Example 1
Solve the following DE
−2
+
=0
Solution
Introduce the trial solution = exp(Y )
Plugging the trial solution into the DE, we have
= Y exp(Y ) and
= Y exp(Y )
Y exp(Y ) − 2Y exp(Y ) + exp(Y ) = 0
(Y − 2Y + 1) exp(Y ) = 0
Hence we get the C-Eq
Y − 2Y + 1 = 0
(Y − 1) = 0
Y( = Y = 1
( = exp( )
and we have
124
(3.19)
(3.20)
3.3 Constant Coefficient Homogeneous DEs
= exp( )
= 1( exp( ) + 1
So we have the G.S.
exp( )
Example 2
Suppose the C-Eq of the given DE is
(Y − Y( ) = 0
Find the G.S. of the DE.
Solution
(Y − Y( ) = 0 means that the C-Eq has three identical roots, thus, we have
the following
= exp(Y( )
= exp(Y( )
( = exp(Y( )
and the G.S. is
= (1( + 1 + 1 ) exp(Y( )
CASE 3: (
−4
< 0)
The C-Eq (3.16) has two complex roots
Y( = v + rœ
Y = v − rœ
where
v=−
2
and œ =
√4
2
−
are both real number. Now we can form our trail solution
(
= exp$ v + rœ
% and
= exp$ v − rœ
%
It is easy to check that for œ ≠ 0, ( and are L.I. (You do not need to
bother with Wronskian since there are only two functions here).
(
=
exp$ v + rœ
%
exp$ v − rœ %
= exp 2rœ ≠ constant
125
Chapter 3 Linear DEs of Higher Order
So now the G.S. is
= 1( exp$(v + rœ) % + 1 exp$(v − rœ) %
(3.21)
Since the DE is raised in real domain, we also want to write the solutions
with the real coefficients. The trick here is to manipulate the constants in
(3.21). We can rewrite (3.21) as
= exp(v ) (1( exp(rœ ) + 1 exp(−rœ ))
From the most well-known Euler’s formula
exp(±rœ ) = cos œ ± r sin œ
we have
Let
= exp(v ) $1( (cos œ + r sin œ ) + 1 (cos œ − r sin œ )%
= exp(v ) $(1( + 1 ) cos œ + (r1( − r1 ) sin œ %
(
= 1( + 1
We have the G.S. in the real form
= exp v
Example 1
Solve the IVP
|
and
( cos œ
= r1( − r1
+
sin œ
−4 +5 =0
(0) = 1,
(0) = 5
(3.22)
Solution
After seeing many examples, now we are able to find the C-Eq directly
from the DE.
Y − 4Y + 5 = 0
(Y − 2) + 1 = 0
This C-Eq has two roots 2 + r and 2 − r. Hence we have the G.S.
( ) = exp(2 ) ( ( cos + sin )
To find the P.S., we plug the I.C. into it. First, we have
126
3.3 Constant Coefficient Homogeneous DEs
For ′, we have
( ) = 2 exp(2 ) (
and
(
cos +
(0) =
(
=1
sin ) + exp(2 ) (−
(0) = 2 ( + = 5
=
1
into
above
equation,
we can get
(
=3
So the solution to the IVP is
( ) = exp(2 ) (cos + 3 sin )
( sin
+
cos )
Plugging
Example 2
Find the G.S. of the following DE.
(Â + 6Â + 13)
Where D is the differential operator
Solution
The C-Eq of the given DE is
That is
Â=
=0
(Y + 6Y + 13) = 0
((Y + 3) + 4) = 0
= −3 + 2r and
=
= −3 − 2r
( =
Hence the G.S. of the given DE is
( ) = exp(−3 ) ( ( cos 2 + ( sin 2 ) + exp(−3 ) (
+ sin 2 )
and gives
cos 2
SUMMARY
For homogeneous 2nd-order constant coefficient DEs
We have trial solution
and the C-Eq
+
+
=0
( ) = exp(Y )
127
Chapter 3 Linear DEs of Higher Order
∆
õé
ö÷ø
∆
0
∆G 0
∆I 0
Y(
ùè , ùé
Y + Y+ =0
√
2
4
√
2
and Y
L.I. Solution
Y( ) Y
Real roots
Y( Y
Y
Real root
Y( v rœ
Y
v rœ
Complex
roots
exp Y(
G.S.
, exp Y(
exp Y
exp$ v
1( exp Y(
, exp Y
exp$ v
rœ
rœ
4
1(
1
1( exp$ v
%
1 exp Y
exp Y
rœ %
1 exp$ v rœ %
exp v
1( cos œ
1 sin œ
%
Theorem 4
If a homogeneous constant coefficient DE has = repeated and identical
roots Y( Y
⋯ Y
Y, then there are = L.I. solutions
exp Y
The G.S. for such DE is
1( exp Y
Or equivalently,
Proof
1(
1
, exp Y
1
exp Y
⋯
Suppose we have DE in the form of
Â
Y(
where  is the differential operator
128
,…,
>(
⋯
1
1
>(
0
exp Y
>(
(3.23)
exp Y
exp Y
(3.24)
(3.25)
(3.26)
3.3 Constant Coefficient Homogeneous DEs
Â≡
We prove that the functions in (3.23) are the solutions to DE (3.26) by
mathematical induction.
First, we have that for
(
(Â − Y( )
= exp(Y( )
(
=Â
=
(
− Y(
(
exp(Y( ) − Y( exp(Y( )
= Y( exp(Y( ) − Y( exp(Y( )
=0
This means ( is a solution of (Â − Y( ) = 0, and hence a solution of
(Â − Y( ) = 0.
Now supposed for / < =,
e
=
e>(
exp(Y( ) is a solution of
(Â − Y( )e = 0
and hence is a solution of (3.26). For / + 1, we prove
is also a solution.
eú(
=
e
(Â − Y( )eú( eú(
= (Â − Y( )e (Â eú( − Y( eú( )
= (Â − Y( )e $ e Y( exp(Y( ) + /
= (Â − Y( )e / e>( exp(Y( )
= /(Â − Y( )e e
=0
This means
eú(
exp(Y( )
e>(
exp(Y( ) − Y(
e
exp(Y( )%
is a solution of
(Â − Y( )eú( = 0
129
Chapter 3 Linear DEs of Higher Order
To prove that the functions in (3.23) are L.I., we can use the conclusion
of Problem 3.3.17 and directly apply the definition of L.I. on them. The
detail of this part of proof is left as a homework problem for you.
Example 1
Find the G.S. of the following DE.
+
+
+
=0
Solution
The DE can be written in terms of differential operator as
(Â + Â + Â + 1) = 0
Thus, the C-Eq is
Y +Y +Y+1=0
Y (Y + 1) (Y 1) = 0
(Y + 1)(Y 1) = 0
Y( = r, Y = −r, Y = −1
Therefore,
exp
, exp r and exp r
are solutions to the original DE and they are L.I. Thus, the G.S. is
( ) = 1( exp
1 cos
1 sin
Example 2
Find the G.S. of the following DE.
9 (9) − 6
( )
( )
=0
Solution
The DE can be written in terms of differential operators
(9Â 9 − 6Â
 ) =0
So the C-Eq is
9Y 9 − 6Y
Y =0
Y 9Y
6Y 1) = 0
Y 3Y 1) = 0
1
Y( Y
Y
0, Y
Y9
3
( ) = 1(
130
1
1
(1
1
19 ) exp E F
3
3.3 Constant Coefficient Homogeneous DEs
Example 3
Find the G.S. for the following DE.
(Â − 5Â + 6)(Â − 1) (Â + 1) = 0
Where
is the differential operator.
Solution
The C-Eq is
Â=
(Y − 5Y + 6)(Y − 1) (Y + 1) = 0
(Y − 2)(Y − 3)(Y − 1) (Y + 1) = 0
Y( = 2, Y = 3, Y = Y = 1, Y9 = r, Yû = −r
Hence the G.S. is
( ) = 1( exp(2 ) + 1 exp(3 ) + (1 + 1 ) exp( ) + 19 cos + 1û sin
Problems
Problem 3.3.1 Use the quadratic formula to solve the following equations
(Note in each case that the roots are not complex).
+r +2 = 0
(1)
(2)
− 2r + 3 = 0
Problem 3.3.2 (a) Use Euler’s formula to show that every complex number
can be written in the form Y exp(rü), where Y ≥ 0 and −8 < ü ≤ 8.
(b) Express the numbers 4 , −2 , 3r , 1 + r and −1 + r√3 in the form of
Y exp(rü).
(c) The two square roots of Y exp(rü) are ±√Y exp š ›. Find the square roots
of the numbers 2 − 2r√3 and −2 + 2r√3.
ëý
Problem 3.3.3 Compute the Wronskian of the three L.I. solutions ( ( ,
of the following 3rd order DE.
+6
+7
+ =0
,
)
Problem 3.3.4 Find the highest point on the solution curve in
131
Chapter 3 Linear DEs of Higher Order
+3 +2 =0
(0) = 1
(0) = 6
Z
Problem 3.3.5 Find the third quadrant point of intersection of the solution
curves with different I.C..
+3 +2 =0
Z ( (0) = 3, ( (0) = 1
(0) = 0, (0) = 1
Problem 3.3.6 Solve the IVP.
3
S
R
Q
+2 =0
(0) = −1
(0) = 0
(0) = 1
Problem 3.3.7 Find a linear homogeneous constant-coefficient DE with the
given G.S..
( ) = (A + ž + 1 ) cos 2 + (Â + þ + M ) sin 2
Problem 3.3.8 Solve the IVP
Z
=
(0) = 1
(0) = (0) = 0
( )
=
Z (0) =
Problem 3.3.9 Solve the IVP
( )
+
+ +2
(0) = (0) = 0
(0) = 30
+ (sin ) = 0
has the discontinuous coefficient function
1,
>0
sin = m
−1,
<0
Show that this DE nevertheless has two L.I. solutions
for all such that
Problem 3.3.10 The DE
132
((
) and
( ) defined
3.3 Constant Coefficient Homogeneous DEs
• Each satisfies the DE at each point ≠ 0
• Each has a continuous derivative at = 0
• ( (0) = (0) = 1 and B (0) = (0) = 0
Problem 3.3.11 Find the G.S. of the following DE.
 ( − 2)( + 3)( + 1) = 0
where  = ⁄
Problem 3.3.12 Solve the following IVP.
−2 +2 =0
|
(0) = 0, (0) = 5
Problem 3.3.13 Solve the following IVP.
+9 =0
|
(0) = 3, (0) = −1, (0) = 2
Problem 3.3.14 Find the G.S. of the following DE.
(Â − 1) (Â − 2) (Â − 3)(Â + 9) ( ) = 0
where  = ⁄ .
Problem 3.3.15 Find the G.S. of the following DE.
− − 15 = 0
Problem 3.3.16 Find the G.S. of the following DE.
9 − 12 + 4 = 0
Problem 3.3.17 For a homogeneous equation with repeated real roots, please
do the following:
(1) For homogeneous equation ( − Y( )eæ ( ) = 0 where  =
derivative operator, Y( is a real constant and /( is a positive integer. Find the
G.S. for the equation.
(2) Compose the G.S. of the following homogeneous equation
(Â − Y( )eæ (Â − Y )eÍ ⋯ (Â − Y )e = 0
is the usual
133
Chapter 3 Linear DEs of Higher Order
where Y( , Y , … , Y are known distinct real constants while /( , / , … , / are
known positive integers. You may use existing theorems without deriving
them, but indicate which one(s) you use.
Problem 3.3.18 Find the G.S. of the following DE:
where
134
E
− vF
( )=
> 0, v = constant and = = positive integer.
3.4 Cauchy-Euler DEs
3.4
Cauchy-Euler DEs
Linear DEs of higher orders with variable coefficients are usually not
solvable analytically even though they are linear. But one class of DEs of
special coefficients can be solved and this class is called Cauchy-Euler
DEs. The N-th order Cauchy-Euler DEs can be expressed as
ê
( )
ÜB
=0
A special case is the 2nd-order Cauchy-Euler DEs:
ê
( )
ÜB
=
+
B
′+
(
′′ = 0
We may use this special case to demonstrate the solution methods and, in
fact, higher order can be solved in the similar fashion. The real geist of
the solution methods is to transform the coefficients of the DEs from
variable to constant.
Method I: Assuming trial solution
Thus,
′
′′
Then the original DE becomes
Or
+
B
$
B
+
(
(
=
+
=
=
+
>(
−1
>
−1
−1 %
=0
=0
135
Chapter 3 Linear DEs of Higher Order
Since
≠ 0, we must have
B
+
( − 1) = 0
+
(
the C-Eq and we can adopt all ideas in solving constant coefficient DEs.
Method II: Substituting to replace independent variable
= exp( )
With this, we can get
′=
Similarly,
′′ =
E
1
=
F=−
=−
′=
1
1
+
Thus,
+
1
=
1
≡
=
1
=−
1
+
11
′′ = − +
This formula can be further generalized.
With the above, we transform our original DE to
Or
B
B
+
+
+
(
(
−
− +
+
=0
=0
a constant coefficient DE and all methods discussed before can be
adopted.
136
3.4 Cauchy-Euler DEs
Example:
Find the G.S. of the following DE
′′ +
− =0
Solution
Method I: Assuming trial solution
( )=
Thus,
>(
′
=
′′
=
−1 >
Then the original DE becomes
( − 1) + − 1 = 0
whose roots are
= −1
( = 1,
The G.S. is
>(
( )= ( (+
Method II: Substituting to replace independent variable
= exp( )
We generate a new DE:
′′ − = 0
whose G.S. is
( ) = ( exp( ) + exp(− )
Back substitute leads to
>(
( )= ( (+
Problems
Problem 3.4.1 Find the G.S. of the following DE.
(
Â+ Â+ ) =0
where  = / , and , and are constants.
Problem 3.4.2 Find the G.S. of the following DE.
+
−9 = 0
Problem 3.4.3 Find the G.S. of the following DE.
+
+
=0
137
Chapter 3 Linear DEs of Higher Order
Problem 3.4.4 Solve the following IVP.
′′ − 2 ′ + 2 = 0
|
1 = 3,
1 =1
Problem 3.4.5 Find the G.S. of the following DE.
+ 2 ′′ − + 2 ′ +
=0
Problem 3.4.6 Find the G.S. of the following DE.
− 2 ′ − 10 = 0
Problem 3.4.7 Find the G.S. of the following DE.
+
−
+ =0
where > 0.
138
3.5 Inhomogeneous Higher Order DEs
3.5
Inhomogeneous Higher Order DEs
Theorem 1
If ¡ is a G.S. to … = 0 and ¢ is a P.S. to … = ( ), then
is the G.S. to the inhomogeneous DE … = ( ).
Proof
Since
and
¢
¡
=
¡
+
¢
is a G.S. to … = 0, we have
…
¡
is a P.S. to … = ( ) gives
Then
Therefore
…$
¡
+
¢
…
¡
+
¢
=0
= ( )
¢%
=… ¡+… ¢
=0+ ( )
= ( )
is a G.S. to … = ( ).
Now let us discuss the way to find the P.S..
Suppose that RHS ( ) = cos / + sin / , then it is reasonable to
expect a P.S. of the same form:
¢
= A cos / + ž sin /
which is a linear combination with undetermined coefficients A and ž.
The reason is that any derivative of such a linear combination of cos /
and sin / has the same form. Therefore, we may substitute this form of
¢ in the given DE to determine the coefficients A and ž by equating
coefficients of cos / and sin / on both sides. The technique of finding
139
Chapter 3 Linear DEs of Higher Order
a P.S. for a inhomogeneous DE is rather tricky and tedious; one must
remember several rules.
Let us consider a few examples.
Example 1
Find the P.S. of
+3
+4 = 3 +2
Solution
Since ( ) = 3 + 2, let us guess that
¢ =A +ž
Therefore
¢ = A and ¢ = 0
Substituting these in the given DE, we have
¢ +3 ¢ +4 ¢ = 3 +2
0 + 3A + 4(A + ž) = 3 + 2
3
1
A = and ž = −
4
16
Therefore, we have a P.S.
3
1
−
¢ =
4
16
Example 2
Find the P.S. of
− 4 = 2 exp(3 )
Solution
Since any derivative of exp(3 ) is a constant multiple of exp(3 ), let us try
¢ = A exp(3 )
¢ = 9A exp(3 )
Substituting it in the given DE, we have
9A exp(3 ) − 4A exp(3 ) = 2 exp(3 )
2
A=
5
Therefore we find a P.S. of the given DE.
2
¢ = exp(3 )
5
Therefore we now have
140
3.5 Inhomogeneous Higher Order DEs
Example 3
Find the P.S. of
3
+ ′ − 2 = 2 cos
Solution
Though our first guess might be ¢ = A cos , the presence of ′ on the LHS
shows us that we will have to include a term of sin as well. So let our trial
solution be
¢ = A cos + ž sin
Therefore
and ¢ = −A cos − ž sin
¢ = −A sin + ž cos
Substituting these into the given DE, we have
3(−A cos − ž sin ) + (−A sin + ž cos ) − 2(A cos + ž sin )
= 2 cos
Equating the terms of cos and sin , we have
−5A + ž = 2
m
−A − 5ž = 0
Solving the above two simultaneous equations we have
5
A=−
13
1
ž=
13
Hence the P.S. is
5
1
cos + sin
¢ =−
13
13
Next, let us introduce several rules for popular cases.
Rule 1
( ) do not satisfy … = 0 (i.e., ( ) and its
If ( ), ( ) and
derivatives are not solution of … = 0), then the trial P.S. ¢ can be
linear combination of all terms in ( ).
Meaning of the Rule
Suppose that no term appearing in either ( ) or any of its derivatives
satisfies the associated homogeneous DE … = 0. Then take as a trial
141
Chapter 3 Linear DEs of Higher Order
solution for ¢ a linear combination of all such L.I. terms and their
derivatives. Then determine the coefficients by substitution of this trial
solution into the inhomogeneous DE … = ( ).
Necessary Conditions
We check the supposition made in Rule 1 by first using the C-Eq to find
the complementary function ¡ , and then write a list of all terms
appearing in ( ) and its successive derivatives. If none of the terms in
this list duplicates a term in ¡ , then we proceed with Rule 1.
Example
Try to find the G.S. of the following DE using undetermined coefficient
method
− 3 − 4 = 15 exp(4 )
Solution
First let us find the G.S. to the homogeneous DE.
′′ − 3 ′ − 4 = 0
The C-Eq is
Y − 3Y − 4 = 0
Y = −1
Y( = 4,
Therefore
¡ = 1( exp(4 ) + 1 exp(− )
Now let us try ¢ = A exp(4 ) as the P.S. to the inhomogeneous DE. Note
that here we have ( ) = 15 exp(4 )
¢ = 4A exp(4 ) and ¢ = 16A exp(4 )
Plugging this back to the DE, we have
LHS = ¢ − 3 ¢ − 4
= 16A exp(4 ) − 3(4A exp(4 )) − 4(A exp(4 ))
= (16 − 12 − 4)A exp(4 )
=0
While
RHS = 15 exp(4 )
Hence LHS ≠ RHS ∀A.
142
3.5 Inhomogeneous Higher Order DEs
No matter what value of A we choose we will never be able to find the
solution. So after reading the Rule mentioned above and the fact that
exp(4 ) term in ¡ ( ¡ = 1( exp(4 ) + 1 exp(− ) ) also appears in
( ) (15 exp(4 )), we cannot use the term A exp(4 ) as the trial P.S.,
as the P.S. should be L.I. with all terms present in ( ) (that is the P.S.
should not be a constant multiple of the terms present in ( )). So now
the question arises of how to solve the given DE. Consider Rule 1A.
Rule 1A
If ( ) is a solution to … = 0 , then one way of composing a trial
solution is to multiply the solution by .
Example (continued)
Now we introduce a new trial solution to the DE ′′ − 3 ′ − 4 =
15 exp 4 :
¢ = A exp(4 )
Hence
¢ = A exp(4 ) + 4A exp(4 ) and ¢ = 8A exp(4 ) + 16A exp(4 )
Plugging these back to the DE, we have
LHS = ¢ − 3 ¢ − 4 ¢
= 8A exp(4 ) + 16A exp(4 ) − 3(A exp(4 ) + 4A exp(4 ))
− 4A exp(4 )
= 5A exp(4 )
= RHS = 15 exp(4 )
This gives
A=3
Therefore we have a P.S.
¢ = 3 exp(4 )
and the G.S. is
Example 2
Find the P.S. of
= 1( exp(4 ) + 1 exp(− ) + 3 exp(4 )
+4 =3
143
Chapter 3 Linear DEs of Higher Order
Solution
We can find the solution to the homogeneous part of the above DE as
sin 2
¡ = ( cos 2 +
The function ( ) = 3 and its derivatives are constant multiples of the
L.I. functions , , and 1. Because none of these appears in ¡ , we try
+ž +1 +Â
¢ =A
Therefore
+ 2ž + 1 and ¢ = 6A + 2ž
¢ = 3A
Substituting the above in the DE, we have
(6A + 2ž + 4 A + ž + 1 + Â) = 3
Equating the coefficients of the like terms, we have
4A 3
4ž 0
€
6A + 41 0
2ž + 4Â 0
These give
3
9
A= ,
ž 0,
1
,
 0
4
8
Therefore, we have the P.S.
3
9
−
¢ =
4
8
Example 3
Find the general form of a P.S. for
+9
sin +
Solution
The C-Eq is
exp 2 )
Y + 9Y 0
3r
Y( 0, Y = 3r, Y
So the G.S. to the homogeneous DE is
¡ = 1( + 1 cos 3 + 1 sin 3
The derivatives of the inhomogeneous RHS involves
the terms
cos , sin , cos , sin
and
exp 2 ) , exp 2 ) , exp 2 )
Since there is no duplication with the terms of the G.S., the trial solution
takes the form
144
3.5 Inhomogeneous Higher Order DEs
= A cos + ž sin + 1 cos + Â sin + þ exp 2 ) + M exp 2 )
+ ‡ exp 2 )
Upon substituting ¢ in the DE and equating the coefficient terms we get
seven equations determining the seven coefficients A, ž, 1, Â, þ, M and ‡
¢
Rule 2 (The case of duplication)
If
( ) contains terms such as h¨ ( ) exp Y ) cos /
h¨ ( ) exp Y ) sin / , then we can select trial solution
¢(
)=
q
or
$(AB + A( + ⋯ + A¨ ¨ exp Y ) cos /
+ (žB + ž( + ⋯ + ž¨ ¨ exp Y ) sin / %
where h¨ is a polynomial of order K and ³ is the smallest non-negative
integer which does not allow duplication of the above solution as the
solution of the homogeneous DE. Then determine the coefficients in ¢
by substituting ¢ into the inhomogeneous DE.
h¨ ( ) =
cos /
( )
B
+
exp Y
exp Y
sin /
h¨
h¨
(
¨
sin /
cos /
exp Y
sin /
+⋯
cos /
¨
q
AB
q
exp Y
exp Y
A cos /
ç
A(
A
A cos /
exp Y
A cos /
AB
A(
⋯
ž sin /
A¨
ž sin /
⋯
ž sin / AB
A¨ ¨
A¨
A(
¨
¨
⋯
145
Chapter 3 Linear DEs of Higher Order
The above table lists the functions that can be chosen as the trial
solutions for a given RHS.
Example 1
Find the particular and G.S. for
Solution
We have the C-Eq
= 3 exp
4
Y
Y =0
1
Y( = Y = 0, Y
The complementary solution is
= 1( + 1 + 1 exp
¡
We now try to form our P.S. as
Ae
ž 1
Â
3 exp
4
q
¢
exp
ž 1
Â
Since we have 1( and 1 in complementary solution, we will have
duplication in ž 1
 terms in ¡ . Therefore ³ ) 0. If we chose ³ =
 will still duplicate the term. Now let ³ = 2, ž
1, ž
1
1
 will not duplicate any terms in ¡ . Therefore, we have ³ = 2.
The part A exp
corresponding to 3 exp
does not duplicate any part
of the complementary function, but the part ž 1
 must be
multiplied by to eliminate duplication. Hence we have
ž
1
Â
¢ = A exp
A
exp
2ž
31
4Â
¢
A exp
2ž 61
12Â
¢
A exp
61 24Â
¢
Substituting these back to the given DE gives
A exp
61 24Â
A exp
2ž 61
12Â = 3 exp
4
4
2A exp
2ž 61
61 24Â
12Â = 3 exp
Equating the like terms, we have
2A = 3
2ž 61 = 0
€
61 24Â = 0
12Â 4
Solving the above simultaneous equations gives us
146
3.5 Inhomogeneous Higher Order DEs
A=
A P.S. is
So the G.S. is
= 1( + 1
3
,
2
¢
ž = 4,
3
= exp
2
4
4
1
1=− , Â=
3
3
3
exp
2
+ 1 exp
4
3
1
3
4
4
3
1
3
Example 2
Determine the appropriate form for a complementary and P.S. if given the
. The roots are Y = 2, 2, 2 and 2 p 3r
roots of the C-Eq and
and
= exp 2
sin 3 .
Solution
Since we are already given the roots of the C-Eq, let us compose the
complementary function. So the complementary function is
= (1( + 1 + 1 ) exp 2
exp 2
1 cos 3
19 sin 3
¡
As a first step toward forming the P.S., we examine the sum
A ž
1
exp 2
exp 2 $ Â þ cos 3
M ‡ sin 3 %
To eliminate duplication with terms of
, the first part corresponding
to
exp 2
must be multiplied by
, and the second part
corresponding to sin 3 must be multiplied by . Hence we would take
= (A + ž + 1 9) exp 2
¢
exp
Example 3
2
9
$ Â
=
sin
þ
cos 3
M
‡
sin 3 %
exp 2
Solution
First, let us find the G.S.. We have the C-Eq
Y
9Y = 0
Y = 3r,
Y = −3r
Y( 0,
The G.S. is
1 sin 3
¡ = 1( + 1 cos 3
Now let us find the P.S. from observing the RHS. We compose our trial.
A cos
A sin
A cos
A9 exp 2
¢ = A( sin
A
exp 2
Aû exp 2
147
Chapter 3 Linear DEs of Higher Order
Problems
Problem 3.5.1 Find the G.S of the following DE.
2
8 = exp 4
Problem 3.5.2 Find the G.S. of the following DE.
′′′′ − = 1
Problem 3.5.3 Find the G.S. of the following DE.
′′′′ − = 4 exp
Problem 3.5.4 Find the G.S. of the following DE.
2 = 18
9
Problem 3.5.5 Find a P.S. of the following DE.
=
+
Problem 3.5.6 Find a P.S. of the following DE.
+
+ + = 1 exp
exp 2
exp 3
Problem 3.5.7 Find the G.S. of the following DE.
= ( + 1 sin
Problem 3.5.8 Find the G.S. of the following DE with the following two
conditions: ) B and = B .
′′ +
= exp r B
Problem 3.5.9 Find the G.S. of the following DE.
9 = cos 3
Problem 3.5.10 Find the G.S. of the following DE.
2
= 5 sin
cos
148
3.5 Inhomogeneous Higher Order DEs
Problem 3.5.11 Find a P.S. of the following DE.
= 1 cos
sin 2
exp
Problem 3.5.12 Find the G.S. of the following DE.
′′ − 4 ′ + 6 =
Problem 3.5.13 Find the G.S. of the following DE.
û ′′ + 2 9 ′ − 12
=1
Problem 3.5.14 Find the G.S. of the following DE.
2 ′ − 6 = 72
9
Problem 3.5.15 Find the G.S. of the following DE.
4
4
3 =8
Problem 3.5.16 Find the G.S. of the following DE.
= ln
Problem 3.5.17 Find the G.S. of the following DE.
=1
Problem 3.5.18 Find the G.S. of the following DE.
′′ + ′ − 4 =
+
>
Problem 3.5.19 Given the three L.I. solutions of the homogeneous portion of
( ) + v ( ) + œ (() + µ = ( ) as
,
, find the G.S. for
( ( ),
the inhomogeneous equation in terms of v, œ, µ, and the given functions.
Problem 3.5.20 Please use any method of your choice to find a P.S.
the following DE
= ( )
¢
for
149
Chapter 3 Linear DEs of Higher Order
where is a constant and
is a given function for the following three
situations. You may express the P.S. ¢
in terms of the given
in some
integral form if you do not have enough information to integrate it out.
(1) Function
is a general form;
(2) Function
is a specific function
= exp(r );
(3) Function ( ) is a specific function ( ) = exp( ).
150
3.6 Variation of Parameters
3.6
Variation of Parameters
Consider an example
= tan . Since
= tan has infinitely
2 sec ,
many L.I. derivatives sec , 2 sec tan , 4 sec tan
…, etc. Therefore, we do not have available a finite linear combination to
use as a trial solution and we do not have corresponding rules to follow
for solving DEs of this type. We have to introduce a new method, i.e.,
variation of parameters.
For … = ( ), the homogeneous DE … = 0 has = solutions
.
(,
, …,
(1) The G.S. of homogeneous DE is ¡ = 1( ( + 1
+⋯ 1
(2) Let us propose to change the 1( , 1 … 1 in the complementary
for composing a trial
function to variables ( ,
…
P.S. for the inhomogeneous DE
¢
((
=
=ê
ëÜ(
=[
Example
=
(
C
)
ë(
(
( )
+
) ë( )
⋯
+⋯
(
« ⋮
Consider a 2nd-order inhomogeneous DE
′′ + h( ) ′ + Ÿ( ) = ( )
(3.27)
Compose a formula for the P.S. for the above DE.
Solution
1) Assume the solutions to the homogeneous DE are
us
(
and
. This gives
151
Chapter 3 Linear DEs of Higher Order
|
(
¢
2) Compose a trial P.S.
h
+ h( )
¢
Therefore we have
¢
=(
as
=
((
( (
( (
( (
+
(
)
Ÿ
+ Ÿ( )
(
+
( (
)′
(
=0
=0
( )
(3.28)
(3.29)
( (
In the above process, we introduced two “free” parameters (
and
. With such, we can impose two constraints. Let us force the
following condition (although one may impose many other conditions).
( (
Therefore, we have
¢
¢
=
From (3.28) we have
¢
=( ( (+
= ( (+
( (
(
Substitute these into
152
=
( (
+
= −h
= −h
+
( (
( (
+
=0
)′
+
+
(
−Ÿ
−Ÿ
(3.30)
(3.31)
+
( (
+
(
(3.32)
3.6 Variation of Parameters
( (
= ( (−h( ) ( − Ÿ( ) ( )
+ (−h( ) − Ÿ( ) )
) − Ÿ(
= −h( ( ( +
= −h ¢ − Ÿ ¢
Finally, plugging back (3.33) into (3.32)
¢
¢
= −h
+h
¢
¢
−Ÿ
So we have
So from (3.30) and (3.35)
+
+Ÿ
k
+Ÿ
But according to (3.27), we have
k
¢
+h
( (
k
+
(
¢
(
(
=
( (
( (
)
+
(3.33)
+
+
(3.34)
= ( )
(3.35)
( (
= ( )
=
0
( )
In theory, one may easily GENERALIZE the above formulation for
higher-order DEs.
( )
(
(
⋮
( >()
(
+ h(
¡
¢
⋮
( >()
=
=
( >()
( (
( (
⋯
⋯
⋱
⋯
+ ⋯+ h
+ ⋯+
+ ⋯+
⋮
( >()
= ( )
(
⋮
=
0
⋮
0
( )
153
Chapter 3 Linear DEs of Higher Order
(
Next, let us derive a formula for
(3.30), we have
(
and
=−
=
where ×
(
=
(,
(
(E
Ý
(
×
From
(3.36)
F+
(
(
¢.
Ý
(,
is the Wronskian. Thus we have
=
=
×
×$
(
(
=−
(
and
(
=−
(,
(
back to (3.36), we have
Plugging
′
(
and plugging (3.36) into (3.35) gives
=−
and, ultimately, for
×
×$
,
%
(3.37)
(,
(
,
%
(3.38)
Plugging (3.37) and (3.38) back into (3.29), we find the trial solution
¢
154
=
( (
=−
(
+
×$
(
,
%
+
×$
(
(
,
%
3.6 Variation of Parameters
¢(
where
=
)=
( ) (( ) −
((
×$ ( ( ),
) ( )
( )
( )%
H( , ) ( )
( ) (( ) −
H( , ) =
×$ ( ( ),
((
) ( )
( )%
is called the kernel mathematically. The kernel is also called the Green’s
function in physics.
SUMMARY
Consider solving a 2nd-order DE with constant coefficients
+
+
= ( )
STEP 1: Find the C-Eq for the homogeneous portion
That is
… =0
Y + Y+
=0
and compose the
STEP 2: Find the two solutions ( and
complementary solutions ¡ of the homogeneous DE.
¡
=
+
( (
STEP 3: Compose the trial P.S. of the inhomogeneous DE … = ( )
where
((
)≠
¢( )
=
((
)
(
+
( ) and are function of .
STEP 4: Solve the equation to find
( (
+
(
( )
and
=0
155
Chapter 3 Linear DEs of Higher Order
( (
OR: Solve the matrix below to find
(
(
(
(
OR: We can find the values of
=−
(
where ×
(,
¢
and
¢
and
=
((
from the formulas given below
( ) ( )
(
and
.
by substituting the values of
=
( (
+
by using the equation below
)=
STEP 6: Now find the G.S.
=
( ) (( ) −
×$ ( ( ),
( (
%
) ( )
×$ ( ( ),
%
=
¢
¢(
0
×$ ( ( ),
is the Wronskian of
STEP 5: Find
below
OR: Find
(
= ( )
+
+
((
(
and
) ( )
%
( (
+
Example 1
Solve
′′ − 3 ′ − 4 = 15 exp 4
by variation of parameters
Solution
STEP 1: The C-Eq is
156
in the equation
3.6 Variation of Parameters
Y
3Y
4=0
Y( = 4, and Y = −1
The solution to the homogeneous DE is
= exp 4
(
= exp
and
STEP 2: The complementary solution is
=
¡
( exp
4
STEP 3: Composing the trial P.S.
(
STEP 4: Find
¢
and
=
exp 4
4 exp 4
( exp
(
exp
(
5 exp 4
(
5
exp
=
=
4
exp
=0
exp
(
(
exp
= 15 exp 4
or, we can find
ת « = _
¢
(3.40)
= 15 exp 4
=3
3
(3.39)+(3.40)
=3
= 15 exp 4
(3.40)-4(3.39)
= −3 exp 5
−3 exp 5
STEP 5: We now find the P.S.
= 3 exp 4
¢
(3.39)
3
= − exp 5
5
3
exp 4
5
directly using the formula given below
=
¢
((
((
)
)
(
( ) (( ) −
×$ ( ( ),
= exp 4
and
( )
exp 4
_=_
( )
exp 4
((
)
%
( )
= exp
exp
exp
_ = −5 exp 3
157
Chapter 3 Linear DEs of Higher Order
¢
=
$
( ) (( ) − (( ) ( )
( )
×[ ]
exp
3
(
exp
3 exp
exp 4
(
exp 5
%
exp 4
1
E exp 5 F
5
×
exp
exp 4
3 exp 4
3
exp 4
3 exp 4
5
which is the same as the ¢ found in the previous page.
STEP 6: We now find the G.S. to the problem
= ¡+ ¢
1( exp 4
1 exp
1 exp 4
3 exp 4
1 exp
15 exp 4
5 exp 3
3 exp 4
3
exp 4
5
3
5
is the final solution to this problem which is same as the solution we
solved using the undetermined coefficients method.
where
1 = 1( −
Example 2
Solve the DE
Solution
Method I
Here we have
Hence
Plugging these into
158
= tan
= tan . The C-Eq to the homogeneous DE is
1=0
Y
Y(, = ±r
1 sin
¡ = 1( cos
=
cos
and
= sin
(
(
= − sin
( (
and
= cos
=0
3.6 Variation of Parameters
Solving for
( and
(
¢
=
=
(cos
sin
sin
cos
+
( (
sin
ln|sec
cos ln | sec
Method II
H
sec
tan |
ln|sec
tan | cos
tan |
cos sin
= cos
= sin
( ) (( ) − (( ) ( )
=
×[ ]
sin cos
cos sin
1
sin
(
,
¢
Finally, we have
= − sin tan
cos
sec
= cos tan
sin
(
Therefore
=
= ( )
( (
we have
=
H( ,
sin
tan
cos ln | sec
= ¡+ ¢
= (1( − ln|sec
tan | cos
tan |
1 sin
159
Chapter 3 Linear DEs of Higher Order
Problems
Problem 3.6.1 For the following 2nd-order, linear, constant coefficient,
inhomogeneous DE
′′ − (Y( + Y ) ′ + Y( Y = ( )
where Y( and Y are two different real constants while ( ) is a real function,
please find
(1) The two L.I. solutions, ( ( ) and ( ), for the homogeneous portion of
DE in terms of Y( and Y .
( ) that form the P.S. ¢ ( ) =
(2) The trial functions ( ( ) and
( ) ( ) by method of variation of parameters.
(( ) (( ) +
(3) The G.S. for the original DE.
Problem 3.6.2 For a given variable coefficient homogeneous DE
+ h( )
+ Ÿ( ) = 0
one solution is given as (
) 0. Find the G.S. of this DE by finding another
= ¹( ) ( ( ). Please note that ( , h
linearly independent solution
and Ÿ
are given functions while ¹
is not.
Problem 3.6.3 Find the P.S. for
′′ +
= cot
Problem 3.6.4 Find the P.S. for
′′′ + ′′ + ′ +
where ( ) is a well-defined function.
Problem 3.6.5 Find a P.S. for
= ( )
′′ + 9 = sin tan
Problem 3.6.6 Find a P.S. for
′′ +
= sin
tan
Problem 3.6.7 Find a P.S. (with constants , ) for
′′ +
= tan
160
3.6 Variation of Parameters
= 2sin
please find
(1) The two L.I. solutions for the homogeneous portion of DE.
(2) The trial functions ( ( ) and ( ) that form the P.S. ¢ ( ) by method of
variation of parameters.
(3) The G.S. for the original DE.
Problem 3.6.8 For inhomogeneous DE
Problem 3.6.9 You can verify by substitution that ¡ = 1( + 1 >( is a
complementary function for the inhomogeneous second-order DE.
′′ + ′ − = 72 9
But before applying the method of variation of parameters, you must first
divide this DE by its leading coefficient
to rewrite it in the standard form.
1
1
′′ +
′−
= 72
= 72
in the DE
…
= ′′ + h
Now processed to solve the DE in
Thus
and thereby derive the P.S.
|
(
+
+
(
( (
¢
′+Ÿ
=3
=0
=
=
9
Problem 3.6.10 Find the G.S. of the following DE by two methods,
′′ − 3 − 4 = 15 exp 4 + 5 exp −
(1) Trail solution method.
(2) Variation of parameters.
Problem 3.6.11 Find a P.S. of
′′′ + h(
′′ + h
′+h
=
,
whose complementary solution is given as
= ( (
+
+
.
¡
Your solution should be expressed in terms of the given functions.
161
Chapter 4
Systems of Linear DEs
4.1
Basics of DE systems
Given below is a general 1st-order DE.
, ,
Here is the independent variable and
=0
(4.1)
is the dependent variable.
The general form of a system of two 1st-order DEs with two dependent
variables ( and is
|
where
,
- ,
(,
(,
,
,
(,
(,
162
=0
=0
(4.2)
4.1 Basics of DE systems
•
•
•
= independent variable
st
( = function of =1 dependent variable,
= function of =2nd dependent variable.
Example 1
is a Coupled ODEs.
|
(
+2
(+
Example 2
The following two DEs
|
are coupled because ( and
is no in the first DE.
Example 3
While DEs
(
+3 (+4 = ( )
+ ( + = -( )
( = ( )
+ (+
= -( )
are mixed in the second DE although there
3 (+5 ( = ( )
+ 2 + 4 = -( )
are two independent (uncoupled) DEs because ( and are in no way
related by the above two DEs. Solving the two DEs, independently, should
yield solutions to the two DEs.
Generalization
A generalization of (4.2) to =
can be written as
(
|
(
¦
order, K DEs and K dependent variables
;
(, (, (
,…,
( )
( ;
,
,
( ;
(, (, (
,…,
( )
( ;
,
,
,…,
( )
;…
,…,
( )
;…
…;
( )
¨, ¨, ¨, … , ¨ )
=0
…;
( )
¨, ¨, ¨, … , ¨ )
=0
163
Chapter 4 Systems of Linear DEs
;
164
(, (, (
…;
,…,
( )
( ;
,
,
( )
¨, ¨, ¨, … , ¨ )
(4.3)
,…,
=0
( )
;…
4.2 First-Order Systems and Applications
4.2
First-Order Systems and Applications
Example 1
Consider a system of a particle of mass m and a spring with constant / that
moves under the influence of a force field M where is the distance by
which it is stretched from the natural length of the spring.
/
K
Figure 4.1 A block-spring system with mass K and spring constant /.
Solution
According to Newton’s 2nd law,
M=K =K
We know the force on the block by the spring is M = −/ , and then we
have
K = −/
That is
/
=−
K
ë
Define
= as constant, we have
=−
or
So the C-Eq is
Thus
and we have G.S.
By the I.C.
Y
=0
=0
Y(, = ±r
= 1( cos(
|
) + 1 sin(
0) = B
(0) = LB
)
165
Chapter 4 Systems of Linear DEs
We can get
•
Finally, we can get the solution
=
B
1( =
1 =
cos
B
LB
+
LB
sin
Example 1A (Modified)
In this example we are studying the motion of the block under the
influence of a force
which is applied to the block.
/
K
Figure 4.2 A block-spring system with an external force
.
The equation of motion should be
K = −/ + ( )
( )
/
=−
+
K
K
Let
/
=
K
and
( )
A
=
K
We have
=A
We can now solve the above DE using variation of parameters learned
previously
= ¡( ) + ¢( )
166
4.2 First-Order Systems and Applications
Obviously one can complicate the system by tangling it with more spring
or more blocks or both. Let us consider next a setup with two vibrating
blocks.
Example 2
We move block 1 by
We move block 2 by
/1
(.
So spring 1 is extended by
.So spring 2 is extended by
1
(.
/2
K1
(
2
.
K2
Figure 4.3 A system of two blocks connected by two springs.
Solution
Force on block 2 is
/
(
/( ( /
(
The DEs for the system are
K
= −/( ( + / ( − ( )
| ( (
= −/ ( − ( )
K
From the first equation, we have
/(
/
+
( − ()
( =−
K( ( K(
And the second equation gives
/
=−
( − ()
K
Force on block 1 is
Let
(
=
eæ
¨æ
,
=
eÍ
¨Í
and
(
=
eÍ
¨æ
(coupled frequency), we can simplify
the Des of the system as following
=− ( (+
ô (
=− ( −
(
(
()
−
()
Example 2A (A Variation of Example 2)
We study the motion of the two blocks under the influence of a force
.
167
Chapter 4 Systems of Linear DEs
1
/1
2
/2
K1
K2
( )
Figure 4.4 A system of two blocks connected by two springs and an external force on
block-2.
while the force on K( is still
(
/( ( /
(
The DEs for the system now become:
K
= −/ ( − ( ) + ( )
| (
K ( = −/( ( + / ( − ( )
This gives
/(
/
( − ()
(+
( =−
K(
K(
( )
/
( − () +
=−
K
K
/
ž
(
K
Solution
The force on K is
where ž
Let
(
=
eæ
¨æ
=
,
£( )
¨Í
=
eÍ
¨Í
/
and
ô
(
(
=
=−
=−
eÍ
¨æ
(coupled frequency), we can have
( (
(
+
−
( − ()
)
( + ž( )
(
Example 3: Kirchhoff circuits
According to Kirchhoff’s voltage law, the left-hand loop of the network has
þB = …
-(
+ (-( − - )X
1
Ÿ + - X( − (-( − - )X = 0
1
The right-hand loop of the network has
168
(4.4)
(4.5)
4.2 First-Order Systems and Applications
Capacitor
Inductor
-1
þ0
-2
X2
X1
Figure 4.5 An example of Kirchhoff circuit.
à
Circuit Element
Voltage Drop
Inductor
X1
Ÿ
1
…
Resistor
Capacitor
Table: Voltage drops across common circuit element.
Since
ௗொ
yields
= - , differentiation of each side of (4.4) and substitution of
1 Ÿ
+ (X( + X )
−X
1
-
This gives
And from (4.5) we have
-
Substitution of
=
æ
=
X
X( + X
-(
=−
-(
−
-(
ௗொ
=0
1
1(X( + X )
X
X
þB
- + - +
… (
…
…
(4.6)
(4.7)
X
X
X
þB
1
E− -( + - + F −
X( + X
…
…
…
1(X( + X )
in (4.6), we have
169
Chapter 4 Systems of Linear DEs
-
=−
X
1
X
1
X
þB
-( +
\ − ]- +
)
…(X( + X
X( + X …
1
X! + X …
(4.8)
From (4.7) and (4.8), we have
X
X
þB
−
…
…
-(
-(
…
=
+
X
þB
X
X
1
1
−
\ − ]
)
X
+
X
…
…(X( + X
X( + X … 1
(
Since, …, 1, þB , X( and X are all constants, we can write the above DEs as
the following
= A- + þ
where - =
-(
, A, and þ are matrices with constant elements.
-
Coupled DE Can be Solved Using the Following Methods
(1) Substitution method
(2) Operator method
We will discuss each method in detail in the following sections.
Problems
Problem 4.2.1 For the following setting with typical assumptions that the
springs are mass-less, and the surface the two massive blocks are placed is
frictionless, please do the following.
(1) Construct the DE that governs the motion of the two blocks.
(2) Find the G.S. for the blocks if we break the middle spring.
(3) Find the G.S. for the blocks if /( = / = / for all cases, you may assume
arbitrary I.C..
170
4.2 First-Order Systems and Applications
/1
/1
K1
K1
1
/2
/2 (
2
−
1)
/2 (
/3
K2
2
−
1)
K2
Figure 4.6 The block-spring system for Problem 4.2.1.
/3
2
Problem 4.2.2 Two particles each of mass m move in the plane with coordinates ( ( ),
under the influence of a force that is directed toward the
origin and has an inverse-square central force field of
/
.
Show that
where Y = V
K
+
.
€
K
/
Y
/
=−
Y
=−
Problem 4.2.3 Suppose that a projectile of mass K moves in a vertical plane in
the atmosphere near the surface of the earth under the influence of two forces:
a downward gravitational force of magnitude K-, and a resistive force M[ that
is directed opposite to the velocity vector L and has magnitude /L (where L =
|L| is the speed of the projectile). Show that the DEs of motion of the projectile
are
K = −/L
|
K = −/L − KProblem 4.2.4 Two massless springs with spring constants /( and / are
hanging from a ceiling. The top end of the spring-1 is fixed at the ceiling, while
the other end is connected to a block-1. The top end of the spring-2 is
connected to block-1, while the other end is connected to block-2. The mass for
block-1 is K( , and that for block-2 is K . Initially, we hold the blocks so that
171
Chapter 4 Systems of Linear DEs
the springs will stay at their nature lengths. At time = 0, we will release both
blocks to let them vibrate under the influence of the spring and gravity with
gravitational constant -. Please find the motion of the blocks.
Problem 4.2.5 Three massless springs with spring constants /( , / and / are
hanging on a ceiling. The top end of the spring-1 is fixed at the ceiling while
the other end is connected to a block-1. The end of the spring-2 is connected to
block-1 while the other end is connected to block-2. The top end of the spring3 is connected to block-2 and the other end is connected to block-3. The mass
for blocks are K( , K and K .
Initially, we hold the blocks so that the springs will stay at their nature lengths.
At time t=0, we will release the blocks to let them vibrate under the influence
of the spring and gravity with gravitational constant -. Please find the motion
of the blocks.
Problem 4.2.6 One end of a massless spring of constant k is attached to a
ceiling and the other end to a small ball of mass K. Normally, the spring
stretches down a little, due to gravity, but stays perfectly vertical. Now,
someone moves the ball up a little and the spring recoils to its natural length
of …B . If the ball is moved a little sideway from the natural state, without
stretching or squeezing the spring, the spring stays on a line about üB degrees
from the vertical line. After being released, the ball starts some magic motion.
Compute its motion. (Hint: Since the move is so small that ü ≈ sin ü.)
Figure 4.7 The pendulum system for Problem 4.2.6.
172
4.3 Substitution Method
4.3
Substitution Method
Example 1
Solve the following two coupled DEs with I.C., i.e., system of DEs.
( = (−8
Z = − (+3
(0) = 2
( (0) =
Solution
STEP 1: Solve for one of the two dependent variables
given two DEs.
The second DE can be written as
(
=3
=3
3 −
Thus, we get a 2nd-order DE for
Therefore
−
or
from the
(4.9)
−
( =3
and ( into the first DE, we get
3 − ) = (3 − ) − 8
Differentiating it gets
STEP 2: Plugging
(
(
4
−
−5
−8
=0
(4.10)
STEP 3: Solving the resulting single-variable (decoupled) DE by C-Eq
method, we have
Y
4Y − 5 = (Y − 5)(Y + 1) = 0
Thus, the solution for is
= 1( exp(5 ) + 1 exp(− )
STEP 4: Back substitute to solve for the other dependent variables, from
(4.9) we get,
−
( =3
3(1( exp(5 ) + 1 exp(− )) − (1( exp(5 ) + 1 exp(− ))
31( exp(5 ) + 31 exp(− ) − (51( exp(5 ) − 1 exp(− ))
21( exp(5 ) + 41 exp(− )
Thus, we have the G.S. for the system of DEs.
= −21( exp(5 ) + 41 exp(− )
| (
= 1( exp(5 ) + 1 exp(− )
STEP 5: Solve for the I.C.
173
Chapter 4 Systems of Linear DEs
|
(
0) = 2 = −21( + 41
(0) = 2 = 1( + 1
1 =1
| (
1 =1
This gives
The final solution is
|
(
= −2 exp(5 ) + 4 exp(− )
= exp(5 ) + exp(− )
Solution Steps
STEP 1: Express one dependent variable by the other by solving the two
DEs, i.e., decoupling variables.
STEP 2: Substitute the above variable to one DE to obtain a singledependent variable DE.
STEP 3: Solve the resulting single-dependent variable DE from the
above STEP 2:.
STEP 4: Back substitute to solve for other dependent variables.
Example 2
Solve the following two DEs.
€
′=4 −3
′=6 −7
0 =2
0 = −1
Solution
STEP 1: Solve the first DE for in terms of and ′, or solve the second DE
for in terms of and ′. (There is no written rule on which variable is
better to eliminate first. This depends highly on experience and
mathematical perception. Fortunately, many times, it would not matter
which order to take.)
From the first DE = 4 − 3 , we get
4
1
=
−
3
3
and
174
4.3 Substitution Method
STEP 2: Plugging
4
1
−
3
3
and into the second DE, we get
1
4
1
4
−
= 6 − 7E −
F
3
3
3
3
=
3 − 10 = 0
STEP 3: Solving the resulting single-dependent-variable DE by C-Eq
method, we have
3Y − 10 = 0
Y
Solve this C-Eq gives
Y( = −5 and Y = 2
We can get the solution for
= 1( exp(−5 ) + 1 exp(2 )
STEP 4: Back-substitute to solve for
.
4
1
=
−
3
3
1
4
(1 exp(−5 ) + 1 exp(2 )) − (1( exp(−5 ) + 1 exp(2 ))
3
3 (
4
1
(1 exp(−5 ) + 1 exp(2 )) − (−51( exp(−5 ) + 21 exp(2 ))
3 (
3
2
31( exp(−5 ) + 1 exp(2 )
3
Therefore, we have the G.S. to the original system
= 1( exp(−5 ) + 1 exp(2 )
2
•
= 31( exp(−5 ) + 1 exp(2 )
3
STEP 5: Solve for the I.C.
0) = 2 = 1( + 1
2
•
(0) = −1 = 31( + 1
3
This solves
1 = −1
| (
1 =3
Thus, the final solution is
= − exp(−5 ) + 3 exp(2 )
|
= −3 exp(−5 ) + 2 exp(2 )
or
Example 3
Solve the following two DEs.
175
Chapter 4 Systems of Linear DEs
Solution
We can easily find
=
|
2
2
−
=3
=1
|
2
2
−
=3
=1
= 1, thus we get (very strange solutions)
= + 1(
|
= +1
Example 4
Solve the following two DEs.
Solution
STEP 1: From the first DE, we get = 3 − 2 ′ and that gives
= (3 − 2 ) = −2
STEP 2: Substituting the above to the second DE, we get
4 + = −1
This can solve
€
= 1( cos + 1 sin − 1
2
2
= −1 cos + 1( sin + 3
2
2
Problems
Problem 4.3.1 Solve the following system of DEs.
= 2 exp(− )
|
+
− =0
Problem 4.3.2 Solve the following system of DEs.
=−
S
= 10 − 7
R (0) = 2
Q (0) = −7
Problem 4.3.3 Use substitution method to solve the following system of 1storder linear equations.
176
4.3 Substitution Method
S
R
Q
=
+2
=2 −2
(0) = 1
(0) = 2
Problem 4.3.4 Find the G.S. of the system given below with the given masses
and spring constants.
K
= −(/( + / ) ( + /
| ( (
K
= / ( − (/ + / )
where K( = K = 1, /( = 1, / = 4 and / = 1
/1
1
K1
/2
2
K2
/3
Figure 4.8 The block-spring system for Problem 4.3.4.
Problem 4.3.5 Solve the following system of DEs.
= ( +2 +
| (
=2 (+ +
Problem 4.3.6 Solve the following system of DEs.
=
|
= −9 + 6
Problem 4.3.7 Solve the following system of DEs with given I.C..
= −2
S
R 0) = 1
Q (0) = 2
Problem 4.3.8 Solve the following system of DEs with given I.C..
177
Chapter 4 Systems of Linear DEs
S
R
Q
= 3 +4
=3 +2
(0) = 1
(0) = 1
Problem 4.3.9 Solve the following system of DEs.
= +2 +Ù
=3 +2
Z
Ù = + +Ù
178
4.4 Operator Method
4.4
Operator Method
In this method we introduce a new operator
Properties of Operators
Â≡
(1) Â
=Â +Â
(2) Â(v ) = vÂ
(3) Â(v + œ ) = v + œÂ
where v and œ are constants.
Example 1
Solve the following system of ODEs by Operator Method. (This problem
was solved earlier in the Substitution Method).
( = (−8
Z = − (+3
(0) = 2
( (0) =
Solution
Let us rewrite the DEs by using the operator
 = (−8
| (
 =− (+3
That is
 1) ( + 8 = 0
|
=0
( + (Â − 3)
Solve DEs (A) and (B) by A
 1) × (B).
$ Â 3)(Â − 1) − 8% = 0
That is
4Â − 5) = 0
Â
whose C-Eq is
Y
4Y − 5 = 0
Y( = 5, Y = −1
Therefore, we solve
= 1( exp(5 ) + 1 exp(− )
Substituting into (B) to solve for ( , we get
(A)
(B)
179
Chapter 4 Systems of Linear DEs
= −21( exp(5 ) + 41 exp(− )
After solving for the I.C. you will have 1( = 1 and 1 = 2.
Finally, we can get the answer
= −2 exp(5 ) + 4 exp(− )
| (
= exp(5 ) + exp(− )
(
Example 2
Let us solve Example 2 of the Substitution Method using the Operator
Method
′=4 −3
′=6 −7
€
0 =2
0 = −1
Solution
We can rewrite the system by using the operator
 =4 −3
|
 =6 −7
Thus, two DEs can be written as
 4) + 3 = 0
|
6 − (Â + 7) = 0
Operating on the two DE by $3 × (ž)% + $(Â + 7) × (A)%, we can get
$ Â
whose C-Eq is
7)(Â − 4) + 18% = (Â + 3Â − 10) = 0
3Y − 10 = 0
Y( = 2 and Y = −5
Y
= 1( exp(2 ) + 1 exp(−5 )
Back-substitute in Eq-(A) to solve for .
2
= 1( exp(2 ) + 31 exp(−5 )
3
The G.S. for the system
= 1( exp(2 ) + 1 exp(−5 )
2
•
= 1( exp(2 ) + 31 exp(−5 )
3
Plugging the I.C., we have
0) = 1( + 1 = 2
2
•
(0) = 1( + 31 = −1
3
This gives
180
(A)
(B)
4.4 Operator Method
1 =3
| (
1 = −1
And the final solution for the problem is
= 3 exp(2 ) − exp(−5 )
|
= 2 exp(2 ) − 3 exp(−5 )
This gives
One application of solving system of DEs is that one can consider
solving DEs of higher order by converting them into a system of 1storder DEs.
Given a function
Let
,
,
, , )=0
(
Thus, the DE becomes
( ,
=
=
=
=
,
,
(,
)=0
Coupling with the dependent variable introduced earlier, we can form a
system of 1st-order DEs
S
R
Q
( ,
,
(
,
=
=
=
(,
)=0
Solving this system of DEs can help solve the original DE of higher
order.
181
Chapter 4 Systems of Linear DEs
Problems
Problem 4.4.1 Find the G.S. of the following system of DEs.
= +2 +Ù
Z ′=6 −
Ù′ = − − 2 − Ù
Problem 4.4.2 Find the G.S. of the following system of DEs.
 +  +  = 2 exp −
|
 −1 +  − =0
Problem 4.4.3 Find the G.S. of the following system of DEs.
′=3 +9
Z ′=2 +2
0 = 0 =2
Problem 4.4.4 Find the G.S. of the following system of DEs.
= ( +2 +
| (
=2 (+ +
Problem 4.4.5 Find the G.S. of the following system of DEs.
′ = + Ù + exp −
Z = +Ù
Ù = +
Problem 4.4.6 Solve the following system of DEs by (1) substitution method
and (2) operation method.
= +
|
=6 −
Problem 4.4.7 Use operational determinant method to solve the following
system of DEs.
 + 2 +  − 3 = exp −2
|
 − 2 +  + 3 = exp 3
182
4.4 Operator Method
Problem 4.4.8 Find the G.S. of the following Des system.
13 − 4 = 6 sin
|
2 −
+9 =0
Problem 4.4.9 Solve the following system of DEs by (1) substitution method
and (2) operator method.
′=3 −2
|
=2 +
183
Chapter 4 Systems of Linear DEs
4.5
Eigen-Analysis Method
All systems of linear DEs can be expressed in terms of matrices and
vectors, which are widely used in linear algebra and multivariable
calculus.
For example, we can express a system of 3 1st-order DEs below
(
=
((
(
(
(
(
(
by a general (dimensionally implicit) notation
= # + M( )
where
=
(
,
#=
((
(
(
(
(
As in the discussions of single DEs, if
0
M( ) = 0 = $
0
,
((
)
(
)"
+!
( )
(( )
M( ) = ! ( )"
( )
the system is said to be homogeneous. O.W., it is inhomogeneous.
Similar to the solution of single dimension DEs, homogeneous systems
are much simpler to solve than the inhomogeneous system with the nonzero inhomogeneous term.
For a homogeneous system
=#
we may assume a trial solution in the form of
= T exp( )
184
4.5 Eigen-Analysis Method
where T is a vector and
matrix #
is a scalar and both are to be determined by the
′ = T exp
Plugging it into the above original homogeneous system, we get
or
T exp
= #T exp
# − % T exp
=$
Since exp
≠ 0, we must have # − % T = $. Considering T = 0
gives only trivial solution, we are only interested in the case where
det # − % = 0
That is, is the eigenvalue and Tis the associated eigenvector of the
matrix #. The solution of the homogeneous system can be written as
= ê ë T& exp
ë
Now, for three different cases of the eigenvalues, we have three different
solution methods.
Case I:
Non-repeated real eigenvalues. The above linear
combination for composing the G.S. holds.
Case II:
Non-repeated complex eigenvalues. The above linear
combination for composing the G.S. holds except that the term
involving exp ë or the eigenvector '& or both may be complex.
Case III:
Repeated real eigenvalues. Composing the second or
third or more solutions is a much trickier process which is
demonstrated by the following Example 2.
For inhomogeneous systems, we need to find a P.S. and then add it to the
solution of the homogeneous portion of the system. To find the P.S., we
185
Chapter 4 Systems of Linear DEs
can use the rule-based trial method as well as the variation of parameters
method.
Example 1
Let us solve Example 2 of the Substitution method using the EigenAnalysis method
Solution
Rewrite the system of DEs in the matrix notation
=
(
The coefficient matrix is
#=
4
6
4
6
Its eigenvalues can be found by solving
which are
det # − % = det
(
= 2,
and the associated eigenvectors are
T( =
Therefore, the solution is
(
=
(
3
,
2
−3
−7
−3
−7
4−
6
(
−3
−7 −
=0
= −5
T =
3
exp(2 ) +
2
1
3
1
exp(−5 )
3
Example 2
Solve the following system using the Eigen-Analysis method
(
=
1 −3
3 7
Solution
The system has two identical eigenvalues
186
(
4.5 Eigen-Analysis Method
(
=4=
=4
(
= T( exp( )
We can find the associated eigenvector T( for the eigenvalue
solution is
(
and its
For the second solution, we need to compose the L.I. solution by
introducing
= (T( + T ) exp( )
We plug this proposed solution to the system. We get
so
= (T( ) exp( ) + (T( + T ) exp( )
= (T( ) exp( ) + (T( + T ) exp( ) = #(T( + T ) exp( )
Additionally, since A
#
#
% T exp( ) = T( exp( )
#
% T = T(
% T( = 0, we have
% #
% T = (# − %)T( = 0
#
% T =0
Now, we can find T with which we can compose the second solution and
then the entire solution.
Now, we return to our original example,
#
% T =
−3 −3 −3
3
3
3
−3
0 0
T =
T =0
3
0 0
which means any solution T will satisfy the above equation. For
convenience, we may select
0
T =š ›
1
with which we construct the eigenvector T(
T( = (# − %)T =
One can also solve
−3
3
−3 0
−3
=
3 1
3
187
Chapter 4 Systems of Linear DEs
3
3
−3
T =0
3 (
=
−3
exp(4 )
3
to find the eigenvector T(.
Finally, we have the two L.I. solutions
=š
−3
3
+
(
−3
0
› exp(4 ) =
exp(4 )
3 +1
1
So, finally, the G.S. for the original DE is
=
( (
+
=
(
−3
exp(4 )
3 +1
−3
exp(4 ) +
3
Example 3
Solve the following system of DEs.
′=3 +4
|
′=3 +2
Solution
The original system can be written as
#=
= # where
3 4
3 2
3−
4
det # − % = det Ý
Ý=0
3
2−
= 6.
with ( = −1,
Next, we find the associated eigenvectors.
For ( = −1, we get
3 − (−1)
4
(
0
=
3
2 − (−1) L(
0
Thus, ( L( = 0, we may select
(
1
L( = −1
So, we get
1
exp(− )
( =
−1
We can easily find its eigenvalues by
For
= 6, we get
Thus, 3
188
3−6
4
3
2−6 L
− 4L = 0, we may select
=
0
0
4.5 Eigen-Analysis Method
L
So, we get
Finally, the G.S. is
=
=
(
=
4
3
4
exp(6 )
3
1
exp(− ) +
−1
4
exp(6 )
3
Example 4
Solve the following system of DEs.
(
1 2 (
=
+
2 1
Solution
First, we try to solve the homogeneous DEs.
(
1 2 (
=
2 1
It can be written as
′=#
where
1 2
#=
2 1
1−
2
det # − % = det Ý
Ý= 1−
−4=0
2
1−
So the eigenvalues are ( = 3,
= −1
For ( = 3, we get
(
0
2 2
=š ›
0
2 −2 L(
We can select
(
1
L( = 1
For ( = −1, we get
0
2 2
=
0
2 2 L
We can select
−1
L = 1
So the G.S. to homogeneous DEs is
(
1
−1
=
= (
exp(3 ) +
exp(− )
1
1
The P.S. may take the following form
A
A
A
(
=
+ ( + B
l =
ž
ž(
žB
189
Chapter 4 Systems of Linear DEs
2A
+
2ž
Plugging them to the original system, we get
2A
A
1 2 A
+ ( =
E
+
2ž
ž(
2 1 ž
43
SAB = −
27
38
žB =
27
11
A( =
9
Rž = − 16
(
9
2
A =−
3
1
Q ž =3
Then the G.S. to the original DEs is
(
=
=
(
(
=
A(
ž(
A(
ž(
+
1 −2
1
−1
exp(3 ) +
exp(− ) +
1
1
3 1
1 −43
+
27 38
Example 5
Find the G.S. of the following.
5
( ) = −6
Ù
6
5
2
−6 −5 ( )
6
5 Ù
Solution
The original DEs can be written as
 5
−5
−2
( )=0
6
Â+6
5
−6
−6
Â−5 Ù
190
AB
F+
žB
+
1 11
9 −16
4.5 Eigen-Analysis Method
Â
6
−6
5
−5
−2
6
Â+6
5
⇒ Â−5
Â+6
5
−5
−2
−6
Â−5
−6
−6
Â−5
6
Â+6
5
⇒ 6(Â − 5) −30 −12
0
Â
Â
6
Â+6
5
⇒ 0 −30 + (Â + 6)(5 − Â) −12 + 5(5 − Â)
0
Â
Â
6
Â+6
5
⇒ 0 Â(−Â − 1) 13 − 5Â
0
Â
Â
6
Â+6
5
Â
Â
⇒ 0
0 Â(−Â − 1) 13 − 5Â
6 Â+6
5
Â
Â
⇒ 0
0
0
Â(Â + 1) + 13 − 5Â
6 Â+6
5
⇒ 0
Â
Â
0
0
 − 4 + 13
Â
So we can have
4Â + 13)Ù( ) = 0
Y
4Y + 13 = 0
Y( = 2 + 3r,
Y = 2 − 3r
C-Eq:
Ù
= ( exp(2 ) cos(3 ) + exp(2 ) sin(3 )
Ù′
= 2 ( + 3 exp 2 cos 3 + 2 − 3 ( exp 2 sin 3
From the second row of the matrix, we can have
Ù =0
= −Ù′
= (−2 ( − 3 ) exp(2 ) cos(3 ) + (−2 + 3 ( ) exp(2 ) sin(3 )
So
So
=
( )
=−
Ù( )
= −Ù( ) +
= − ( exp(2 ) cos(3 ) − exp(2 ) sin(3 ) +
From the first row of the matrix, we can have
6 + ′ + 6 + 5Ù = 0
So
191
Chapter 4 Systems of Linear DEs
− ( ) − 6 ( ) − 5Ù( )
6
− (
(+
exp(2 ) cos(3 ) +
exp(2 ) sin(3 ) −
=
2
2
Then the solution is
− (
(+
=
exp(2 ) cos(3 ) +
exp(2 ) sin(3 ) −
2
2
= − ( exp(2 ) cos(3 ) − exp(2 ) sin(3 ) +
Ù
= ( exp(2 ) cos(3 ) + exp(2 ) sin(3 )
=
Problems
Problem 4.5.1 Solve the following system of DEs by (1) substitution method
and (2) operator method.
′=3 −2
|
=2 +
Problem 4.5.2 Solve the following system using Eigen-Analysis method.
1 −3
′
=
3 7
Problem 4.5.3 Solve the following system using Eigen-Analysis method.
3 exp
1 2
=
+
2 1
−
Problem 4.5.4 Solve the following system using Eigen-Analysis method.
1 3
1
=
+
exp −2
1 −1
0
Problem 4.5.5 Solve the following system using Eigen-Analysis method.
1 2
−1
=
+
4 −1
1
Problem 4.5.6 Solve the following system using Eigen-Analysis method.
′−4 +2 =0
Z ′ + 4 − 4 + 2Ù = 0
Ù′ + 4 − 4Ù = 0
192
Chapter 5
Laplace Transforms
5.1
Laplace Transforms
Why do we learn Laplace Transforms for a course with focus on ODEs?
In Chapters 3 and 4, we discussed how to solve linear DEs of systems of
them. In certain cases, the methods in Chapter 3 can be quite tedious and
such methods as Laplace transform method can provide much
convenience. The Differentiation Operator  can be viewed as a
transform which, when applied to the function
, yields the new
function Â{
}=
. Similarly, the Laplace transformation ℒ
involves the operation of integration and yields a new function
} = M ³ of a new independent variable. This new function
ℒ{
usually can be manipulated much more easily. This is the key reason that
the Laplace Transform is introduced here.
193
Chapter 5 Laplace Transforms
After learning how to compute the Laplace transform M ³ of a
function
, we will learn how the Laplace transform converts a DE
with the unknown function
and its derivatives into an Algebraic
Equation in M ³ . Because algebraic equations are usually easier to solve
than DEs, this method simplifies the problem of finding the
solution
.
Definition
Given a function
∀ ≥ 0, the Laplace Transform of the function
( ) is another function M(³) defined as follows.
M(³) = ℒ{ ( )} =
•
B
exp(−³ ) ( )
for all values of ³ for which the improper integral converges.
194
5.2 Properties of Laplace Transforms
5.2
Properties of Laplace Transforms
Computing Laplace transforms requires understanding of the key
properties of the transforms. In the following, we present a few such
properties.
The first and the most important property is that of linearity.
Linearity Property
ℒ{
and
}=
ℒ{-( )} =
•
( ) exp(−³ )
= M(³)
-( ) exp(−³ )
= ‡(³)
B
•
B
The linearity of the transform states that,
ℒ{v ( ) + œ-( )} = vM(³) + œ‡(³)
where v and œ are constants.
Proof
LHS = ℒ{v ( ) + œ-( )}
=
=
•
$v ( ) + œ-( )% exp(−³ )
B
•
B
v ( ) exp(−³ )
= vM(³) + œ‡(³)
= RHS
+
•
B
œ-( ) exp(−³ )
195
Chapter 5 Laplace Transforms
5.2.1
Laplace Transforms for Polynomials
Polynomials are a class of functions of form
k
=
B
+
(
+
+ ⋯+
From the linearity, we know the Laplace transform h (³) = ℒ{k ( )}
has form
h (³) =
B ℒ{1} +
} + ⋯+
( ℒ{
ℒ{ }
Now let us compute ℒ{ }. We start from the simplest case ℒ{1}
ℒ{1} =
=
•
1 × exp(−³ )
B
•
B
exp(−³ )
•
1
= − exp(−³ )_
³
B
1
1
= − exp(−³ )_
+ exp(−³ )_
³
³
→•
1
= lim exp(−³ )
→B ³
→B
It is obvious that the integral diverges if ³ ≤ 0, resulting in meaningless
transform. We only consider ³ > 0 for which we have
ℒ{1} =
1
³
Thus, the Laplace transform of 1 is 1⁄³.
Next, we consider ℒ{ }. Starting from the definition,
ℒ{ } =
•
B
Using integration by parts, we have
196
exp(−³ )
5.2 Properties of Laplace Transforms
1 •
(exp(−³ ))
³ B
•
1
1
= − exp(−³ )_ − \−
³
³
B
•
1
= − \−
exp(−³ ) ]
³
B
1
= …{1}
³
ℒ{ } = −
From the previous transform ℒ{1} =
ℒ{ } =
(
q
•
B
exp(−³ )
]
∀³ > 0, we then have
1
∀³ >0
³
Alternatively, we can derive this directly by taking derivative of
ℒ{1} =
That is
LHS =
³
=−
ℒ{1} =
•
B
³
³
•
B
exp(−³ )
ℒ{1} =
1
E F
³ ³
exp(−³ )
RHS =
By equating both sides, we get
1
³
= −…{ }
=
•
B
¬
(exp(−³ ))
¬³
1
1
E F=−
³ ³
³
−ℒ{ } = −
Or finally, we get the intended transform
1
³
197
Chapter 5 Laplace Transforms
1
³
ℒ{ } =
Iteratively using this method, we can easily derive
•
B
That is
³
1
E F
³ ³
ℒ{ } =
¬
[ exp(−³ )]
¬³
•
B
exp(−³ )
ℒ{ } =
2
³
⋯⋯
=−
2
³
=
2
³
∀³ > 0
It is easy to derive the general formula for a power function (and
naturally for a polynomial)
ℒ{ } =
³
=!
ú(
∀³ > 0
which can be proved by induction as follows.
Proof
For = = 0, we have already discussed.
Suppose the formula is true for =. This means
=!
ℒ{ } = ú( ∀³ > 0
³
from which we have
³
198
ℒ{ } =
E
³ ³
=!
ú(
F
5.2 Properties of Laplace Transforms
•
B
That is
¬
(
¬³
•
B
ℒ{
ú(
which finishes the proof.
=! ⋅ (= + 1)
³ ú
(= + 1)!
=
³ ú
exp(−³ ))
=−
exp(−³ )
ú( }
=
(= + 1)!
∀³ > 0
³ ú
Example
If @ v = ℒ{ - }, then prove
³
$@ v % = −@(v + 1)
Solution
To prove this, we simply derive the formula directly
LHS =
³
³
ℒ{ - }
•
B
•
B
¬
¬³
•
B
ℒ{
5.2.2
Given ℒ{
-
exp(−³ )
-
exp(−³ ))
-ú( exp(−³
-ú( }
= RHS
)
The Translator Property
} = M(³), compute ℒ{ ( ) exp( )}.
ℒ{ ( ) exp( )} =
•
B
( ) exp(
−³ )
199
Chapter 5 Laplace Transforms
=
By substitution
B
= ³ − , we have
ℒ{ ( ) exp
That is
Similarly, we can get
•
}=
( ) exp$
•
B
³ %
( ) exp
= M( )
ℒ{ ( ) exp
ℒ{ ( ) exp
Combining the two, we have
ℒ{ ( ) exp p
} = M(³ − )
} = M(³ + )
} = M(³ ∓ )
This formula is the translator property of Laplace transform.
}
Example 1
Compute ℒ{exp
Solution
= 1, we know
ℒ{exp
} = ℒ{exp
1
,
³
Thus, by the translator property, we have
Let
M ³ = ℒ{ ( )} =
ℒ{exp
Example 2
Compute ℒ{ exp
}
Solution
Let
= , we have
200
} = M(³ − ) =
³
Ê 1}
1
³G0
,
³G
5.2 Properties of Laplace Transforms
M ³ = ℒ{ ( )} = ℒ{ } =
By the translator property, we have
} = M(³ − ) =
ℒ{ exp
Example 3
Compute ℒ{
Solution
Let
=
³
1
,
³
1
, we have
M ³ = ℒ{ ( )} = ℒ{ } =
By the translator property, we have
} = M(³ − ) =
exp
In general, we can prove
ℒ{
,
³G
}
exp
ℒ{
³G
}=
=!
(³ − )
ℒ{exp r
}=
exp
³
ú(
2
,
³
2
,
,
³G
³G
³G
From the above Example 1, we can generalize the formula to complex
plane
ℒ{exp
Example 4
Compute ℒ{cos
} and ℒ{sin
Solution
From Euler’s formula, we have
exp r
|
exp r
we have
cos
}=
r
=
}
³
³
= cos
= cos
exp r
1
1
r
r
r sin
r sin
exp
2
r
201
Chapter 5 Laplace Transforms
exp r
/
2
1
1
}
}
ℒ{exp r
ℒ{exp r
2
2
1
1
1
1
E
F
E
F
2 ³ r
2 ³ r
1
1
1
E
F
2 ³ r
³ r
³
³
Therefore, we have obtained one of the most important transforms,
³
ℒ{cos } =
³ +
Similarly, for sin , we have
exp r
exp r
sin
=
2r
and thus
1
} ℒ{exp r
}
ℒ{sin } =
ℒ{exp r
2r
1
1
1
E
F
³ r
2r ³ r
Thus,
ℒ{cos
exp r
} =ℒô
³
Therefore, we have obtained another one of the most important
transforms,
ℒ{sin
Example 5
Compute ℒ{cosh
202
³ +
}
Solution
From definition, we know that
Thus
}=
cosh
ℒ{cosh
=
exp
exp
} = ℒô
1
1
E
2 ³
exp
2
³
exp
2
1
F
/
5.2 Properties of Laplace Transforms
=
³
³ −
Similarly, we can have
ℒ{sinh
Example 6
Compute ℒ{exp
}
Solution
Let
= cos
cos
}=
³ −
. From previous discussion, we know that
³
M ³ = ℒ{cos } =
³ +
Therefore, by the translator property, we know that
³−
ℒ{exp
cos } = M(³ − ) =
(³ − ) +
Example 7
Compute ℒ{3 exp 2
Solution
5.2.3
ℒ{3 exp 2
2 sin 3 }
2 sin 3 } = ℒ{3 exp 2
1
3
³ 2 ³ ³
3³
144³
³ ³ 2 ³
1 cos 6 }
³
36
72
36
Shifting Property
Given a unit step function defined as follows and has a graph given
below. (Figure 5.1)
=m
0,
1,
I0
≥0
203
Chapter 5 Laplace Transforms
1
0
Figure 5.1 The graph of the unit step function
.
Since
= 1 for ≥ 0 and because the Laplace transform involves
only the values of a function for ≥ 0 we see immediately that
1
ℒ{ ( )} = ,
³
³≥0
Furthermore, for any function ( ), M(³) = ℒ{ ( )}, we know that
ℒ{ ( ) ( )} = M(³)
The graph of a unit step function w ( ) = ( − ) appears in the
following figure. Its jump occurs at = rather than at = 0. So we
have
w(
)= ( − )=m
0,
1,
<
≥
1
0
Figure 5.2 The graph of the unit step function
204
.
5.2 Properties of Laplace Transforms
w
To calculate the Laplace transform of
ℒ{
} = ℒ{ ( − )}
w
=
=
•
B
w
B
=−
=
( − ) exp
0 ⋅ exp
exp
exp
³
³
³
³
w
}=
³
³
•
,
x
exp
³
exp
w
•
Üw
Thus, we get one important transform
ℒ{
, we start from the definition.
³ G 0,
³
G0
∀³,
G0
Example 1
Suppose M ³ = ℒ{ ( )}, compute ℒ{
}
Solution
From the definition of Laplace, we have
•
M ³ = ℒ{ ( )} =
On the other hand, we have
}=
ℒ{
•
B
w
B
Let 0 = − , we have
ℒ{
0
exp
³
( ) exp
B
³
( − ) ( − ) exp
•
³
} = exp
exp
exp
w
•
w
³
( − ) exp
•
B
³
³
exp$ ³
(0) exp
³ ℒ{ 0 }
³ M ³
³0
0
%
205
Chapter 5 Laplace Transforms
Example 2
Compute ℒ{
Solution
Let
= , we know that
}
M ³ = ℒ{ ( )} =
1
³
Using the result of example 1, we have
} = ℒ{ ( − ) ( − )}
ℒ{
exp(− ³) M(³)
exp(− ³)
³
5.2.4
The t-multiplication property
The -multiplication property, also called the frequency differentiation, is
} = M(³) , when taking
another important property. Given ℒ{
derivative WRT ³, we have
³
\
•
( ) exp(−³ )
B
]=
³
M(³)
By exchanging the derivative and integral operation, we have
•
B
−
Thus,
( ( ) exp(−³ ))
•
B
…{ ( )} =
Similarly, we can have
206
³
( ) exp(−³ )
•
B
ℒ{
= M (³)
= M (³)
( ) exp(−³ )
( )} = M (³)
= −M (³)
5.2 Properties of Laplace Transforms
Generally,
} = (−1 M
ℒ{
Example 1
Compute ℒ{ exp
³
}
Solution
Method I: By -multiplication property
Let
= exp
, we know from shifting property,
1
}=
M ³ = ℒ{exp
³
By applying the -pultiplication property, we have
} = −M (³)
ℒ{
1
E
F
³
1
³
Method II: By shifting property
Let
= , we know that
1
M ³ = ℒ{ } =
³
By applying the shifting property, we have
} = M(³ − )
ℒ{exp
1
³
Method III: By differentiation property
Consider
= exp
and denote M ³ = ℒ{ ( )}, we know that
= exp
exp
Perform Laplace transform on both sides, we have
} = ℒ{exp
}
ℒ{
}
ℒ{
By applying differentiation property, we have
1
³M ³
0
M ³
³
Solving this algebraic equation, we can get
1
0
M ³ =
³
³
Since 0
0, we final have
1
M ³ =
³
207
Chapter 5 Laplace Transforms
Example 2
Compute …{ sin
}
Solution
Method I: By -multiplication property
Let
= sin , then we know
M ³ = ℒ{ ( )} =
³ +
Applying the -multiplication property, we have
} = −M (³)
ℒ{
š
That is
ℒ{ sin
³
2 ³
(³ + )
}=
›
2 ³
(³ + )
Method II: By differentiation property
Let
= sin
and denote M ³ = ℒ{ ( )}. By taking derivative, we
have
= sin
+
cos
and
= 2 cos
−
sin
, we can get
Perform Laplace transform on
} = 2 …{cos } − ℒ{ sin }
ℒ{
Applying the differentiation property, this becomes
2 ³
³ 0) − (0) =
− M(³)
³ M ³
³ +
It is easy to see that
0) = (0) = 0. Thus, we can get
2 ³
M ³ =
(³ + )
Remarks
Similarly, we will have,
208
ℒ{
sin
}=
³
š
³ +
6 ³ −2
(³ + )
›
5.2 Properties of Laplace Transforms
Example 3
Compute ℒ{ cos
} and ℒ{
Solution
Similar to example 2, we have
cos
³
³ +
³
2³
³
} = −š
ℒ{ cos
and
5.2.5
ℒ{
}
›
³
›
³ +
2³
6³
³
}=š
cos
Periodic Functions
with period @ that can be defined as
For a periodic function
= ( + @) ∀ ≥ 0
The Laplace transform can be obtained by the original definition
ℒ{
}=
=
=
=
•
B
C
B
C
B
C
B
( ) exp
( ) exp
C
C
( ) exp
C
B
( ) exp
³
³
( ) exp
³
( + 2@
³
³
C
C
B
1
C
( ) exp
…
³
( + @) exp$ ³
exp$ ³
exp
³@
2@ %
exp
…
@ %
³2@
⋯
209
Chapter 5 Laplace Transforms
=
1
1 − exp(−³@)
C
( ) exp(−³ )
B
Therefore, for periodic functions, one only needs to compute the
transform for the first period and then multiply the transform by a
period-dependent factor given above.
5.2.6
Differentiation and Integration Property
In order to utilize Laplace transforms to solve DEs, we need to be able to
perform Laplace transforms on differentiation and integration.
Let us first consider ℒ{ ( )} , where M(³) = ℒ{ ( )} . By using
integration by part, we have
ℒ{ ( )} =
•
B
( ) exp(−³ )
= exp(−³ ) ( )|•
B −
•
B
( ) (exp(−³ ))
= exp(−³ × ∞) (∞) − exp(−³ × 0) (0)
since M(³) = ´B
•
+³
•
B
( ) exp(−³ )
( ) exp(−³ )
, we have
ℒ{ ( )} = ³M(³) − (0)
Starting from this result, we can derive the Laplace transform for higher
order derivatives.
For ℒ{
210
( )}, we have
ℒ{
( )} = ℒ m$ ( )% 1
= ³ℒ{ ( )} − (0)
= ³[³M(³) − (0)] −
(0)
5.2 Properties of Laplace Transforms
Thus
Similarly,
ℒ{
ℒ{
} = ³ M(³) − ³ (0
} = ³ℒ{ ( )} − (0
³ M ³
³
0
³
In the same manner we can generalize for
ℒ2
3 = ³ M(³) − ³
(0
>(
Laplace transform for integration, e.g. ℒ m´B
0
⋯
complicated.
Consider
and we denote
and
=
B
0
0
>(
0
0
01, is a little more
(0) 0
‡(³) = ℒ{-( )}
M(³) = ℒ{ ( )}
From previous discussion, we know that
Since This gives
ℒ{- ( )} = ³‡(³) − -(0
= ( ), this means
M ³
³‡(³) − -(0
‡ ³
M ³
³
- 0
211
Chapter 5 Laplace Transforms
It is easy to calculate
B
- 0) =
Hence, we have
ℒô
where
Similarly, we have
and
B
(0) 0 = 0
B
(0) 0/ =
M(³)
³
M(³) = ℒ{ ( )}
ℒ |4 ( ( )
ℒ |6 ( ( )
(
5=
M(³)
³
5=
(
M(³)
³
Problems
Problem 5.2.1 The graph of a square wave function -( ) is shown in Figure
5.3. Express -( ) in terms of the function
•
and hence deduce that
( ) = ê(−1)
ℒ{-( )} =
212
ÜB
( − =)
1 − exp(−³)
1
³
= tanh
³(1 + exp(−³)) ³
2
5.2 Properties of Laplace Transforms
1
0
1
2
3
4
5
−1
Figure 5.3 The square wave function for Problem 5.2.1.
Problem 5.2.2 Compute the Laplace Transform
exp(v ) sin(
ℒ|
))5
Problem 5.2.3 Compute the Laplace Transform
ℒ •exp( )
√
B
exp(−0 ) 0 7
Problem 5.2.4 Compute the Laplace Transform of the following function.
)
( ) = sin( ( ) + exp(v ) cos(
Problem 5.2.5 Compute the Laplace Transforms of the following un-related
functions.
1
∞
( ) = ê ( − =)
( )= −8 9
2
In both cases above, > 0 and 8 9 is the floor function of and ( − =) is the
==0
usual step function.
Problem 5.2.6 Laplace transform is defined as
213
Chapter 5 Laplace Transforms
M ³ =
•
( ) exp(−³ )
B
Performing Laplace transform on the following functions:
(1) ( ) = 9
(2) ( ) = exp( )
(3) ( ) = sin( )
(4) ( ) = exp( ) sin( )
Problem 5.2.7 Compute the Laplace transform of the following triangular
wave function shown in Figure 5.4.
Hint: Obviously, the function during the 1st peorid can be written as
2A E F ,
@
( )=€
2A E1 − F ,
@
( )
0≤ ≤
@
2
@
≤ ≤@
2
A
−2@
−@
0
@
Figure 5.4 The triangular wave function for Problem 5.2.7.
214
2@
5.3 Inverse Laplace Transforms
5.3
Inverse Laplace Transforms
We introduce the inverse Laplace transform
ℒ >( {M ³ } = ( )
where
M(³) = ℒ{ ( )}
Theoretically, we have to calculate the Bromwich integral (also called
the Fourier-Mellin integral) to find the inverse Laplace transform.
( ) = ℒ >( {M(³)} =
1
28r
:úë⋅•
:>ë⋅•
exp(³ ) M(³) ³
where µ > X(³¢ ) for every singularity ³¢ of M(³).
To actually calculate this integral, one needs to use the Cauchy residual
theorem. However, for most cases, we can find out the inverse Laplace
transform by simply looping up in the Laplace transform table.
Example 1
Compute
Solution
Example 2
Compute
1
1
ℒ >( | −
5
³
³−
1
1
1
1
ℒ >( | −
5 = ℒ >( | 5 − ℒ >( |
5
³
³−
³
³−
exp( )
1
ℒ >( |tan>( 5
³
215
Chapter 5 Laplace Transforms
Notice that the derivative of tan>( is a simple rational function
q
(
Solution
1
1
tan>( = −
³ +1
³
³
We can find the inverse Laplace transform by applying the t-multiplication
property.
Denote
1
M ³ = tan>( and ³ = …>({M(³)}
³
Since
} = −M (³)
ℒ{
We know that
1
= − ℒ >( {M (³)}
1
sin
Example 3
Compute
ℒ >( |−
1
ℒ >( |
5
³ (³ − )
Solution
Method I: Using the integration property
From the integration property, we have
That is
where
Now let
216
1
5
³ +1
ℒô
B
ℒ >( ô
(0) 0/ =
M ³
/=
³
B
M(³)
³
(0) 0
M ³ = ℒ{ ( )}
1
³ (³ − )
1
‡ ³ =
³(³ − )
M ³ =
5.3 Inverse Laplace Transforms
„ ³ =
We know that
and
-
ℎ
= …>( {„(³)}
exp( )
= ℒ >( {‡(³)}
„ ³
/
ℒ >( ô
³
B
B
1
1
Finally, we have
1
³−
ℎ 0
0
exp( 0) 0
exp( 0)_
B
(exp( 0) − 1)
= ℒ >( {M(³)}
‡ ³
ℒ >( ô
/
³
1
(exp( 0) − 1) 0
B
1
(exp( ) −
− 1)
Method II: Using partial fractions
Let
1
A ž
1
= + +
³ (³ − ) ³ ³
³−
We have
1 = (A³ + ž)(³ − ) + 1³
ž A ³ ž
A 1 ³
This gives
1
1
1
A = − , ž = − and 1 =
which means
1
1 1 1 1
1
1
=− ⋅ − ⋅ + ⋅
³ (³ − )
³
³
³−
217
Chapter 5 Laplace Transforms
By linearity, we have
1
1
1
1
1
1
1
ℒ >( |
5 = − ℒ >( | 5 − ℒ >( | 5 + ℒ >( |
5
³ (³ − )
³
³
³−
exp( )
1
− +
exp( ) −
−1
Problems
1
1
(sin / − / cos / )
ℒ >( |
5=
(³ + à )
2/
Problem 5.3.1 Prove that
If given
³ −/
(³ + / )
/
ℒ{sin / } =
³ +/
ℒ{/ cos / } =
1
exp š− ›
³ Ä = ; $2 %
Z
B √
³
Problem 5.3.2 Show the relationship
ℒ
>(
where ;B ( ) is the so-called Bessel’s function defined as
∞
;B $2√ % = ê
ÜB
(−1)
$√ %
2 ⋅ =!
Problem 5.3.3 Perform the following inverse transform.
2³
ℒ >( |
5
(³ − 1)
Problem 5.3.4 Perform the following inverse transform.
exp(− ³)
ℒ >( ô
/
³(1 − exp(− ³))
218
5.3 Inverse Laplace Transforms
Problem 5.3.5 The following equation is one of the so-called Volterra Integral
Equations containing obviously the unknown function
:
= sin( ) + 2
B
cos( − 0) (0) 0
Use Laplace Transform to solve this equation by completing the following
steps.
(1) Use Laplace Transform to convert the above equation to an algebraic
equation containing unknown function in Laplace space (³) with given
formula
³
(³)
ℒ ô cos( − 0) (0) 0/ =
³+1
B
(2) Solve the algebraic equation for (³).
(3) Inverse transform to obtain the unknown function ( ) as a function of t.
Problem 5.3.6 Solve the following Volterra Integral Equations.
( ) = 2 exp(3 ) −
B
exp$2( − 0)% (0) 0
219
Chapter 5 Laplace Transforms
5.4
The Convolution of Two Functions
and -
Consider two functions
⨂-
=
. Let us define a binary operation
B
(0)-( − 0) 0
which is called the convolution of the functions ( ) and -( ). This
operation has elegant yet useful property under Laplace transform. This
property is called Convolution Theorem.
Convolution Theorem
Denoting M(³) = ℒ{ ( )} and ‡(³) = ℒ{-( )}, we have
Proof
ℒ{ ( )⨂-( )} = ℒ{ ( )}ℒ{-( )} = M(³)‡(³)
( )⨂-( ) =
=
=
(0)-( − 0) 0
B
(0)-( − 0) 0 +
B
•
B
•
0 × (0)-( − 0) 0
( − 0) (0)-( − 0) 0
where ( − 0) is the unit step function. Thus, we have
ℒ{ ( )⨂-( )} =
=
•
B
•
B
Introducing a new variable
ℒ{ ( )⨂-( )} =
220
•
B
•
exp(−³ )
•
B
(
•
B
B
( − 0) (0)-( − 0) 0
exp(−³ ) ( − 0) (0)-( − 0)
= − 0 to replace , we have
exp[−³(
(
+ 0)] ( ( ) (0)-( ( ) 0
0
(
5.4 The Convolution of Two Functions
=
=
Conversely,
•
exp(−³ ( ) ( ( )-( ( )
B
•
B
exp(−³ ( ) -( ( )
= M(³) × ‡(³)
(
•
B
(
exp(−³0) (0) 0
ℒ >( {M(³) × ‡(³)} = ( )⨂-( )
Thus we can find the inverse transform of the usual product of two
functions M(³)‡(³) provided we can evaluate the convolution of ( )
and -( ).
Example 1
Compute cos )⨂(sin )
Solution
cos ⨂ sin =
B
cos( − 0) sin 0 0
Applying the trigonometric identity
sin(A ± ž) = sin A cos ž ± cos A sin ž
These can be written as
1
cos A sin ž = [sin(A + ž) − sin(A − ž)]
2
Thus,
1
(sin − sin( − 20)) 0
cos ⨂ sin =
2 B
1
1
E0 sin − cos( − 20)F_
2
2
=ÜB
1
sin
2
So,
1
sin
cos ⨂ sin =
2
Naturally,
cos ⨂ sin ≠ cos × sin
221
Chapter 5 Laplace Transforms
Example 2
Compute
1
ℒ >( |
5
³ (³ − )
1
1
1
5 = ℒ >( | ⋅
5
ℒ >( |
³ (³ − )
³ ³−
1
1
ℒ >( | 5 ⨂ℒ >( |
5
³
³−
⨂ exp( )
Solution
B
B
exp( 0) ( − 0) 0
exp( 0) − 0 exp( 0)) 0
B
exp( 0) 0 −
B
0
(exp( 0))
0
1
E exp( 0) − exp( 0) + exp( 0)F_
1
(exp( ) −
− 1)
=ÜB
Consistent with results obtained earlier in Example 3 of the previous
section.
Problems
Problem 5.4.1 Use Convolution Theorem to prove the following. Identify
1
/ = exp( ) erf$√ %
ℒ >( ô
(³ − 1)√³
where ℒ >( denotes an inverse Laplace Transform and erf( ) is the so-called
“error function” defined by
2
exp −
erf
≡
√8 B
(Hint: substitute = √ .)
222
5.4 The Convolution of Two Functions
Problem 5.4.2 Solve the following DE.
= cos(3 ) −
B
exp(0) ( − 0) 0
223
Chapter 5 Laplace Transforms
5.5
Application of Laplace Transforms
Laplace transforms can help solve some of the linear DEs (and some
special nonlinear DEs) much more conveniently. Starting from Section
5.2.4, we demonstrate such values.
First, we must use these key formulas
ℒ2
( )
} = ³ℒ{ ( )} − (0)
ℒ{
(
)}
= ³ ℒ{ ( )} − ³ (0) −
ℒ{
⋯⋯
( )3 = ³ ℒ{ ( )} − ³
>(
= ³ ℒ{ ( )} − ê ³
ëÜ(
(0)
(0) − ⋯ −
>ë (ë>()
(0)
( >()
(0)
Thus, for a linear DE of the following form
ê
ëÜB
ë
(ë)
( ) = >( )
applying Laplace transforms on both sides of the equation gives
ê
ëÜB
ë ℒ2
(ë)
( )3 = ℒ{>( )}
By applying the differentiation property, we can write the equation as
where
(e)
ℒ{ ( )} =
ℒ{>( )} + ∑ëÜB ë ∑ë@Ü( ³ ë>@
∑ëÜB ë ³ ë
(0) are I.C..
(@>()
(0)
Finally, the inverse Laplace transform yields the solution to the original
DE.
224
5.5 Application of Laplace Transforms
=ℒ
Example 1
Compute
>(
…{>( )} + ∑ëÜB ë ∑ë@Ü( ³ ë>@
ô
∑ëÜB ë ³ ë
6 = 0 for
0) = 2 and
(@>()
(0)
/
0) = −1
Solution
Applying Laplace transforms to both sides of the DE, we can get
ℒ{
6 } = …{0}
ℒ{ } …{ } 6…{ } = 0
³
³ 0) − (0)% − $³ (³) − (0)% − 6 (³) = 0
$³
Plug in the I.C., and we can get
³
2³ + 1 − ³ (³) + 2 − 6 (³) = 0
³
³ 6) − 2³ + 3 = 0
³ ³
2³ − 3
³ =
³ −³−6
By partial fraction decomposition we have,
1
3
1
7
+ ⋅
³ = ⋅
5 ³+2 5 ³−3
Thus, we can find the solution of the original DE by inverse Laplace
transform:
= ℒ >( { (³)}
7
1
3
1
+ ⋅
5
ℒ >( | ⋅
5 ³+2 5 ³−3
3
7
exp(−2 ) + exp(3 )
5
5
Example 2
Compute
4 = sin 3 for
0) =
(0) = 0
Solution
Such a problem arises in the motion of a mass and spring system with
external force as shown in Figure 5.5.
Apply Laplace transforms on both sides of the DE
ℒ{
4 } = ℒ{sin 3 }
225
Chapter 5 Laplace Transforms
/
K
Figure 5.5 A block-spring system with an external force
$³
³
³ 0
0 %
Plugging in the I.C., we can get
³ ³
Finding the partial fraction
³ =
³ =
Solving for A, ž, 1 and  gives
A = 1 = 0,
4 =
³
3
4
³
4 ³
A³ + ž
³ +4
ž
3
1³
³
³ =
9
³
3
.
3
9
Â
9
3
3
, and  = −
5
5
1
3
1
3
⋅
⋅
4 5 ³
9
5 ³
3
2
1
3
⋅
⋅
10 ³
2
5 ³
3
Finally, inverse Laplace transform generates the solution for the original
DE
= ℒ >( { (³)}
3
2
1
3
ℒ >( | ⋅
⋅
5
10 ³
2
5 ³
3
2
1 >(
3
3 >(
ℒ |
5
ℒ |
5
³
2
5
³
3
10
3
1
sin 2
sin 3
10
5
Thus, we get
226
³ =
5.5 Application of Laplace Transforms
Example 3
Solve IVP
2 = −6
2
2
40 sin 3
Z ′′ = 2
0
′(0
0
′ 0
0
which can be concluded from the following spring system. (Figure 5.6)
/1 = 4
K1
/2 = 2
K2
Figure 5.6 A system of two blocks connected by springs and an external force on
.
Solution
From the I.C. we can get
} = ³ (³)
ℒ{
} = ³ (³)
ℒ{
Performing Laplace transform on both sides of the original DEs, we have,
2³
³ = −6 ³
2 ³
3
³
³ =2 ³
2 ³
40 E
F
³
9
where we have used
3
…{sin 3 } =
³
9
Resulting two algebraic equations defined in the ³-space:
3 ³
³ =0
³
120
Z
2 ³
³
2 ³ =
³
9
Substituting the first equation
³ = (³ + 3 ³
To the second, we get
120
³ =
9 $ ³
2 ³
3
2%
³
120
1 ³
4 ³
9
³
Plugging this back to the first equation, we have
227
Chapter 5 Laplace Transforms
Next, express
we let
³ and
120(³ + 3)
(³ + 1)(³ + 4)(³ + 9)
³ using partial fractions, for example for
³ =
³
A
ž
1
120
=
+
+
(³ + 1)(³ + 4)(³ + 9) ³ + 1 ³ + 4 ³ + 9
Substitution of ³ = −1 in the above equation gives A = 5. Similarly, we
can get ž = −8 and 1 = 3. Thus,
8
3
5
−
+
³ =
³ +1 ³ +4 ³ +9
2
3
1
−4⋅
+
5⋅
³ +4 ³ +9
³ +1
Inverse Laplace transform produces
= ℒ >( { (³)}
5 sin − 4 sin 2 + sin 3
Similarly,
= 10 sin + 4 sin 2 − 6 sin 3
Example 4
Solve IVP for the Bessel’s equation of order 0.
=0
|
(0) = 1 and
0) = 0
Solution
The I.C. give us
} = ³ (³) − 1
ℒ{
} = ³ (³) − ³
ℒ{
Because and ′′ are each multiplied by , by applying the -multiplication
property, we can get the transformed equation as
³
³
³
³ ³
1) − $ (³)% = 0
³
³
The result of differentiation and simplification is the DE
³
1) (³) + ³ (³) = 0
Since the DE is separable we have,
³
³
=−
³ +1
³
and its G.S. is
1
³ =
√³ + 1
228
5.5 Application of Laplace Transforms
Remarks
(1) Here the constant 1 in the G.S. of
Let ³ = 0, we have
³ is actually not an arbitrary number.
(0) = 1
On the other hand, from definition of Laplace transform,
(0) =
•
B
( )
This means 1 is an normalization factor.
(2) The solution to the original DE, i.e., the inverse Laplace transform for
1
(³) =
√³ + 1
is a function called zero order Bessel function of first kind. A general form
of v order Bessel function is defined as
•
;B ( ) = ê
Example 5
Solve IVP.
¨ÜB
|
(−1)¨
š ›
K! Γ(K + v + 1) 2
¨ú-
4 = 2 exp( )
(0) = (0) = 0
Solution
Take Laplace Transform on both sides of the DE
ℒ{
4 } = ℒ{2 exp( )}
From the I.C., we have
2
³
4) (³) =
³−1
2
1
³ =E
FE
F
³−1 ³ +4
Therefore, we know that the solution to the original DE is
= ℒ >( { (³)}
1
2
FE
F5
ℒ >( |E
³−1 ³ +4
1
2
ℒ >( |
5 ⨂ℒ >( |
5
³−1
³ +4
exp( ) ⨂ sin 2
Since
exp( ) ⨂ sin 2 =
B
exp( − 0) sin 20 0
229
Chapter 5 Laplace Transforms
Let
= exp( )
-=
B
B
exp(−0) sin 20 0
exp(−0) sin 20 0
By applying integration by part repeatedly, we have
-=−
B
sin 20 (exp(−0))
exp(−0) sin 20|B + 2
exp(−0) sin 2 − 2
B
B
exp(−0) cos 20 0
cos 20 (exp(−0))
exp(−0) sin 2 − 2 exp(−0) cos 20 |B + 4
This gives
B
exp(− ) sin 2 − 2 exp(− ) cos 2 + 2 − 4-
and finally
exp(−0) sin 20 0
1
2
2
- = − exp(− ) sin 2 − exp(− ) cos 2 +
5
5
5
= exp(− ) ⋅ 2
1
2
exp(− ) − sin 2 − cos 2
5
5
5
Problems
Problem 5.5.1 Apply the convolution theorem to derive the indicated solution
of the given DE with I.C. 0) = (0) = 0.
1
( − 0)sin20 0
+ 4 = ( ); ( ) =
2 B
Problem 5.5.2 Solve the following integral-DE and you may not leave the
solution at a convolution form.
Z
230
( )+2 ( )−4
B
exp( − 0) ( 0) 0 = sin
(0) = 0
5.5 Application of Laplace Transforms
Problem 5.5.3 Find the G.S. of the following DEs.
′′ = −4 + sin
|
′′ = 4 − 8
Problem 5.5.4 Use Laplace Transform to find the P.S. of the following IVP.
′′′ + ′′ − 6 ′ = 0
0 =0
•
′ 0 = ′′ 0 = 1
′′ + 4 = −5¼ − 3
0 =1
′ 0 =0
where ¼ − is the so-called Delta-function which is defined as
0,
≠
¼ −
=m
∞,
=
with given properties
Problem 5.5.5 Solve the IVP
Z
•
and
•
•
-
>•
¼
¼
−
−
=1
=-
Problem 5.5.6 Solve the following IVP.
′′ + 6 ′ + 8 = −¼
Z
0 =1
′ 0 =0
Problem 5.5.7 Solve the following IVP.
′′ + 2 ′ + = ¼
−¼
Z
0 =0
′ 0 =0
−2
−2
Problem 5.5.8 Solve the following IVP.
231
Chapter 5 Laplace Transforms
S ′′ +
R
Q
•
= ê¼
ÜB
0 =0
′ 0 =0
− 2=
B
Problem 5.5.9 Solve the following IVP.
′′ + 2 ′ + =
− +¼
Z
0 =0
0 =0
−
Problem 5.5.10 Solve the following system of IVP.
= +2
S
′=2 −2
R 0 =1
Q 0 =0
Problem 5.5.11 Solve the following system of IVP.
= −2 ( + + ¼ − 0
S (
= ( − + ¼ − 20
( 0 = ( 0 =0
R
0 =
0 =0
Q
Problem 5.5.12 Solve the following systems of IVP.
+4
+6
⊗ exp
= sin
|
0 =0
Problem 5.5.13 Solve the following system of DEs.
2 1
=
+ exp
1 2
Problem 5.5.14 Use Laplace Transform to find the P.S. of the following
systems.
′′ − 6 + 8 = 2
|
0 =
0 =0
Problem 5.5.15 Solve the following DE.
232
5.5 Application of Laplace Transforms
|
′′ + 2 − 1 ′ − 2 = 0
0 = ′ 0 =0
Problem 5.5.16 Solve the following DE.
′′ + ′ + = 0
• 0 = v = constant
0 =0
Problem 5.5.17 Solve the following DE.
′′ + 2 ′ + =
|
0 = ′ 0 =0
Problem 5.5.18 Use Laplace Transform to find the P.S. of the following IVP.
′′ + 4 ′ + 13 = exp −
Z
0 =0
′ 0 =2
Problem 5.5.19 Use Laplace Transform and another method to solve the
following DE and compare the results.
′′ − ′ − 12 = sin 4 + exp 3
|
0 = 2, ′ 0 = 1
Problem 5.5.20 Solve the following DE (where B , LB and
constants) by: (1) Variation of Parameters; (2) Laplace Transform.
=
′′ +
Z
0 = B
′ 0 = LB
are given
Problem 5.5.21 Solve the following IVP
+ ( = sin
|
0 = 0, 0 = 0
by
(1) Any method of your choice except Laplace Transform.
(2) The method of Laplace Transform.
Problem 5.5.22 Solve the following DE using two different methods:
233
Chapter 5 Laplace Transforms
=1− −1
−1
0 =0
where
− 1 is the so-called Step Function defined as
0, ≤ 1
−1 =m
1, ≥ 1
(1) Laplace Transform method.
(2) Any other method of your choice.
|
′−
Problem 5.5.23 Solve the following IVP
′′ + 4
= $1 −
− 28 %cos 2
Z
0 =0
0 =0
′′
+
= −1 C D
Z
0 =0
0 =0
where C D denotes the greatest integer not exceeding C D , e.g., C0.918D =
0; C1.234D = 1; C1989.64D = 1989. Your final answer may not contain the
convolution operator.
Problem 5.5.24 Solve the following DE by your favorite method:
Problem 5.5.25 The motion of a particle in the plane can be described by
+
= B sin B
|
+
= B cos B
where , B , B are all constants. Initially, the particle is placed at the origin at
rest. Find the trajectories of the particle for
(1) = B
(2) ≠ B
Problem 5.5.26 A block of mass K is attached to a mass-less spring of spring
constant /, and they are placed on a horizontal and perfectly smooth bench.
During the first 0 time, we add a constant force f to the block from left to
(
right. During the second 0 time, the force direction is reversed but its constant
(
magnitude retains. Repeat this process till eternity. Find the displacement of
the block as a function of time. The initial displacement and speed can be set to
zero.
234
5.5 Application of Laplace Transforms
K
Figure 5.7 The block-spring system for Problem 5.5.26 & Problem 5.5.27.
Problem 5.5.27 A block is attached to a mass-less spring of spring and they
are placed on a horizontal and perfectly smooth bench. We add a force
=
cos 2 to the block during ∈ [0, 28] and remove it at all other times (The
spring is still there). The equation of motion of the block is
′′ + 4 =
|
0 = ′ 0 =0
Solve the equation.
Problem 5.5.28 A block of mass m is attached to two mass-less springs of
spring constants /( and / and they are placed on a horizontal and perfectly
smooth bench. During 0/2 time, we add a constant force to the block from
left to right. At the end of the 0/2 time, immediately, we reverse the direction
of the force (but keep the magnitude) and act on the block for another 0/2
time. Then, we reverse the force and keep repeating this process forever. Find
the displacement of the block as a function of time.
/1
/2
Figure 5.8 The block-spring system for Problem 5.5.28.
Problem 5.5.29 Two blocks (A & B) of the same mass K are attached to three
identical mass-less springs (as shown). The assembly is placed on a horizontal
and perfectly smooth bench. Initially, springs stay at their natural lengths and
both blocks are at rest. During two brief moments, two forces were added to
the blocks respectively:
∈ ª0, 0/2]
∈ ª0/2, 0]
B
= mB
and G
º( ) = m
0
O. W.
0
O. W.
235
Chapter 5 Laplace Transforms
Compute the displacements of the blocks as a function of time. All mentioned
parameters, including the spring constant /, are given. You may not leave the
solution at a convolution form
/
/
A
/
ž
Figure 5.9 The block-spring system for Problem 5.5.29.
Problem 5.5.30 Two blocks (A & B) of the same mass m are attached to two
identical mass-less springs (as shown). The assembly is placed on a horizontal
and perfectly smooth bench. Initially, springs stay at their natural lengths and
both blocks are at rest. During two brief moments, two forces were added to
the blocks respectively:
∈ ª0, 0/2]
∈ ª0/2, 0]
=m B
= mB
and G
º
0
O. W.
0
O. W.
Compute the displacements of the blocks as a function of time. All mentioned
parameters, including the spring constant, are given. You may not leave the
solution in convolution form.
/
A
/
ž
Figure 5.10 The block-spring system for Problem 5.5.30 & Problem 5.5.31.
Problem 5.5.31 Two blocks (A & B) of the same mass K are attached to two
identical mass-less springs (as shown). The assembly is placed on a horizontal
and perfectly smooth bench. Initially, springs stay at their natural lengths, and
both blocks are at rest. During two instances = B and 2 B, two forces were
added to the blocks respectively. The equation of motion of the two blocks can
be written as
K = −/ ( + /( − ( ) + B ¼( − B )
S (
K = −/( − ( ) + B ¼( − 2 B )
( (0) = ( (0) = 0
R
(0) = (0) = 0
Q
236
5.5 Application of Laplace Transforms
Compute the displacements of the blocks as a function of time. All parameters,
including K, /, B , B , are given. You may leave the solution at a convolution
form if you need to.
Problem 5.5.32 Consider a system of two masses K( and K (from left to
right) connected to three massless springs whose spring constants are /( , / ,
and / (from left to right). The entire space is placed on a frictionless leveled
surface as shown below.
/1
1
K1
/2
2
K2
/3
Figure 5.11 The block-spring system for Problem 5.5.32.
(1) Derive the DEs of motion for the two masses.
(2) Solve the DEs you derived above with the following I.C. and simplified
parameters K( = K = 1, /( = 1, / = 2, / = 3,
(0) = 0
( (0) =
(0)
=
0
Z
(
(0) = L
237
Appendix A
Solutions to Selected Problems
Chapter 1 First-Order DEs
1.1 Definition of DEs
= − sin
2 sin 2
=
−
cos
4 cos 2
(
4 cos 2
cos
cos 2
…HS
( = − cos
(
= 3 cos 2 = RHS
is
a
solution
of
the DE.
(
= cos
2 sin 2
= sin
cos 2 ⇒ |
= − sin
4 cos 2
LHS=
+ = − sin
4 cos 2
sin
cos 2
= 3 cos 2 =RHS
is a solution of the DE.
Problem 1.1.1
(
= cos
Problem 1.1.2
cos 2 ⇒ |
(
(
=
238
cos ln
Appendix A - Chapter 1
LHS
⇒Z
(
= cos ln
sin ln
=−
(
sin ln
cos ln
2 (
= − (sin ln
cos ln
cos ln
2 cos ln
= 0 = RHS
( is a solution of the DE.
= sin ln
= sin ln
cos ln
⇒Z
cos ln
sin ln
=
(
(
LHS
2
= − (sin ln
cos ln
2 sin ln
= 0 = RHS
is a solution of the DE.
cos ln
Problem 1.1.3
From the orthogonal relation, for two function curves
that are orthogonal to each other, we have
⋅ - = −1
Here we have
( )=
+/
( )=2
Thus, we have
1
=−
2
1
1
= −
= − ln
1
2
2
sin ln
sin ln
and
Problem 1.1.4
By observation, we can find that
239
Appendix A Solutions to Selected Problems
=
and
This gives
That means
Problem 1.1.5
Guess
=
(
+
) =3
=
=
is a solution.
+
=RHS
=LHS
= cos
= − sin
= − cos
LHS
= − cos
cos = 0 = RHS
cos is a solution of the DE.
1
1+
Problem 1.1.6
=
⇒
=−
LHS
is a solution of the DE.
)
2
1
1+
= 0 = RHS
( ) = 1 exp
⇒
= −31 exp
3 1 exp
LHS
3
= −31 exp
= 0 = RHS
is a solution of the DE.
1 exp
Plugging 0) = 7 into the G.S., we have
1 exp 0) = 7
gives 1 = 7
Therefore
= 7 exp
Problem 1.1.7
240
2
1+
Appendix A - Chapter 1
Problem 1.1.8
1 sin
= ( + 1) cos ⇒ = cos
LHS cos
1 sin
1 cos tan
= cos
1 sin
1 sin
= cos = RHS
Plugging 8
0 into the G.S., we have
(8) = (8 + 1) cos 8
= −(8 + 1) = 0
gives 1 = −8
Therefore
( ) = ( − 8) cos
= (tan
Problem 1.1.9
Formula
and chain rule give
On the other hand
=3
=3
=3
=3
(tan
=
sin
\
cos
sin
′=
cos
1) ′
1
3
1
1
1
cos
1
cos
1 ′
1
1]
cos
1
1
cos
1
So given function satisfies the DE.
If
= tan
1 then
0) = 1 gives the equation
tan 1 = 1. Hence the value of 1 is 1 = 8/4 (as is this value
plus any integral multiple of 8).
241
Appendix A Solutions to Selected Problems
1 9 1
4
31
5
=
4
5 9 31 3 9 31
LHS =
4
4
2 9 = RHS
Plugging (2 = 1 into the G.S., we have
1
1
=1
2 = Ê 29
2
4
1 = −56
Problem 1.1.10
=
Problem 1.1.11
(
Problem 1.1.12
Plugging
Problem 1.1.13
242
= 3 exp 3
= −3 exp
0
3
= exp 3
=9 (
( = 9 exp 3
= exp 3
,
= 9 exp 3 = 9
(
,
= ln
LHS = exp ln
=
= ( + 1) ⋅
1
1
1
1
1
⋅
1
1
= 1 = RHS
1
0 into the G.S., we have
0
ln 1 = 0
1 1
1 exp
= −= >( 1 exp
Appendix A - Chapter 1
LHS
Plugging
=
= >( 1 exp
1 exp
0 RHS
0
2014 into the G.S., we have
0
1 2014
>(
243
Appendix A Solutions to Selected Problems
1.2 Mathematical Models
Problem 1.2.1
From the given condition, we have the acceleration function
= −/, where / is a positive constant, thus
L
= −/ ⇒ L( ) = −/
L( ) = −/ + 1
Plug the condition L(0) = 88, we can get 1 = 88. Thus, we
have velocity function
L( ) = −/ + 88
The car will stop skidding once the velocity is zero. That is
L( ) = −/ + 88 = 0
This gives
88
stop =
/
The car will stop skidding when = 88⁄/. On the other hand
L( ) =
⇒ ( )=
L
/
+ 88 + 1
2
Considering (0) = 0, we can have
1=0
Thus, we have the displacement function
/
( )=−
+ 88
2
Plugging the condition that
$ stop % = 176
gives
/ 88
88
− E F + 88
= 176
2 /
/
This gives
244
( )=−
Appendix A - Chapter 1
That is / = 22 ft/s and stop = 4 sec.
So the car provides a constant deceleration of 22ft/s2 and the car
continues to skid for 4 seconds.
L
Problem 1.2.2
= ( ) = 50 sin 5
L( ) =
50 sin 5
= −10 cos 5 + 1
L(0) = −10 ⇒ −10 + 1 = −10 ⇒ 1 = 0
L( ) = −10 cos 5
( )=
=
= L( )
L( )
−10 cos 5
= −2 sin 5 + 1
Problem 1.2.3
L( ) =
Hence
( )
1
( + 1)
1
1
⋅
=−
= − 1 ( + 1)
=
L(0) = 0 gives
(0) = 8 ⇒ 1 = 8
( ) = 8 − 2 sin 5
1=
1
=−1
>(
+1
245
Appendix A Solutions to Selected Problems
and
( )=
L( ) =
L( )
1
1
E1 −
( + 1)
=−1
>(
F
1
1
E1 −
F
( + 1) >(
=−1
1
1
1
+
⋅
+1
=
(
(= − 1)(= − 2)
+ 1) >
=−1
(0) = 0 gives
1
1=−
(= − 1)(= − 2)
Thus
1
1
1
( )=
+
⋅
=−1
(= − 1)(= − 2) ( + 1) >
1
−
(= − 1)(= − 2)
=
Problem 1.2.4
(1) Establish the DE of draining process
A( )
= −/V
Z
( = 0) = B
A( ) = 8Y = 8(X − (X − ) ) = 8 (2X − )
A( )
8 (2X − )
=
= −/V
8
8V (2X − )
B
[
V (2X − )
= −/
=−
B
/
2 9 B
F_ = −/
5
[
9
4
2
8X E0 − (2X) F − 8 E0 − (2X) F = −/
3
5
246
4
8E X
3
−
Appendix A - Chapter 1
28√2 8 9
X
15 /
(2) If we double the radius of the container,
A( ) = 8(4X − (2X − ) ) = 8 (4X − )
A( )
8 (4X − )
=
= −/V
=
B
8
[
V (4X − )
=−
C
B
/
9
2
8
8X E0 − (4X) F − 8 E0 − (4X) F = −/@
5
3
128 8 9
@=
X
64 /
Problem 1.2.5
The equation can be written as
8 (2X − )
= −/V
Z
( = 0) = 2X
Solving this IVP, we get
(
/
E − 2X F
=
8
Thus,
(
/
E − 2X F
=
8
[
B
We have
9
Now, we insert
8 4(2X)
+2
H
15
4
= X to find @( ,
9
8 4(2X)
@( = H
+2
15
/
2X
E − FI =
5
3
9
9
2X
8 4(2X)
14(X)
E − FI = Þ
−
ß
15
15
5
3
/
247
Appendix A Solutions to Selected Problems
Next, we find @ by setting
=0
9
8 4(2X)
@ = Þ
ß
/
15
Finally, we get @ > @( .
Problem 1.2.6
Let be the vertical distance from the hole. Then we have
19
F
Y =X E +
20 10„
Thus
19
8X E +
F = −/V
20 10„
19
8X E +
F
= −/V
20 10„
1
19
š + 10„ ›
/
20
=
−
J
8X
V
B
=0
19
10
(
+
1
15„
−
29√2 (
/
„ =−
30
8X
−
1
15„
−
14√2 (
/
„ =−
15
8X
29√2„8X
30/
Now turn the cup upside down
19
Y =X E −
F
20 10„
19
8X E −
F = −/V
20 10„
19
1
š −
/
20 10„ ›
=
−
J
8X
V
B
When
19
10
248
(
=
Appendix A - Chapter 1
When
=0
=
14√2„8X
15/
Problem 1.2.7
For draining the upper half:
T
A( )
= −/V =
Y + ( − X) = X
A( ) = 8Y = 8(X − ( − X) ) = 8 (2X − )
8 (2X − )
= −/V
8
8V (2X − )
[
[
E2X
(
−
F
9
= −/
[
= −/
B
æ
2X
− ßó = −/ (
3
5
2
2
[
8 4
2 9 [
− E X −
F_ = (
/ 3
5
[
9
2 9
4
2
8 4
− \ X EX F − X − E X (2X) − (2X) F] =
5
3
5
/ 3
8 14 − 16√2 9
X ]= (
− \
15
/
For draining the lower half:
Y + (X − ) = X
A( ) = 8Y = 8(X − (X − ) ) = 8 (2X − )
8V (2X − ) = −/(
8
8Þ
B
[
E2X
(
−
F
= −/(
B
(
Í
249
Appendix A Solutions to Selected Problems
9
We want
B
2X
8Þ
− ßó = −/(
3
5
2
2 [
2 9 B
8 4
F_ =
− E X −
5
/( 3
[
8
4
2 9
− \− E X EX F − X F] =
/(
3
5
8 4 9 2 9
E X − X F=
5
/( 3
8 14 9
E X F=
/( 15
= (.
8 14 − 16√2 9
8 14 9
E X F=− \
X ]
15
/
/( 15
14
$14 − 16√2%
=−
/(
/
14
/( = −
14 − 16√2
7
/ ≈ 1.623/
/( =
8√2 − 7
Problem 1.2.8
Torricelli’s law tells us that
T
= −/V
T = −/V
A( )
= −/V
A( )
=
−
/V
Then the time it takes to drain the container from full (
to a certain height ¨ is
250
= 2X)
A( )
= ¨
/V
B
We seek the ¨ such that the time it takes to drain the container
from height 2R to height ¨ is equal to the time it takes to drain
the container from height ¨ to height 0, i.e.,
B
K
A( )
@
A( )
=
−
=
−
2
/V
/V
[
K
K
[
−
Appendix A - Chapter 1
=
K
where @ is the time to drain the container fully. As shown in the
textbook, cross-section area for any liquid surface height is
A( ) = 8Y = 8(X − (X − ) ) = 8 (2X − )
Thus we may find @:
B
C
8 (2X − )
=
−
/V
[
B
B
8
−
E2XV − F
=@
/ [
9
8 4
2 9 B
8 4
2
− E X −
F_ = @ = E X(2X) − (2X) F
/ 3
5
/ 3
5
[
8 2 9 2 9
82 9
X
@= Þ X − X ß=
5
/
/ 3
L
And we want
K
[
−
¨
such that
8 (2X − )
/V
−
B
=
B
K
−
8 (2X − )
V (2X − )
/V
=2 X
9
=
@ 82 9
=
X
/
2
9
4
2 9 K
E X −
F_ = 2 X
3
5
B
9
9
4
2
X ¨− ¨=2 X
3
5
K
251
Appendix A Solutions to Selected Problems
Then the volume of the leftover liquid is given by
B
K
A( )
=
B
K
8(2X −
)
Problem 1.2.9
(1) Then total mass for a boat with = boaters is given by +
=K. The total force propelling the boat is = B . Water resistance
is †B L, which acts in the opposite direction. Thus, we have the
following DE
L
( + =K)
= = B − †B L
(2) This is a separable equation.
Hence we have
(
(
+ =K)
L
=
= B − †B L
+
+ =K)
ln(= B − †B L) = +
†B
With L( = 0) = 0, one can get
( + =K)
ln(= B )
=−
†B
Plugging in this value of in the solution, we get
= B
†B
L( ) =
E1 − exp E−
FF
†B
+ =K
(3) The total travel time @ follows the following equation,
C
C
= B
†B
…=
L( ) =
E1 − exp E−
FF
+ =K
B
B †B
+K
†B
+ =K
= B
E@ +
exp E−
F−
F
…=
†B
†B
+ =K
†B
Differentiate @ WRT = (assuming real variable instead of integer
−
for convenience) and found out that
C
< 0 and thus the more
boaters, the shorter the time. The answer is then “To win the
race, put as many boaters as practical.”
252
Appendix A - Chapter 1
Alternate argument, one can compute the boat speed WRT =.
L
†B
B
= E1 − exp E−
FF
= †B
+ =K
=K B †B
†B
+
exp E−
F
†B ( + =K)
+ =K
Now
†B
B
E1 − exp E−
FF ≥ 0 ∀ ≥ 0
†B
+ =K
and
=K B †B
†B
exp E−
F>0
(
+ =K)
†B
+ =K
L
>0∀ ≥0
=
In other words L is an increasing function of = . So, at any
instant, the more boaters, the faster the boat travels. “The added
weight of the boater has not slowed down the boat, as long as the
added boater does not pedal backward.”
Thus
253
Appendix A Solutions to Selected Problems
1.3 Separation of Variables
Problem 1.3.1
This is a separable DE, and by separation of variables, we have
3
3
=−
⇒
=−
+1
+1
−2
−2
3
ln( + 1) = −ln( − 2) + 1
2
1
=2+
( + 1)
Problem 1.3.2
>
3
2
>
(
(
=4
( (
=4
=
=3
= E2
(
4
(
+ 1(
+ 1F
Problem 1.3.3
We can rewrite the DE as
= (1 + )(1 + )
and this is a separable DE. Hence we have
1+
1+
254
= (1 + )
=
(1 + )
ln|1 + | =
Problem 1.3.4
Since
we have
=
exp(− )
=
=−
+
Appendix A - Chapter 1
1
2
+1
exp(− )
=
. That is
exp(− )
(exp(− ))
= − exp(− ) +
exp(− )
= − exp(− ) − exp(− ) + 1
From the I.C. (0) = 1, we can get
−1 + 1 = 1
This gives 1 = 2, and the solution to the DE is
( ) = − exp(− ) − exp(− ) + 2
Problem 1.3.5
=
= −2 cos 2
−2 cos 2
= − sin 2 + 1
I.C. (0) = 2014 gives
− sin 0 + 1 = 2014
1 = 2014
Problem 1.3.6
sec
⋅
=
1
2√
255
Appendix A Solutions to Selected Problems
sec
2√
= √ +1
tan
The G.S. is
Plug the I.C. (4) =
=
M
= tan>( $√ + 1%
in to the G.S., we can get
8
= tan>((2 + 1)
4
1 = −1
= tan>( (√ − 1)
The P.S. of the DE is
Problem 1.3.7
=
ln
=2 +3
(2 + 3
=
exp(
+ exp(
exp(
)
))
)+1
= exp( + exp( ) + 1)
Given the I.C. (0) = 5, we have
5 = exp(0 + 1 + 1) ⇒ 1 = ln 5 − 1
Thus the P.S. for the DE is
= exp( + exp( ) + ln 5 − 1)
= 5 exp( + exp( ) − 1)
The G.S. is
Problem 1.3.8
+1
+1
256
= cos
=
cos
1
4
Appendix A - Chapter 1
(
+ 1)
=
+1
1
ln(
4
+ 1) = sin + 1
Problem 1.3.9
The DE can be written as
=5
which is a separable DE. Thus,
−
Problem 1.3.10
=2
1
=
=
+3
(5
9
This is also separable equation
−
Then substitution of
So 1 = −1 and
1
cos
=
=
−4
−2
−4 )
=
(2 + 3
+
+1
+1
(2 + 3
)
)
−1
+ +1
= 1 gives
−1
−1 =
1+1+1
=
=
−1
+ −1
257
Appendix A Solutions to Selected Problems
cos
Problem 1.3.11
cos
Problem 1.3.12
sin
=
=
=
( )=
(
tan
258
sin
2
+1
cos
8
8
= 1 sin
2
2
8
1=
2
The explicit P.S. is
Problem 1.3.14
=
cos
sin
ln | | = ln | sin | + 1
= 1 sin
8
8
š ›=
2
2
gives
Problem 1.3.13
=2
8
sin
2
+ 1) tan
=
1
− ln|cos | = ln(
2
( + 3) E
=
+1
F = ( − 2)
+ 1) + 1
Appendix A - Chapter 1
= ( + 3)>
( − 2)
( − 2)
Problem 1.3.15
=
( + 3)>
−
=
=
=3
+3
( + 3)
1
−( − 2)>( = E− F ( + 3)> + A
2
1
>(
( − 2) = E F ( + 3)> + ž
2
1
+ 2ž( + 3)
( − 2)>( =
2( + 3)
( − 2)(1 + 1( + 3) )
1=
2( + 3)
2( + 3)
= −2
1 + 1( + 3)
2( + 3)
=2+
1 + 1( + 3)
1
E FE
1
E F
F=
=
+3
( + 3)
( + 3)
+ 1(
2
( + 3)
= exp \
1( ]
2
( + 3)
= 1 exp \
]
2
(1 + 3)
(1) = 1 = 1 exp \
] = 1 exp(8)
2
1 = exp(−8)
ln
=
259
Appendix A Solutions to Selected Problems
= exp \
( + 3)
− 8]
2
Problem 1.3.16
The equation can be written as
$(1 + ) % = cos
Integrate on both sides,
(1 + ) =
cos
+ 1 = sin + 1
sin + 1
+1
We find 1 = 1 by plugging in the I.C.. Finally,
sin + 1
=
+1
Thus,
Problem 1.3.17
So the G.S. is
260
=
=
œ (v − ln )
ln
=
œ(v − ln )
− (v − ln )
=
œ(v − ln )
− ln(v − ln )
=
œ
− ln(v − lny)
=
œ
ln(v − ln )
= +1
œ
v − ln = exp$−(œ + 1( )%
= exp$v − exp$−(œ + 1( )%%
−
If (0) =
If (∞) =
If (œ < 0)
Appendix A - Chapter 1
B
•
•
v − ln
B
= exp(−1( )
= exp(∞) (œ > 0) ⇒ v = ln
=0
•
•
261
Appendix A Solutions to Selected Problems
1.4 Linear First-Order DEs
Problem 1.4.1
Divided (
+ 1) by both sides of the DE, we have
3
6 exp š− 2 ›
3
+
=
+1
+1
Let
]
+1
3 ( + 1) − 3
= exp \
]
+1
3
3
− ln( + 1)F
= exp E
2
2
3
>
= ( + 1) exp E
F
2
We have
Thus,
3
( ) = exp \
=
=(
1
( )
( ( ) ) =6 (
6 (
+ 1)>
+ 1) exp E−
3
2
9
+ 1)>
F E−2(
9
+ 1)> + 1F
3
3
F + 1( + 1) exp E−
F
2
2
which is the G.S.. Given (0) = 1, we have
1 = −2 + 1 ⇒ 1 = 3
Hence the P.S. of the DE is
3
3
= −2 exp E−
F + 3( + 1) exp E−
F
2
2
= −2 exp E−
Problem 1.4.2
(a) We know that
262
Appendix A - Chapter 1
¡
= \1exp E−
= 1 exp E−
= −1 exp E−
h( )
h( )
h( )
F]
F E−
h( )
F h( )
= −h( ) ¡
It is easy to see that
¡ + h( ) ¡ = −h( ) ¡ + h( )
This proves it is a G.S. of the original DE.
(b) We have that
¢( )
= \|exp E−
= Eexp E−
+ exp E−
h( )
h( )
h( )
= −h( ) exp E−
F
¡
=0
F E Ÿ( ) exp E h( )
FF E Ÿ( ) exp E h( )
F E Ÿ( ) exp E h( )
h( )
+ exp E−
F
F
F
F
F E Ÿ( ) exp E h( )
h( )
F5 ]
F
F
F Ÿ( ) exp E h( )
= −h( ) ¢ ( ) + Ÿ( )
Now it is easy to see
LHS = ¢ h( ) ¢
= −h( ) ¢ + Ÿ( ) + h( ) ¢
= Ÿ( ) = RHS
h
= 0, we know that
(c) Since ¡
is a G.S. of
¡ = −h( ) ¡
On the other hand, we know that
¢ = Ÿ( ) − h( ) ¢
Then for = ¡ + ¢ , we have
F
F
263
Appendix A Solutions to Selected Problems
= $ ¡ + ¢%
= ¡+ ¢
= −h( ) ¡ + Ÿ( ) − h( )
= −h( )$ ¡ + ¢ % + Ÿ( )
= −h( ) + Ÿ( )
+ h( ) = Ÿ( )
which means it is a G.S. to the DE.
That means
Problem 1.4.3
The DE can be written as
Let
We have that
That gives
+
2
= exp(2 ln )
=
( ( ) ) =7
=
=
Given (2) = 5, we have
264
=7
( ) = exp E
=
Thus the P.S. is
2
>
7
4
>
5=7+
=
7
4
E 7
E
7
4
+1
F
F
+ 1F
>
1
⇒ 1 = −8
4
−8
>
¢
Appendix A - Chapter 1
Problem 1.4.4
Let
We can have
which gives
( ) = exp E cot
= exp(ln sin )
= sin
( ( ) ) = sin cos
1
E sin cos
sin
sin + 1
=
2 sin
= 21
( ) = exp E −3
Thus
F
=
Problem 1.4.5
Method I
This DE can be written as
−3
which is a linear DE. Let
We know that
F
(exp(−
= exp(
= exp(−
) ) = 21
)
21
= −7 + 1 exp(
Method II
The DE can be written as
)
)
F
exp(−
exp(−
)
)
=3
+7
which is a separable DE. Thus, we have
265
Appendix A Solutions to Selected Problems
=
+7
ln( + 7) =
= 1 exp(
Problem 1.4.6
The DE can be written as
Let
We can have
That gives
−
2
3
+1
)−7
=
cos
( ) = exp E −
= exp(−2 ln )
= >
( ( ) ) =
=
=
1
( )
cos
2
cos
cos
F
( )
( )
= (sin + 1)
Thus, the G.S. to the original DE is
= sin + 1
Problem 1.4.7
Let
We have
266
+
1
2 +1
(
= (2 + 1)
( ) = exp E
2 +1
(
= (2 + 1)
F
Appendix A - Chapter 1
Thus
E(2 + 1)
(
= (2 + 1)>
F =2 +1
(
(
= (2 + 1)> (
(2 + 1)
+
+ 1)
Problem 1.4.8
Method I: Separation
The equation can be written as
2 ( + 1)
=
+1
2
=
+1
+1
ln( + 1) = ln( + 1) + 1
=1 +1−1
Method II: Linear 1st-order
Rewriting the equation in the following form, we find that it is
linear
2
2
−
=
+1
+1
Let
2
( ) = exp E −
F
+1
1
=
+1
Thus we know,
2
1
⋅ F =
E
( + 1)
+1
2
= ( + 1)
( + 1)
1
+ 1F
= ( + 1) E−
+1
267
Appendix A Solutions to Selected Problems
= −1 + 1(
+ 1)
Problem 1.4.9
To solve this DE, we can regard as the dependent variable and
as the independent variable. Thus, we have
(1 + 2
)
=1+
1+
1+2
1+2
=
1+
2
1
−
=
1+
1+
This is a 1st-order Linear DE of WRT . Let
2
( ) = exp E −
F
1+
(1 + )
= exp \−
]
1+
= exp[− ln(1 + )]
1
=
1+
We have that
1
( ( ) ) =
( )
1+
which gives
1
1
( )
=
( ) 1+
1
= (1 + )
(1 + )
1
1
= (1 + ) E tan>( +
+ 1F
2
21 +
268
=
Appendix A - Chapter 1
Note: Let = tan ü , we have 1 +
= sec ü and
sec ü ü. Thus, the last integral becomes
1
1
=
sec ü ü
(1 + )
sec ü
=
cos ü ü
1
(1 − cos 2ü) ü
2
ü sin 2ü
+1
= −
4
2
1
= (ü − sin ü cos ü) + 1
2
tan ü
1
F+1
= Eü −
sec ü
2
1
1
= tan>( + ⋅
+1
2
2 1+
=
Problem 1.4.10
We can write the DE as
Let
+
1
2
=5
1
2
( ) = exp E
>
(
1
= exp E ln F
2
Thus we have
This gives
=
(
( ( ) ) =5
=
=
1
( )
5
>
>
(
(
F
( )
( )
269
Appendix A Solutions to Selected Problems
(
(5 + 1)
1
= 5√ +
√
=
>
Problem 1.4.11
The DE can be written as
−
exp( )
=
which can be considered as a DE of
WRT . Thus, Let
( ) = exp E −1
We can have
which gives
= exp(− )
( ( ) ) =
exp( ) ( )
1
( )
=
= exp( )
exp( ) ( )
= exp( ) E
=
Problem 1.4.12
Let
Thus
270
1
2
F
1
2
+ 1F
exp( ) + 1 exp( )
+
2
+1
( ) = exp E
= ( + 1)
(( + 1)
=3
2
+1
F
) = 3( + 1)
Appendix A - Chapter 1
Problem 1.4.13
Converting the DE
= ( + 1)> (( + 1) + 1)
1
= +1+
( + 1)
The integrating factor is
+
1
1+
( ) = exp E
Then,
F=1+
sin
1+
$(1 + ) % = sin
(0) =
The solution is
Substitute L =
sin
1+
( ( ) ) = ( )
(1 + ) =
Since (0) = 1,
1
1+
=
=
sin
= − cos + 1
− cos + 1
1+
1−1
=1−1=1
1
1=2
=
Problem 1.4.14
2 − cos
1+
L = ′′
L +3 L =4
3L
L +
=4
h( ) =
3
,
Ÿ
=4
271
Appendix A Solutions to Selected Problems
( ) = exp E
3
F=
( ( )L) = ( )Ÿ( )
L=
Since L =
272
L=
1
=
6
4
6
=
4
6
−
(4
+
+
>
2
)
+
(
Appendix A - Chapter 1
1.5 Substitution Methods
Problem 1.5.1
We can write the DE as
=
Let
We have
Thus,
=
+
Problem 1.5.2
Rewrite the DE as
=
Let
We have
=
=
and
−
That is
+š ›
1
=
=
+
=
= ln
=−
+
1
ln
1
+ Ë4 š ›
=
and
=
1
+
273
Appendix A Solutions to Selected Problems
Thus, we have
+
1
ln š2
2
š2
Þ2
2
Problem 1.5.3
Since L = ln , we have
=
√4
V4
+ V4
1
V4
=
1› = ln
(
1› = 1
Ë4 š ›
1(
(
1ß = 1
V4
= exp L
1
=1
= exp L
and
L
Put this back to the DE, we have
exp L L
h
exp L = Ÿ( )L exp L
It is easy to see that exp L G 0, so we can have
h
= Ÿ( )L( )
L
Problem 1.5.4
E
274
5
F 5
=
5
5
=
2
=2
2
=2
2
Appendix A - Chapter 1
1
5
>(
E F =
2
2
L=
⇒ L =2
(
(
1
= E F L> L
=L ⇒
2
(
( >(
1 >(
1
5
E FL L
E F L = E F EL F
2
2
2
L
= 5
L
1
= exp E E F
F = exp ln
( L ) =5
L=
L=
L=
=
Problem 1.5.5
Let L = ln , we have
exp L L
5 9
5
=E
+
1
1
+
1
1
(
+ F
= exp L and = exp L L′. Thus,
4 exp L
2 exp L L = 0
2
L
L=4
which is a linear DE. Let
We have
5
=
= exp E
L=
>
2
4
F=
275
Appendix A Solutions to Selected Problems
=
= exp
Thus, we have
K
Problem 1.5.6
K
Substitute
2ln = 0
¨
=
2
, then
The DE becomes
=L ,
F=
= (K + 1
(K + 1
¨ú(
¨ú(
K + 1 ¨ú
K 2
K+1 ¨
=
K 2
K+1 ¨
ln =
K 2
K 1 ¨
= exp E
K 2
=
æ
Ì
>
2 ln = 0
¨
= exp E
=
276
1
1
− (K + 1 ¨ 2 = 0
2
= (K + 1 ¨
+2
Let L =
1
>
ln ,
Using 1st-order linear
Problem 1.5.7
+1
>
>
>
= 3L L′
F
Let
2
L − L=4
= exp E −
= (2
>
1
and consider L as a function of . Now we have
L
L
L
=
=
=L
Thus, the DE becomes
L
L
L
= 3L
=
3L
L = 1(
That is
−
Problem 1.5.9
Method I: Separable
This DE can be write as
Thus
F=
(L ) = 4
1
L > =2
1
L=2
Finally,
Problem 1.5.8
Let L =
2
Appendix A - Chapter 1
1
2
= 1(
= 1( + 1
−
=
277
Appendix A Solutions to Selected Problems
1
ln
2
>
1
1
>
1
2
1 = 1 exp
= 1 exp
ln
=
1(
Method II: Bernoulli
Let L = > . Thus L = −2 > ′. The DE becomes
1
L=
− L
2
L − 2 L = −2
This is a linear DE, so let
= exp E −2
F
= exp
Thus, we know that
exp
L = −2 exp
exp
1
L = exp
= 1 1 exp
Put L = > back, we have
>
− 1 = 1 exp
Problem 1.5.10
Let
We have
=
Thus, the DE becomes
That is
278
+
=
and
=−
=−
=
(2
2
2
+
+
1
Appendix A - Chapter 1
ln
1
ln
ln
(
+
+
which gives
Problem 1.5.11
The DE can be written as
Let
We have
The DE becomes
=
=
− exp
The solution of the DE is
exp š
+ . We have
=
)
1
=1
+ exp š ›
=
=
+
=
1
=1
and
exp
Problem 1.5.12
Let =
1
1 = − ln
=
+
+ exp
=
= ln
› = ln
1
1
− 1. Thus
=1
1
=
1
279
Appendix A Solutions to Selected Problems
1
+
That is
1
− ln
− ln
− ln
=
1 =
+1
1 = +1
1 =1
Problem 1.5.13
Let L = sin , then L′ = ′ cos . The DE becomes
2 LL′ = 4 + L
1
L −
L = 2 L >(
2
Let = L . Then ′ = 2LL′
1
=4
−
We have
( ) = exp E −
(
Then,
Problem 1.5.14
Let L = ′. Then
Let
280
=
sin
F=
)′ = 4
=4 +1
=4 +1
>(
Solution is
1
>(
=4
−1
>(
= L and the DE becomes
L = ( + L)
+ L. Then L = − 1
−1=
+1
=
Appendix A - Chapter 1
= arctan + 1(
= tan( − 1( )
Thus
Finally,
=L= −
= tan( − 1( ) −
= − ln(cos( − 1( )) −
Problem 1.5.15
Let L = exp( ). Thus
1
2
= ln L and we have
1
= L′
L
+1
1
( + L) L = − 1
L
L
( + L)L = − L
Let = + L. Then we have L = − 1 and
− =− +2
=2
1
=
+ 1(
2
=2 +1
Put back = + L gives
L − +2 L =1
and finally
exp(2 ) − + 2 exp( ) = 1
Thus the DE becomes
Problem 1.5.16
We can rewrite the DE as
=E
>
−9
(
F
which is a Bernoulli’s equation. Let L =
>(
and we have
281
Appendix A Solutions to Selected Problems
L =−
Thus, the DE becomes
It is easy to get
Therefore
(
L =9
−
+2
L=6
+2
= E6
Problem 1.5.17
We can rewrite the DE as
1
+
′
>
>
>
>
(
(
>(
+ 1F
1
=
+1
>
√1 +
This is a Bernoulli’s equation. Let L = , we have
3
3
L + L=
√1 +
Let
3
( ) = exp E
F
Thus,
(
L=
That is
282
=
=
>
>
L) =
3
3
√1 +
√1 +
3
E V1 +
2
+ 1F
3
= E V1 + + 1F
2
+ 1( = 1 exp( )
Appendix A - Chapter 1
Problem 1.5.18
Let L = exp( ), L = exp( ) ′. Thus, the DE becomes
2
L − L = 2 exp(2 )
Let
( ) = exp E −
(
>
L=
F=
L) = 2 exp(2 )
>
2 exp(2 )
(exp(2 ) + 1)
= 2 ln + ln(exp(2 ) + 1)
Thus
Problem 1.5.19
Let
=
2
L = ln
⁄
⇒
= exp(L) ⇒
= exp(L)
L
and by the above, we have
L
exp(L)
exp(L)
− 4 L exp(L) + 2
L =0
Substituting
⇒
L
exp(L)
2
−4 L =− L
which is Bernoulli’s equation. Let
= L (>
We have
− 4(1 − =)
= −(1 − =)
2
This is a 1st-order DE of u WRT , the integrating factor is
( ) = exp E −4(1 − =)
F = exp(2(= − 1)
)
283
Appendix A Solutions to Selected Problems
)E
( ) = exp(−2(= − 1)
+ 1F
2(= − 1)
exp(2(= − 1)
Back substitute twice to get the solution ( ).
Problem 1.5.20
Let L =
)
and consider L as a function of . Now we have
L
L
L
=
=
=L
Thus, the DE becomes
L
L
L
+L = L
+L =
( L) =
1
+1
2
1
1
+
L=
2
L=
Put L =
back, we have
ln(
Problem 1.5.21
It is easy to observe that
rewrite the DE as
=
2
1
1
+
2
=1
+ 1(
+ 1( ) = + 1̅
= 0 is a solution. For
6 š › + 2 š › + \9 + 8 š ›]
284
≠ 0, we can
=0
Appendix A - Chapter 1
Let
=
= +
. Thus,
(9
)=0
6 + 2 + + 8 )( +
9+8
5
=−
3 +2
which is a separable DE. We can get
3 ln + ln(3 + 2 ) = −5 ln + 1
(3 + 2 ) = 1 >9
Put =
back, we have
3
+2
=1
We know that
=
Problem 1.5.22
(1) From
>(
+ , we can have
1
=− +
and ′ in the original DE,
1
1
1
1
3
− +
+ 7E + F − 3E + F =
Substituting
−
1
+
+
7
+
7
+
′+
−3E
−3
>(
1
+
2
=0
+
F=
3
=3
in terms of , where h( ) =
which is a Bernoulli equation of
>(
, Ÿ( ) = 3, = = 2.
(2) Let L = (> = (> = >( , we have
L
+ (1 − =)h( )L = (1 − =)Ÿ( )
L
L − = −3
This is a 1st-order linear DE, where integrating factor is
285
Appendix A Solutions to Selected Problems
Then,
L=
1
E
( ) = exp E −
( )(−3)
1
+ 1F =
F=
E−
= (−3 ln + 1) = −3 ln + 1
Substituting back L with >( and with −
1
1
−3 ln + 1 = =
−
1
1
=
+
−3 ln + 1
Problem 1.5.23
Let L = ′, thus
1
3
>(
+ 1F
, we have
>(
= L′
and the original DE becomes
L +3 L =4
( L) = 4
L = 2 + 1(
2 1(
L= +
Put L = ′ back, we have
=
=
2
+
1(
2 1(
E + F
= 2 ln −
1(
+1
2
Problem 1.5.24
Convert the DE to the form of Bernoulli DE.
>(
− 1000 LLL = −2
>(
>
Let L =
,L = −
′. Plug into the above DE,
286
Appendix A - Chapter 1
−
L − 1000 LLL L = −2
L + 1000 LLL L = 2 >(
The integrating factor is
= exp E 1000
LLL
(BBB )
2
F = exp(
Multiply the integrating factor to both sides,
(exp( (BBB ) L) = 2 >( exp(
L = exp(−
= exp(
(BBB )
E 2
>(
(BBB )
(BBB )
>( exp( (BBB )
>(
exp(
(BBB )
>(
F
( + )
=
( + )
(when + ≠ 0, dividing the equation by + )
This equation is simply homogeneous and separable.
Introducing substitution
Problem 1.5.25
=
L=− ,
=L+ L
Plug this into original DE, we have
L
1
L + =
L
which is a Bernoulli’s equation with = = −1. Let
= L (> = L
So that
(
1 (
L = u> ′
L= ,
2
Then
L
1
2
2
L + =
,
+
=
L
2
( ) = exp E k( ) F = exp E
F = exp(2ln ) =
287
Appendix A Solutions to Selected Problems
= exp E−
=
=
1
1
2
E E
(
k( )
+ 1)
F E Ÿ( ) exp E k( )
F
+ 1F
After back substitution
=š ›
One gets the G.S.
Problem 1.5.26
F + 1F
+
1
=
+
1
2
+1
>(
=0
(Dividing the equation with 2
)
1
1 >(
+
=−
2
Let
L = (> =
So that
(
1 (
= L > L′
=L ,
2
Plug these to equation
2
+2
+1 = 0
We have
L +2 L +1 = 0
(dividing the equation with )
2
1
2
1
L + L =− ,
h( ) = ,
Ÿ( ) = −
( ) = exp E k( )
F = exp E
2
L = E Ÿ( ) exp E k( )
288
F = exp(2ln ) =
F + 1F
=
=
1
1
Therefore the G.S. is
E E−
1
(− + 1)
=
Appendix A - Chapter 1
1
F
+ 1F
(− + 1)
Problem 1.5.27
Method I: Separation of variables
(
+ 1)
= 2 ( + 1)
2
1
=
+1
+1
ln( + 1) = ln( + 1) + ln(1)
y = 1( + 1) − 1
Method II: Integration factor
2
2
−
=
+1
+1
Find the integration factor,
2
1
( ) = exp E−
F=
+1
+1
Multiply the integration factor and take an integral,
2
š
›=
( + 1)
+1
2
1
=
š
›=
=−
+C
( + 1)
+1
+1
+1
= 1( + 1) − 1
Problem 1.5.28
Using substitution L = / , we get
L′ = − V1 + L
Solving this equation by separation of variables ( and L), we
get
289
Appendix A Solutions to Selected Problems
(>Ô
=
\š ›
(úÔ
−š ›
]
2
< 1, the solution can be zero (answer to (4)).
= 1, we have
(1) If
(2) If
E1 − š › F
2
Thus, at = 0, we have = /2.
(3) If > 1, blows up easily.
=
Problem 1.5.29
Let
−
1
+
+
Let
L=
1
=
+ ,
=−
2
,
=3
L
2
L − L = −3,
1
+
F
1
−
=−
2
2
+3
+ 3
(Bernoulli equation)
å( ) = 3,
==2
+ (1 − =)h( )L = (1 − =)å( )
L( ) =
=
=
290
=−
1
8 1
4
= 3E + F − E + F+
2
h( ) = ,
>(
E
( ) = exp E −
1
( )
>( (
−
)
( ) (−3)
3
+
2
F=
>
=
Therefore
Appendix A - Chapter 1
1
( )−
( )=
1
1
3 +1
+
1
Problem 1.5.30
The original DE can be converted if we let
=œ+Ù
= (H( ) + + œ)( − œ)
= (H( ) + 2œ + Ù)Ù
= (H( ) + 2œ)Ù + Ù
or
Ù′ − (H( ) + 2œ)Ù = Ù
which is a perfect Bernoulli equation with
h( ) = −(H( ) + 2œ)Ù
Ÿ( ) = 1
==2
− =œ
+œ
−œ
=œ +
( −œ )=œ +
Problem 1.5.31
=
Let
Then
= ∙
šœ + ›
œ +
=
œ
−œ
1−
+ ,
=
+
=
œ+
1−œ
291
Appendix A Solutions to Selected Problems
œ+
œ+œ
− ,
=
1−œ
1−
1−œ
=
œ+œ
1
1
−
= ln + 1
1+
œ 1+
1
1
arctan( ) + ln|1 + | = ln + 1
2
œ
œ
+ (
arctan( ) = ln
2 1+
=
œ
+ (]
= tan \ ln
2 1+
Back substituting, one can get the final solution
œ
= tan \ ln
+ (]
2 1+
So
Problem 1.5.32
For = = 0
If
292
>0
=
=
+
1
š
+
+
=
+ ›
+ \Ó ]
=
=
Appendix A - Chapter 1
1
1
E F S U tan>( S U =
Ó
Ó
R T
R T
1
tan>( Ë
√
tan>( Ë
Ë
If
<0
=√
= tan š√
= Ó tan š√
1
1
2√−
1
S
R
1
For = = −2
ln íí
− √−
+ √−
′=
L=
+A
( + A)
( + A)›
( + A)›
− \Ó− ]
− Ó−
2√−
=
1
−
=
1
+ Ó−
− Ó−
+ Ó−
íí =
+Â
U
T
= M exp$2√−
+
+A
>
L =−
=
+Â
+þ
%
>
293
Appendix A Solutions to Selected Problems
1
= −L > L
L
−L > L = L > + >
L =− − > L
This is now a simple homogeneous equation and one can solve it
easily by a substitution.
L
Ù=
=
( Ù) = − − Ù
Ù = − Ù −Ù−
1
1
Ù=
− Ù −Ù−
Case (1): 1 − 4 > 0
Let Ù( , Ù be the roots of the equation −cz − z − b = 0. Then
we have
1
1
1
1
E
−
F Ù=
(Ù − Ù( ) Ù − Ù( Ù − Ù
1
Ù − Ù(
ln _
_ = ln | | + ‡(
(Ù − Ù( ) Ù − Ù
where G( is a constant.
Case (2): 1 − 4 < 0
1
1
Ù=
1
4 −1
− šÙ + 2 › + −4
Let
1
4 −1
, =Ù+
A = − ,ž =
2
−4
Then by assumption, we have Až > 0. So
We have
1
√Až
1
Ë
Až
294
1
A
\ Óž ] + 1
A
Ë
=
ž
1
A
=>( ÞË ß + ‡ = à=| |
ž
Appendix A - Chapter 1
A
1
Ë arctan Ë + ‡ = ln | |
ž
Až
where G is another constant.
Case (3): 1 − 4 = 0
Then
1
1
1
−
Ù=
1
šÙ + 2 ›
or
2
= „ exp E
F
2 +
The G.S. can be expressed differently depending on the values of
1−4 .
295
Appendix A Solutions to Selected Problems
1.6 The Exact DEs
Problem 1.6.1
From the DE, we have
We can evaluate
and
, ) = 1 + ln(
) and
( , )=
( , )=
1
( , )=
1
= . This means the DE is exact. Now we can
which gives
solve the DE by exact DE method.
M( , ) =
=
and
gives
=
( , )
ln + -( )
¬M
= ( , )
¬
ln + - ( ) = 1 + ln(
- ( ) = 1 + ln
-( ) = + ln −
= ln
M( , ) = ln + ln
and finally, we have the G.S. to the DE
ln + ln = 1
Thus,
296
)
Appendix A - Chapter 1
Problem 1.6.2
(1) Multiplying the integrating factor on both side of the original
DE
exp E h( )
F A( , )
+ exp E h( )
( , ) = exp E h( )
( , ) = exp E h( )
¬ž ¬A
¬
−
=
exp E h
¬
¬
¬
F−
− žh exp E h
F ž( , )
F A( , )
F ž( , )
¬ž
exp E h
¬
F
¬A ¬ž
−
− žhF exp E h
¬
¬
=0
The new DE is an exact DE.
=E
(2) According to š
Xº( , )
XG( , )
− X › /ž(
X
=0
F
F
, ) = h( ),
¬A( , ) ¬ž( , )
−
F −2 −
E
3
¬
¬
)
h( =
=
=−
ž( , )
The integrating factor is
3
( ) = exp E h( ) F = exp E −
F= >
The new exact DE is
> (2
− ) +
>
( , )=2
− >
,
=0
( , )=
1 >
M( , ) =
+ -( ) = −2 >( +
2
With ¬M/¬ = , we have
¬M
= > +- ( )= = >
¬
-( )=0
>
>
+ -( )
297
Appendix A Solutions to Selected Problems
The solution of the DE is
2
Problem 1.6.3
(1)
If
If
=
, check if
+
is a function of pure , or
is a function of pure .
Choose the easiest
or
+
1
2
= 1(
>
v( ) + $œ( ) + µ( )% = 0
= œ( ) + µ( ),
= v( )
(i.e., the DE is exact), then the solution is
M( , ) =
≠
>(
-( ) = 1
−
−
−
°
E
F±
= ( )
= -( )
( ) = exp E
( ) = exp E−
( )
-( )
F
F
Let ® = ( ) and ® = ( ) , replace the and
solution above with ® and ®.
(2)
µ
œ
œ
=−
,h
=
′+
v
v
v
and
µ
Ÿ
=−
v
298
=1
in the
Appendix A - Chapter 1
The solution is
= exp E−
h
F
Ÿ
+
Problem 1.6.4
(1) Check if this DE is Exact.
, = $h
− Ÿ %,
, = 1
¬
,
¬
,
h
=
≠
=0
¬
¬
So, this DE is not an Exact DE.
(2) If not, convert it to an Exact DE.
For converting given DE into Exact DE, we need to calculate
two ratios.
¬
,
¬
,
−
¬
¬
=h
=
,
Because
is a function of , we have the integration factor
= exp E
F = exp E k
F
By multiplying the integration factor into given DE, we can
obtain the Exact DE.
−Ÿ % +
=0
$h
exp E k
F $h
,
So, we can get new
,
exp E k
−Ÿ
and
= exp E k
,
%
,
+ exp E k
F $h
= exp E k
Fh
=
¬
¬
,
as following
−Ÿ
F
=
¬
¬
%
F
=0
,
299
Appendix A Solutions to Selected Problems
= exp E k
Fh
(3) Solve the DE using the Exact Equation method and your
solution may be expressed in terms of the functions h
and Ÿ .
¬M ,
=
, = exp E k
F $h
−Ÿ %
¬
¬M ,
=
, = exp E k
F
¬
By integrating first equation, we can get
M
,
=
=
,
exp E k
F
−
+-
Ÿ
exp E k
FE k
+By integrating second equation, we can get
M
,
=
,
+
=
exp E k
F+
By comparing the two results, we get the solution.
M
,
=
(4) If h
exp E k
−
Ÿ
= and Ÿ
(
F
exp E k
=
¶·¸
FE k
F
, get the specific solution.
We can apply the result of previous problem.
M
=
=
300
,
=
exp E k
+1
1
exp E
F−
− sin + 1
cos
F
F−
exp E
Ÿ
1
exp E k
F
+1
+1
F
Appendix A - Chapter 1
M=
− sin + 1
Problem 1.6.5
First, let us check if the DE is exact. We have
and ( , =
, =2 −
= −2 and
=
≠ , the DE is not exact. Since
−
= −3
We have
−
3
=−
which is a single-variable function. Let
3
= exp E −
We know that
is exact.
2
M
and
gives
=
2
,
>
=0
>
=
=
-
2
¬
M=
¬
=−
>
-( ) = −
Therefore, the solution to the DE is
F
2
+- ( )
301
Appendix A Solutions to Selected Problems
Problem 1.6.6
Method I: Substitution
The DE can be written as
2
=−
Let
We know that
The DE becomes
Plug
=
+
ln 2
2
2
2
back, we have
4
2š ›
2
=
=1
3
4
=
=−
2
+
3
4
3
=−
1
1 = −3 ln
1=1 >
1=1
2
1(
>
=1
Method II: Exact DE
The DE can be written as
(3
4
=0
2
Thus we have
3
2
and = 4
Since
= 4 and
=4
We know that it is an exact DE. Hence we have
302
Appendix A - Chapter 1
M
,
=
=2
Thus, the G.S. of the DE is
2
2
We have
-
¬M
=
¬
On the other hand
=3
=3
=
2
=1
Problem 1.6.7
Method I: Exact DE
We can write the DE as
( +3
where
3 and
3
=0
=3
. Thus, we have
=
=3
Hence the DE is exact. Now we can get
M
Put it back, we have
Thus, we have G.S.
,
=
=
1
2
( ,
3
¬M
¬
3
=3
=−
1
-( ) = −
2
=
-
303
Appendix A Solutions to Selected Problems
1
2
Method II: Substitution
We can write the DE as
Let L = . Thus
3
= L and
1
=
1
2
=1
3
3
= L + L . This gives
1 3L
L+ L =
L 3
L 3
1
L =
1 6L L
1
ln 1 6L L = ln
2
1 6L L = 1 >
Put L = / back, we have
1
Problem 1.6.8
We have
( ,
6
+6
=
š › =1
and
1(
>
=1
( ,
=2
= 1 and
=2
we know the DE is not exact. Since
−
= −1
We can find
−
1
=−
=
By checking
which is a single-variable function of . Let
1
= exp E
−
F=
304
=
Appendix A - Chapter 1
Now we can have an exact DE
+ (2
To solve this, we have
M
,
=
=
Since
we can get
2
-
¬M
=
¬
-
Thus, the solution to the DE is
=0
9
+ -( )
=2
1
=
6
û
6
û
9
=1
Problem 1.6.9
Rearrange the original equation to yield
(2
1 % =0
$ 2
Then two parameter functions are given below.
, = (2
( , = −2
1
Calculate the difference.
−
= (−2
2 =0
This DE is exact.
Expand the original equation. Using the formula
L L to trace back the solution.
2
2
=0
One more step,
=0
or
L
305
Appendix A Solutions to Selected Problems
=0
=1
or
Problem 1.6.10
Multiplying the DE by
makes the equation
+(
−1
=0
The DE can be made exact by multiplying both sides by the
integration factor
3
= exp \
] = > .
,
M( ,
( ,
= ,
=
( ,
Problem 1.6.11
( ,
¬
¬
=
It is an exact equation since
M ³,
306
+ -( )
=
¬M
= +- ( )=
¬
-( )=− > ⇒The solution is then found to be
M( ,
=0
= −
>
=
+
= cos
,
XY
X
1
−
,
X
X
=
=1
+ exp
=
1
−
ln
¬
1
=
¬
= 0.
-
>
>(
>
Appendix A - Chapter 1
=
(cos
ln
-
= sin + ln + -( )
¬M
= + - = + exp( )
¬
- = exp( )
-( ) = exp( )
Thus, the solution is M( , ) = sin + ln + exp( ) = 1.
Problem 1.6.12
Method I: Exact DE
(exp( ) + cos ) + ( exp( ) + sin ) = 0
= exp( ) + cos
= exp( ) + sin
= exp( ) + cos
= exp( ) + cos
¬M
= exp( ) + cos
=
¬
M( , ) =
= exp( ) + sin + A( )
¬M
= exp( ) + sin + A ( ) = = exp( ) + sin
¬
exp( ) + sin + A ( ) = exp( ) + sin
A( )=0
A( ) = 1
M( , ) = exp( ) + sin + 1 = 1
exp( ) + sin = 1(
Method II: Substitution
exp( ) + cos + ( exp( ) + sin ) = 0
L = exp( ) + sin
L = exp( ) + exp( ) + sin + cos
L = ( exp( ) + sin ) + exp( ) + cos
L =0
L=1
exp( ) + sin = 1
307
Appendix A Solutions to Selected Problems
1.7 Riccati DEs
Problem 1.7.1
Since
We have
=
=
+
(
(
−
1
L
1
L′
L
Put these back to the DE, we have
1
2 ( 1
1
L = A( ) E ( +
+ F + ž( ) E ( + F + 1( )
(−
L
L
L
L
Since ( is a solution of the DE, we know ( = A( ) ( +
ž( ) ( + 1( ). This gives
A( ) ž( )
1
(
+
− L = 2A( ) +
L
L
L
L
That is
L + (2A ( + ž)L = −A
Problem 1.7.2
Let
We have that
=
+
=1−
1
L
1
L′
L
Thus, the DE becomes
1
2
1
1− L + +
+
=1+
L
L L
That is
L −2 L =1
Let
( ) = exp E −2
308
F
Appendix A - Chapter 1
)
= exp(−
( ( )L) = ( )
We have that
This gives
)
L = exp(
)\
= exp(
Finally, we can have
=
+
1
L
exp(−
√8
erf
2
)
+ 1]
√8
= + exp −
\ erf
2
Note: The error function is defined as
2
erf
=
exp −
√8 B
Problem 1.7.3
Substitute
L
Integrate
L
−
=1−
L
1
=
L
⇒
−
+ 1]
=1−
1
1
= 1 + E− F
L
4
L
L
1
−
=
⇒ −4 L =
L
4L
1−
−4
−4L =
L=
+ ⇒ L=−
L
>(
L
+
4
309
Appendix A Solutions to Selected Problems
Since
1
=
L
−
1
−
⇒ L=
=−
4
+
+
=−
Problem 1.7.4
Substitute
and
+
4
=
+
1
−
1
L
=1−
L
L
=
1
Ù
The original DE becomes
1
L
1
1 − − 13 \ + E + F ] + 26 E + F = 1
L
L
L
13
L
+
= 0,
L = −13,
L = −13 + 1
L
L
Put this back, and the solution to the original DE is
1
= +
1 − 13
Problem 1.7.5
Substituting
+
1 Ù
Ù
Substituting into the original DE, we get
We get
310
=−
Ù
Appendix A - Chapter 1
+
+2 Ù+1=0
which is a 1st-order linear DE and solvable with the integrating
factor method
= exp E
+2
F
Apply the integrating factor to the both sides of the Ùfunction,
and we will have:
Ù
+
+2
Ù+
=0
L =
Using the formula
whose G.S. is
Ù=−
1
L+L
Ù =−
Finally, the G.S. for the original DE is
1
= + = −
Ù
´
where
= exp E
Problem 1.7.6
′=
+4
−
=
, we get
+1
+2
−1
F
+4
+ §, we have
(
1 L
1
1
=E + F +4 E + F−
+4
L
L
L
Simplifying it, we have
L′ + 2 + 4 L = −1
This is a 1st-order linear DE, where
h
= 2 + 4 ,Ÿ
= −1
= exp$´ 2 + 4
% = exp 2 +
This is a Riccati DE. Substitution
−
(1)
311
Appendix A Solutions to Selected Problems
Multiplying on both side of Eq-(1) by
, we have
L ′=−
exp ´ 2 +
+1
L
=−
exp 2 +
So the G.S. for original DE is
1
exp 2 +
= + = −
L
exp ´ 2 +
+1
Problem 1.7.7
The DE is a Ricatti Equation with the given solution (
sin , So the necessary substitution is
1
1
= ( ( ) + = sin
L
L
L
′ = cos −
L
Substituting back into the original DE gives
1
2 cos − sin
šsin
›
L
L
cos −
=
L
2cos
2 sin
1
sin
2 cos − sin
L
L
=
2cos
1
tan
L
= cos +
+
L
2cos
The DE can then be rewritten as
1
L + tan L = − sec
2
Multiply both sides by the integration factor
( ) = exp E tan
F = exp(ln|sec |) = sec
1
sec L + sec tan L = − sec
2
312
=
(sec L)
=−
1
2
Appendix A - Chapter 1
sec
1
sec L = − tan + 1
2
1
1 cos
L = − sin
2
Back substituting for
L = ( − sin >(
gives
1
1
= − sin
1 cos
2
− sin
2
sin
=
cos − sin
Problem 1.7.8
It is Riccati DE and we know one P.S. is çè =
1
1
+
= (+ =
L
L
1 L
′=2 −
L
The original DE becomes
1
1
1 L
= E + F + v( ) E + −
2 −
L
L
L
or
L
+ (v( ) + 2 )L = −1
If v( ) ≡ −2
the solution is
If v( ) ≠ −2
, for any , then
§
( ) = exp$´ (v( ) + 2
+
)
. We substitute:
F+2 −
= −1, L( ) = − + 1 . So
1
− +1
, the integrating factor is
=
é
% = exp E´ v( )
+
2
3
F
313
Appendix A Solutions to Selected Problems
Multiplying on both sides by the integrating factor, we have
2
2
Eexp E´ v( ) +
F LF = − exp E´ v( ) +
F
3
3
2
exp E´ v( ) +
FL
3
2
= − exp E´ v( ) +
F
+1
3
2
2
L = − exp E−´ v( ) −
F exp E´ v( ) +
F
3
3
2
F
+ 1 exp E−´ v( ) −
3
Substituting back, we have
=
+ E− exp E−´ v( )
−
2
3
F
exp E´ v( )
+ 1 exp E−´ v( )
−
2
3
+
2
3
>(
FF
Problem 1.7.9
The DE is another Riccati DE with a given solution of
=
Using the substitution
1
+
=
L
L
=2 −
L
The DE then becomes
1
1
L
\2 − ] + E + F − E + F = 2
L
L
L
2
1
2 − L + + − −
−
=2
L
L
L
L
314
F
Appendix A - Chapter 1
−
L
L −
L +
L
L
=−
=
1
1
L
.
Multiplying both sides by the integration factor
1
( ) = exp E
F=
gives
L +L =−
( L)
Back substitute
L=
1
L=
+
=−
1
1
1
+1
=
1+1
L = ( − )>(
1+1
1
=
−
=
+
1
.
1+1
315
Appendix A Solutions to Selected Problems
Chapter 2 Mathematical Models
2.1 Population Model
Problem 2.1.1
From the population model, we can get
h
= (œ − ¼)h
= (/( h − / h)h
= (/( − / )h
where /( and / are both constant. Let / = /( − / , which is
also a constant.
(1) From the problem, we know that
/(
/
œ=
and ¼ =
√h
√h
where /( and / are both constant. Thus
h
= (œ − ¼)h
/(
316
−
/
Fh
√h √h
= (/( − / )√h
Let / = /( + / which is also a constant. We have
h
=/
√h
2√h = / + 1(
1
h( ) = E / + 1F
2
Plugging h(0) = hB into it, we can get 1 = hB . Thus
1
h( ) = E / + VhB F
2
(2) Given hB = 100 and h( = 6) = 169, we can get
=E
Appendix A - Chapter 2
1
169 = E × 6/ + √100F
2
This gives / = 1. Thus, it is straightforward to evaluate h(12) =
16 = 256.
Problem 2.1.2
(1) From the problem, we can get
œ = /( h and ¼ = / h
h
= /h
Solve this separable DE, we can get
1
= −/ + 1
h
That is
1
h( ) =
1−/
Plugging h(0) = hB into the solution, we can get
1
hB =
1
Thus, the population is
1
h( ) =
1
hB − /
hB
=
1 − /hB
(2) When → 1⁄/hB we know that /hB → 1 . Thus, (1 −
/hB ) → 0 and
hB
→∞
h( ) =
1 − /hB
(3) Given hB = 6 and h(10) = 9, we can get
6
9=
1 − / ∙ 6 ∙ 10
1
/=
180
317
Appendix A Solutions to Selected Problems
Thus, the doomsday occurs at
1
=
= 30 months
/hB
(4) Suppose œ < ¼, we know that / < 0. Thus, we know that
>0
1 − /hB > 0,
and
h( ) → 0,
→∞
Problem 2.1.3
According to the discussion in textbook, we know that we can
always write the logistic equation into form
h
= /h( − h)
where
= /, and
=
Since for / = > 0, the limiting population will be
easily calculate that
žB ⁄hB žB hB
= =
=
ÂB
ÂB⁄hB
, we can
Problem 2.1.4
(1) This is a 1st-order separable DE.
h
= / +1
h(ln − ln h)
Let ln h = ¹, then h ¹ = h. Substitute h by ¹ and ¹,
h ¹
¹
=
= / +1
h(ln − ¹)
ln − ¹
− ln(ln − ¹) = / + 1
ln(ln − ln h) = −/ − 1
ln − ln h = H exp(−/ )
h( ) = exp(ln − H exp(−/ ))
h( = 0) = hB
318
Appendix A - Chapter 2
H = ln − ln hB
h( ) = exp(ln − (ln − ln hB ) exp(−/ ))
(2) Since hB > ,
ln − ln hB < 0
h
h( )
Figure A.1 If hB G
h
, population will decrease and approach value
as → ∞.
h( )
Figure A.2 If hB <
, population will increase and approach value
as → ∞.
319
Appendix A Solutions to Selected Problems
h
(ln
− ln hB ) exp(−/ ) < 0
= /h(ln
− ln hB ) exp(−/ ) < 0
This means h( ) is a monotonically decreasing function of .
When → ∞, h( ) → .
(3) Since hB < ,
ln − ln hB > 0
(ln − ln hB ) exp(−/ ) > 0
h
= /h(ln − ln hB ) exp(−/ ) > 0
This means h( ) is a monotonically increasing function of .
When → ∞, h( ) → .
Problem 2.1.5
The DE is
h
= œB exp(−v ) h
which is a separable DE. We have
h
= œB exp(−v )
h
œB
ln h = − exp(−v ) + 1
v
Since h 0) = hB , we can get
œB
1 = ln hB +
v
Thus,
œB
ln h = ln hB + (1 − exp(−v ))
v
œB
h( ) = hB exp ° (1 − exp(−v ))±
v
Now it is clear that for → ∞, exp(−v ) → 0 and hence
œB
h( ) → hB exp E F
v
320
Appendix A - Chapter 2
Problem 2.1.6
Since / G 0,
> 0 and h G 0, for the DE
h
= /h( − h)
we consider the following two cases.
(1) If G h, we have
h
= /h( − h) > 0
Thus the function h( ) is monotonically increasing WRT .
h
/ increases
/ increases
Figure A.3 Function h( ) is close to value
as / increases.
Since h/ is the slope of the graph, as / increases, the slope
of the curve increase, then it will go closer to .
(2) If
< h, we know
h
= /h( − h) < 0
We know h( ) is a monotonically decreasing function in terms
of ., see fig.1(Blue lines). As / increases, the absolute value
of h⁄ , say | h⁄ | increases, and the slope of the curve is
larger, thus the curve goes closer to .
321
Appendix A Solutions to Selected Problems
From above, we can conclude, as / increases, the curve goes
closer to the limiting population .
Problem 2.1.7
Birth rate: œ = 0 , and death rate: ¼ = /⁄Vk , where / is a
constant
By general population function, we have
h
= (œ − ¼)h
/
Fh
√h
= −/√h
with I.C. h(0) = hB . Solving this 1st-order separable DE, we
have
h
= −/
√h
2√h = −/ + 1
h(0) = hB ⇒ 1 = 2VhB
Thus
/
h
= E−
VhB F
2
From the I.C., we know h 0) = hB = 900, and h(6) = 441 ,
thus
6/
441 = E−
√900F ⇒ / = 3
2
Therefore
3
h( ) = E−
30F
2
All the fish being dead means h( ) = 0, then we have
3
30F = 0 ⇒ = 20 weeks
E−
2
= E0 −
322
Appendix A - Chapter 2
Problem 2.1.8
From the I.C., we have
ž(0)
8
Â(0)
6
=
=
and =
=
h(0) 120
h (0) 120
To solve this DE, we should write the DE into standard logistic
equation form
h
= /h( − h)
where
/ = , and
=
Since we know the solution to the logistic equation should be
hB
h( ) =
(
hB +
− hB ) exp(−/ )
Thus, for h( ) = .95 , we can solve for
hB
0.95 =
(
hB +
− hB ) exp(−/ )
gives = 27.69
Problem 2.1.9
With v = /( h, œ = / h and / = / − /( , population model is
given by
h
= ( œ − v)h = (/ − /( )h = /h
Take the anti-derivative and gives
1
− =/ +1
h
(1) Coupling with I.C., where ( œ − v) = /h > 0, and locate the
solution as
>(
1
h =E −/ F
hB
(2) Calculate the doomsday formula.
The denominator is zero, which satisfies the definition of
doomsday. So, time is given below as
323
Appendix A Solutions to Selected Problems
1
hB /
(3) Calculate the doomsday time.
First, calculate the difference between the birth rate and death
rate.
@=
h(12) = šl − ( œ − v) ›
(
D
>(
=š
>(
(
−
12/›
B((
>9
= 4027
/ ≈ 2.0745 ∙ 10
Second, calculate the time.
1
1
=
≈ 23.970
@=
hB / 2011(2.0745 ∙ 10>9 )
(4) Calculate the population.
Due to (v − œ) = −/h > 0, we have (−/) > 0. So,
>(
1
h = E − / F → 0, → +∞
hB
As time goes to infinity, the population goes to zero.
Problem 2.1.10
The population modeling is given by
(
/
h
= (œ − v)h = E0 −
F h = −/h
√h
The solution is followed by
2√h = −/ + 1
When we couple with the I.C., we get
1 = 2√4000
2√h = −/ + 2√4000
When we plug in the population value of 2014 WRT time 11, the
unknown coefficient is
2√4000 − 2√2014
≈ 3.3396
/=
11
So, the time for all the fish in the lake to die (let h( ) = 0)
1
@ = = 37.876
/
324
Appendix A - Chapter 2
If the population does not change with time, / = 0. So we need
4000 rather than 2014.
325
Appendix A Solutions to Selected Problems
2.2 Acceleration-Velocity Model
Problem 2.2.1
(1) This is a separable DE which can be written as
L
= −/
L
ln L = −/ + 1
L = 1 exp(−/ )
From the I.C., we know that L(0) = LB . This gives 1 = LB .
Hence the velocity function is
L( ) = LB exp(−/ )
Also, we know
L=
= LB exp(−/ )
LB
= − exp(−/ ) + 1
/
Since (0) = B , we have
LB
1= B+
/
Thus
LB
( ) = B + (1 − exp(−/ ))
/
(2) As → ∞, we know that e>e → 0. Thus
LB
→ B+ ,
→∞
/
thus
Problem 2.2.2
§
§D L
326
L
L
⁄
= −/L
=
B
−/
Appendix A - Chapter 2
(
>
(
−2L > + 2LB = −/
(
2L > =
L( ) =
and
VLB / + 2
$VLB / + 2%
= L( ) =
( )=−
VLB
4LB
4LB
$VLB / + 2%
4VLB
+1
/$2 + / VLB %
Putting the I.C. (0) = B into it, we can have
2VLB
1= B+
/
Thus
( )=
B
+
As → ∞, we know that
2
2VLB
2
\1 −
]
/
/ VLB + 2
→ 0,
→∞
/ VLB + 2
Thus, it is easy to know the body only goes a finite distance
before it stops.
Note
As the power is greater than 3/2, the unknown will be found in the numerator
part of function ( ), which means as → , → . You can check it by
yourself.
∞
Problem 2.2.3
Let
L=
∞
Y
327
Appendix A Solutions to Selected Problems
and consider L to be an function of Y. We can get
Y
L
L Y
L
=
=
⋅
=L
Y
Y
Thus, the DE becomes
‡ ‰
‡ ¨
L
=−
+
L
(Î − Y)
Y
Y
By separation of variables, we have
‡ ‰
‡ ¨
L L = E−
+
F
(Î − Y)
Y
‡ ‰ ‡ ¨
L
=
+
+1
2
Y
Î−Y
Since Y (0) = L(0) = LB and Y(0) = 0 , we know L(Y = 0) =
LB . This gives
LB ‡ ‰ ‡ ¨
−
−
1=
2
X
Î−X
Hence the velocity function is
L = Ë2‡
1 1
− F + 2‡
Y X
‰E
¨
1
1
E
−
F + LB
Î−Y Î−X
The earth and moon attractions balance at the point where the
right-hand side in the acceleration function vanished, which
means
‡ ¨
‡ ‰
+
=0
−
(Î − Y)
Y
solve this equation for Y, we have
Y=
V
‰Î
V ‰−V ¨
If we substitute this value of r, ¨ = 7.35 × 10 (kg), Î =
3.844 × 10Z (m), ‰ = 5.975 × 10 (kg), X = 6.378 × 10û
(m), and set L = 0(to just reach the balancing point), we can
solve the resulting equation for LB = 11109 m/s. Note that this
328
Appendix A - Chapter 2
is only 71 m/s less than the earth escape velocity of 11180 m/s,
so the moon really does not help much.
Problem 2.2.4
(1) The DE for this motion is
L
K
= −K- − /L
L=
⇒K
L
=K
= KL
L
= −K- − /L
/
L
K
This is a separable DE, and by separation of variables, we have
L L
=
+1
/
−- − K L
Let
Ko + KK
/
⇒ L=−
o
o = −- − L ⇒ L = −
/
/
K
Ko + K- K
K
o=š ›
š1 + › o
/o
/
/
o
K
= š › (o + -à=|o|)
/
K
/
/
⇒ š › E−- − L + - à= _−- − L_F = + 1
/
K
K
K
/
/
(0) = 0
⇒ 1 = š › E−- − LB + - à= _−- − LB _F
|
L(0) = LB
/
K
K
K
/
K- + /L
⇒ ( ) = š › E− (L − LB ) + - à= _
_F
/
K
K- + /LB
At highest point ¨w we have L = 0
K
/
K_F
¨w = š › E LB + - à= _
/
K
K- + /LB
(2)
L
L
/L
K
= −K- − /L ⇒
= −- −
K
⇒L
L
L
= −- −
329
Appendix A Solutions to Selected Problems
which is a separable DE, we have
L
=
/
−- − L
K
/
K
−K- − /L
K
+ 1 = − à= _−- − L_ = − à= _
_
K
/
K
/
K
−K- − /LB
L(0) = LB ⇒ 1 = − à= _
_
K
/
K
−K- − /L
−K- − /LB
K
( ) = − à= _
_ + à= _
_
/
K
/
K
K
K- + /LB
= à= _
_
/
K- + /L
At highest point, we have L = 0
K
K
K- + /LB
/LB
à= _
_ = à= _1 +
_
¤¢ =
K
K/
/
Problem 2.2.5
(1) The DE is
L
= −/L
ln L = −/ + 1
L = 1 exp(−/ )
Since L(0) = 40 and L(10) = 20, we can solve that
ln 2
1 = 40 and / =
10
which gives
L = 40 exp(−/ )
and similarly,
gives
B
330
=
=L
B
40 exp(−/ )
Appendix A - Chapter 2
when → ∞, we have
(2) The DE becomes
This gives
=
40
(1 − exp(−/ ))
/
=
400
≈ 577 ft
ln 2
L
= −/L
1
=/ +1
L
1
/ +1
Given L(0) = 40 and L(10) = 20, we can get
1
1
and / =
1=
400
40
Thus
40
L=
1 + 40/
And finally
40
= L( ) =
1 + 10
which is
We can get
L=
= 400 ln E1 +
Since (0) = 0, we get 1 = 0. Thus
F+1
10
( ) = 400 ln E1 +
and (60) = 400 ln š1 + (B› ≈ 778 ft.
ûB
F
10
331
Appendix A Solutions to Selected Problems
Problem 2.2.6
L
K
M‡ = KMX = /L
Figure A.4 A system of moving up object.
(1) The total force on this object is M = −(M[ + MÅ ) , by
Newton’s law, M = K = K
§
thus M = −K- − /L = K
§
which is a 1st-order separable DE.
Solving this DE with I.C. L(0) = LB , we get
K/
K› Ï k E−
F−
L( ) = šLB +
/
K
/
(2) At the point it reverses, the velocity of the object vanishes at
L( ) = 0, thus
K/
K› exp E−
F−
=0
šLB +
/
K
/
K
K= − ln
/ K- + /LB
It depends on the initial velocity LB (the bigger it is, it takes
longer to reverse the direction).
(3) When the object keeps a constant speed, the acceleration
¨¥
vanishes. Thus, = −K- − /L = 0 ⇒ L = −
, which is
independent on the initial velocity LB .
332
e
Appendix A - Chapter 2
L = −LB cos(v)
Problem 2.2.7
gives
cos(v) =
V
+
L = −LB sin(v) − o
and
gives
sin(v) =
+
−LB sin(v) − o sin(v)
o
=
=
+
−LB cos(v)
cos(v) LB cos(v)
=
=
Substitute
=
+
+
=
V
o V
°
LB
o
+
LB
+
\
Ó1 +
o
Ë1 +
LB
=
=
+
=
+
+
±
]
o
V1 +
LB
o
V1 +
LB
o
ln$ + √1 + % = − ln( ) +
LB
=
333
Appendix A Solutions to Selected Problems
when o ≥ o¡
If o > LB
When o < o¡
…=
When o = 0
( )=
2
( )=
Y
B
V
( )=
2
(>
°š ›
(>
lim š ›
→B
2
(>
°š ›
…=
−š ›
[
§D
→∞
[
§D
(ú
−š ›
Y
=
B
§D
(ú
[
§D
(>
°š ›
+
[
§D
B
[
§D
Ë1 + E
(ú
−š ›
B
§D
±
±
F
±=0
Problem 2.2.8
Let ( be the time you accelerate, be the time you decelerate.
= , which is the time you travel from one stop sign to
(+
another. Note that
( ( =
Since your initial velocity and final velocity are both zero.
So
1
1
=…
( ( +
2
2
plug in
and we have
334
=
(
(
Appendix A - Chapter 2
which gives
(
=Ë
1
2
(
2
( (
…
+
+
(
1
2
and
(
(
=…
=Ë
(
2
(…
+
), the result stays the same.
If L¨ ≥ ( ( (=
If L¨ < ( ( , then you should accelerate to L¨ (need time ( ).
Then travel at a constant velocity L¨ for time
. Finally
decelerate from L¨ to velocity 0 (need time ).The total time
is = ( + + . Here
L¨
L¨
gives …( =
( =
2 (
(
L¨
L¨
=
gives … =
2
and
L
L
…− ¨ − ¨
2 ( 2
=
… = … − …( − … gives
L¨
Finally
L
L
…− ¨ − ¨
L¨ L¨
2 ( 2
=
+
+
L¨
(
Problem 2.2.9
From the problem, we know that with closed parachute, we have
L
K
= K- − /L
This is a linear DE and we have
/
K+ 1( exp E− F
L=
K
/
Given L(0) = 0, we know that
K1( = −
/
335
Appendix A Solutions to Selected Problems
Hence
And we can get
L=
=
L
=
L
K/
E1 − exp E− FF
/
K
K /
K+
exp E− F + 1
/
K
/
Since (0) = 0, we have
K 1 =−
/
Thus
K
/
K
K( )=
E + exp E− F − F
/
K
/
/
Suppose he open the parachute at = ( , we know that
KK
/(
K
E ( + exp E−
F− F
( =
/
/
/
K
and
K/(
L( =
E1 − exp E−
FF
K
/
Starting from ( , the parachute will be open and thus, the DE
becomes
KL = K- − =/L
which we can get
K=/
L=
+ 1 exp E−
F
=/
K
where from L( ( ) = L( , we have
(= − 1)/ (
= − 1 K=/ (
K1 =
exp E
F−
exp \
]
K
K
= /
/
For , we have
=
336
Appendix A - Chapter 2
K1 K
=/
−
exp E−
F+1
=/
=/
K
where 1 satisfies ( ( ) = ( . After finding 1 , we can have
=„− (
Solving for
such that ( ) =
and make L( ) = LB , we
can find the optimal ( .
=
Problem 2.2.10
We start with the usual force equation
L
L
L
L
‡ K
K
=K
=K
= KL
=−
( + X)
where M is the mass on the Earth, m is the mass of the projectile,
y is the height from the earth’s surface and Y = + X.
Hence we will separate this equation and integrate
§
‡
L L=
−
( + X)
§D
B
to obtain
L = LB − 2‡
We found the velocity to be
L
= ËLB − 2‡
1
E
X
X
1
1
E
X
X
1
F
F = ËLB − 2‡
The velocity will become imaginary when
1 1
LB I 2‡ E
F
X Y
1
E
X
1
F
Y
Problem 2.2.11
Consider L to be a function of . Thus,
L
L
L
=
=
=L
and the original DE can be written as
337
Appendix A Solutions to Selected Problems
‡
− œ exp(− )
( + X)
‡
− œ exp(− )F
L L = E−
( + X)
‡
L
=
+ œ exp(− ) + 1
+X
2
Plugging I.C. (0) = 0 and (0) = LB , we have
LB ‡
=
+œ+1
X
2
LB ‡
1=
œ
2
X
2‡
2œ exp
1
L = LB −
X X
max occurs when L = 0, which gives the implicit solution:
2‡ max
2œ exp
max = LB − 2œ
X X
max
For the total time @, we should consider
L
and it gives
@=
ËLB −
max
B
L
=−
2‡
°LB −
L=
X X
2‡
Problem 2.2.12
Before opening
In vertical ( -) direction:
338
2œ exp
X X
K
L
2œ exp
= K-
1 =
1 ±
>
(
Appendix A - Chapter 2
§\
B
1
= 2
⇒
L =
=- ,
„ 1
= 2 2
B
B
,
(
(
,L = -
=
B
„
= Ë ,L
-
In horizontal ( -) direction:
There is no force, constant velocity LB
1
= LB ⇒ = , = LB
2
=
So the length
…( =
B
æ
Ë1
1 LB „
= Ë
2
LB
E
„
After opening
In vertical ( -) direction:
K
K L
=
§\æ K- − vL
§\
,
F
L
B
(
= LB
=
-
(
=-
(
= V-„
1
= 2 LB
„
= LB Ë
-
J
§D Ó
¥
B
(
Ë1
LB
V-„
ln Þ
2LB
E
LB
-„
Ë
LB
F
1ß
= K- − vL
,L
KKv
+ šV-„ −
› exp E− F
v
v
K
KKv
=
+ šV-„ −
› exp E− F ,
v
v
K
B
KKv
=
E
+ šV-„ −
› exp E− FF
v
v
K
B
KK
Kv
K
K=
− šV-„ −
› exp E− F + (V-„ −
)
v
v
v
K
v
v
=
339
Appendix A Solutions to Selected Problems
„ KK
Kv
K
K=
− šV-„ −
› exp E−
F + šV-„ −
›
K
2
v
v
v
v
v
We could get implictitly.
In horizontal ( -) direction:
L
K
= −vL
K
vL
v
F
K
§D
B
v
v
= LB exp E− F ,
=
LB exp E− F
K
K
B
B
KLB
v
KLB
K
v
=−
exp E− F +
⇒ = − ln(1
K
v
KLB
v
v
K
Kv
K
KK− šV-„ −
› exp š− › + šV-„ −
›
=
v
v
K
v
v
v
K v
K
Kv
=−
ln E1
F
šV-„
› 1
v
KLB
v
v
KLB
KK
šV-„
›
v
v
K1
K=
+ šV-„
›
v
LB
v
vLB − K
KLB
v
KLB
=−
exp E−
F+
v
K
v
So the length
−
… =
=
B
B
Í
Í
§]
Ë1
S
^1
E
L =
F
Kv
vLB K
R
S
, L = LB exp E−
1
šV-„
LB
R
The length of the trajectory is … = …( + …
(
KU
›U _
v
T
T
Problem 2.2.13
From the problem, we know that the resistance in vector form is
340
X̀Ñ = −v|LÑ|
LÑ
|LÑ|
Appendix A - Chapter 2
= −vLÑ
Now it is clear to see the resistance in and directions are
independent from each other. Thus, we only need to consider the
axis. Before the parachute opens, we have
L
= −-
L = −- + 1(
given L (0) = 0, 1( = 0 we have
L = −1
1
=− 2
Plugging in 0) = „, 1 = „
1
=„− 2
(
If the man opens the parachute at = „, the time it takes from
which gives
start to open the parachute is
1
1
„=„− 2
2
(
=Ë
„
-
(
L = −V-„
After opening the parachute, the equation of motion becomes
L
v
= −- − L
K
which is a separable DE. The G.S. is
K
v
− ln š L + -› = + 1
v
K
and at that time,
341
Appendix A Solutions to Selected Problems
Given L (0) = −V-„, we have
K
v
1 = − ln š- − V-„›
v
K
Thus,
v
K - − K V-„
ln
=
v
-+KL
v
› − K- = vL
$K- − vV-„% exp š−
K
Kv
KL =š
− V-„› exp š−
›−
v
K
v
and
v
KK K− V-„› exp š−
›−
+1
=− š
K
v
v v
Given (0) = 0, we have
K - KV-„
−
v
v
(
and solve for ( ) = „ gives an implicit equation
1 =
„ K
v
1 v„
= ÞË − ß šexp š−
› − 1›
- v
K
2 K-
The total time is =
(
+
.
In the case he opens the parachute at
(
and at that time
3„
=Ë
2-
= „, we have
(
3-„
L ( ( ) = −Ë
2
The equation after opening the parachute then becomes
342
Appendix A - Chapter 2
L =Þ
K3
v
− Ë -„ß exp š
›
v
2
K
and the implicit expression for
3„
3 v„
= ÞË
24 K-
is
K
v
ß šexp š
v
K
›
Kv
1›
Problem 2.2.14
This is a problem of Newtonian mechanics M¥ M = K . Total
weight of the man and his parachute m with acceleration ,
gravitational force M¥ and a drag force M . Let be the distance
above the earth’s surface, with positive direction downward, =
§
, where L =
is the velocity and M¥ = −K-.
Since the drag force is assumed to be proportional to velocity,
M = −/L (with closed paracute) and M = −=/L (with open
parachute).
The deployment occurs at time B ,
During initial fall with closed parachute,
L
K
= K- − /L, L(0) = 0
This is an IVP problem. Solve this separable DE. We obtain,
/L
L
=-−
K
or
L
=
/L
-− K
/L
K š- − K ›
−
=
/ š- − /L ›
K
343
Appendix A Solutions to Selected Problems
K
/L
ln E- − F = + 1
/
K
/L
/
K
/
-−
= 1( exp E− F ⇒ L( ) = E- − 1( exp E− FF
K
K
/
K
= 0, L(0) = 0
K
/
1( = = L = - E1 exp E
FF
/
K
K
K
/
= - E + exp E− FF + 1
/
/
K
K
(0) = 0 ⇒ 1 = −
/
So the distance of close parachute fall is
K
K
/ B
K
- E B + exp E−
FF −
( =
/
/
K
/
At B ,
/ B
K
E1 exp E
FF
T B =
K
/
Starting with the opening parachute B time, the DE becomes
L
K
= K- − =/L
−
K
whose solution is
Since
K
L( ) =
L
L
= K- − =/L
L = K- − =/L
K
=/
E- − 1 exp E−
FF
=/
K
L( B ) =
K
- E1
/
=/ B
1
F =- E
1 = exp E
=
K
K
=
E- − 1
=/
344
/ B
FF
K
/ B
1 exp E
FF
K
=/
exp E−
FF
K
exp E
Appendix A - Chapter 2
=
K
K
/
-E +1
exp E− FF + 1
=/
=/
K
( B) = (
which means
K
K
/ B
-E B +1
exp E−
FF + 1
=/
=/
K
K
K
/ B
K
= - E B + exp E−
FF −
/
/
K
/
One can easily find the constant 1
„= (+
K
K
/(
=
-E ( +1
exp E−
FF + 1
=/
=/
K
K
K
/ B
K
= „ − - E B + exp E−
FF −
K
/
/
/
The falling time t( is in principle solvable from the above
equation, but it is tedious and may require a numerical solution
technique. (Reaching this point without the actual solutions for
1 and ( is sufficient to gain full marks.)
Therefore the total falling time is ( + B , the speed he hits the ground
is
K
=/
L( ( ) =
E- − 1 exp E−
FF
=/
K
We can adjust the B to enable the quickest fall and lightest hit on
the ground.
Problem 2.2.15
Suppose the speed of the jet is LB . And the jet needs to know
which angle to go in, call it v. Then, without the wind the jet’s
velocity will be [LB cos(v) , LB cos(v)]. With the wind the jet’s
velocity can be [LB cos(v) , LB cos(v)L[ ].
345
Appendix A Solutions to Selected Problems
12
Wind
11
L =
(2)
⇒
(1)
L =
L2
v
Figure A.5 The jet landing model.
= LB cos v = LB
= LB sin(v) + L[ = LB
=
L0
V 2+ 2
L0
+ Lo
V 2+ 2
+
V
⇒
V
=
Problem 2.2.16
According to Newton’s Law, we have
L
K
= vL + œL
Since
then
Plugging this back
346
L=
L
=
=L
L
L
L
+
+
+ L[
(1)
(2)
2
Lo
Ë1 + š ›
L0
KL
K
L
L
Appendix A - Chapter 2
= vL + œL
= v + œL
K
L=
v + œL
K
ln(v + œL) = + 1
œ
At = 0, it gives L = LB . So,
K
1 = ln(v + œLB )
œ
Therefore,
K
v + œL
(L) = ln E
F
œ
v + œLB
The maximal distance is
v
K
ln E
F
¨ (L = 0) =
v + œLB
œ
If we double the initial speed,
K
v + œL
(L) = ln E
F
œ
v + 2œLB
K
v
ln E
F
¨ (L = 0) =
œ
v + 2œLB
Problem 2.2.17
Assume the following speeds.
LB and L( are entering and exiting Medium-1, respectively.
L( and L are entering and exiting Medium-2, respectively.
L and L are entering and exiting Medium-3, respectively.
For Medium-1: the equation of motion is
K
§
= −/L or K
§
§D
B
§æ
Integrating both sides, we get:
L=
a
−
= KL
/
K
§
= −/L
347
Appendix A Solutions to Selected Problems
/
…
K
For Medium-2, treat it using the same strategy for Medium-1.
L
= −/L
K
Thus, the solution is given by
L( = LB −
or
K
L
= KL
Integrating both sides, we get
§æ
1
L=
§D L
a
B
L
−
= −/L
/
K
Thus, the solution is given by
/
/
/
L = L( exp E− …F = (LB − …) exp E− …F
K
K
K
For Medium-3, we know easily (from Medium-1)
/
L =L − …
K
Force L = 0 and we have
/
L =L − …=0
K
or
/
L = …
K
Trace back and find the final solution
Problem 2.2.18
s$ =
b
b
d Eè + efg E dFF
c
c
Method I:
For all three cases, we have the following DE.
L
L
L
K
=K
=K
L = −/L Z
L( = 0) = LB
348
Appendix A - Chapter 2
B
We can easily get the following.
§D
L (>- L = −
B
h
/
K
The bullet stops at the max distance Y .
K §D (>K LB >=
L
L
=
Y
/ B
/ 2−v
For Case 1, v = 1,
K LB(
=
Y
/ 1
For Case 2, v = 3/2,
(/
K LB
K (/
=
= 2 LB
Y
/ 1/2
/
For Case 3, v = 2,
Y →∞
Method II:
(i)
So we have
K
L
= −/( L
L
/(
= −E F
L
K
/(
ln L = − E F
K
/(
+ 1F
L = exp E−
K
/(
F
L = 1( exp E−
K
L(0) = LB = 1(
/(
F
K
/(
= LB exp E−
F
K
/(
F
= LB exp E−
K
L( ) = LB exp E−
349
Appendix A Solutions to Selected Problems
=
LB exp š−
−
/(
K
/(
›
K +1
/( LB
/(
exp E−
F+1
K
K
/( LB
(0) = B = −
+1
K
/( LB
1 = B+
K
=−
/( LB
/(
E1 − exp E−
FF
K
K
e
As goes to infinity, exp š− ¨æ › goes to 0. Thus x goes to
/( LB
B+
K
So the total distance travelled is x at infinity minus B , or
/( LB
K
(ii)
L
K
= −/ L
So we have
( )=
B
+
L>
L
(
/
L
K
/
L=−
K
=−
/
L>
=−
+1
1
K
−2
(
/
−1
+
L> =
2
2K
>
/
L = E1( +
F
2K
L(0) = LB = 1(>
1
1( =
VLB
350
Appendix A - Chapter 2
1
/
L=\
+
]
VLB 2K
=
=
1
VLB
>
L
+
/
2K
/
=
2K
2K
+
=
/
2K
>
/
>(
/
2K 1
+
=−
] +1
\
/ VLB 2K
Putting in I.C. (0) = B :
2K
VLB + 1
B =−
/
>(
2K 1
/
2K
( )=
+
\
] + B+
VLB
/ VLB 2K
/
So we have
=
/ VLB
2K
2K
( )=
+
\
]
/ 2KVLB 2KVLB
>(
+
B
+
2K
VLB
/
2KVLB
2K
2K
\
]+ B+
VLB
/
/ 2K + / VLB
So as goes to infinity, goes to
2K
VLB
B+
/
Thus the total distance is this value minus B , or
2K
VLB
/
(iii)
( )=
351
Appendix A Solutions to Selected Problems
K
L>
L
= −/ L
L=−
−L >( = −
/
K
/
+1
K
>(
/
+1 F
K
1
L(0) = LB =
1
/
1 >(
L( ) = E
+ F
K
LB
L=E
=
L
/
1
K
ln E
+ F+1
LB
K
/
K 1
K
(0) = B = ln + 1 = − ln LB + 1
/
LB
/
K
/
1
K
( ) = ln E
+ F + B + ln LB
/
K
LB
/
As goes to infinity, ( ) goes to infinity. So in this situation the
bullet does not stop.
=
Problem 2.2.19
The motion equation is given below.
L
v(œ + L )
=−
K
(1) Suppose the distance position function is given by , which
is a function of , only.
Rearrange the motion equation.
v(œ + L )
L
=−
K
We know the fact that
=L
352
Appendix A - Chapter 2
So, the rearranged motion equation becomes
L
L
v(œ + L )
E F
=L
=−
K
Use the separation of variable to get
2v
+1
ln(L + œ) = −
K
Coupled with I.C. to locate the unknown coefficient 1.
ln(LB + œ) = 1
So, the expression for velocity and position is given by
2v
ln(L + œ) = −
+ ln(LB + œ)
K
When the velocity decreases to zero, the bullet reaches the
farthest.
2v
ln œ = −
+ ln(LB + œ)
K
The maximum distance is
K (LB + œ)
=
ln
2v
œ
(2) Coupling the I.C. with original motion equation to give
L
v(œ + L )
=
−
Z
K
L( = 0) = LB
Using the integral table to gain
1
L
v
tan>(
=1−
K
Vœ
Vœ
v
›]
K
Apply the I.C. to locate the unknown coefficient.
The explicit expression of velocity is
L = Vœtan \Vœ š1 −
L( = 0) = LB = Vœtan$Vœ1%
The explicit expression of C is
353
Appendix A Solutions to Selected Problems
1=
L
tan>( \ B ]
Vœ
Vœ
When the velocity becomes zero, time reaches the maximum.
v
@=
K
Finally, time is
@=
tan>( \
LB
]
Vœ
Vœ
L
K ∙ tan>( \ B ]
Vœ
vVœ
As long as time is in the interval ∈ [0, @], the bullet is in
motion.
Problem 2.2.20
(0, 0)
( , )
v
Figure A.6 The ferryboat model.
( , 0)
(1) Establish the Des for the boat’s trajectory,
L =
€
L =
354
= −TB cos v
= ×( ) − TB sin v
(1)
(2)
Appendix A - Chapter 2
Ssin v =
We know that
V
+
Rcos v =
V +
Q
Eq-(2) divided by Eq-(1), we get
×( ) − TB sin v
×( )
=
=
+ tan v
TB cos v
−TB cos v
oB ( − )
oB ( − )
Ë1 + š › +
=−
+ =−
TB cos v
TB
=−
( − )
Ë1 + š › +
1
oB
TB
(2) By substitution L = / , we can transform the equation as
( − )
VÙ + L + L
L+ L =−
−
V1 + L
L =−
where
=
By separation of variables, we have
L
=−
√1 + L
−
−2
+1
2
With the I.C. (v) = 0, we find 1 = / 2 ). Finally, we have
1 −
ln šL + V1 + L › = š
›
2
( − )
L + V1 + L = exp \
]
2
Substituting back L with / ,
ln šL + V1 + L › =
355
Appendix A Solutions to Selected Problems
( − )
]
2
(3) If the river flows slowly relative to the boat,
result becomes → 0, which is a straight line.
+ Ë1 + š › = exp \
Problem 2.2.21
The net force
M = K- −
K-
→ ∞, the
− μL
(positive sign means the force is downwards).
So
K- K1
L M K- š1 − › − μL μ š μ − μ − L›
= =
=
K
K
K
K
L=
K- Kμš μ −
−
L›
μ
μt K- KK- KL( ) = exp E− F E
−
F+E
−
F
K μ
μ
μ
μ
We also know that
= L( ) ⇒
J
B
=
C
B
L( )
K- KK
K- Kμ
@› E
−
F E− F + E
−
F@
μ
μ
μ
μ
μ
K
K- KK
−E
−
F E− F
μ
μ
μ
Once we solve for @, which is the time that the egg hits the
bottom, we can just plug @ into L( )to get the impact speed (at
time @).
Alternative Method
L
L
KL
=K
=K
L = K- −
− †L
K
„ = exp š−
356
Appendix A - Chapter 2
Or
Define
L
1
†
L = - E1 − F − L
K
1
v = - E1 − F ,
We have separable equation
L
L = v + œL
L L
=
B v + œL
where L£ is the impact speed. So
§j
Next, solve the equation
œ=−
†
K
J
B
L vln(v + œL) L£ L£ vln$v + œL£ % vln(v)
−
+
_ = −
œ
œ
0
œ
œ
œ
You can solve for L£ now.
„=
357
Appendix A Solutions to Selected Problems
2.3 An example in Finance
Problem 2.3.1
(1) Establish the DE
¹ = ¹Y
¹
Which gives
− ŸB
= ¹Y − ŸB
I.C.: ¹ 0
¹B
(2) We can rewrite the DE
¹′ ¹Y
ŸB
which is a linear DE. Thus
¹ exp Y )) = −ŸB exp(−Y )
¹ exp
¹
Plug in the I.C. ¹ 0
Finally,
exp Y
exp Y
]
Y
Y
ŸB ŸB
+
+ 1 exp(−Y )
Y
Y
¹B , we have
ŸB
¹B
+1
Y
ŸB
1 ¹B
Y
Y ) = −ŸB \
¹ )=
ŸB
ŸB
ŸB
+
+ E¹B − F exp(−Y )
Y
Y
Y
Problem 2.3.2
(1) The change of loan
Thus, we have an equation
358
after time interval
= Y −Â
= Y−Â
1
Appendix A - Chapter 2
with (0) = ¹.
(2) The above DE can be solved easily by separation of variables.
Y−Â
=
+1
1
ln( Y − Â) = + 1
Y
1
1
( ) = $Â + exp$Y( + 1)%%
7 ⇒ 1 = ln(Y¹ − Â)
Y
Y
(0) = ¹
1
( ) = (Â + exp(Y + ln(Y¹ − Â)))
Y
1
= (Â + (Y¹ − Â) exp(Y ))
Y
(3) Obviously, if Y¹ − Â > 0 , the loan will grow with time.
O.W., it will decrease. Thus, the critical point is Y¹ − Â = 0, i.e.,
 = Y¹ (The Youngs only pays off the interest).
(4) Pay off in days,
1
( ) = (Â + (Y¹ − Â) exp(Y )) = 0
Y
1
Y¹
= − ln E1 − F
Y
Â
(5) Pay off in /2 days,
1
Y¹
= − ln E1 − F
2
Y
þ
where þ is the daily payment to pay off loan in half time. We
also know
1
Y¹
= − ln E1 − F
Y
þ
2
=−
1
Y¹
1
Y¹
ln E1 − F = − ln Ë1 −
2Y
þ
Y
þ
Comparing it with previous equation, we have
359
Appendix A Solutions to Selected Problems
Let
Ë1 −
Y¹
Y¹
=1−
Â
þ
= Y¹/Â and Y¹ = Â, then
√1 −
þ=
=1−
Â
1 − √1 −
Â
þ
Problem 2.3.3
(1) The equation can be written as ¹ ¹Y − ×
be solved easily by separation of variables
á
¹
= Y
×
áD ¹ −
B
Y
Õ
Õ
We can then get ¹ ) = + š¹B − › exp(Y )
Ç
which can
Ç
Now, the pay-off time is when
×
×
¹(@) = + E¹B − F exp(Y ) = 0
Y
Y
Y¹B
1
F
@ = − ln E1 −
×
Y
(2) According to the result above,
@(
@ š¹B , ×, Y → 0›
1
Y¹B
1
Y¹B
¹B
à= E1 −
FF = − E−
F=
→B
×
×
×
Y
Y
×
2¹B
¹B
@ = @ E¹B , ×, Y =
F=
à= 2 = (2 à= 2).
×
×
2¹B
360
àrK E
Appendix A - Chapter 2
(3) With equation in (1) we have
1
Y¹B
@
à= E1 −
F
Y
×
Now to pay in half time, we need to have
1
Y¹B
@
= − à= E1 −
F
×
Y
2
Solving these two algebraic equations, we get
× = × + V×(× − Y¹B )
The additional amount for each payment is now
× = × + V×(× − Y¹B )
Problem 2.3.4
Mr. Wyze
Borrow = ÙB
Payment = oB
Interest rate = Y[ = Y
Ù[ ( )= amt owed to bank at time
Ù[ = decrement of loan due to payments made
Ù[ = Ù[ ( )Y − oB
Ù[ = Ù[ Y − oB
Ûk
Ù[
Y
oB =
ÛD Ù[ −
ÜB
Y
oB
oB
Ù[ ( ) =
+ šÙB − › exp(Y )
Y
Y
Mr. Fulesch
Borrow = ÙB
Payment = oB
Interest rate = Y£ = Y(1 + )
9
(
Ù£ ( )= amt owed to bank at time
Ù£ = decrement of loan due to payments made
Ù£ = Ù£ ( )Y£ − oB
361
Appendix A Solutions to Selected Problems
Y
(1)
Ù£ = Ù£ (1 + ) − oB
5
It is in the form of a linear equation.
+ h( ) = Ÿ( )
Y
h( ) = − (1 + )
5
Y
Y
( ) = exp \
(1 + ) ] = 1 exp °− \ + ]±
5 B
2
5
Ÿ( ) = −oB
$Ù£ ( ) ( )% = Ÿ( ) ( )
Ù£ ( ) = ÙB − oB ( )
Solve for with Eq-(1) and Eq-(2).
B
( )
(2)
Problem 2.3.5
With the new periodic payment 1 + v)×B ,
Ù = ÙY − (1 + v)×B
Ù
=Y
(1 + v)×B
Ù−
Y
B
Cl
Ù
=
Y
B
áD Ù − (1 + v)×B
Y
(1 + v)×B
−
Y
ß = Y@ln Þ
(1 + v)×B
¹B −
Y
(1 + v)×B
1
@- = ln \
]
(1 + v)×B − ¹B Y
Y
Similarly, we can find the payoff time with periodic payment ×B
is
1
×B
@
ln E
F
×B − ¹B Y
Y
362
Appendix A - Chapter 2
At periodic payment ×B, the total interests paid to the bank is
×B
F
¹B Y@ ¹B ln E
×B − ¹B Y
At periodic payment 1 + v)×B , the total interests paid to the
bank is
(1 + v)×B
¹B Y@- = ¹B ln \
]
(1 + v)×B − ¹B Y
Problem 2.3.6
(1) With double periodic payment, we have
Ù = ÙY − 2×
Ù
=Y
2×
Ù− Y
B
CÍk
Ù
=
Y
2×
B
áD Ù −
Y
2×
−
Y ß = Y@
ln Þ
[
2×
¹B − Y
1
2×
F
@ [ = ln E
Y
2× − ¹B Y
(2) With double rate, we have
Ù = 2ÙY − × Ç
Ù
= 2Y
×Ç
Ù − 2Y
B
Cæ
Ù
=
2Y
×Ç
B
áD Ù −
2Y
×Ç
− 2Y
ln Þ
ß = 2Y@(
×
¹B − Ç
2Y
363
Appendix A Solutions to Selected Problems
×Ç=
2¹B Y
1 − exp(−2Y@( )
Problem 2.3.7
(1) With periodic payment =×, we have
Ù = ÙY − =×
Ù
=Y
=×
Ù− Y
B
Ck
Ù
=
Y
=×
B
áD Ù −
Y
=×
− Y
ß = Y@[
ln Þ
=×
¹B − Y
1
=×
@[ = ln E
F
Y
=× − ¹B Y
(2) With interest rate Y, we have
Ù = ÙY − ×Ç
Ù
= Y
×
Ù− Ç
Y
B
C
Ù
=
Y
×
áD Ù − Ç
B
Y
×Ç
−
Y ß = Y@
ln Þ
×
¹B − Ç
Y
Y¹B
×Ç =
1 − exp(− Y@)
Problem 2.3.8
364
× = ×Y
− ×B
Appendix A - Chapter 2
×
=
áD ×Y − ×B
B
and
or
If payment is vY
and
or
Let
where
CD
B
= @B
1
Y¹B
F
@B = − ln E1 −
Y
×B
× = ×vY
×
=
áD ×vY − ×B
B
@( = −
− ×B
Cæ
B
= @(
1
vY¹B
ln E1 −
F
v×B
vY
1
Y¹B
@(Y) = − ln E1 −
F = (Y)-(Y)
Y
×B
(Y) =
1
Y
Y¹B
_
×B
@ is always greater than zero, we have @ = (Y)-(Y) ≥ 0
1
Y¹B
Y¹B
(Y) = > 0 ⇒ -(Y) = −ln _1 −
_ > 0 ⇒ _1 −
_<1
×B
×B
Y
@B = @(Y) = (Y)-(Y)
@( = @(vY)-(vY)
Let
¹Y
A=
×B
Then
and
-(Y) = −ln _1 −
365
Appendix A Solutions to Selected Problems
(1) If
>1
ln|1 − A|
Y
€
ln|1 − vA|
@( = −
vY
@B = −
vY > Y ⇒
1
1
< ⇒ (vY) < (Y)
vY Y
(I)
One case.
-(Y) = −ln 1 − A) and -(vY) = − ln(1 − vA)
1 > vA > A > 0 ⇒ 0 < 1 − vA < 1 − A < 1
− ln(1 − αA) > −ln 1 − A) > 0
(II)
-(vY) > -(Y)
Another case.
-(Y) = −ln A 1) and -(vY) = −ln vA 1)
Then
1 > Av − 1 > A − 1 > 0
0 < − ln(vA − 1) < −ln A 1)
(III)
⇒ -(vY) < -(Y)
If (I) and (II) are satisfied, it is difficult to get a result.
If (I) and (III) are satisfied, then @( < @B .
So the function should be
1
¹Y
@B = − ln E − 1F
Y
×B
and
1
¹vY
@( = − ln E
1F
vY
×B
This means ¹Y > ×B
(2) If 0 < v < 1
With the similar method, we have @( > @B
(3) If v = 0, the DE becomes
× = −×B
and I.C. is ×(0) = ¹B , which is a 1st-order separable DE.
366
Appendix A - Chapter 2
B
Solving it, we can get
áD
×=
Cæ
B
−×B
⇒ @( =
¹B
×B
367
Appendix A Solutions to Selected Problems
Chapter 3 Linear DEs of Higher Order
3.1 Classification of DEs
Problem 3.1.1
For
For
Problem 3.1.3
For (
Thus
+
n
1
LHS = −
1
1
1
+ E F = 0 = RHS
1
1(1 − 1)
+E F =
≠0
, we know = 3
LHS =
and
×6
6 = RHS
where 1 ≠ 1 we have
LHS = 1 × 61
61
≠6
6 . Thus
1, we know that ( = 0 and (
0. Thus
LHS = ( ( + ( ( )
0 RHS
= √ , we know that
1
1
=
and
=−
2√
4 √
1
F+E
F
4 √
2√
RHS
LHS = √ E−
0
368
=−
= where 1 ≠ 1 and 1 ≠ 0, we have
Problem 3.1.2
From
For
LHS =
1
Appendix A - Chapter 3
However for
(
+
, we have
LHS = $√ + 1% E−
1
4 √
≠0
1
4 √
F+E
2√
Problem 3.1.4
(a) ¡ is the G.S. of the homogeneous DE, thus
¡ +k ¡ +å ¡ =0
¢ is a P.S. of the inhomogeneous DE, then
¢ +k ¢+å ¢ = ( )
) = ¡ + ¢ , we have
Let
LHS = $
¡
+
¢%
+ k$
¡
+
¢%
1
+ å$
F
¡
+
¢%
+ k ¡ + å ¡ ) + $ ¢ + k ¢ + å ¢%
0 + ( ) = RHS
(b) From part (a), we know
= ¡ + ¢ = 1 + 1( cos + 1 sin
is a solution of the given DE, and
(0) = −1 = 1 + 1( and
0
1=1
gives 1( = −2 and 1 = −1. The P.S. is
( ) = 1 − 2 cos − sin
¡
369
Appendix A Solutions to Selected Problems
3.2 Linear Independence
Problem 3.2.1
Part 1
…( … ( ) = …( $( + v  + œ ) %
+v
+œ )
…(
 + v(  + œ( )( + v
+œ )
( )
+ (v( + v )
+ (v( v + œ( + œ )
+ (v( œ + v œ( ) + œ( œ
… …( ( ) = … $  + v(  + œ( ) %
… ( + v( + œ( )
 + v  + œ )( + v( + œ( )
( )
+ (v( + v )
+ (v( v + œ( + œ )
+ (v( œ + v œ( ) + œ( œ
…( … ( ) = … …( ( )
Part 2
,
… ≡ Â+1
…( ≡ Â
…( … ( ) = …( $( Â + 1) %
+ )
…(
 + )(
+ )
(
+
+2
+
)
(
)
+
= … $(Â
… …(
%
… ( + )
 + 1)( + )
+ ( + 1) + ( + 1)
…( … ( ) ≠ … …( ( )
Problem 3.2.2
Note that
cosh =
370
exp( ) − exp( )
exp( ) + exp( )
; and sinh =
2
2
Appendix A - Chapter 3
cosh + sinh = exp( )
That means we are able to find a linear combination of the three
functions is identically zero. Hence the three functions are L.D..
we know that
Problem 3.2.3
Let ( = 1,
=
= 0, then we have
sin + exp( ) = 0
(⋅0+
So these functions are L.D..
Note: 0 is always L.D. with any other function.
Problem 3.2.4
Let ( = 15,
= −16, = −6, then we have
15 ⋅ 2 − 16 ⋅ 3 − 6(5 − 8 ) = 0
So these functions are L.D..
Problem 3.2.5
) = 0 , and -( ) = cos
( ) = 8 gives
gives - ( ) = 0. Hence the Wronskian
×( , -) = _
_
8 1
Ý
Ý=0
0 0
Thus ( ) and -( ) are not L.I..
Problem 3.2.6
(1) Since both
we know that
and
(
and
A( )
A( )
A( )
The Wronskian of
(
(
+ sin
=1
are the solution to the DE
+ ž( ) + 1( ) = 0
and
= −ž( )
= −ž( )
is
(
− 1( )
− 1( )
(
371
Appendix A Solutions to Selected Problems
)=Ý
×( ( ,
Thus, we have
A( )
×
(
(
(
= A( )(
(
−
Ý
(
−
A )( ( −
) − (A
( A
, we have
Substituting A ( and A
×
A( )
= ( (−ž
−1
)−
()
)
()
(
(−ž
(
−1
()
ž ( −
()
ž )×( )
(2) To solve this equation, we notice it is a separable DE
ž
×
=−
A
×
ž( )
ln × = −
+H
A( )
ž( )
× = H exp \
]
A( )
Problem 3.2.7
For ë ) =
ë
( ) = r! and
Thus, the Wronskian
ë
(ë)
, we know that
ë
@
) = 0,
o>r
⋯
⋯
(
×( B , ë , … , ) = í ⋮B
⋮
⋱
⋮ í
( )
( )
( )
⋯
B
(
is actually an upper triangular matrix with diagonal elements
Thus, we have
372
ë
(ë)
B
= r!
(
Appendix A - Chapter 3
×( B , ( , … ,
We know functions
B, (, … ,
) = p r! ≠ 0
ëÜB
are L.I.
Problem 3.2.8
Suppose ( , … ,
are linear independent solutions for the
homogeneous DE.
⋯
( )
( )
(( )
⋯
(
)
(
)
( )
(
×( ) = í
í
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>( ( )
⋯
(
Taking the derivative of × ) we get
⋯
( )
( )
(( )
× )
⋯
( )
( )
(( )
= í
í
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>( ( )
⋯
(
⋯
( )
′( )
( ′( )
⋯
(
)
(
)
(
)
í
í (
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>( ( )
⋯
(
( )
⋯
( )
(( )
⋯
( )
( )
( )
+í (
í+⋯
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>( ( )
⋯
(
(
)
(
)
⋯
( )
(
⋯
(
)
(
)
( )
+í (
í
⋮
⋮
⋮
⋱
( ) ⋯
( )
(( )
All the terms except the last one are zero.
( ) ⋯
( )
(( )
× )
⋯
( )
( )
( )
=í (
í
⋮
⋮
⋮
⋱
( ) ⋯
( )
(( )
Plugging ( , … , back to the homogeneous DE, we have
373
Appendix A Solutions to Selected Problems
(
= h( ( )
= h( ( )
= h( ( )
Put this into the DE wrt W
×( )
=
í
((
)
′
( ( )
(
>(
>(
⋮
>(
( )
− ⋯− h ( )
−⋯− h ( )
(
− ⋯− h ( )
′( )
⋯
⋯
⋱
⋯
( )
′( )
í
⋮
⋮
( )
( )
( )
( )
⋯
(( )
′
′
′
⋯
( )
( )
( ( )
×í
í
⋮
⋮
⋮
⋱
−k( ( ) ( >( ( ) −k( ( ) >( ( ) ⋯ −k( ( ) >( ( )
⋯
( )
( )
(( )
⋯
(
)
(
)
( )
(
+í
í+…
⋮
⋮
⋮
⋱
−k ( ) ( > ( ) −k ( ) > ( ) ⋯ −k ( ) > ( )
( )
⋯
( )
(( )
⋯
( )
( )
(( )
…+í
í
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>(
( )
−k ( ) (
−k ( )
⋯ −k ( )
Now everything except the first term cancels out
×( )
⋯
)
( )
( )
⋯
( )
( )
(( )
= í
í
⋮
⋮
⋮
⋱
>(
>(
( ) ⋯ −h( ( ) >( ( )
−h( ( ) ( ( ) −h( ( )
( ) ⋯
( )
(( )
⋯
( )
( )
( )
= −h( ( ) í (
í
⋮
⋮
⋮
⋱
>( ( )
>( ( )
>( ( )
⋯
(
((
374
⋮
(( )
Appendix A - Chapter 3
= −h( ( )×( )
×( )
= −h( ( )×( )
Thus, ×( ) is a function of h( ) . Easily, we can solve the
above equation to get
×( ) = 1 exp E−
h( ( )
F
Problem 3.2.9
First of all, we know that exp$ Y( + Y + Y ) % > 0, ∀ , Y( , Y
and Y . Notice that
1 1 1
Y ó = (Y − Y( )(Y − Y( )(Y − Y )
ó Y! Y
Y( Y Y
is a Vandermonde determinant which we know that for distinct
Y( , Y and Y , it is non-zero. Therefore, the Wronskian
1 1 1
Y
)
× = exp$(Y( + Y + Y % ó ! Y Y ó ≠ 0
Y( Y
Y
and hence exp(Y( ), exp(Y ) and exp(Y ) are L.I..
Problem 3.2.10
First, if ∃ 1 ≤ r, o ≤ =, r ≠ o such that Yë Y@ , from the property
of determinants we know that T 0 . Now suppose Yë ’s are
mutually distinct. Some simple calculations suggest
T
p $Yë
(r@sër
Y@ %
We will prove this by induction. It is easy to see this is true
for = 2. Suppose it is true for = /, then for = / + 1, we
have
375
Appendix A Solutions to Selected Problems
1
Y(
=í ⋮
Y(e
1 ⋯
1
Y ⋯ Yeú(
Teú(
⋮ ⋱
⋮ í
e
Y e ⋯ Yeú(
0
0
⋯
0
1
Y( − Yeú( Y − Yeú( ⋯ Ye − Yeú( Yeú(
í
⋮
⋮
⋯
⋱
⋮ í
e
e
e
e
Y(e − Yeú(
Y e − Yeú(
⋯ Yee − Yeú(
Yeú(
Similarly without changing the value of T, we can subtract, in
order, Yeú( times row / from row / + 1, Yeú( times row / 1
from row /, and so on, till we subtract Yeú( times row 2 from
row 3.
Teú(
0
(Y( − Y/+1 )Y01
í
⋮
Y1 − Y/+1 )Y/−1
1
e
1) Te p Yë
e
e
ëÜ(
Te p Yeú(
ëÜ(
p
(r@sëreú(
Problem 3.2.11
We can get the Wronskian
376
,…,
)=í
Yeú(
Y@ %
$Yë
Here finishes the proof.
×( ( ,
Yë
⋯
0
⋯ (Y/ − Y/+1 ) Y0/
⋱
⋮
⋯ (Y/ − Y/+1 )Y/−1
/
(
(
⋮
( >()
(
⋮
( >()
⋯
⋯
⋱
⋯
⋮
( >()
í
1
0
í
⋮
0
Appendix A - Chapter 3
exp(Y )
⋯
exp(Y )
exp Y( )
Y exp(Y )
⋯
Y exp(Y )
Y( exp(Y( )
í
í
⋮
⋮
⋱
⋮
Y( >( exp(Y( ) Y >( exp(Y ) ⋯ Y >( exp(Y )
1
1
⋯
1
Y
⋯
Y
Y(
exp$ Y( + Y + ⋯ + Y ) % í
í
⋮
⋮
⋱
⋮
Y( >( Y >( ⋯ Y >(
Since exp$ Y( + Y + ⋯ + Y ) % > 0, and the second part is a
Vandermonde determinant, we know that for Y( , Y , … , Y that are
distinct, the Wronskian is non-zero. Hence ë are L.I..
377
Appendix A Solutions to Selected Problems
3.3 Constant Coefficient Homogeneous DEs
Problem 3.3.1
(1) We have
This gives ( = r and
(2) We have
This gives
(
(,
= 3r and
(,
−r ± √r − 8
2
= −2r
=
=
2r ± V(2r) − 12
2
= −r
Problem 3.3.2
(a) We know that any complex number z can be written as Ù
+ r , where and are the real part and imaginary part
respectively, and this number can also be written in polarcoordinate as
Ù = Y(cos ü + r sin ü), where Y = V
+
and ü = tan>(
By Euler’s law exp(rü) = cos ü + r cos ü, so we proved that any
complex number can be written in the form Y exp(rü).
(b)
4 = 4 r0 ⇒ 4 = 4 exp r0)
−2 = −2 r0 = 2 exp r8
8
3r = 0 + 3r = 3 exp šr ›
2
8
8
1 r ⇒ Y = √2 and ü = ⇒ 1 r = √2 exp šr ›
4
4
2
28
1 r√3 ⇒ Y = 2 and ü = 8 ⇒ 1 r√3 = 2 exp Er F
3
3
(c) We have
8
28
2 2r√3 = 4 exp š r › and 2 2r√3 = 4 exp Er F
3
3
378
Appendix A - Chapter 3
Hence
2
8
2r√3 = šp2 exp š r ›› and
6
8
= šp2 exp šr ››
3
Problem 3.3.3
Let = exp( ), then
=
¤
=
=
=
−3
−
2r√3
+2
+2
6
=0
7
+3 +3 + =0
Y + 3Y + 3Y + 1 = 0
(Y + 1 = 0
Y( = Y = Y = −1
Therefore, the G.S. is
exp(− )
= 1( exp(− ) + 1 exp(− ) + 1
>(
>(
>(
( ) = 1(
+1
ln + 1
ln
>(
>(
,
=
ln ,
= >( ln
( =
×( ( , , ) =
>(
− >
2 >
>(
>
>
1 ln
2 ln
3
ln
ó
>(
>
>
2 ln
ln
2 6 ln
2 ln
ln
1
ln
ln
>( > >
=
ó
ó 1 1 ln
2 ln
ln
2 2 ln
3 2 6 ln
2 ln
Here
−3
= ln
2
379
ó
Appendix A Solutions to Selected Problems
1 ln
=
ó0
1
0 −3
= >û (2 6 ln
>û
ln
2 ln
2 6 ln
6 ln
×
(,
ó=
,
=2
>û
>û
=
2
_
1
2 ln
3 2 6 ln
û
Problem 3.3.4
This is a 2nd-order DE with constant coefficients, C-Eq is
Y
3Y 2 = 0
(Y + 1 Y 2 = 0
which gives
Y = −2
Y( = −1,
Thus the G.S. is
= 1( exp(− ) + 1 exp(−2
Given I.C. 0) = 1, 0) = 6, we know that
1( + 1 = 1
|
1( 21 = 6
This gives
1 =8
| (
1 = −7
Therefore
( ) = 8 exp(− ) − 7 exp(−2
At the highest point, the velocity vanishes, which means
= −8 exp(− ) + 14 exp 2 = 0
Solving this, we get
7
= ln
4
and
7
16
Eln F =
4
7
Thus, the highest point is
380
_
7 16
Eln , F
4 7
Appendix A - Chapter 3
Problem 3.3.5
This is a 2nd-order DE with constant coefficients, C-Eq is
3Y 2 = 0
Y
(Y + 1 Y 2 = 0
which gives
Y( = −1,
Y = −2
Thus the G.S. is
= 1( exp(− ) + 1 exp(−2
From the given I.C., we have for (
1( 1 = 3
|
−1( − 21 = 1
This gives
1 =7
| (
1 = −4
2
( = 7 exp(− ) − 4 exp
Similarly, we have for
1( 1 = 0
|
−1( − 21 = 1
This gives
1 =1
| (
1 = −1
= exp(− ) − exp(−2
To find their intersection, let
= ( )
(
This gives
7 exp(− ) − 4 exp 2 = exp(− ) − exp(−2
= − ln 2
Thus
ln 2 = −2
(,
Hence the intersection is
381
Appendix A Solutions to Selected Problems
ln 2 , 2
Problem 3.3.6
The C-Eq is
It is easy to solve
3Y
Y( = Y = 0,
Thus, we have G.S.
= 1( + 1
Given the I.C., we can get
S
1
R
which gives
2Y = 0
Q
1(
Y =−
+ 1 exp E−
1 = −1
2
1 =0
3
4
1 =1
9
S1( = −
3
1 =
2
R
9
Q1 = 4
Therefore, the P.S. is
1
= E 13
4
6
2
3
2
F
3
13
4
9 exp E
2
FF
3
Problem 3.3.7
From the solution we know that the C-Eq has 3 repeated roots at
both 2r and 2r. That means
Y 2r Y 2r = 0
û
48Y
64 = 0
Y + 12Y
Therefore, the DE is
382
Appendix A - Chapter 3
(û)
Problem 3.3.8
The C-Eq is
+ 12
Y
We can get
Thus, the G.S. is
Y( = 1, Y
From the I.C., we have
64 = 0
Y =1
Y 1 =0
1 Y
= 1( exp( ) + 1 exp \
This gives
48
1
=
,
2
S
1 √3r
1
1(
2
R
−1 √3r
1
+
1
(
Q
2
−1 p √3r
2
√3r
1 exp \
]
1( 1
1 =1
1 √3r
1 =0
2
1 √3r
1 =0
2
1( = 1 = 1 =
1
2
√3r
]
1
3
Hence we have
1
1
1 √3r
1
1 √3r
= exp
exp \
exp \
]
]
3
3
2
3
2
which can be written as
2
1
1
√3
exp E
F cos
= exp
3
2
3
2
Problem 3.3.9
For the 4th order DE with constant coefficients, the C-Eq is
Y
Y
Y
Y 2=0
(Y + 1 Y 2 Y
1 =0
383
Appendix A Solutions to Selected Problems
The G.S. is
1 cos
1 sin
= 1( exp(− ) + 1 exp(2
From the given I.C., we have
1( 1
1 =0
−1( + 21
1 =0
€
1 =0
1( + 41
−1( + 81 − 1 = 30
Solving for the I.C. we have,
1( = −5, 1 = 2, 1 = 3 and 1 = −9
Thus, the P.S. is
= −5 exp
2 exp 2
3 cos
9 sin
Problem 3.3.10
If G 0, then the DE is
+ = 0 with the G.S. = A cos +
ž sin
− = 0 with G.S. = 1 exp( ) +
If I 0 the DE becomes
 exp(− )
To satisfy the I.C. ( (0) = 1 and ( 0) = 0, we have
1
A=1
1+Â =1
and m
⇒1=Â=
m
ž=0
1 Â=0
2
Thus,
cos ,
0
=m
cosh ,
<0
You can try the similar ways to get ,
sin ,
≥0
=m
sinh ,
I0
Problem 3.3.11
This is a higher order DE with constant coefficients, the
corresponding C-Eq is
Y (Y − 2 Y 3 Y
1 =0
Thus we have
Y( = Y = Y = 0, Y = 2, Y9 = −3, Yû, = pr
384
Appendix A - Chapter 3
So the solution is
= 1( + 1 + 1
Problem 3.3.12
The C-Eq is
+ 1 exp(2
1 cos
Y
19 exp
3
1û sin
2Y 2 = 0
Y(, = 1 p r
1 cos
= exp( ) (1( sin
From the given I.C., we have
1 = 0, 1( = 5
Thus, we have
= 5 exp
sin
The G.S. is
Problem 3.3.13
The C-Eq is
Y
9Y = 0
Y( = 0, Y , = p3r
= 1( + 1 sin 3
1 cos 3
From the given I.C., we have
1( 1 = 3
31 = −1
•
91 = 2
This gives
29
S1( =
9
1
1 =−
3
R
2
Q1 = − 9
Thus, the solution is
The G.S. is
385
Appendix A Solutions to Selected Problems
=
Problem 3.3.14
The C-Eq is
29
9
1
sin 3
3
2
cos 3
9
9 =0
Y 1 Y 2 Y 3 Y
Y( = Y = Y = 1, Y = Y9 = 2, Yû = 3, Y ,Z = p3r
Thus, the G.S. is
= (1( + 1 + 1 ) exp( ) + (1 + 19 ) exp(2
1û exp 3
1 cos 3
1Z sin 3
Problem 3.3.15
This is a 2nd-order DE with constant coefficients, the C-Eq is
Y 15 = 0
Y
Solve it, we have
1 √61
1 √61
Y( =
,
Y =
2
2
1 √61
1 √61
= exp \
],
]
( = exp \
2
2
Thus the G.S. is
1 √61
1 √61
= ( exp \
exp \
]
]
2
2
Problem 3.3.16
The C-Eq is
(
386
9Y
12Y 4 = 0
(3Y
⇒
−2 =0
2
⇒Y=
3
2
2
( ) = exp E F
= exp E F ;
3
3
2
2
= ( exp E F
exp E F
3
3
Appendix A - Chapter 3
Problem 3.3.17
1 This is a kth order DE with constant coefficients, and the CEq is Y Y( eæ = 0 . Since Y = Y( has /( repeated roots, we
have
( ) = $1( + 1 + 1
+ ⋯ + 1eæ eæ >( % exp(Y( )
2 The C-Eq for this DE is
Y Y( eæ Y Y eÍ … (Y − Y )e = 0
Thus Y = Y( is /( repeated root, Y = Y is / repeated root, …,
Y = Y is / repeated root.
( ) = $1(( + 1( + ⋯ + 1(e eæ >( % exp(Y( )
+ $1( + 1
+ ⋯ + 1 eÍ eÍ >( % exp(Y )
+⋯
+ $1 ( + 1
where 1ë@ means constant.
+ ⋯+1
e
e >(
% exp(Y )
Problem 3.3.18
First find the G.S. of corresponding homogenous DE:
Let
− vF
E
= exp( ), we have
Eexp( )
exp( )
E
− vF
( )=0
− vF
=0
=0
The C-Eq is(Y − v) = 0. There are n identical real roots.
So the G.S. for the homogenous DE is
+ ⋯ + 1 >( )
¡ ( ) = exp(v ) (1( + 1 + 1
= exp(v ) ê 1e
eÜ(
e>(
Then plugging x back, we have
387
Appendix A Solutions to Selected Problems
¡(
)=
-
ê 1e (ln )e>(
eÜ(
Finding the P.S. consists of three cases:
Case I: v = 0
The original DE becomes
( )
= exp( )
whose P.S. is
¢ = exp( )
The G.S. is
+ ⋯ + 1 >( ) + exp( )
= ¡ + ¢ = (1( + 1 + 1
Or in terms of
= (1( + 1 ln + 1 (ln ) + ⋯ + 1 (ln ) >( ) +
Case II: v = 1, we select the trial P.S. as
exp( )
¢ =A
Thus,
LHS = E
=E
=E
=E
=E
=E
=E
− 1F
− 1F
− 1F
− 1F
− 1F
− 1F
− 1F
A
>(
>(
>(
>
>
>
E
E
exp
A=
1F A
A
E
E
exp
>(
exp
A=
>(
1F A=
A= =
exp
1
>(
exp
>
= ⋯ = A=! exp( )
Plugging in to the original DE, we get
Or
388
E
− 1F
A
exp
A
exp
exp
exp
A=
= exp( )
F
>(
exp
F
Appendix A - Chapter 3
A(=!) exp( ) = exp( )
Thus,
A=
The G.S. for case v = 1 is
= ¡+ ¢
= exp( ) (1( + 1
+1
1
=!
+ ⋯+1
>( )
+
1
=!
Or in terms of
= ¡+ ¢
= (1( + 1 ln + 1 (ln ) + ⋯ + 1 (ln )
1
+
ln
=!
Case III: v ) 1
We select trial P.S.
¢ = A exp( )
Thus,
LHS = E
− vF
=E
− vF
=E
− vF
=E
Thus
− vF
= ⋯ = (1
E
A(1
Or
i.e.,
The G.S. is
=
¡
+
¢
>(
>(
>
>
E
− vF A exp( )
E
− vF (1
(1
(1
v A exp
v A exp
v A exp
vF A exp
v
A=
exp
1
1
v
exp
>( )
v A exp
= exp( )
= exp( )
389
Appendix A Solutions to Selected Problems
= exp(v ) (1( + 1
Or in terms of
= ¡+ ¢
= - (1( + 1 ln
+1
+⋯+ 1
1
exp
+
1 v
>( )
+ 1 (ln ) + ⋯ + 1 (ln ) >( )
1
+
1 v
One can see that this case covers the simple case of v = 0.
390
Appendix A - Chapter 3
3.4 Cauchy-Euler DEs
Problem 3.4.1
The original DE can be written as the following form
+
Let = ln , then
=
E
+
=0
= exp(− )
F
Eexp(− ) exp( )
(exp(− ))
exp(− )
exp(−2 )
F
+ exp(− )
+ exp(− )
+ exp(−2 )
exp(−2 ) \
−
E
E
F
F
]
Plugging these back to the DE, we have
exp(2 ) °exp(−2 ) \
+
=0
−
+( − )
]± + exp( ) Eexp(− )
+
F
=0
which is a linear DE WRT . The C-Eq is
Y + ( − )Y + = 0
If ( − ) − 4
± V( − ) − 4
2
> 0, we have two distinct real roots.
Y(, =
−
391
Appendix A Solutions to Selected Problems
Put = ln
If ( − )
Put = ln
If ( − )
( ) = 1( exp(Y( ) + 1 exp(Y )
back, we have
( ) = 1( Çæ + 1 ÇÍ
− 4 = 0, we have repeated root Y, thus
( ) = (1( + 1 ) exp(Y )
back, we have
( ) = (1( + 1 ln ) Ç
− 4 < 0, we have two complex conjugated roots
Y(, = Y exp(±rü)
( ) = (1( sin ü + 1 cos ü exp Y
Put = ln back, we have
= 1( sin ü ln + 1 cos ü ln
Thus
Problem 3.4.2
We use the substitution L = ln , we get the DE
L = ln
L = ln ⇒ L =
L
=
=−
1
⇒
=−
L
+
L
=
1
+\
\
L
L
L
−9 § = 0
Y −9 = 0
Y = ±3
=
( = exp 3
= exp −3 = >
+1 >
¡ = 1(
§
392
]
]
Ç
Appendix A - Chapter 3
Problem 3.4.3
We use the substitution
L = ln
L = ln ⇒ L =
L
=
=−
§
C-Eq:
(
+
Y +
1
⇒
=−
−1
L
+
L
1
+\
§
+
\
L
L
]
]
L
=0
−1 Y+ =0
−1 ±V −1 −4
2
− −1 +V −1 −4
= exp °
L±
2
Y=
−
−
= exp °
Problem 3.4.4
Let = ln , then
−1 −V −1
2
= exp −
= exp −2 \
Thus, the DE becomes
Its C-Eq is
=
−3
−
+2 =0
−4
L±
]
393
Appendix A Solutions to Selected Problems
Y − 3Y + 2 = 0
Y( = 1, Y = 2
Thus
= 1( exp
+ 1 exp 2
= 1( + 1
From the given I.C., we can get
1 +1 =3
| (
1( + 21 = 1
This gives
1 =5
| (
1 = −2
Thus the IVP has solution
=5 −2
Put
back, we have
Problem 3.4.5
Let =
+ 2 , we have
=
+ 2 >(
= −1
+2 >
Substitute into DE
+2
−2 +1 =0
−1 = 0
=1
( = 1,
The G.S. is
= 1( + 2 + 1
+ 2 ln| + 2|
Problem 3.4.6
Let =
Thus,
394
,
′=
>(
−1
, ′′ =
−1
>
>( − 10 = 0
−2
− 1 − 2 − 10 = 0
− 3 − 10 = 0
>
Appendix A - Chapter 3
There are two real and different roots
So the G.S. is
= 1( 9 + 1
Problem 3.4.7
Substitute
we have
=Y
Then
Ç>(
,
=Y Y−1
=
Ç>
,
Ç
(
= 5,
>
= −2
=Y Y−1 Y−2
Y Y−1 Y−2 +Y Y−1 −Y+1 =0
Y − 2Y + 1 = 0
Y−1 Y −Y−1 =0
which has three distinct roots, resulting three L.I. solutions.
1 + √5
1 − √5
,
Y = 1,
2
2
=
(
(
+
(>√9
+
Ç>
(ú√9
395
Appendix A Solutions to Selected Problems
3.5 Inhomogeneous Higher Order DEs
Problem 3.5.1
This is a 2nd-order DE with constant coefficients, and the
corresponding C-Eq is
Y − 2Y − 8 = Y + 2 Y − 4 = 0
Y( = −2,
Y =4
Thus, ¡
= 1( exp −2 + 1 exp 4
= exp 4
Trial solution is
= A q exp 4
¢
Compare terms in ¡ and ¢ , we notice ³ = 1.
+ A exp 4
¢ = A4 exp 4
= A exp 4 ⟹ ô
¢
+ A16 exp 4
¢ = A8 exp 4
Plugging into original DE,
A8 exp 4 + 16 exp 4 − 2 A4 exp 4 + A exp 4
− 8A exp 4 = exp 4
A6 exp 4 = exp 4
1
A = ⟹ ¢ = exp 4
6
6
So the G.S. is
= 1( exp −2
+ C exp 4
+ exp 4
6
Problem 3.5.2
C-Eq for homogeneous portion of the DE is
Y −1 = 0
Y( = 1, Y = −1, Y = r, Y = −r
= ( exp
+ exp − + cos +
¡
Trial P.S.
¢ =A
So
¢ =0= ¢ = ¢ = ¢
396
sin
Appendix A - Chapter 3
After substitution, we get
The G.S. is
= ¡+ ¢
= ( exp
+
0−A = 1
A = −1
¢ = −1
exp −
+
cos +
sin − 1
Problem 3.5.3
We first consider the homogeneous equation
− =0
The C-Eq is
Y −1= Y +1 Y −1
= Y+r Y−r Y+1 Y−1
=0
The roots are Y( = −r , Y = r, Y = −1, Y = 1.
Then the complementary solution can be written as
+ 1 exp −
¶ = 1( cos + 1 sin + 1 exp
Next we assume the P.S. yŠ has the form
¢ = A exp
Its derivatives are
+ A exp
¢ = exp
+ A exp
¢ = 2A exp
+ A exp
¢ = 3A exp
+ A exp
¢ = 4A exp
Substituting them in the inhomogeneous ODE, we have
4A exp
= 4 exp
Solving the equation gives us
A=1
¢ = exp
Finally the G.S. of the original ODE is
397
Appendix A Solutions to Selected Problems
=
¢
+
¡
=
exp
+ 1( cos + 1 sin + 1 exp
+ 1 exp −
Problem 3.5.4
Solve the corresponding homogeneous DE. The C-Eq is
Y −Y −Y −Y−2 = 0
Y+1 Y +1 Y−2 = 0
Y( = −1, Y = 2, Y , = ±r
The G.S. to the homogeneous DE is
+ 1 exp 2 + 1 sin + 1 cos
¡ = 1( exp −
From the RHS
= 18 9 , we can assume the P.S.
q
AB + A( + A
+A
+A
+ A9 9
¢ =
Since there has no polynomial in the homogeneous solution, we
can let ³ = 0. Thus
+A
+A
+ A9 9
¢ = AB + A( + A
and
+ 3A
+ 4A
+ 5A9
¢ = A( + 2A
+ 12A
+ 20A9
¢ = 2A + 6A
+ 60A9
¢ = 6A + 24A
= 24A + 120A9
Plugging this into the DE, we have
¢
− ¢ − ¢ − ¢ − 2 ¢ = 18 9
Equating the terms, we have
24A − 6A − 2A − A( − 2AB = 0
S120A − 24A − 6A − 2A − 2A = 0
9
(
−60A9 − 12A − 3A − 2A = 0
−20A9 − 4A − 2A = 0
R
−5A9 − 2A = 0
−2A9 = 18
Q
Solving this gives
398
¢
135
4
135
A( =
2
135
A =
2
R
A = 45
45
A =
2
Q A9 = −9
Appendix A - Chapter 3
SAB =
Thus, the solution is
135 135
135
+
+
¢ =
4
2
2
Finally, the G.S. is
+ 45
+
45
2
−9
9
135
135
+
2
2
+ 45
+
45
2
−9
9
= 1( exp −
+
+ 1 sin + 1 cos + 1 exp 2
+
135
4
Problem 3.5.5
We can assume the P.S. has form
q
AB + A( + A
+A
¢ =
To determine ³, we have to find the G.S. to the homogenous
equation. The C-Eq is
Y +Y +Y+1=0
Y+1 Y +1 =0
Y( = −1, Y , = ±r
This means the polynomials are not part of the ¡ . Hence ³ = 0.
Now we have
+A
¢ = AB + A( + A
and
+ 3A
¢ = A( + 2A
¢ = 2A + 6A
399
Appendix A Solutions to Selected Problems
¢
= 6A
Plugging this back to the DE
+
¢ + ¢ + ¢+ ¢ =
Equating the terms, we have
AB + A( + 2A + 6A = 0
A( + 2A + 6A = 0
€
A + 3A = 1
A =1
which gives
AB = 0
A = −2
€ (
A = −2
A =1
Thus, we have a P.S. of the DE
+
¢ = −2 − 2
Problem 3.5.6
We can assume the P.S. has form
+ ž exp 2 + 1 exp
+Â
¢ = A exp 3
Thus
+ 2ž exp 2 + 1 exp
¢ = 3A exp 3
and
+ 4ž exp 2 + 1 exp
¢ = 9A exp 3
+ 8ž exp 2 + 1 exp
¢ = 27A exp 3
Putting this back to the DE, equating the terms and we can get
27A + 9A + 3A + A = 1
8ž + 4ž + 2ž + ž = 1
€
1+1+1+1 =1
Â=1
this gives
400
1
40
1
ž=
15
R
1
1=
4
QÂ = 1
Appendix A - Chapter 3
SA =
Thus, the P.S. is
1
exp 3
¢ =
40
+
1
exp 2
15
1
+ exp
4
+1
Problem 3.5.7
The C-Eq of the homogeneous DE is
Y + Y =0
Y( = Y = 0, Y , = ± r
Thus the G.S. for the homogeneous DE is
+ A sin
+ A cos
¡ = A( + A
Since sin
and cos
are part of the G.S., we can assume the
P.S. has form
sin
+1
cos
+ 1 sin
¢ = 1(
+ 1 cos
1 +
1 −
1( sin
¢ = 21( − 2 1 − 4
+ 21 + 2 1 + 4 1( −
1 −
1 cos
¢
= 41
− 121(
+1
+ 81
sin
+ 1(
+ −41
− 121
+1
− 81(
+1
cos
Putting these back and equating the terms, we can get
401
Appendix A Solutions to Selected Problems
S
1( = 0
1 =
4
1
R1 = −
Finally, we have
Q1 =
= ¡+ ¢
1
4
5
Problem 3.5.8
The homogeneous G.S. is
+ sin
If =
¡ = ( cos
¢ = ž exp r
and find ¢ ′′
¢,
Plug
¢ ′′
to
So, the G.S. is
=
+
¡ +¢ = ( cos
¢
=
= exp r
1
2r
+
If ) B , suppose the P.S.
Plug ¢ , ¢ ′′ to
+
then
So, the G.S. is
=
402
¡+ ¢
=
¢
( cos
=
B
¢
A=
1
−
+
B,
B
B
exp r
sin
suppose the P.S.:
, then find ž =
B
+
= A exp r
= exp r
B
1
−
B
exp r
sin
B
B
+
1
2r
B
B
1
−
exp r
and find
B
(
.
ëtD
B
exp r
¢ ′′
B
Appendix A - Chapter 3
Problem 3.5.9
The C-Eq of the homogeneous DE is
Y +9 = 0
Y(, = ±3r
Thus the G.S. for the homogeneous DE is
¡ = A( cos 3 + A sin 3
Since sin 3 and cos 3 are part of the G.S.,
we can assume the P.S. has form
1( sin 3 + 1 cos 3
¢ =
Thus,
¢ = 1( sin 3 + 1 cos 3
+3 1( cos 3 − 3 1 sin 3
and
¢ = 61( − 9 1 cos 3 + −61 − 9 1( sin 3
Putting these back and equating the terms, we can get
1
1( = ,
1 =0
6
Thus, the P.S. is
1
sin 3
¢ =
6
and the G.S. for the original DE is
1
= A( cos 3 + A sin 3 +
sin 3
6
Problem 3.5.10
First we find the complementary solution, i.e., the solution to
+2 + =0
The C-Eq is Y + 2Y + 1 = 0 whose two identical roots Y =
−1. So
= ( exp − +
exp −
¡
403
Appendix A Solutions to Selected Problems
Now we must solve for the P.S., ¢ . Because we see that the
inhomogeneous term is a sin and cos , we will use the method
of undetermined coefficients with our guess of ¢ as,
¢ = A cos + ž sin
¢ = −A sin + ž cos ,
¢ = −A cos − ž sin
Substituting these values into the DE, we obtain that
−A cos − ž sin − 2A sin + 2ž cos + A cos + žsin
= 5 sin + 5 cos
Hence we find that
−2A sin + 2ž cos = 5 sin + 5 cos
Therefore
5
5
A=− ,
ž=
2
2
The final solution is then
5
= ¢ + ¡ = sin − cos
2
+ ( exp − +
exp −
Problem 3.5.11
The C-Eq is
Y +Y +Y+1=0
Y , = ±r
Y( = −1,
¡ = 1( exp(− ) + 1 sin + 1 cos
We can assume the particular has form
¢ = A + (ž cos + 1 sin ) + Â cos 2
+þ sin 2 + M exp(− )
=
ž
cos
+ 1 sin + (−ž sin + 1 cos )
¢
−2Â sin 2 + 2þ cos 2 + M exp(− ) − M exp(− )
¢ = −2ž sin + 21 cos − (ž cos + 1 sin )
−4Â cos − 4þ sin 2 − 2M exp(− ) + M exp(− )
¢ = −3(ž cos + 1 sin )
404
Appendix A - Chapter 3
+ (ž sin − 1 cos ) + 8Â sin 2 − 8þ cos 2
+ (3 − )M exp(− )
Plugging this back and comparing the terms, we have
A=1
S 21 − 2ž = 1
−2ž − 21 = 0
−3Â
− 6þ = 0
R
−3þ + 6Â = 1
Q
2M = 1
Solving this gives
A=1
S
1
ž=−
4
1
1=
4
2
RÂ = 15
1
þ=−
15
1
QM = 2
The P.S. is
1
1
2
cos +
sin + cos 2
¢ =1−
4
4
15
1
1
exp(− )
− sin 2 +
2
15
Problem 3.5.12
Let = exp( ), then = ln
−
−4
=
=
−
+ 6 = exp(3 )
405
Appendix A Solutions to Selected Problems
−5
+ 6 = exp(3 )
C-Eq of homogeneous portion:
Y − 5Y + 6 = 0
(Y − 3)(Y − 2) = 0
Y =2
Y( = 3,
¡ ( ) = 1( exp(3 ) + 1 exp(2 )
Assume the P.S. has form
¢ = A exp(3 )
′ = A exp(3 ) + 3A exp(3 )
′′ = 6A exp(3 ) + 9A exp(3 )
¢
¢
Plugging these back, we have
A exp(3 ) = exp(3 )
Thus,
A=1
¢ = exp(3 )
The G.S. is
( ) = 1( exp(3 ) + 1 exp(2 ) + exp(3 )
( ) = 1( + 1
+ ln
Problem 3.5.13
Use the trial function = - and plug into the homogeneous
equation first.
û( - )
+ 2 9 ( - ) − 12
= (v(v − 1) + 2v − 12) ú= (v + v − 12) ú- = 0
(v + 4)(v − 3) = 0
v = −4,3
Thus,
>
+
n( ) = (
To find P.S. use trial function
>
ln
l =A
406
Appendix A - Chapter 3
û (A >
ln ) + 2 9 (A > ln ) − 12 (A > ln ) = 1
û (20A >û ln − 9A >û )
+ 2 9 (−4A >9 ln + A >9 )
− 12A ln = 1
(20 − 8 − 12) ln + A(−9 + 2) = 1
1
A=−
7
The G.S. is then
1 >
= ( > +
−
ln
7
Problem 3.5.14
Let = exp( ), then = ln
C-Eq:
=
=
−
− 6 = 72 exp(5 )
+
Y +Y−6=0
(Y + 3)(Y − 2) = 0
Y =2
Y( = −3,
¡ ( ) = 1( exp(−3 ) + 1 exp(2 )
Assume the P.S. has form
¢
¢
¢
= A exp(5 )
= 5A exp(5 )
= 25A exp(5 )
Plugging these back, we have
25A exp(5 ) + 5A exp(5 ) − 6A exp(5 ) = 72 exp(5 )
Thus,
A=3
¢ = 3 exp(5 )
407
Appendix A Solutions to Selected Problems
The G.S. is
( ) = 1( exp(−3 ) + 1 exp(2 ) + 3 exp(5 )
( ) = 1( > + 1
+3 9
Problem 3.5.15
We use the substitution = ln , we have
=
= −
The equation becomes
4
4 − 4 − 4 + 3 = 8 exp E F
3
4
4 − 8 + 3 = 8 exp E F
3
which is a constant coefficient inhomogeneous linear DE WRT .
The C-Eq to the associated homogeneous DE is
4Y − 8Y + 3 = 0
3
1
Y =
Y( = ,
2
2
3
1
F + 1 exp E F
¡ ( ) = 1( exp E
2
2
Assume the P.S. has the form ¢ ( ) = A exp š ›, then
64
4
32
4
4
4
A exp E F − A exp E F + 3A exp E F = 8 exp E F
9
3
3
3
3
3
72
A=−
5
Thus,
3
1
72
4
( ) = 1( exp E F + 1 exp E F − exp E F
2
2
5
3
That is
(
72
( ) = 1( + 1
−
5
408
Appendix A - Chapter 3
Problem 3.5.16
Using Euler’s Equation approach.
= exp( ) ,
=
,
−
+
+
= ln
=
+
=
=
−
The homogeneous portion of the solution
¡ ( ) = 1( cos + 1 sin
Try P.S.
¢( ) = A + ž
We get A = 1, ž = 0. The final solution is
( ) = 1( cos + 1 sin +
( ) = 1( cos(ln ) + 1 sin(ln ) + ln
Problem 3.5.17
We use the substitution = ln , we have
=
= −
=2 −3 + u
The equation reduces to
u − 2 + 2 − = 1 + exp( ) + exp(2 )
The C-Eq for the homogeneous equation is
Y − 2Y + 2Y − 1 = 0
(Y − 1)(Y − Y + 1) = 0
1 ± r√3
Y( = 1, Y , =
2
Thus, the G.S. to the homogeneous DE is
¡(
) = 1( exp( ) + \1 cos
1
√3
√3
+ 1 sin
] exp E F
2
2
2
409
Appendix A Solutions to Selected Problems
Since e is part of the G.S., we can assume the P.S. has form
¢ ( ) = A( exp( ) + A exp(2 ) + A
Thus,
¢ = A( ( + 1) exp( ) + 2A exp(2 )
¢ = A( ( + 2) exp( ) + 4A exp(2 )
u¢ = A( ( + 3) exp( ) + 8A exp(2 )
Plugging these into the DE, we have
A( exp( ) + 3A exp(2 ) − A = exp( ) + exp(2 ) + 1
This gives
1
A = ,
A = −1
A( = 1,
3
Thus,
1
¢ = exp( ) + exp(2 ) − 1
3
and
1
( ) = exp( ) + exp(2 ) − 1 + 1( exp( )
3
1
√3
√3
+ 1 sin
+ \1 cos
] exp E F
2
2
2
Finally, we have
1
( ) = ln +
− 1 + 1(
3
√3
√3
+ \1 cos \ ln ] + 1 sin \ ln ]] √
2
2
Problem 3.5.18
Plug in test solution = - to solve homogeneous equation:
[v(v − 1) + v − 4] - = 0
v −4=0
v = ±2
>
is the homogeneous solution. For
So n = 1( + 1
inhomogeneous part, the trial solution can be
410
Appendix A - Chapter 3
= A ln + ž > ln
where A and ž are constants. Plugging into DE gives
(A ln + ž > ln ) + (A ln + ž > ln )
− 4(A ln + ž > ln )
= $A(2 ln + 3) + ž > (6 ln − 5)%
+ $A (2 ln + 1) + ž > (−2 ln + 1)%
− 4(A ln + ž > ln )
= 4A − 4ž >
1
1
A = ,ž = −
4
4
The G.S. is then
1
= 1( + 1 > + ( − > ) ln
4
l
Problem 3.5.19
, which imply
Given two L.I. solutions ( ( ) and
+
v
+
œ
=
0
(
(
| (
+v
+œ
=0
1 × − 2 × ( gives
= y ( −
v
− (
(
The Wronskian
×
This gives
=Ý
(
(
×
v
=
×
(
Ý =
−
=−
×
=−v
×
ln × = − v
× = exp E−
(
v
(
−
(1)
(2)
(
×
F
This is the required relation between the Wronskian and v
.
411
Appendix A Solutions to Selected Problems
The variation of parameters formula gives:
y(
− y( y
y
¢ =
×
Substitute this value of × in the above variation of parameters
formulate find a P.S.
− (
(
¢ =
ñ
exp š− ´ v
›
Since ( and are the two L.I. solutions of the homogeneous
portion of " + v
′ + œ( ) = ( ), the G.S. is
(
( )
¡ ) = 1( ( ( ) + 1
( ) = ¡ ( ) + ¢( )
y ( )y( ( ) − y( ( )y ( )
( )
= 1( ( ( ) + 1 ( ) +
ñ
exp š− ´ v( ) ›
Problem 3.5.20
C-Eq of homogeneous portion:
+
=0
= 0 → Y( , Y = ±r
Y +
) + sin( )
¡ = ( cos(
(1) Function ( ) is a general form
( )=
where
=
¡
+
¢
( cos(
¢(
H( , 0) =
)=
)+
cos
H( , 0) (0) 0
sin(
)+
=
=D
H( , 0)
sin 0 − cos 0 sin
(0) 0
(2) Function ( ) is a specific function ( ) = exp(r ) .
Naturally, ( )is a solution of the homogeneous equation and
thus, the trial solution should be
)
¢ ( ) = A exp(r
412
Appendix A - Chapter 3
) = A exp(r ) + A r exp(r )
exp(r ) + 2Ar exp(r )
¢ ( ) = −A
Thus, we have
exp(r ) + 2Ar exp(r ) + A
exp(r )
−A
= exp(r )
From this, we get
1
A=
2r
Thus, the P.S. should be
1
exp(r )
¢( ) =
2r
1
( ) = ¡ + ¢ = ( cos( ) + sin( ) +
exp(r )
2r
Thus
¢(
(3) Function ( ) is a specific function
≠0
)
¢ = A exp(
¢ = A exp(
exp(
¢ =A
Since
+
= exp( )
A exp( ) + A exp(
1
A=
( )=
¡
Remarks
When = 0
+
¢
=
¢
=
( cos(
1
exp(
)+
( ) = exp(
)
)
) = exp(
)
sin(
)+
). For
)
1
exp(
)
=1
−1=0
413
Appendix A Solutions to Selected Problems
(Y − 1)(Y + 1) = 0
Y( = 1, Y = −1
( ) = ¡ = ( exp( ) + exp(− )
414
Appendix A - Chapter 3
3.6 Variation of Parameters
Problem 3.6.1
(1) This is a 2nd-order DE with constant coefficients, the C-Eq is
Y − (Y( + Y )Y + Y( Y = 0
whose solution are Y( and Y . Thus, the two solutions are
= exp(Y )
( = exp(Y( ),
(2) Let ¢ = ( ( +
, and the trial solutions are
( )
exp(Y )
( ) =−
( )
( =−
×( )
×( )
exp(Y( )
(( )
( ) =−
( )
=−
×( )
×( )
where
exp(Y( )
exp(Y )
×( ) = _
_
Y( exp(Y( ) Y exp(Y )
= (Y − Y( ) exp$(Y( + Y ) %
(3) The G.S. for the whole equation
¢ = ( (+
1
=
$exp$Y ( − )% − exp$Y( ( − )%% ( )
Y − Y(
Thus the G.S. is
= 1( exp(Y( ) + 1 exp(Y )
1
+
$exp$Y ( − )%
Y − Y(
− exp$Y( ( − )%% ( )
Problem 3.6.2
First we must take all necessary derivatives of ( ),
( ) = ¹ ( + ¹′ (
( ) = ¹ ( + 2¹ ( ′ + ¹′′ (
Plugging these values into the original DE, we obtain
415
Appendix A Solutions to Selected Problems
¹ ( + 2¹′ ( + ¹′′ ( + h( )(¹ ( + ¹ ( ) + Ÿ( )¹ ( = 0
Now note that we can collect terms to get
¹( ( + h( ) ( + Ÿ( ) ( ) + ¹ ( + ¹ (2 ( + h( ) ( ) = 0
Since ( is a solution
¹( ( + h( ) ( + Ÿ( ) ( ) = 0
We get
¹ ( + ¹ (2 ( + h( ) ( ) = 0
Now we will perform the following substitution to cast the
equation into a 1st-order DE form
= ¹ =>
= ¹′′
We obtain
′ + °2
+ h( )± = 0
(
(
Using the integrating factor method, we find that
( ) = exp \−
Thus,
°2
= exp E−2 ln
( )=
Now we must solve for Z.
¹( ) =
The second L.I. solution is
( )=
((
(
(
(
−
+ h( )±
exp(− ´ h( )
(( )
exp(− ´ h( )
(( )
)
)
)
exp(− ´ h( )
(( )
Problem 3.6.3
The corresponding homogeneous DE is
+ =0
416
h( )
F
)
]
Appendix A - Chapter 3
Y +1 = 0
Y(, = pr
So the G.S. to the homogeneous DE is
1 cos
¡ = 1( sin
By using the variation of parameters method, denote
= cos and = cot
( = sin ,
Then
The C-Eq is
×( ( ,
)=Ý
¢
=−
(
= sin
+
×
(csc − sin
= sin ln|csc
Problem 3.6.4
The homogeneous equation is
Let
Then
¢
(
= exp(− ) ,
=(
cot |
cos
Ý
sin
(
×
cos
( (
+
¢
=
( (
+
cos
=0
Y +Y +Y+1=0
(Y + 1 Y
1 =0
Y( = −1, Y , = pr
The C-Eq is
We have
(
sin
=Ý
cos
= −1
and we can get the P.S.
Ý
(
= sin ,
+
)+(
+
( (
= cos
+
+
)
417
Appendix A Solutions to Selected Problems
Since
Thus
¢
is not unique, we can assume
+
( (+
¢
=
( (
+
+
=0
Similarly, we have
)+( ( (+
)
+
+
¢ =( ( ( +
and we can also assume
+
=0
( (+
Thus,
+
¢ = ( ( +
Finally,
)+( ( ( +
)
+
+
¢ =( ( ( +
and
are solution of the corresponding
Since ( ,
homogeneous DE, thus
( =− ( − (− (
=− − −
Z
=− − −
Plugging these into the equation for ¢ , we can get
)
+
¢ =− ¢ − ¢− ¢+( ( ( +
Therefore, from the original DE, we have
+
= ( )
( ( +
Now we have
+
=0
( (+
+
=0
Z ( (+
+
= ( )
( ( +
or we can write them in the matrix form
(
0
(
! "= 0
(
( )
(
Solving this, we can get
418
(
×( , ) ( )
×( ( , , )
×( , ( ) ( )
×( ( , , )
×( ( , ) ( )
×( ( , , )
=
=
=
Problem 3.6.5
Appendix A - Chapter 3
The homogeneous equation is
+9 =0
whose C-Eq is
Y +9 = 0
Y(, = p3r
Thus, we have
and
= sin 3
( = cos 3
The P.S. has the form
¢ = ( (+
where ( and
satisfies
=0
( (+
= ( )
( (+
where
( ) = sin tan
and we know that the Wronskian
cos 3
sin 3
×( ( , ) = Ý
Ý
−3 sin 3
3 cos 3
=3
We get
(
=−
1
=−
3
×
sin 3 sin tan
419
Appendix A Solutions to Selected Problems
and
1
= − Esin
3
=
+ ln cos F
1
cos 3 cos cot
3
1
= (− sin cos + 2 cos sin − )
3
=
The P.S. is
(
×
1
+ sin
2
¢
=
( (
+
Problem 3.6.6
= 0 where = is a P.S..
If = 0, we have
Now consider
≠ 0 . It is easy to get the G.S. for the
homogeneous equation
+ 1 sin
¡ = 1( cos
Thus, we have
( ) (( ) − (( ) ( )
( )
¢ =
×$ ( ( ), ( )%
, ( ) = sin
, ( ) = sin tan
where ( ( ) = cos
and the Wronskian × = .
We can get
1
( sin
+ 2 cos
ln cos )
¢ =
2
Problem 3.6.7
The C-Eq of the homogeneous equation is
Y + =0
which gives
Y( , Y = ±r√
The G.S. for the homogeneous equation is
% + 1 sin$√
¡ = 1( cos$√
420
%
Appendix A - Chapter 3
Thus, we have
¢
=
( ) (( ) −
×$ ( ( ),
where ( ( ) = 1( cos(√
tan( ) and
×$ ( ( ),
),
((
) ( )
( )%
( )
( ) = 1 sin(√
( )% = Ý
(
(
Ý=√
), ( ) =
Problem 3.6.8
(1) The associated homogeneous DE is
+ = 0, and the CEq for this equation is
Y +1 = 0
¡ ( ) = 1( sin + 1 cos
The two L.I. solutions for the homogeneous DE is
= cos
( = sin ,
and ( ) = 2 sin .
(2) The Wronskian is
(
sin
cos
×( ) = Ý
Ý=Ý
Ý = −1
cos
− sin
(
( ) ( )
cos ∙ 2 sin
cos
=−
=−
( =−
×( )
−1
2
sin ∙ 2 sin
sin
(( ) ( )
=
=
=
−
×( )
−1
2
¢( ) = ( ( +
cos 2 sin
sin 2 cos
=−
+
− cos
2
2
(3) The G.S. of the DE is
( ) = ¡ + ¢ = 1( sin + 1 cos + sin − cos
Problem 3.6.9
With ( = ,
= 1/ , and ( ) = 72
parameters takes the form
, the variation of
421
Appendix A Solutions to Selected Problems
€
(
(
+
=0
−
= 72
Upon multiplying the second equation by
readily solve first for (
( = 36
And then
(
=
=
=−
36
−36
(
9
=9
= −36
and then adding, we
9
= −6
Then it follows that: a P.S. can be written as
¢ = ( (+
1
= ⋅9 − ⋅6 û
=3
û
9
Problem 3.6.10
The homogenous portion is
′′ − 3 − 4 = 0
C-Eq:
Y − 3Y − 4 = 0
Y( = 4,
Y = −1
¡ ( ) = 1( exp(4 ) + 1 exp(− )
Method I: Trail Solution
Trial P.S. is ¢ ( ) = A exp(4 ) + ž exp(− )
Since each term in the trial P.S.
¢ ( ) depends on terms
in ¡ ( ), we modify terms in ¢ ( ), i.e., multiply each term in
¢ ( ) with t we get
¢ ( ) = A exp(4 ) + ž exp(− )
422
Appendix A - Chapter 3
Evaluating
¢ = A exp(4 ) + 4A exp(4 ) + ž exp(− ) − ž exp(− )
¢ = 8A exp(4 ) + 16A exp(4 ) − 2ž exp(− ) + ž exp(− )
Substituting and equating the coefficients
− 3 − 4 = 15 exp(4 ) + 5 exp(− )
Produces A = 3, ž = −1
( ) = ¡ ( ) + ¢( )
= 1( exp(4 ) + 1 exp(− ) + 3 exp(4 ) − exp(− )
Method II: Variation of Parameters
From ¡ ( ) we have ( ( ) = exp(4 ) , ( ) = exp(− ) and
( ) = − exp(− )
( ( ) = 4 exp(4 ) ,
Evaluate the Wronskian of ( , .
exp(4 )
exp(− )
(
×( ) = Ý
Ý=_
_ = −5 exp(3 )
)
4 exp(4
− exp(− )
(
We have
( ) ( (³) − ( ( ) (³)
(³) ³
¢ =
×(³)
where
(³) = exp(−³) , ×(³) = −5 exp(3³)
( (³) = exp(4³) ,
We get
3
1
¢ = 3 exp(4 ) + exp(4 ) − exp(− ) − exp(− )
5
5
( ) = ¡( ) + ¢( )
3
= 1( exp(4 ) + 1 exp(− ) + 3 exp(4 ) + exp(4 )
5
1
− exp(− ) − exp(− )
5
= 1 exp(4 ) + 1 exp(− ) + 3 exp(4 ) − exp(− )
where 1 = 1( + , 1 = 1 −
9
(
9
Problem 3.6.11
The Wronskian is
423
Appendix A Solutions to Selected Problems
×
×( ( ,
)=w
,
Taking the derivative, we have
=
=ó
(,
=w
=w
and
(
(
−k(
w
(
(
(
w
(
(
ó+w
(
(
w
(
(
w
(
(
(
w+w
(
(
−k(
=w
−k(
(
+w (
−k
= −k( w
(
(
(
(
(
(
w
(
(
(
w+w (
−k
−k
w
(
−k
−k
w
w = −k( ×
× = H exp E −k( ( ) F
424
(
are solutions of the DE
( + k( ( + k ( + k ( = 0
+ k( + k
+k
=0
+ k( + k
+k
=0
( = −k( ( − k ( − k ( =
= −k( − k
−k
=
⇒
= −k( − k
−k
=
×
(
(
−k
w
Appendix A - Chapter 3
×( ( ,
(
(
Ý=
(
−
(
,
×( , ( ) =
(− (
×( , ) ( )
(( ) =
×( ( , , )
×( , ( ) ( )
( )=
⇒
×( ( , , )
×( ( , ) ( )
( )=
×( ( , , )
Finally, substituting ( ( ), ( ), ( ), ( ( ) , ( ), ( ) into
( ) ( )+ ( ) ( )
¢ = (( ) (( ) +
Similarly
×( ,
)=
)=Ý
−
425
Appendix A Solutions to Selected Problems
Chapter 4 Systems of Linear DEs
4.2 First-Order Systems and Applications
Problem 4.2.1
(1) Consider the forces on the first object: the spring /( is
stretched ( units and spring / by − ( units. Stretched force
is pointed away from the object. By Newton’s law, we have
K ( = / ( − ( ) − /( ( = −(/( + / ) ( + /
For the second object, the spring / is stretched by ( − ( )
units. Again, by
units, and the spring / is compressed by
Newton’s law, we have
K = −/ ( − ( ) − /
= / ( − (/ + / )
(2) If the spring / is broken, then / = 0, the DEs are
K = −/( (
| (
K = −/
(3) If /( = / = / = /
K (
K ( = −2/ ( + / ⇒
=
Z
/+2 (
K = / ( − 2/
K (
/+2
( )
(
( )
(
K
/+2
(
=/
=/
(
(
− 2/ \
− 2K
K (
]
/+2 (
(
− 4/
(
2/ + 2/K
3/
=0
( +
K (
K
which is a 4th order DE with constant coefficient. Solving the CEq, and after we find the characteristic roots, we can easily get
the generally solution of ( ( ). ( ) can be attained with the
similar method.
( )
(
426
+
Appendix A - Chapter 4
Problem 4.2.2
The force can be show as the figure on the top of the next page.
From Newton’s law, we have:
Force in direction: M = −M cos ü = K
ü
M
( , )
Y
Figure A.7 The particle moves under the influence of force M.
Force is direction: M = −M sin ü = K
From the given condition, we know:
Y=V
+
sin ü =
and M =
/
+
and cos ü =
Y
Y
By substitution of sin ü, cos ü, M, we have
/
K =−
Y
€
/
K =−
Y
=
/
Y
427
Appendix A Solutions to Selected Problems
Problem 4.2.3
Forces on this object can be shown in the following figure. By
Newton’s law:
Forces in x-direction: M = −M cos ü = K
Forces in y directions: M = −M sin ü − MÅ = K
Also from the given condition, we have
L
cos ü =
=
L
L
L
sin ü =
=
L
L
MÅ = K- and M = /L
By substitution, we have
K = −/L ′
|
K = −/L − KL = L′
MX
ü
L
L = ′
M‡
Figure A.8 A projectile model moving with speed L.
Problem 4.2.4
Assuming the displacements of blocks from the nature positions
( ). The DEs of motion for the two blocks are
( ( ) and
428
Appendix A - Chapter 4
− ( ) + K( ( = −/( ( + / (
K
= −/ ( − ( ) + K By substitution, we get
K
| (
+ [(/( + / )K + / K( « − /( /
= (/( + / )K This is a constant coefficient inhomogeneous DE that can be
solved with C-Eq method.
( ) = ( cos ( + sin (
+ cos
+ sin
+ l( )
where ( and
are two characteristic frequencies. Back
substitute can generate the solution for ( ( ).
K( K
( )
Problem 4.2.5
Let ë ( ) be the displacement of block-i from its initial position,
r = 1,2,3.
The DEs for the system are
K( ′′( = −/( ( + / ( − ( ) + K( ZK ′′ = −/ ( − ( ) + / ( − − ( ) + K K ′′ = −/ ( − − ( ) + K Also, we have the I.C.
(0) = ′ (0) = (0) = ′ (0) = 0
( (0) = ′( (0) =
The above system of DEs together with the I.C. describes the
motion of three blocks.
Let
/(
/
/
/
/
o( =
,o =
,o =
, o( =
,o
=
,
K(
K
K
K(
K
We can have
+( ′′ = −(o( + o( ) ( + o(
S
= −(o + o ) − o
+
o
+(
= −o
+o (+o
+R
Q ( (0) = ′( (0) = (0) = ′ (0) = (0) = ′ (0) = 0
429
Appendix A Solutions to Selected Problems
Problem 4.2.6
On the ball, the force can be decomposed into two components,
one along the direction of the spring and another ü along
which we can establish the following two DEs.
K = K- − /
(1)
|
(2)
)«
K[ü( + …B
= −K-ü
Along the extension of the spring direction, the gravity is
K- cos ü ≈ K- × 1
Because the angle is small so we can make the approximation
given above.
Eq-(1) is trivial to solve, one can write the solution as
= sin( ) +
and
¥
tÍ
can be absorbed to …B where
=
ë
. Inserting such
solution to Eq-(2) results in
ü = −-ü
ü ( + …B ) + 2ü
+ ü sin
= −-ü
ü (sin
+ …B ) + 2ü cos
Now, we combine with I.C. to get
+ ü sin
= −-ü
ü (sin
+ …B ) + 2ü cos
|
ü( = 0) = üB , ü ( = 0) = 0
We can now perform Laplace transform on the above equation to
solve it or we can use other methods.
430
Appendix A - Chapter 4
4.3 Substitution Method
Problem 4.3.1
Sum up two equations,
= exp(− )
( ) = exp(− ) + 1( + 1
Substitute into second equation
exp(− ) +
− exp(− ) − 1( − 1 = 0
= 1( + 1
1(
1
=
+ 1
+1 +1
6
2
whose solution is
Problem 4.3.2
From the first equation = − , we know that = − ′.
Substituting this into the second DE, we get
(− ′) = 10 + 7 ⇒ ′′ + 7 ′ + 10 = 0
This is a 2nd-order DE with constant coefficients and the C-Eq is
Y + 7Y + 10 = 0
Y( = −5, Y = −2
Therefore,
= 1( exp(−5 ) + 1 exp(−2 )
and thus, we can have
=−
= 51( exp(−5 ) + 21 exp(−2 )
From the given I.C., we know that
1 +1 =2
| (
51( + 21 = −7
Solving this, we can get
11
17
1 =
1( = − ,
3
3
Finally, the solutions to the DEs are
431
Appendix A Solutions to Selected Problems
11
17
exp(−5 ) + exp(−2 )
3
3
€
55
34
= − exp(−5 ) + exp(−2 )
3
3
=−
Problem 4.3.3
From the first DE
−
2
− ′
=
2
Plugging into the second DE, we have
−
−
=2 −2
2
2
+ −6 =0
This is a 2nd-order DE with constant coefficients, and the C-Eq
is
Y + Y − 6 = (Y + 3)(Y − 2) = 0
Y( = −3, Y = 2
( ) = 1( exp(−3 ) + 1 exp(2 )
−
1
( )=
= −21( exp(−3 ) + 1 exp(2 )
2
2
Insert I.C., we get 1( = −3/5 and 1 = 8/5. Thus the G.S. of
the system is
8 exp(2 ) − 3 exp(−3 )
( )=
5
€
4 exp(2 ) + 6 exp(−3 )
( )=
5
=
Problem 4.3.4
As per the information given in the above problem we have,
K
= −(/( + / ) ( + /
| ( (
K
= / ( − (/ + / )
Given K( = K = 1, /( = 1, / = 4 and / = 1, we have
432
= −5 ( + 4
=4 ( 5
From the second equation, we have
5
( =
4
4
Put it back to the first equation, we can get
|
\
4
Appendix A - Chapter 4
(
5
] = −5 \
4
4
10
5
]
4
+9
4
=0
Y + 10Y + 9 = 0
Y , = ±r
Y(, = ±3r,
This gives the G.S. for :
= 1( sin 3 + 1 cos 3 + 1 sin + 1 cos
and put it back for ( :
5
( =
4 4
= −1( sin 3 − 1 cos 3 + 1 sin + 1 cos
The C-Eq is
Problem 4.3.5
From
We can have
(
=
Substitute into
We can have
C-Eq:
−
2
=2
−
(
−2
=
,
(
(
+
+
+2
−3
(
=
+
=−
Y
2Y − 3 = 0
Y( = 3, Y = −1
−
2
−2
+4
433
Appendix A Solutions to Selected Problems
= ( exp(3 ) + exp(− )
Then we can have the P.S. like the following form
+ A( + AB
¢ =A
+ A(
¢ = 2A
¢ = 2A
Substitute into the original equation.
2A − 2(2A + A( ) − 3(A
+ A( + AB ) = − + 4
Then
38
16
1
, A( = −
,A =
AB =
27
9
3
So the G.S. is
1
16
38
= ( exp(3 ) + exp(− ) +
−
+
3
9
27
2
16
= 3 ( exp(3 ) − exp(− ) +
−
3
9
Substitute into
− −
( =
2
We can have
2
11
43
exp(− ) −
+
−
( = ( exp(3 ) −
3
9
27
So the solution is
2
11
43
exp(− ) −
+
−
( = ( exp(3 ) −
3
9
27
1
16
38
= ( exp(3 ) + exp(− ) +
−
+
3
9
27
¡
Problem 4.3.6
434
=
= = −9 + 6
−6 +9 =0
Y − 6Y + 9 = 0
Y( = Y = 3
Appendix A - Chapter 4
Hence
Problem 4.3.7
( ) = (A + ž ) exp(3 )
( ) = (3A + ž + 3ž ) exp(3 )
|
From Eq-(2)
Plugging this into Eq-(1)
+
The C-Eq is
= −2
= −
=
=
(1)
(2)
+
+
= −2
= + −2
+ =0
Y +1 = 0
Y(, = ±r
( ) = 1( cos + 1 sin
Put this back, and
( ) = (1 − 1( ) sin + (1( + 1 ) cos
Plugging in the I.C., we have
1 = 1( + 1
|
2 = 1(
1 =2
| (
1 = −1
Thus
( ) = cos − 3 sin
|
( ) = 2 cos − sin
Problem 4.3.8
From Eq-(2)
|
=3 +4
=3 +2
(1)
(2)
435
Appendix A Solutions to Selected Problems
−2
3
−2
3
=
Plugging this into Eq-(1)
−2
3
=
=
5
−2 +4
−6 =0
Y − 5Y − 6 = 0
(Y + 1)(Y − 6) = 0
Y( = −1, Y = 6
( ) = 1( exp(− ) + 1 exp(6 )
( ) = −1( exp(− ) + 61 exp(6 )
Put this back, and
4
( ) = −1( exp(− ) + 1 exp(6 )
3
Plugging in the I.C., we have
4
1 = −1( + 1
•
3
1 = 1( + 1
1
1( =
7
€
6
1 =
7
Thus
8
1
( ) = − exp(− ) + exp(6 )
7
7
€
1
6
( ) = exp(− ) + exp(6 )
7
7
The C-Eq is
Problem 4.3.9
Using Eq-(1)
436
Ù=
−
−2
Appendix A - Chapter 4
Ù =
−
−2 ′
Substitute into Eq-(3)
− −2 = + + − −2
− 2 = − + 2 (4)
Using Eq-(2)
−2
=
3
−2
=
3
−2
=
3
Substitute above DEs to Eq-(4)
−2
2( − 2 )
−
=− +2 ′
3
3
−4
2 +3 =0
Find the C-Eq
Y − 4Y
2Y + 3 = 0
5 + √13
5 − √13
Y( = −1,
Y =
,
Y =
2
2
( ) = ( exp(− ) + +
5 + √13
5 − √13
( ) = − ( exp(− ) +
+
2
2
Using Eq-(2)
−2 =3
5 + √13
5 − √13
+
−2 (−2 −2
− (+
2
2
=3 (+3 +3
( = −3 (
6 − 6√13
=−
12
6 + 6√13
=−
12
( ) = ( exp(− ) + +
437
Appendix A Solutions to Selected Problems
( )=−
( exp(−
)+
5 + √13
2
+
5 − √13
2
− −2 =Ù
5 + √13
5 − √13
+
− (− − +2
− (+
2
2
12 − 12√13
12 + 12√13
+
+
12
12
= (+ +
5 − √13
5 + √13
=
,
=
( = 0,
2
2
( ) = ( exp(− ) + +
6 − 6√13
6 + 6√13
( ) = −3 ( exp(− ) −
−
12
12
5 + √13
5 − √13
+
Ù( ) =
2
2
Using Eq-(1)
438
(
Appendix A - Chapter 4
4.4 Operator Method
Problem 4.4.1
From the problem
(Â − 1) − 2 − Ù = 0
Z−6 + (Â + 1) = 0
+ 2 + (Â + 1)Ù = 0
Â−1
−2
−1
0
( )= 0
−6
Â+1
0
1
2
Â+1 Ù
0
The operational determinant is:
Â−1
−2
−1
w −6
Â+1
0 w = Â(Â + 4 Â 3)
1
2
Â+1
So , , Ù all satisfy a third-order homogeneous linear DE with
C-Eq
Y(Y + 4 Y 3) = 0
the corresponding G.S. are
= ( + exp(−4
exp(3 )
exp(3 )
Z = ( + exp(−4
exp(3 )
= ( + exp(−4
If we substitute ( ) and ( ) in the second DE = 6 − ,
and collect the coefficients of like terms, we have:
3 exp(3 )
−4 exp(−4
= 6 ( + 6 exp(−4
6 exp(3 ) − ( − exp(−4
exp(3 )
This gives
3
= −2 , =
( = 6 (,
2
Similarly, we can substitute ( ), ( ) and Ù( ) in the first or
third DE, and compare the coefficients of like terms, we have
=− , =−
( = −12 ( ,
gives
439
Appendix A Solutions to Selected Problems
thus the final G.S. are:
= (+
€ =6
(
−2
Ù = −12
(
exp(−4
−
exp(−4
exp(−4
exp(3 )
3
exp(3 )
2
exp(3 )
Problem 4.4.2
By the following operations
(Â − Â) × Eq-(1) −Â × Eq-(2)
we have:
(Â − Â)(Â + Â) − Â (Â − 1) = (Â − Â)2 exp(− )
which gives
0 = 4 exp(− )
This is impossible, so the DEs have no solution.
Problem 4.4.3
( − 3) − 9 = 0
−2 + (Â − 2) = 0
Â−3
−9
⇒Ý
Ý =0
−2
Â−2
⇒ (Â − 5Â − 12) = 0
Y − 5Y − 12 = 0
5 ± √73
Y(, =
2
( ) = A( exp(Y( ) + ž( exp(Y )
( ) = A exp(Y( ) + ž exp(Y )
Apply the I.C. (0) = (0) = 2 gives
A( + ž( = A + ž = 2
The C-Eq is
Problem 4.4.4
The original DEs can be written as
Â−1
−2
−2
Â−1
440
(
=
Â−1
−2
−2
Â−1
So we can have
Â
Appendix A - Chapter 4
−2
Â−1
Â−1
−2
−2
Â−1
⇒
2(Â − 1)
−4
2
−2
Â
−
1
⇒
0 Â − 2Â − 3 −
⇒
2Â − 3)
=−
+4
+4
Y
2Y − 3 = 0
Y( = 3, Y = −1
=
exp(− )
¡
( exp(3 ) +
Then we can have the P.S. like the following form
+ A( + AB
¢ =A
+ A(
¢ = 2A
¢ = 2A
Substitute into the original equation
+ A( + AB ) = − + 4
2A − 2(2A + A( ) − 3(A
Then
38
16
1
AB =
, A( = −
,A =
27
9
3
So the G.S. is
1
16
38
= ( exp(3 ) + exp(− ) +
−
+
3
9
27
From the first row of the matrix
−2 ( + (Â − 1) =
− −
( =
2
2
11
43
= ( exp(3 ) − exp(− ) −
+
−
3
9
27
So the solution is
2
11
43
exp(− ) −
+
−
( = ( exp(3 ) −
3
9
27
C-Eq:
441
Appendix A Solutions to Selected Problems
=
Problem 4.4.5
( exp(3
)+
exp(− ) +
1
3
−
 − 1 − 1Ù = exp(− )
−1 + Â − 1Ù = 0
−1 − 1 + ÂÙ = 0
16
38
+
9
27
 −1 −1
w−1 Â −1w = Â(Â − 1) + 1(−Â − 1) − 1(1 + Â)
−1 −1 Â
=Â −Â−Â−1−1−Â
= Â − 3Â − 2
= (Â + 1) (Â − 2) = 0
( ) = ( exp(2 ) + exp(− ) + exp(− )
( ) = ( exp(2 ) + exp(− ) + exp(− )
Ù( ) = ( exp(2 ) + exp(− ) + exp(− )
1
= 1( exp(2 ) − (1 + 1 ) exp(− ) ,
S
3
1
1
= − exp( ) + 1( exp(2 ) + 1 exp(− ) ,
2
3
R
1
1
)
)
)
Q Ù = − 2 exp( + 3 1( exp(2 + 1 exp(− .
Problem 4.4.6
Method I: Substitution
By the first equation, we have = − , and =
− .
Return it to the second equation, we have
− =6 −
( − ).
= 7 . The C-Eq is Y = 7. Y( = √7, Y = −√7. Therefore
( ) = ( exp$√7 % + exp$−√7 %
ô
( ) = (√7 − 1) ( exp$√7 % − (√7 + 1) exp$−√7 %
Method II: Operator
The DEs are written as
442
|
(Â − 1) − = 0
−6 + (Â + 1) = 0
Appendix A - Chapter 4
By (Â + 1) ×(3)+(4), we have
(Â + 1)(Â − 1) − 6 = 0
(Â − 7) = 0
The C-Eq is
Y −7 = 0
Y( = V7, Y = −√7
Therefore
( ) = ( exp$√7 % + exp$−√7 %
ô
( ) = (√7 − 1) ( exp$√7 % − (√7 + 1)
(3)
(4)
exp$−√7 %
Problem 4.4.7
We add and subtract the two DEs respectively, and we have
1
(1)
Â( + ) = (exp(−2 ) + exp(3 ))
2
1
(2)
2 − 3 = (exp(−2 ) − exp(3 ))
2
(
From (2) we have = (exp(−2 ) − exp(3 )) + .
We return to (1) and
5
1
1
F = (exp(−2 ) − exp(3 )).
E exp(−2 ) − exp(3 )) +
2
2
4
Then
2
1
= exp(−2 ) + exp(3 )
5
2
and
1
1
( ) = − exp(−2 ) + exp(3 ) + 1( .
5
6
(
Therefore ( ) = − B exp(−2 ) + 1(.
The solution is
443
Appendix A Solutions to Selected Problems
€
3
1
exp(−2 ) + 1(
2
20
1
1
( ) = − exp(−2 ) + exp(3 ) + 1(
5
6
( )=−
Problem 4.4.8
The DE can be rewritten into operator form
(Â − 4
13Â = 6 sin (1)
ô
2Â + (−Â + 9) = 0 (2)
From the equation
Eq-(1) × 2Â − Eq-(2) × (Â − 4
we get
Â
13Â + 36 = 12 cos
C-Eq:
Y + 13Y + 36 = 0
Y(, = ±2r,
Y , = ±3r
¡ = 1( sin 2 + 1 cos 2 + 1 sin 3 + 1 cos 3
Consider the DE has only even number of derivatives, we can
assume
¢ = A cos
This gives
= −A cos ,
A=
1
2
( )
¢
= A cos
1
( ) = 1( sin 2 + 1 cos 2 + 1 sin 3 + 1 cos 3 + cos
2
Put this back into (2) and we can get
1
= ( −9 )
2
1
= (−131( sin 2t − 131 cos 2t − 181 sin 3t
2
− 181 cos 3t − 5 cos )
So
444
¢
Appendix A - Chapter 4
1 13
13
= E 1( cos 2t − 1 sin 2t + 61 cos 3t − 61 sin 3t
2 2
2
+ 5 sin F
Problem 4.4.9
Method I: Substitution
From the second equation
−
2
−
=
2
Substitute these into the first equation
−
−
=3
−2
2
2
−4
7 =0
C-Eq:
Y − 4Y 7 = 0
So
( )=
=
Y( = 2 + √3r,
Y = 2 − √3r
) cos(√3 ) + exp(2 ) sin(√3 )
( ) = $2 ( + √3 % exp(2 ) cos$√3 %
( )=
−
2
1
= $
2
(
( exp(2
+ $2
− √3 ( % exp(2 ) sin$√3 %
+ √3 % exp(2 ) cos(√3 )
1
+ $
2
− √3 ( % exp(2 ) sin(√3 )
( ) = ( exp(2 ) cos(√3 ) +
Method II: Operator
The original DEs can be written as
So
exp(2 ) sin(√3 )
445
Appendix A Solutions to Selected Problems
Â−3
2
=0
−2
Â−1
Â−3
2
−2
Â−1
−2
Â−1
⇒
⇒
−2
Â−1
Â−3
2
2Â − 6
4
−2
Â−1
⇒
0 4
 1)( − 3)
So we can have
 1)( − 3)% ( ) = ( − 4 7) ( ) = 0
$4
C-Eq:
Y − 4Y 7 = 0
Y( = 2 + √3r,
Y = 2 − √3r
) cos(√3 ) + exp(2 ) sin(√3 )
( ) = $2 ( + √3 % exp(2 ) cos$√3 %
So
( )=
( exp(2
+ $2 − √3 ( % exp(2 ) sin$√3 %
From the first row of the matrix, we can have
−2 + − = 0
−
( )=
2
1
1
= cos(√3 ) + sin(√3 )
2
2
( ) = ( exp(2 ) cos(√3 ) + exp(2 ) sin(√3 )
446
Appendix A - Chapter 4
4.5 Eigen-Analysis Method
Problem 4.5.1
det
1−
3
−3
7−
= (1 − )(7 − ) + 9
=( −4 =0
The system has two identical eigenvalues
=4
( =
We can find the associated eigenvector T( for the eigenvalue (
and its solution is
( = T( exp( )
For the second solution, we need to compose the L.I. solution by
introducing
= (T( + T ) exp( )
We plug this proposed solution to the system, we get
= (T( ) exp( ) + (T( + T ) exp( )
so
= (T( ) exp( ) + (T( + T ) exp( )
= A(T( + T ) exp( )
or
(A − )T exp( ) = T( exp( )
or
(A − )T = T(
(A
Additionally, since − )T( = 0, we have
(A − )(A − )T = (A − )T( = 0
or
(A − ) T = 0
Now, we can find T with which we can compose the second
solution and then the entire solution.
Now, we return to our original example,
(A − ) T = (A − 4 T
−3 −3 −3 −3
=
T
3
3
3
3
447
Appendix A Solutions to Selected Problems
0 0
T =0
0 0
which means any solution T will satisfy the above equation. For
convenience, we may select
0
T =
1
with
−3 −3 0
−3
=
T( = (A − )T =
3
3 1
3
which we construct the eigenvector T( and solve
−3 −3
T =0
3
3 (
=
to find the eigenvector T( .
Finally, we have the two L.I. solutions
−3
exp(4
( =
3
−3
−3
0
=š
+
› exp(4 =
exp(4
3 +1
3
1
So, finally, the G.S. for the original DE is
3
−3
= (
= ( (+
exp(4
exp(4
3 +1
3
3 exp( )
1 2 ( )
+
2 1
−
First, we try to solve the homogeneous DEs
1 2 (
(
=
2 1
where
1−λ
2
det
= (1 − λ) − 4 = 0
2
1−λ
So the eigenvalues are λ( = 3, λ = −1
For ( = 3, we get
0
−2 2
(
=
0
2 −2 L(
We can select
Problem 4.5.2
448
=
For
(
= −1, we get
2
2
1
(
L( = 1
2
2 L
=
Appendix A - Chapter 4
0
0
−1
L = 1
So the G.S. to homogeneous DEs is
1
−1
(
=
= (
exp(3 ) +
exp(− )
1
1
The P.S. may take the following form
A
A
A
(
=
∙ 3 exp( ) − ( ∙ + B
l =
ž
ž(
žB
2A
A
(
=3
exp( ) − 2 (
2ž
ž(
Plugging them to the original system, we get
2A
A
exp( ) − 2 (
3
2ž
ž(
A
A
A
3 exp( )
1 2
exp( ) − (
+ B F+
=
E3
ž
ž(
žB
2 1
−
We can select
Solve for AB , A( , A , žB , ž( , ž and plug them into ( ) =
( )=
(
=
1
−1
exp(3 ) +
exp(− )
1
1
A
A
A
+
3 exp( ) − (
+ B
ž
ž(
žB
(
Problem 4.5.3
Eigen-Analysis method for the homogeneous portion:
1 3
=
1 −1
3
Ý
Ý = (1 − )(−1 − ) − 3 = − 4 = 0
1 −1 −
whose roots are
= −2
( =2
Find eigenvectors. For ( = 2
(
449
Appendix A Solutions to Selected Problems
For
So
−1 3 0
Ý ⇒− +3 =0
1 −3 0
3
T( =
1
= −2
¡
= 1(
3 3 0
Ý ⇒ 3 +3 =0
1 1 0
−1
T =
1
3
−1
exp(2 ) + 1
exp(−2 )
1
1
Let P.S. be
¢ = ž exp(−2 ) + þ exp(−2 )
¢ = −2ž exp(−2 ) + þ exp(−2 ) − 2þ exp(−2 )
Plugging into the original system, we have
−2ž exp(−2 ) + þ exp(−2 ) − 2þ exp(−2 )
1
= A(ž exp(−2 ) + exp(−2 )) +
exp(−2 )
0
Grouping like terms, we get
1
−2ž + þ = Až +
ô
0
−2þ = Aþ
Solve the 2nd equation
Ï
þ=
−Ï
where Ï is a constant.
Then the first equation becomes
c
1
−2ž +
= Až +
−c
0
which allows us to construct the following two equations for ž
vector
or
Thus
450
|
−4
(
|
ž=
(
+ 3 = −2 ( + Ï − 1
= −2 + 2 ( − 2Ï + 1
(
+ = Ï/3 − 1/3
= Ï − 1/2
(+
Ï 1
1
− =Ï−
3 3
2
or
Ï=
Thus
Also
We may choose
Then
=
¢
=
+
And the G.S. is
¡
1
4
þ=
1 1
4 −1
ž=
1 1
4 0
(
+
=−
1
4
1 1
1 1
exp(−2 ) +
4 −1
4 0
¢
Appendix A - Chapter 4
exp(−2 )
3
−1
exp(2 ) + 1
exp(−2 )
= 1(
1
1
1 1
1 1
+
exp(−2 ) +
exp(−2 )
4 0
4 −1
Problem 4.5.4
First find solution to corresponding homogeneous system:
1 2
A=
4
1
To find eigenvalues set det A − - = 0
1−
2
Ý
Ý = (1 − )(−1 − ) − 8 = 0
4
1−
= −3
( =3
Find eigenvector for ( = 3
0
1−3
2
(
=
0
4
1 − 3 L(
−2 ( + 2L( = 0
1
T( = š ›
1
451
Appendix A Solutions to Selected Problems
= −3
1 − (−3)
4
2
0
=
1 − (−3) L
0
2L = 0
4
1
T =
−2
So the complementary solution is:
1
1
exp(3 ) + 1
exp(−3 )
¡ ( ) = 1(
1
−2
The P.S. has form:
For
¢
¢
=ž +Â =
=
+
(
(
(
Plugging back into original system, we get:
1 2
−1
(
=
E ( + ( F+
4
1
1
=
=
=
1
4
(
4
(
4
2
1
(
+2
(
+2
(
We can solve first for B:
(
4
−1
1
(
2
+
1
4
1
1
+
2
1
+
(
4
(
4
+2
(
−1
0
=
1
0
(
+
+2
(
−1
1
+2 =1
| (
= −1
4 (
Adding twice the second equation to the first equation yields:
9 ( = −1
1
( =−
9
i.e.,
452
Appendix A - Chapter 4
Then the first equation gives
=
5
9
From our above equation we have:
1 −1
(+2
= ( =
4 (
9 5
i.e.,
1
=−
(+2
9
€
5
4 (
=
9
Solving this system finds:
1
1
=−
( =
9
9
So the P.S. is:
1 1
1 −1
+
¢ =
5
9 −1
9
And the G.S. is:
= ¡+ ¢
1 −1
1
1
= 1(
exp(3 ) + 1
exp(−3 ) +
1
−2
9 5
+
1 1
9 −1
Problem 4.5.5
This system may be written as
4
2 0
( ) = −4 4
2 ( )
Ù
0 −4 4 Ù
Let us find the eigenvalues and eigenvectors of the coefficient
matrix.
|#
%| = 0
4
2
0
w −4 4
2 w=0
0
−4 4
(4
8
2$−4 4
4
%=0
453
Appendix A Solutions to Selected Problems
( −4
8 =0
= 4, = 8
( = 0,
The corresponding eigenvectors can be found from the equation
ALë = ë Lë , where Lë is an eigenvector corresponding to ë .
Hence,
1
1
1
L( = 2 ,
L = 0 ,
L = −2
2
−2
2
The solution of the system is given by
= ( L( exp( ( ) + L exp( ) + L exp( )
where
Hence
(,
( )=
Ù
454
,
(
are arbitrary constants and
1
2 exp(0 ) +
2
= ( ).
Ù
1
1
0 exp(4
−2 exp(8 )
−2
2
exp(4
exp(8 )
(+
2 ( − 2 exp(8 )
"
( )=!
Ù
2 exp(8 )
2 ( − 2 exp(4
Appendix A - Chapter 5
Chapter 5 Laplace Transforms
5.2 Properties of Laplace Transforms
Problem 5.2.1
From the figure we write the square wave function as
1,
2 ≤ < 2 +1
-( ) = m
,
= 0,1,2, …
−1,
2 +1< ≤ 2 +2
We can write the above function as
-( ) = 2 ( ) − ( ) − 2 ( − 1) + 2 ( − 2) − 2 ( − 3)
+⋯
•
( − =) − ( )
= 2 ê(−1)
ÜB
=2 ( )− ( )
Taking Laplace on both sides we get
LHS = ℒ{-( )}
= ℒ{2 ( ) − ( )}
•
= 2 ê(−1) ℒ{ ( − =)} − ℒ{ ( )}
ÜB
•
= 2 ê(−1)
=
ÜB
•
exp(−=³) 1
−
³
³
1
2
ê(−1) exp(−=³) −
³
³
ÜB
2
1
1
= E
F−
³ 1 + exp(−³)
³
1 1 − exp(−³)
= \
]
³ 1 + exp(−³)
1
³
= tanh š › = RHS
³
2
455
Appendix A Solutions to Selected Problems
Problem 5.2.2
} = ℒ{ exp v ) sin(
M ³ = ℒ{
= ³(−1) M(³) − (0)
Problem 5.2.3
)} = ³M(³) − (0)
F −0
= ³E
(³ − v) +
2(³ − v)
= ³ \−
]
((³ − v) + )
(−2 )( (³ − v) + ) + 8(³ − v)
=³
((³ − v) + )
6 (³ − v) − 2
= ³\
]
((³ − v) + )
6 ³(³ − v) − 2³
=
((³ − v) + )
ℒ •exp( )
√
B
√8 2 exp( )
= ℒ•
2
√8
exp(−0 ) 0 7
√
B
exp(−0 ) 0 7
exp( ) √ 1
= ℒ•
exp(−0 ) 20 0 7
√8 B 0
1
= ℒô
exp( − )
/
B √8
1
√8
ℒ |exp( )
=
5
2
√8
1
= ℒ{exp( )} ℒ |
5
√8
1
=
(³ − 1)√³
456
Appendix A - Chapter 5
2A
³@
tanh
³ @
4
ℒ{ ( )} =
Problem 5.2.4
Apply Laplace transform to the function, and use the property of
Laplace transform to split it into two parts.
)}
ℒ{
} = ℒ{ sin( ( )} + ℒ{exp(v ) cos(
For the first term ℒ{ sin( ( )}, we use the property of tmultiplication
( )} = (−1) M ( ) (³)
ℒ{
In this case, = = 2. ( ) = sin( ( ). Then
ℒ{
sin(
(
M(³) =
(
³ +
(
6 (³ − 2 (
(³ + ( )
)}, we use the property
)} = (−1) M ( ) (³) =
For the second term ℒ{exp(v ) cos(
of translator
)} = M(³ − v)
ℒ{exp(v ) cos(
(
)
). Then
In this case,
= cos(
³
M(³) =
³ +
³−v
)} =
ℒ{exp(v ) cos(
(³ − v) +
Finally, after applying Laplace transform to the function ( ), we
will have
6 (³ − 2 (
³−v
ℒ{ ( )} =
+
(³ + ( )
(³ − v) +
Problem 5.2.5
1)
((
•
) = ê ( − =)
ÜB
457
Appendix A Solutions to Selected Problems
•
By taking the Laplace Transform, we get
ℒ{ ( ( )} = ℒ •ê ( − =)7
ÜB
•
= ê ℒ{ ( − =)}
ÜB
•
=ê
=
So the solution is
=
ℒ{ ( ( )} =
exp(−=³)
³
ÜB
•
1
ê exp(−=³)
s
ÜB
1
³(1 − exp(−³))
1
³(1 − exp(−³))
( )= −8 9
In this case, the floor function can be expressed as
2)
•
8 9 = ê ( − =)
Ü(
•
So, if we perform the same process of (1), then
ℒ{8 9} = ℒ •ê ( − =)7
•
Ü(
= ê ℒ{ ( − =)}
Ü(
•
=ê
Ü(
458
exp(−=³)
³
Appendix A - Chapter 5
•
1
= ê exp(−=³)
s
Ü(
exp(−³)
³(1 − exp(−³))
So, we obtain the following solution
1
exp(−³)
ℒ{ ( )} = ℒ{ } − ℒ{8 9} = −
s
³(1 − exp(−³))
=
(1) M(³) = ℒ{ 9 } = ´B
Let
exp(−³ )
• 9
Problem 5.2.6
= ³ , then = /³ and
•
M(³) =
B
1
³û
1
= û
³
5!
= û
³
=
=
•
B
Γ( ) =
/³.
š › exp(− )
³
B
•
Note
Gamma function is defined as
9
9
û>( exp(−
•
B
exp(− )
exp(− )
)
>(
and for an integer n, we have
Γ(= + 1) = =Γ(=) = =!
(2)
M(³) = ℒ{exp( )}
(3) When ³ > ,
=
=
•
exp( ) exp(−³ )
B
•
B
exp$( − ³) %
459
Appendix A Solutions to Selected Problems
exp$( − ³) %
1
x =
−³
³−
B
)}
M(³) = ℒ{sin(
(4)
•
=
B
=−
1
³
exp( ) exp(−³ )
•
sin(
B
By product rule,
exp(−³ )
sin(
M(³) = −
³
=0−0+
=−
=−
=
=
³
³
³
³
−
−
•
B
³
³
cos(
B
+
) (exp(−³ ))
)x +
B
1
³
•
) exp(−³ )
•
) exp(−³ )Ý
³
B
•
B
•
exp(−³ ) (cos(
exp(− ) sin(
B
M(³)
\1 +
³
)
] M(³) =
M(³) =
³ +
M(³) = ℒ{exp( ) sin( )}
=
=
460
exp(−³ ) (sin(
B
) (exp(−³ ))
cos(
cos(
•
•
³
•
•
B
•
B
exp( ) sin(
))
³
) exp(−³ )
exp(−(³ − ) ) sin(
)
))
Appendix A - Chapter 5
Substitute (³ − ) with ³(, we can change the integral to the
same form as part (3). Thus,
M(³) =
³( +
=
(³ − ) +
Problem 5.2.7
This is a periodic function (of period @ ) whose Laplace
transform can be expressed as
ℒ{ ( )} =
1
1 − exp(−³@)
C
C
B
( ) exp(−³ )
where the integral portion can be carried out as
=
C
B
B
( ) exp(−³ )
2A E F exp(−³ )
@
+
C
C
2A E1 − F exp(−³ )
@
A
³@
2A
³@
2A
= E− exp E− F −
exp E− F +
F
³
2
³ @
2
³ @
2A
A
³@
+E
exp(−³@) + exp E− F
³ @
³
2
2A
³@
−
exp E− FF
³ @
2
2A
³@
=
E−2 exp E− F + 1 + exp(−³@)F
³ @
2
2A
³@
=
E1 − exp E− FF
2
³ @
Therefore, the transform is
ℒ{ ( )} =
2A
³@
1
E1 − exp E− FF
2
1 − exp(−³@) ³ @
461
Appendix A Solutions to Selected Problems
³@
2A 1 − exp š− 2 ›
=
³ @ 1 + exp š− ³@›
2
³@
³@
2A exp š 4 › − exp š− 4 ›
=
Þ
ß
³ @ exp š³@› + exp š− ³@ ›
4
4
³@
2A
³@
2A sinh 4
=
tanh
=
³@
³ @
4
³ @ cosh
4
462
Appendix A - Chapter 5
5.3 Inverse Laplace Transforms
Problem 5.3.1
Taking Laplace on the RHS of the given problem, we get
1
ℒ|
sin / − / cos / 5
2/
1
ℒ{sin / } − /ℒ{ cos / }
=
2/
/ ³ −/
1
/
−
=
\
]
³ +/
2/ ³ + /
1
2/
=
\
]
³ +/
2/
1
=
³ +/
= ℒ{LHS}
Hence the above is proven.
Problem 5.3.2
We have
•
ℒ2;B $2√ %3 = ℒ •ê
•
=ê
ÜB
•
=ê
ÜB
ÜB
−1
$√ % 7
2 ⋅ =!
−1
ℒ{ }
2 ⋅ =!
−1
=! ⋅ ³
ú(
1
•
š− ³ ›
1
= ê
=!
³
=
ÜB
1
exp š− ³ ›
³
463
Appendix A Solutions to Selected Problems
which finishes the proof.
Problem 5.3.3
Let us suppose that the solution has the form of
=
Then we have a Laplace transformation for this function.
2³
}=
} = − ℒ{
ℒ{
³ −1
³
So, we have
1
1 1
1
}=
ℒ{
= E
−
F
³ −1 2 ³−1 ³+1
By taking the inverse Laplace transform, we get
1
= exp − exp −
2
Consequently, the solution is
1
=
= ªexp − exp − «
2
Let k =
Problem 5.3.4
Let
ℒ{
464
and
M ³ =
= −³
= w,
1
1 − exp −³
exp
=
1
1 − exp − ³
1
=
1 − exp − ³
1
=
³ 1 − exp − ³
1
=
³ 1 − exp − ³
}=
¢
exp −³
B
− 1 exp
>wq
B
>wq
B
exp −³
exp
>wq
B
$
exp
š− › E− F
³
³
− 1 exp
%yB
>wq
.
Appendix A - Chapter 5
1
$ − ³ − 1 exp − ³ + 1%
³ 1 − exp − ³
exp − ³
1
−
=
³ 1 − exp − ³
³
=
Since
ℒ |− − -
ℒ >( ô
5=
1
−‡ ³
³
exp − ³
/=
³ 1 − exp − ³
−
Problem 5.3.5
Performing Laplace Transform on both sides of the equation, we
get
= ℒ{sin
ℒô
B
cos
} + 2ℒ ô
−0
B
cos
0
−0
1
2³
+
³ +1 ³ +1
1
³ =
³−1
>( {
=ℒ
³ } = exp
=
Problem 5.3.6
Take Laplace on each side of the equation:
ℒ{ } = ℒ{2 exp 3 } − ℒ ô
Let ℒ{ } =
0/
B
exp$2
0
−0 %
0/
0
0/
³ . By the convolution of two functions, we have
³
}=
ℒ ô exp$2 − 0 % 0 0/ = ℒ{exp 2 }ℒ{
³−2
B
Substitution of the above equations, we have
465
Appendix A Solutions to Selected Problems
This gives
Thus
466
³ =
³ =
2
³
−
³−3 ³−2
1
1
+
³−1 ³−3
= exp
+ exp 3
Appendix A - Chapter 5
5.4 The Convolution of Two Functions
Problem 5.4.1
ℒ >( |
Let
1
³ − 1√³
-
By convolution
5 = ℒ >( |
1
³−1
1
= ℒ >( |
5 = exp
³−1
(
1
= ³>
‡ ³ =
√³
M ³ =
= ℒ >( |³ > 5 =
(
{E FÜ√M
|}}}}}~ -
=
√0
B
Ï k −0
√0
=
=
B
B
=Ï k
2
√8
=Ï k
Ï k
√0
√=
B
1
z š2›
1
√8
Ï k
Ï k
0=
(
−0 - 0
√8
⟹ 0=2
= √0 ⟹ 0 =
Ï k
>
(
ℒ >( {M ³ ‡ ³ } =
Let
1 1
5
³ − 1 √³
−0 ∙
B
,
0
1
√80
Ï k −0
√0
√= Ï
B
Ï k −
k −
0
0
2
ÏY $√0%
467
Appendix A Solutions to Selected Problems
³
³ +o
³
ℒ{cos 3 } =
³ +9
ℒ{cos o } =
Problem 5.4.2
ℒ{
ℒ{ cos 3 } = −
ℒ{exp
B
exp 0
⨂
ℒ{
}=−
M ³
³
³
³ −9
š
›=
³ +9
³ ³ +9
}= ³
−0
} = ℒ{exp
0 = exp
}ℒ{
⨂
}=
1
³−1
³
³ −9
−
³ +9
³−1
Solving the above equation, one gets
1 ³ −9
³ = E1 − F
³ ³ +9
Assuming
³ −9
=‡ ³
³ +9
Inverse transforming the above, one gets
³ −9
³ + 9 − 18
=
³ +9
³ +9
1
18
‡ ³ =
−
³ +9
³ +9
>(
= … {‡ ³ }
1
1
ℒ >( |
sin / − / cos /
5=
³ +/
2/
1
1
ℒ >( |
5 = sin /
³ +/
/
468
³ =
³
1
= sin 3
3
Appendix A - Chapter 5
18
sin 3 − 3 cos 3
54
1
³ =‡ ³ − ‡ ³
³
-
- 0
=-
−
B
- 0
0
‡ ³
³
B
= cos 3
0 sin 30
sin 30
x −
0=
3
3
B
B
ℒô
B
−
- 0
0/ =
0
sin 3
cos 3
1
+
−
3
9
9
cos 3
1
sin 3
−
+
= cos 3 −
9
9
3
=
469
Appendix A Solutions to Selected Problems
5.5 Application of Laplace Transforms
Problem 5.5.1
Let us apply Laplace Transform to the equation. Since I.C. are 0
³ +4 ³ =M ³
³
where M ³ is the Laplace Transform of ³ . We got algebraic
equation for ³
M ³
³ =
³ +4
Because
1
1
ℒ >( |
5 = sin2
³ +4
2
we can rewrite the equation on term of two Laplace Transforms
1
}ℒ | sin2 5
³ = ℒ{
2
Then the Convolution Property yields
1
1
− 0 sin20 0
=
E sin2 F =
2 B
2
Problem 5.5.2
Solve the integral-DE
′
+2
−4
B
exp
−0
0
0 = sin
,
0 = 0.
Taking the Laplace Transform of each side of the equation we
find,
1
}=
$³ ³ − 0 % + 2 ³ − 4ℒ{exp ⊗
³ +1
1
1
F ³ =
³ ³ + 2 ³ − 4E
³ +1
³−1
and solving for ³ , we obtain (and performing partial fraction)
³−1
1 4 − 3s
1
2
³ =
=
E
+
+
F
³ +1 ³ +³−6
25 ³ + 1 s − 2 s + 3
Taking the inverse transform, we get
470
Appendix A - Chapter 5
= ℒ >( {
=
³ }=
1
4sin
25
1
2
1 >( 4 − 3s
ℒ |
+
+
5
25
³ +1 s−2 s+3
− 3 cos
+ exp 2
+ 2 exp −3
Problem 5.5.3
Take Laplace on each side of two DEs, we have
ℒ{ } = −4ℒ{ } + ℒ{sin }
|
ℒ{ } = 4ℒ{ } − 8ℒ{ }
Let ℒ{ } = ³ . We know that
0
ℒ{ } = ³
³ − 0 ³−
Plugging into the first DE we get
³
³ −
0 ³ − ’ 0 = −4
1
³ +1
0 ³ + 0 ³+
0 +1
³ +4 ³ +1
Thus, the G.S. of
is
= ℒ >( { ³ }.
Similarly, for , let ℒ{ } = ³ }. This gives
ℒ{ } = ³
³ − 0 ³−
0
The second DE can be written as
0 =4 ³ −8 ³
³ − 0 ³−
³
This gives
4 ³ + 0 ³+
0
³ =
³ +8
Plugging ³ into the above equation, we can get ³ in terms
of ³, then the G.S. of
is:
= ℒ >( { ³ }
This gives
³ =
0 ³ +
³ +
Problem 5.5.4
Take Laplace on each side of the equation, we have
ℒ{ } + ℒ{ } − 6ℒ{ } = 0
Let ℒ{ } = ³ , then
ℒ{ } = ³ ³ − 0 = ³ ³
471
Appendix A Solutions to Selected Problems
}=³
³ −³ 0 −
³ −1
0 =³
}=³
³ −³ 0 −³ 0 −
0
=³
³ −³−1
By substitution, we have
³ −³−1+³
³ − 1 − 6³ ³ = 0
³
³+2
³ =
³ ³+3 ³−2
1 1 1
1
2
1
=− ⋅ −
⋅
+ ⋅
3 ³ 15 ³ + 3 5 ³ − 2
= ℒ >( { ³ }
2
1 1
= − − exp −3 + exp 2
5
3 15
ℒ{
ℒ{
Problem 5.5.5
Take Laplace on each side of the equation:
ℒ{ } + 4…{ } = −5ℒ{¼ − 3 }
³ − 0 ³−
0 + 4 ³ = −5 exp 3³
³
Given 0 = 1, 0 = 0, we have
³ − ³ + 4 ³ = −5 exp 3³
³
³
exp 3³
³
³ = −5 ⋅
⋅
+
³ +4
³
³ +2
Thus
= −5
Note
It can be proved that
B
cos 2
This means
and
ℒ{¼
− 0 ¼ 0 + 3 0 + cos 2
¼
= ′
−
} = exp − ³
ℒ{¼
}=1
Problem 5.5.6
Apply Laplace transforms on both sides of the DE,
472
Appendix A - Chapter 5
ℒ{ + 6 + 8 } = ℒ{−¼ − 2 }
³ − ³ + 6³ ³ − 6 + 8 ³ = − exp −2³
³
³ + 6 − exp −2³
³ + 6 − exp −2³
³ =
=
³ + 6³ + 8
³+2 ³+4
2
1
1
1
1
=
−
− exp −2³ E
−
F
³+2 ³+4 2
³+2 ³+4
1
1
1
1
2
−
− exp −2³ E
−
F5
= ℒ >( |
³+2 ³+4
³+2 ³+4 2
1
= 2 exp −2 − exp −4 − ¼ − 2 ⊗ exp −2
2
1
+ ¼ − 2 ⊗ exp −4
2
=¼
−¼ −2
0 =0
0 =0
Apply Laplace transform on both sides.
ℒ{¼ } = 1
ℒ{¼ − } = exp − ³ }
ℒ{ + 2 + } = ℒ{¼
−¼ −2 }
0 % + 2$³ ³ − 0 % + ³
³ −³ 0 −
$³
= 1 − exp −2³
0 =0
0 =
We get
1 − exp −2³
1 − exp −2³
³ =
=
³ + 2³ + 1
³+1
1
exp −2³
=
−
³+1
³+1
Since
ℒ{
−0
− 0 } = exp −0³ M ³
Inversely transform leads to
= exp − − − 2 exp$− − 2 %
−2
Problem 5.5.7
+2
+
473
Appendix A Solutions to Selected Problems
} = ℒ{¼
ℒ{¼w
Problem 5.5.8
} + ℒ{
ℒ{
³ +
Note that
ÜB
•
ÜB
•
1
ê
ÜB
> 0,
ℒ{
ℒ{¼
} = exp − ³
•
− 2=
•
B
}
³ = ê exp −2= B ³
ÜB
³ = ê exp −2= B ³
•
³ =ê
ÜB
exp −2= B ³
³ +
−
•
−
exp −2= B ³
exp −2= B ³
7 = ê ℒ >( ô
/
³ +
³ +
= ê ℒ >( ô
Problem 5.5.9
Case (I)
ÜB
ℒ >( {exp − ³ M ³ } =
= ℒ >( •ê
474
0 +
ÜB
•
=
•
} = ê ℒ{¼
³ −³ 0 −
³
−
1
− 2=
>0
−
−
ÜB
exp −2= B ³
/
³ +
}=
}=
=
B
sin
− 2=
exp − ³
³
•
B
•
B
¼
¼
−
−
B
exp −³
exp −³
Appendix A - Chapter 5
= exp − ³
Applying Laplace to both sides of the equation
³
³ −³ 0 −
=
0 + 2$³
³ −
exp − ³
+ exp − ³
³
0 %+
³
exp − ³
+ exp − ³
³
exp − ³
exp − ³
+ exp − ³
exp − ³
³
³
=
+
³ =
³+1
³+1
³+1
Substitute I.C.
³ + 2³
³
= ℒ >( {
= ℒ >( |
³ +
³ =
³ }
1
³ ³+1
exp − ³ 5 + ℒ >( |
1
³+1
exp − ³ 5
1
1
1
−
F5
= ℒ >( |exp − ³ E −
³ ³+1
³+1
+ ℒ >( |
=
−
2
=
−
$1 − exp$−
Case (II)
≤ 0,
−
−
+
+
>0
ℒ{¼
1
³+1
exp − ³ 5
− exp$−
−
−
−
ℒ{
−
exp$−
−
−
−
%−
−
−
−
%
%3
exp$−
−
exp$−
}=
1
³
} = exp − ³
−
%
exp$−
−
%
−
%%
475
Appendix A Solutions to Selected Problems
1
+ exp − ³
³
exp − ³
+
³+1
Applying Laplace to both sides of the equation
³
³ + 2³
³ =
³ +
1
³ ³+1
³ =
1
exp − ³
1
1
= ℒ >( | −
−
5 + ℒ >( ô
/
³ ³+1
³+1
³+1
=
− exp −
+
= 1 − exp −
+
− exp −
−
−
− exp −
−
−
exp$−
exp$−
Problem 5.5.10
Take Laplace on each side of both equations.
ℒ{ } = ℒ{ } + 2ℒ{ }
|
ℒ{ } = 2ℒ{ } − 2ℒ{ }
Given I.C., this gives
³−1 ³ =2 ³ +1
|
2 ³ − ³+2 ³
This gives
1
1
4
1
³ = ⋅
+ ⋅
5 ³+3 5 ³−2
€
2
1
2
1
³ =− ⋅
+ ⋅
5 ³+3 5 ³−2
Finally
4 exp 2 + exp −3
=
5
€
2 exp 2 − exp −3
=
5
476
−
−
%
%
Appendix A - Chapter 5
…{ ( } = −2…{ ( } + …{ } + exp −0³
…{ } = …{ ( } − …{ } + exp −20³
³ + exp −0³
³ ( ³ = −2 ( ³ +
|
³
³ = ( ³ −
³ + exp −20³
³ = ³ + 2 ( ³ − exp −0³
|
³ +1
³ = ( ³ + exp −20³
Insert first equation into second.
³ + 1 $ ³ + 2 ( ³ − exp −0³ % = ( ³ + exp −20³
³ + 3³ + 1 ( ³ = exp −0³ exp −0³ + ³ + 1
Problem 5.5.11
(
=
|
=ℒ
>(
S
exp −0³
− 0 ℒ >(
S
exp −0³
3 + √5
3 − √5
R
\³ + 2 ] \³ + 2 ]
Q
\³ +
Using partial fractions we find
1
Where
and
‚
3 + √5
3 − √5 •
R
\³ + 2 ] \³ + 2 ]
Q
€
+
\³ +
exp −0³ + ³ + 1
³ +1
‚
3 + √5
3 − √5 •
2 ] \³ + 2 ]€
3 + √5
3 − √5
2 ] \³ + 2 ]
ž=
=
1
³ +
ž
3 + √5
2
+
³ +
Â
3 − √5
2
√5
477
Appendix A Solutions to Selected Problems
³ +1
and
\³ +
Â=−
3 + √5
3 − √5
2 ] \³ + 2 ]
where
M=
and
„=
Then
(
=
−0 Þ
−
+
We also have
³ =
³ +1
478
+
(
1
1
√5
=
³ +
M
3 + √5
2
+
³ +
„
3 − √5
2
1 + √5
2√5
√5 − 1
2√5
−0 Þ
2
2
Ë
sin Ë
3 + √5
√5 3 + √5
1
2
2
Ë
sin Ë
ß
3 − √5
√5 3 − √5
1 + √5
2
2
Ë
sin Ë
2√5
3 + √5
3 + √5
√5 − 1
2
2
Ë
sin Ë
ß
2√5
3 − √5
3 − √5
³ + exp −20³
Appendix A - Chapter 5
³ =
1 Sexp −0³ exp −0³ + ³ + 1
³ + 1^
3 + √5
3 − √5
\³ + 2 ] \³ + 2 ]
R
U
+ exp −20³ _
S
= exp −0³ ^
+
R
T
exp −0³
³ + 1 \³ +
\³ +
1
3 + √5
3 − √5
2 ] \³ + 2 ]
3 + √5
3 − √5
2 ] \³ + 2 ]
+
By partial fractions we have
1
³ + 1 \³ +
When
Then
A=
1 U
³ + 1_
T
3 + √5
3 − √5
2 ] \³ + 2 ]
A
ž
1
=
+
+
³ +1
3 + √5
3 − √5
\³ + 2 ] \³ + 2 ]
1
√5 + 1
,ž =
2
5 + √5
,1 =
2
5 − √5
479
Appendix A Solutions to Selected Problems
=
−0 Þ
+
+
+
−
2
−0 Þ
1
√5 + 1
sin
2
3 + √5
Ë
sin Ë
2
5 + √5 3 + √5
2
3 − √5
Ë
sin Ë
ß
2
5 − √5 3 − √5
1
2
2
2
Ë
sin Ë
3 + √5
√5 3 + √5
1
2
2
Ë
sin Ë
ß
3 − √5
√5 3 − √5
Problem 5.5.12
Apply Laplace transform on both sides we have
1
= ℒ{sin }
³ ³ +4 ³ +6 ³
³−1
6
³ E³ + 4 +
F = ℒ{sin }
³−1
³−1
³ = ℒ{sin }
³+1 ³+2
Using partial fractions, we get
³−1
3
2
=
−
³+1 ³+2
³+2 ³+1
Then ³ can be expressed as
2
3
−
F
³ = ℒ{sin } E
³+2 ³+1
Finally apply inverse Laplace transform and use convolution
theorem
480
Appendix A - Chapter 5
ℒ >( {
³ } = sin
= 3 sin
3
2
5 − sin ⨂ℒ >( |
5
³+2
³+1
⨂ exp −2 − 2 sin ⨂ exp −
⨂ℒ >( |
Problem 5.5.13
By applying Laplace transform to given a system of DE, we get
following.
=2
+
+
|
=
+2
+ exp
1
³ ³ − 0 =2 ³ + ³ +
³
€
1
³ ³ − 0 = ³ +2 ³ +
³−1
0 ³−2
³−2
S
=
+
³−1 ³−3
³ ³−1 ³−3
0
1
+
+
³−1 ³−3
³−1 ³−3
0
1
R
³ =
+
³−1 ³−3
³ ³−1 ³−3
0 ³−2
³−2
+
+
³−1 ³−3
³−1 ³−3
Q
We need partial fractions for each term in the above DEs.
0 ³−2
0
1
1
S
=
+
³−1 ³−3
2 ³−1 ³−3
³−2
5
2
1
1
=− −
+
+
³
−
1
³
−
3
³
9³ 3³
2 ³−1
18 ³ − 3
0
0
−1
1
R
=
+
³−1 ³−3
2 ³−1 ³−3
1
1
1
1
=−
−
+
Q ³−1 ³−3
4 ³−1
2 ³−1
4 ³−3
481
Appendix A Solutions to Selected Problems
0 −1
1
+
2 ³−1 ³−3
³−1 ³−3
1
4
1
1
1
=
+
−
+
³ ³−1 ³−3
9³ 3³
2 ³−1
18 ³ − 3
0 ³−2
0
1
1
R
=
+
³−1 ³−3
2 ³−1 ³−3
³−2
1
1
1
=−
+
+
4 ³−1
2 ³−1
4 ³−3
Q ³−1 ³−3
Using results above, we perform an inverse Laplace transform,
or directly use the convolution.
0
S
=
ℒ{ } − 6ℒ{ } + 8ℒ{ } = 2ℒ{1}
From given I.C., we have
2
³ − 6³ ³ + 8 ³ =
³
³
2
³ =
³ ³−2 ³−4
1
1
1
1 1 1
+ ⋅
= ⋅ − ⋅
4 ³ 2 ³−2 4 ³−4
Therefore
1 1
1
= − exp 2 + exp 4
4
4 2
Problem 5.5.14
Problem 5.5.15
−$³
ℒ{
} + 2ℒ{
³ % + 2 m−$³
=
482
−1
} − 2ℒ{ } = 0
³ % −³
³ 1−2
4 ³+1
³
=−
³ ³+2
³
³ = 1 ³ + 2³ >
1
exp −
2
⋅ cosh − sinh
³ =0
Appendix A - Chapter 5
Problem 5.5.16
Perform Laplace Transform on equation
−
ℒô
³
³
/+ℒ|
5 + ℒ{ } = 0
− ³v + ³ − v −
⇒ (³ + 1) ′ + ³ = 0
which is a separable DE. We have
³ ³
=−
³ +1
1
à= = − à=(³ + 1)
2
1
(³) =
√³ + 1
( ) = ℒ >( { (³)} = ℒ >( |
1
=0
√³ + 1
5
ℒ{ } − 2ℒ{ } + ℒ{ } = ℒ{ ( )}
³ (³) − 2³ (³) + (³) = M(³)
M(³)
(³) =
(³ − 2³ + 1)
( ) = ℒ >( { (³)}
Problem 5.5.17
Problem 5.5.18
³
1
(³ + 1)
2³ + 4³ + 13
2 + 1⁄(³ + 1)
(³) =
=
(³ + 1) (³ + 4³ + 13)
³ + 4³ + 13
(³
(³) − 2) + 4³ (³) + 13 (³) =
483
Appendix A Solutions to Selected Problems
=
So
( )=
1
1
5
³+2
E−
+
+
(³ + 2) + 9
50
³ + 1 (³ + 1)
3
+ 32
F
(³ + 2) + 9
1
$(−1 + 5 ) exp(− ) + exp(−2 ) (cos3
50
+ 32sin3 )%
Problem 5.5.19
Method I: Laplace Transform
By applying the Laplace Transform to the given DE, we get
ℒ{ } − ℒ{ } − 12ℒ{ } = ℒ{sin 4 + exp(3 )}
$³ (³) − ³ (0) − (0)% − $³ (³) − (0)% − 12 (³)
4
1
=
+
³ + 16 s − 3
From the I.C., we have
4
1
(³ (³) − 2³ − 1) − (³ (³) − 2) − 12 (³) =
+
³ + 16 s − 3
2³ − 7³ + 36³ − 108³ + 52
(³) =
(³ − 3)(³ + 3)(³ − 4)(³ + 16)
Now, if we apply the partial fraction for the RHS, we get
2³ − 7³ + 36³ − 108³ + 52
(³ − 3)(³ + 3)(³ − 4)(³ + 16)
A
ž
1
³ + þ
=
+
+
+
³ − 3 ³ + 3 ³ − 4 ³ + 16
484
Appendix A - Chapter 5
1
6
1051
B=
A+ž+1+Â =2
S
1050
−A − 7ž − 4Â + þ = −7
65
4A + 28ž + 71 − 9Â − 4þ = 36 ⇒
C=
56
R
R −16A − 112ž + 36Â − 9þ = −108
1
Q−192A + 192ž − 1441 + 36þ = 52
D=
200
28
E
=
−
Q
200
So, we get
2³ − 7³ + 36³ − 108³ + 52
(³) =
(³ − 3)(³ + 3)(³ − 4)(³ + 16)
1051 1
65 1
1 1
F+
E
F+ E
F
=− E
1050 ³ + 3
56 ³ − 4
6 ³−3
1
³
7
4
+
š
›−
E
F
200 ³ + 16
200 ³ + 16
By using the inverse Laplace Transform, we obtain
1051
65
7
1
( )=
exp(−3 ) + exp(4 ) −
sin4 +
cos4
1050
56
200
200
1
− exp(3 )
6
Method II:
For the common solution, we have to solve the C-Eq.
Y − Y − 12 = 0, Y = −3, 4
The common solution is
¡ ( ) = 1( exp(−3 ) + 1 exp(4 )
For the P.S., we get it by considering the RHS of the given DE.
¢ ( ) = A sin 4 + žcos4 + 1 exp(3 )
¢ ( ) = 4Acos4 − 4žsin4 + 31 exp(3 )
¢ ( ) = −16Asin4 − 16žcos4 + 91 exp(3 )
¢ ( ) − ¢ ( ) − 12 ¢ ( )
= (−28A + 4ž)sin4 + (−4A − 28ž)cos4 − 61 exp(3 )
S A=−
485
Appendix A Solutions to Selected Problems
= sin4 + exp(3 )
From this relation we get the system of equations
7
SA = −
200
−28A + 4ž = 1
1
•−4A − 28ž = 0 ⇒ B =
200
R
−61 = 1
1
C
=
−
Q
6
So, the P.S. is
7
1
1
sin4 +
cos4 − exp(3 )
¢( ) = −
200
200
6
The G.S. is
( ) = ¡ ( ) + ¢( )
7
1
= 1( exp(−3 ) + 1 exp(4 ) −
sin4 +
cos4
200
200
1
− exp(3 )
6
Because we have I.C., we can determine the coefficients of
common solution.
1
1
(0) = 1( + 1 +
− =2
200 6
28 3
(0) = −31( + 41 −
− =1
200 6
1
1
1051
(0) = 1( + 1 +
C( =
− =2
200 6
1050
⇒€
€
28 3
65
(0) = −31( + 41 −
− =1
C =
200 6
56
Consequently, we have the G.S. with determined solution
( ) = ¡( ) + ¢( )
1051
65
7
1
=
exp(−3 ) + exp(4 ) −
sin4 +
cos4
1050
56
200
200
1
− exp(3 )
6
486
Appendix A - Chapter 5
Problem 5.5.20
(1) The C-Eq of the homogeneous portion is
=0
Y +
Y = pr
The two L.I. solutions are
,
= sin
( = cos
cos
sin
×=Ý
+ sin
=1
Ý = cos
− sin
cos
The trial functions are
( ) ( )
( )
= − cos
( =−
×( )
(( ) ( )
( )
=
= sin
×( )
The G.S. of the original DE is
= 1( cos
+ 1 sin
+ ( cos
+ sin
(2) Take Laplace on both sides of the DE, we have
ℒ{ ′′} + ℒ{ } = ℒ{ ( )}
Let (³) = ℒ{ }
ℒ{ ′′} = ³ (³) − ³ (0) − ′(0) = ³ (³) − B ³ − LB
By substitution
³ (³) − B ³ − LB + o (³) = ℒ{ ( )}
ℒ{ ( )} + B ³ + LB
(³) =
³ +o
1
o
³
LB
o
= ℒ{ ( )}
+ B
+
o ³ +o
o
³ +o
³ +o
( ) = ℒ >( { (³)}
LB
1
( − 0) sin(o ) + B cos(o ) + sin(o )
=
o B
o
Problem 5.5.21
(1) The corresponding homogeneous DE is:
+ ( =0 which
is a 2nd-order DE with constant coefficients, the C-Eq is
Y + ( = 0 ⇒ Y(, = p ( r
¡ ( ) = 1( cos( ( ) + 1 sin( ( )
487
Appendix A Solutions to Selected Problems
) = cos( ( )
( ) = sin( ( )
sin( ( )
cos( ( )
×( ) = _
_=
)
− ( sin( (
( cos( ( )
Let ¢ ( ) = ( ( ) ( ( ) + ( ) ( ), we have
( ) ( )
sin( ( ) sin(
=−
(( ) = −
×( )
(
1
šcos$( ( + ) % − cos$( ( −
=
2 (
1 sin$( ( + ) % sin$( ( − )
=
−
\
2 (
(+
(−
cos( ( ) sin(
(( ) ( )
( )=−
=−
×( )
(
1
šsin$( ( + ) % − sin$( ( −
=
2 (
1
cos$( ( + ) % cos$( ( −
=
−
\−
2 (
(+
(−
Thus a P.S. can be written as
( ) ( )
¢ ( ) = (( ) ( ( ) +
|
=
cos(
2
(
(
) sin$(
\
+
+
(+
(
sin(
2
(
(
)
) %
\−
−
sin$(
cos$(
−
(−
(
+
(+
(
(
)
) %›
%
]
)
) %›
) %
]
) %
]
) %
− ) %
]
(−
cos( ( )sin$( ( + ) % − sin( ( )cos$( ( + ) %
=
2 (( ( + )
) % + sin( ( )cos$( ( − ) %
− cos( ( ) sin$( ( −
+
2 (( ( − )
sin$( ( + ) − ( % sin( ( − ( ( − ) )
=
+
2 (( ( − )
2 (( ( + )
−
488
((
cos$(
(
Appendix A - Chapter 5
sin(
)
( −
Thus, the G.S. of the DE is
( ) = ¡( ) + ¢( )
=
= 1( cos(
With the given I.C.
(B)Ü † (B)ÜB
Thus
(
) + 1 sin(
|}}}}}}}}}~ 1( = 0, 1 =
(
)−
(( (
sin(
( −
−
sin( ( )
sin(
)
−
) ( (− )
(( ( −
sin( ( ) sin(
=
E
−
(
( −
( )=
)
)
)
F
Note
We applied the following triangle relations.
³r=(v p œ) = ³r= v ‡³ œ p ‡³ v ³r= œ
1
³r= v ³r= œ = − ( ‡³(v + œ) − ‡³(v − œ))
2
1
³r= v ‡³ œ = (³r=(v + œ) + ³r=(v − œ))
2
(2) Take Laplace on each side of the DE, we have
ℒ{x′′} + ω( L{x} = ℒ{sin(ω t)}
Let
(³) = …{ } ⇒ …{ }= ³ ( ) − (0)³ − (0) = ³
Substituting …{ } and …{ } in the DE, we have
(³) =
=
³
(³) +
(³ +
(
(³) =
³ +
)(³ + ( )
1
1
(
−
\
− (
³ +
(³ + (
(³)
]
489
Appendix A Solutions to Selected Problems
Thus
( ) = ℒ >( { ( )} =
is the solution.
ω
sin(ω( t) sin(ω t)
E
−
F
ω(
ω
ω − ω(
Problem 5.5.22
(1) Take Laplace on both sides of the DE.
ℒ{ ′} − ℒ{ } = ℒ{1} − ℒ{( − 1) ( − 1)}
Let (³) = ℒ{ }, then ℒ{ ′} = ³ (³) − (0) = ³ (³).
1
1
ℒ{1} = , ℒ{( − 1) ( − 1)} = Ï k(−³)
³
³
By substitution, we have
³ − Ï k(−³)
1 1
³ (³) − (³) = − Ï k(−³) =
³
³ ³
Note
ℒ{ ( − 1) ( − )} = M(³) exp(− ³) where M(³) = ℒ{ ( )}.
490
Solve the above equation for (³)
³ − exp(−³)
1
1 exp(−³)
(³) =
=
− −
³ (³ − 1)
³ − 1 ³ ³ (³ − 1)
1
1
1
1 1
=
− − exp(−³) E
− − F
³−1 ³
³−1 ³ ³
1 1 exp(−³) exp(−³) exp(−³)
1
− −
+
+
=
³
³
³−1 ³ Ï ³−1
( ) = ℒ >( { (³)}
( − 1)
= exp(−r) − 1 −
+ ( − 1) + ( − 1) ( − 1)
Ï
1
= exp(−r) − 1 + ( − 1) E − F
Ï
(2) We divide the equation in two regions
If ≤ 1, the DE is
− =1
|
(0) = 0
whose solution can be easily found as ( ) = exp( ) − 1.
If > 1, the DE is
Appendix A - Chapter 5
− =2−
(0) = 0
which is a 1st-order Linear DE, and
h( ) = −1, Ÿ( ) = 2 −
|
( ) = exp E−
Solving this DE, we have
1
( )= E
Ÿ + 1F
F = exp( )
= exp( ) E exp(− ) (2 − )
+ 1F
= exp( ) (−2 exp(− ) + exp(− ) + exp(− ) + 1)
= −1 + + 1 exp( )
I.C. yields (0) = 0 ⇒ 1 = 1 ⇒ ( ) = exp( ) + + 1 . Thus
the final solution of this DE is
exp( ) − 1
≤1
( )=|
)
exp( + − 1
>1
Problem 5.5.23
For the original DE, if 0 < < 28 then ( − 28) = 0
Or
( ) + 4 ( ) = cos 2
Thus
( ) + 4 ( ) = cos 2
(0) = 0
Z
(0) = 0
Using Laplace transformation method, we have
(³ + 4) (³) = ℒ{ ‡³2 }
So
1
( ) = sin 2 ⊗ cos 2
2
1
=
sin 20 cos 2( − 0) 0
2 B
491
Appendix A Solutions to Selected Problems
1
sin 2
4
For the original DE, if > 28 then ( − 28) = 1
Or
( )+4 ( ) =0
with starting from = 28.
When = 28, from the condition of the phase I, 0 < < 28, we
know
(28) = 0, (28) = 28
After resetting the origin to = 28, DE system becomes
( )+4 ( )=0
(0) = 0
Z
(0) = 28
whose G.S. for > 28 is
( ) = 8 sin 2
Many also directly evaluate the convolution of the following to
find the G.S.
1
( ) = sin 2 ⊗ $$1 − ( − 28)%cos 2 %
2
=
Problem 5.5.24
Applying Laplace transformation on both sides of DE, we have
³ (³) − ³ (0) − (0) + (³) = ℒ{(−1)C D }
Plugging the IC, we get
(³ + 1) (³) = ℒ{(−1)C D }
Or
1
(³) =
ℒ{(−1)C D }
³ +1
Applying inverse transform, we have
1
1
( ) = ℒ >( ô
…2(−1)C D 3/ = ℒ >( |
5 ⨂(−1)C D
³ +1
³ +1
= sin ⨂(−1)C
492
D
Appendix A - Chapter 5
Now, one can compute the convolution directly by definition.
One may also compute the Laplace transform of the RHS.
ℒ2(−1)C D 3 =
=
•
B
(
B
(−1)C D exp(−³ )
exp(−³ )
+
−
exp(−³ )
(
•
exp(−³ )
…
1 2
= + ° ê(−1) exp(−=³) − 1±
³ ³
ÜB
•
Thus,
1 2
= − + ê(−1) exp(−=³)
³ ³
ÜB
•
1 2
1
(³) =
°− + ê(−1) exp(−=³)±
³ ³
³ +1
ÜB
•
³
2 1
1
+
ê(−1) exp(−=³)
=− +
³ ³ +1 ³ +1³
ÜB
•
( ) = −1 + cos + 2 sin ⊗ ê(−1)
•
sin ⊗ ê(−1)
=
ÜB
B
•
•
( − =)
sin 0 ê(−1)
= ê(−1)
ÜB
•
= ê(−1)
ÜB
ÜB
B
B
( − =)
ÜB
( − 0 − =) 0
sin 0 ( − 0 − =) 0
>
•
sin 0 0 + ê(−1)
ÜB
>
0 sin 0 0
493
Appendix A Solutions to Selected Problems
•
Thus
= ê(−1) (1 − cos( − =))
ÜB
•
( ) = −1 + cos + 2 ê(−1) (1 − cos( − =))
Problem 5.5.25
=
(1) When
( )=
(2) When
B
494
(0) = (0) =
(0) =
B
…{
³
B
(
ÜB
} + …{
(³) +
+³ )
(³) =
=
( ) = …>( ô−
≠
B
³
} = …{
B
2
(³) +
(³) =
B
(
E
(
E
(³) =
1
(0) = 0
B sin(
B
+³ )
B
+³
)}
F (sin( ) −
cos
2
1
B
F/=
sin
+³
2
(³) =
B
B
B
B
B
+³
)( + ³ )
B
B³
(³) =
( B + ³ )( + ³ )
A³ + ž
1³ + Â
B B
=
+
( B + ³ )( + ³ )
+³
B +³
+ ³ ) + (1³ + Â)( B + ³ )
B B = (A³ + ž)(
³ + A³ + ž + ž³ + 1 B ³ + 1³ + Â
B =A
+ ³
+
)
B
Appendix A - Chapter 5
A+1 =0⇒A +1 B =0
ž+Â = 0⇒ ž +Â B = B
1 = −A and  = −ž
A −A B =0⇒A=1 =0
ž −ž B = B B
ž=
…>( ô
Â=
B
B
−
B
−
1
B B
\
]+
− B
B +³
( )=
B
B
)+
B
B
−
B
E
1
F/
+³
sin( )
− B
B −
A³ + ž
1³ + Â
B³
=
+
+³
( B + ³ )( + ³ )
B +³
³ + A³ + ž + ž³ + 1 B ³ + 1³ +  B + ³
B³ = A
A+1 =0⇒A +1 B = B
ž+Â =0⇒ž +Â B =0
A −A B = B
B
sin(
B
B
B
B
A=
( )=
B
−
1=
B
B
−
B
−
ž=Â=0
cos(
B
B
)+
B
B
B
B
B
−
cos(
)
Problem 5.5.26
Observe that the external force is periodic and during the first
period, the external force looks like the following,
495
Appendix A Solutions to Selected Problems
0
∈ [0, )
2
Y( ) = €
0
−
∈ [ , 0)
2
Hence, force as a function of time can be given as
Y( ) = Y( + 0)
We have assumed that the directions towards right are positive.
Let (0) = 0 be the initial displacement and (0) = 0 be the
initial velocity, where ( ) represents the displacement.
According to our sign convention, the force exerted by the spring
on mass m is always in the negative direction.
Hence the DE which governs the motion is
K
= −/ + Y( )
We can write this equation as
where
and
+
å( ) =
= å( )
1
Y( )
K
/
=Ë
K
Take the Laplace transform on both sides. Hence
(³) = Ÿ(³)
³ (³) +
Ÿ(³)
(³) =
³ +
Taking the Inverse Laplace transform of this equation, we get
Ÿ(³)
1
( ) = ℒ >( ô
/ = ℒ >( m
1 ℒ >( {Ÿ(³)}
³ +
³ +
1
= sin( ) å( )
496
( )=
1
sin(
K
Appendix A - Chapter 5
) Y( )
cos 2
∈ ª0,28«
( )=m
0
o. w.
( ) = $1 − ( − 28)% cos 2
= cos 2 − ( − 28) cos 2( − 28)
ℒ{ ( )} = ℒ{cos 2 − ( − 28) cos 2( − 28)}
= ℒ{cos 2 } − ℒ{ ( − 28) cos 2( − 28)}
³(1 − exp(−28³))
=
³ +4
ℒ{ + 4 } = ℒ{ ( )}
³(1 − exp(−28³))
³ (³) − ³ (0) − (0) + 4 (³) =
³ +4
³(1 − exp(−28³))
(³)(³ + 4) =
³ +4
³(1 − exp(−28³))
(³) =
(³ + 4)
³(1 − exp(−28³))
( ) = ℒ >( { (³)} = ℒ >( ô
/
(³ + 4)
³
1
ℒ >( |
sin 2
5=
(³ + 4)
4
1
( ) = sin 2 − ( − 28) ( − 28) sin 2( − 28)
4
4
1
sin 2
∈ ª0, 28«
= €4
1
8 sin 2
o. w.
2
Problem 5.5.27
Problem 5.5.28
By setting the naturally spot at 0 of -axis, we can write the
system as
497
Appendix A Solutions to Selected Problems
K = M − (/( + / )
Applying Laplace transforms on both sides,
K$³ (³) − ³ (0) − (0)% = ℒ{M} − (/( + / ) (³)
The force can be expressed as
•
M=
°2 ê(−1)
ÜB
0
š − =› − ( )±
2
0³
tanh š ›
4
³
With I.C., (0) = (0) = 0, simplify the DE,
0³
K³ (³) = tanh š › − (/( + / ) (³)
³
4
0³
tanh š 4 ›
(³) =
³ K³ + /( + /
>(
The solution is ( ) = ℒ { (³)}.
ℒ{M} =
Problem 5.5.29
The system of DEs
K ( ′′ = /(
K ′′ = /(
Defining o = Ó , we get
ë
The system becomes
( ′′
498
=o (
− 2 () +
−
2 )+
(
− 2 () +
º(
)
(
G )
º(
)
K
G( )
′′ = o ( ( − 2 ) +
K
Applying Laplace transform on the above system yields
(³ + 2w ) ( = o
+ Mº (³)
(³ + 2w ) = o ( + MG (³)
where
º( )
Mº (³) = ℒ |
5
K
MG (³) = ℒ |
G(
K
)
Appendix A - Chapter 5
5
Solving the above DEs, we get
Mº (³ + 2o ) + MG o
( =
³ + 4o ³ + 3o
Mº ³ + (2Mº + MG )o
(³ + o )(³ + 3o )
Mº + MG
1
1
Mº − MG
+
2
2 ³ + 3o
³ +o
Mº o + MG (³ + 2o )
=
³ + 4o ³ + 3o
MG ³ + (2MG + Mº )o
(³ + o )(³ + 3o )
1
Mº − MG
1
Mº + MG
−
³ +o
2
2 ³ + 3o
Then we have
1
1
Mº − MG
M + MG
>( º
+
|
5
( =ℒ
³ +o
2
2 ³ + 3o
1
1 >(
ℒ {Mº + MG } ⊗ sin o +
ℒ >( {Mº − MG }
=
2o
2√3o
⊗ sin√3 o
1
1
{ º ( ) + G ( )} ⊗ sin o +
{ º( ) −
=
2Ko
2√3Ko
⊗ sin √3 o
+
MG
1
Mº − MG
1
M
º
= ℒ >( |
−
5
³ +o
³ + 3o
2
2
1
1 >(
ℒ {Mº + MG } ⊗ sin o −
ℒ >( {Mº + MG }
=
2o
2√3o
=
⊗ sin √3 o
1
1
{ ( ) + G ( )} ⊗ sin o −
{ º( ) −
2Ko º
2√3Ko
⊗ sin √3 o
G(
)}
G(
)}
499
Appendix A Solutions to Selected Problems
The above convolutions are trivial to compute.
Problem 5.5.30
Two driving forces can be expressed as following (Here B is
constant).
0
º ( ) = B \ ( ) − š − ›]
2
0
º ( ) = B š š − › − ( − 0)›
2
Now, we transform the DEs through a Laplace transform.
0
K ( = −/ ( + /( − ( ) + B \ ( ) − š − ›]
2
0
K = −/( − ( ) + B š š − › − ( − 0)›
2
K$³ ( (³) − ³ ( (0) − ( (0)%
K$³
= −2/
(³) − ³
=/
( (³) +
( (³) −
Substitute I.C.
(K³ + 2)
(0) −
/
(0)%
(³) +
( (0)
BÞ
=
(0)
=
(
( (³) −
−kX( (s) + (K³ + /)
500
/
0³
1 exp š− 2 ›
(³) + B Þ −
ß
³
³
/
0³
2 › − exp(−0³)ß
³
³
exp š−
(0) = 0
(0) = 0
0³
1 exp š− 2 ›
(³) = B Þ −
ß
³
³
0³
exp š− 2 › exp(−0³)
(³) = B Þ
−
ß
³
³
Appendix A - Chapter 5
0³
B
šK³ + / − K³ exp š− 2 › − / exp(−0³)›
³
( (³) =
(K³ + /)(K³ + 2/) − /
(³)
0³
B
š/ + (K³ + /) exp š− › − (K³ + 2/) exp(−0³)›
³
2
=
(K³ + /)(K³ + 2/) − /
After performing inverse Laplace transform, we get the solution.
>(
( ( ) = ℒ { ( (³)}
( ) = ℒ >( { (³)}
Problem 5.5.31
Applying Laplace Transform the both equations:
K³ ( (³) = −2/ ( (³) + / (³) + B exp(− B ³)
(³) = / ( (³) − / (³) + B exp(−2 B ³)
K³
Writing these in matrix and using Gaussian Elimination to solve
for (³).
2/
/
B
−
exp(− B ³)
³ +
K
K
( (³)
K
=
/
/
(³)
B
−
³ +
exp(−2 B ³)
K
K
K
2/
/
³ +
−
K
K
( (³)
(³)
/
2/
/
0
E³ + F E³ + F −
K
K
K
=
B
K
exp(− B ³)
2/ B
/ B
F exp(−2 B ³) +
exp(− B ³)
K
K K
3 − √5 /
3 + √5 /
] \³ +
\³ +
] (³)
2 K
2 K
2/ B
/ B
exp(− B ³)
= E³ + F exp(−2 B ³) +
K
K K
E³ +
501
Appendix A Solutions to Selected Problems
(³) =
\(³ +
+
(
B
exp(−2 B ³)
K
=
³ +
( )=
1
B
K
B
K
+
sin
1
+
(
)
B
K
exp(−2 B ³) +
(³ +
( )(³
+
/ B
exp(− B ³)]
K
)
/
exp(− B ³)
K B
(³ + ( )(³ + )
exp(−2 B ³) +
⊗ ¼( − 2 B )
°
B
K
¼( − 2 B ) +
/
K
B ¼(
)
⊗ (sin ( ⊗ sin
where we have substituted
3 ∓ √5 /
(, =
2 K
above, and the substitution
2/
=
( +
K
Now solve for ( (³).
/
2/
B
(³) = exp(− B ³)
E³ + F ( (³) −
K
K
K
( (³)
+
+
502
=
B
/
/
2/
K š³ + K ›
B
ÞK
exp(−2 B ³)
³ +
/
B
K exp(−2 B ³) + K B exp(− B ³)
(³ + ( )(³ + )
exp(− B ³)ß
−
B )±
((
)=
/ B
1
¼( − 2 B ) ⊗
sin
KK
+
/
°
K
⊗
where
+
1
B
K
(
sin
B
K
¼( −
=
¼( − 2 B ) +
(
⊗
B) ⊗
2/
=
K
1
1
(
sin
Appendix A - Chapter 5
/
K
sin
⊗
1
B ¼(
⊗
1
sin
−
sin
B )±
+
Problem 5.5.32
(1) Considering the spring forces acting on the two masses, we
get
= −(/( + / ) ( + /
K
| ( (
K
= / ( − (/ + / )
(2) Plug the parameters into the DEs, we get
= −3 + 2
| (
=2 (−5
From here, we have many ways to solve the equation. For
example, substitution or Laplace method can be used. Let us use
substitution method.
From the second equation, we get
1
+5
( = (
2
Plugging into the first equation, we get
8
11 = 0
whose C-Eq is Y + 8Y + 11 = 0 whose solutions are
Y = −4 p √5
Next, one can find all four solutions ( , ,
,
503
Appendix A Solutions to Selected Problems
Ó 4 p √5,
Ó 4 p √5
So that one can compose the final solution for K( and then for
K
= 1( Ï k( ( ) + 1 Ï k( ) + 1 Ï k( )
+1 Ï k( )
One can find ( and then determine 1( , 1 , 1 , 1 by I.C..
504
Appendix B
Laplace Transforms
Selected Laplace Transforms
(Š) = d>è {‹(Œ)} ‹(Œ) = d{ (Š)}
¼( − )
exp(− ³)
¼ ( )
³
exp(− ³)
³
( − )
, =
1
³
0
³
505
2!
³
=!
ú(
Appendix B Laplace Transforms
sin
cos
sin
cos
exp(− )
exp(− )
exp(− ) sin
exp(− ) cos
³ +
³
³
2³
(³ +
)
³ −
(³
)
1
³+
=!
(³ + )
ú(
(³ + ) +
(³
³
)
Selected Properties of Laplace Transforms
(Š)
((
506
)
( )
( )
‹(Œ)
( )
M(³)
M(³)
M( (³)
M (³)
Appendix B Laplace Transforms
( )
³M(³) − (0)
³ M(³) − ³ >( (0)
−³ > (0) − ³ >
− ⋯ − ( >() (0)
( )
B
M(³)
³
(0) 0
exp(− ) ( )
(0)
M(³ + )
( − 0) ( − 0)
exp(−³0) M(³)
( )⨂-( )
M(³) ⋅ ‡(³)
( )
−
M(³)
³
(−1) M (³)
( )
( )
q
1
( ), > 0
M(³)
³
Mš ›
Remarks
(1) ¼( ) is the Dirac delta-function which is defined as
+∞, = 0
¼( ) = m
0,
≠0
and also satisfied
ú•
(2)
>•
¼( )
=1
( ) is the unit step function which is defined as
507
Appendix B Laplace Transforms
0,
<0
( )=m
1,
0
(3) ⨂ is the convolution operator which is defined as
( )⨂-( ) =
508
B
(0)-( − 0) 0
Appendix C
Derivatives & Integrals
1.
2.
3.
4.
5.
6.
7.
8.
( L) =
¤
L+
M$-( )% =
(
)==
(ln ) =
(
â
¥
>(
∙
§
¥
(Chain Rule)
(exp( )) = exp( )
(sin ) = cos
(cos ) = − sin
(tan ) = sec
509
Appendix C Derivatives & Integrals
9.
L=´
L + ´L
10. ´ exp( )
= exp( ) + 1
(Integration by Parts)
11. ´
12. ´
13. ´
=
+1
(= ≠ −1)
w]
= − cos + 1
15. ´ cos
= sin + 1
= − ln|cos | + 1
16. ´ tan
= ln|sec + tan | + 1
17. ´ sec
18. ´ csc
19. ´ Í
√w >
21. ´ Í
w ú
ú(
L = L − ´L
= Ž• w + 1
14. ´ sin
510
(
ú(
= ln| | + 1
20. ´ Í
w ú
⇒ ´
Í
Í
Í
= ln|csc + cot | + 1
= sin>( + 1
w
= w tan>( w + 1
=
(
(
úw
ln Ý >wÝ +
w
1
Appendix D
Abbreviations
Abbreviation
C
Meaning
Constant, or Const
C-Eq
Characteristic Equation
DE(s)
Differential Equation(s)
Eq
Equation
G.S.
General Solution(s)
I.C.
Initial Condition(s)
IVP
Initial Value Problem(s)
L.D.
Linearly Dependent
LHS
Left Hand Side
511
Appendix D Abbreviations
L.I.
512
Linearly Independent
O.W.
Otherwise
P.S.
Particular Solution(s)
RHS
Right Hand Side
WRT
With Respect To
Appendix E
Teaching Plans
I. Ten homework sets, three quizzes, and one final exam that will
contribute to a total of 100 points for the entire semester. The usual
partitioning of grades:
II. During the semester, you expect 9-10 homework sets, approximately
one per week.
Item
10 HWs
Quiz 1
Quiz 2
Quiz 3
Final
Total
Expected Time
(Hours)
2-3
1.5
1.5
1.5
2.5
513
Points
20
15
15
15
35
100
Appendix E Teaching Plans
(1)
(2)
(3)
(4)
You usually have one week for each assignment.
All homework grades will be counted.
No grade for late homework.
No assignments are given for quiz weeks.
III. Four tests include one final and three in-class quizzes.
(1) All tests except the final will take place in the classroom.
(2) All tests are open-book: calculators, your own lecture notes,
and textbook are allowed during all tests.
(3) The final exam is cumulative.
(4) Scores will be curved.
IV. Student partitioning for the class
20% Like it and need it
15% Like it but don ot need it
60% Do not like it but need it
05% Do not like it and do not need it
Instructor: Prof. Yuefan Deng
Email Address: Yuefan.Deng@StonyBrook.edu
+1 (631) 877-7979
Mobile:
http://www.ams.sunysb.edu/~deng
Web:
514
References
1. Introduction to Numerical Analysis by J. Stoer and R. Bulirsch
(Springer-Verlag, 1980.)
2. Numerical Mathematics and Computing (2nd Ed.) by Ward Cheney
and David Kincaid (Books/Cole Publishing Company, Pacific
Grove, CA. 1985.)
3. Introduction to Numerical Analysis by F. B. Hilderbrand (Dover,
1987. ISBN: 0-486-65363-3)
4. Computer Methods for Mathematical Computations by Forsythe,
Malcolm, and Moler.
5. Numerical Methods for Engineers and Computer Scientists by Paul
F. Hultquist (Benjamin/Cummings, 1988.)
6. Applied Numerical Linear Algebra by William W. Hage (PrenticeHall, N.J., 1988.)
7. Discrete Variable Methods in Ordinary DE by Peter Henrici (John
Wiley \& Sons, New York, 1962.)
8. A Scientist's and Engineer's Guide to Workstations and
Supercomputers by Rubin H. Landau and Paul J. Fink, Jr. (1995)
9. Numerical Initial Value Problems in Ordinary DE by C. William
Gear (Prentice-Hall, N.J., 1971.)
10. Finite Difference Methods in Partial DE by Mitchell and Griffith
515
References
11. C. H. Edwards and D. E. Penney, Differential Equations and
Boundary Value Problems (4th Ed, Prentice Hall, 2007)
516
Index
constant coefficient, 107, 121, 127,
128, 155
variable coefficient, 107
complementary solutions, 155
cooling model
heat conductivity, 10
heat transfer, 9
solution, 11
coupled DE, 170
A
Abel’s formula, 119
analytical methods, 2
analytical solutions, 6
B
Bessel function
zero order Bessel function, 229
α order Bessel function, 229
Bromwich integral (Fourier-Mellin
integral), 215
D
Delta-function, 231, 507
derivative, 5, 6, 7, 8, 29, 30, 33, 35, 323
determinant, 116
Vandermonde, 120
Wronskian, 116, 118, 131, 154, 156,
376
differential equations (DEs), 1
draining model, 13
cross-section area, 14, 15
draining constant, 14, 17
draining equation, 15
C
characteristic equation (C-Eq), 121
classification of DEs, 4
homogenous DE, 4
inhomogeneous DE, 4
order of DEs, 4
coefficient
517
Index
initial height, 15
solution of the draining equation,
16
time needed to empty the
container, 16
E
eigenvalues, 185
eigenvectors, 185
error function, 222
escape velocity, 93
Euler’s formula, 126
exact DE theorem, 56
F
financial model
increase of loan, 101
interest rate, 101, 102, 103
loan equation, 101
payment rate, 101
solution, 102
G
Gamma function, 459
gravitational constant, 93
gravitational force, 93
Green’s function (kernel), 155
H
homogeneous system, 184
homogeneous system
solution of the homogeneous
system, 185
518
I
inhomogeneous system, 184
initial conditions (I.C.), 7
initial value problem (IVP), 17
integrating factor, 33, 34, 35, 55, 56,
57, 61, 63, 68
L
Laplace Transforms
binary operation, 220
convolution theorem, 220
frequency differentiation, 206
linearity property, 195
translator property, 200
Linear combination, 108, 114, 139
linear differential operator, 113
properties of the linear differential
operator, 114
Linearity of Des, 5
Linearity of DEs
linear DEs, 5, 6, 35, 45, 67, 72, 73,
107, 114
nonlinear DEs, 5, 6, 107
M
matrix, 156, 170, 184, 185
motion model
acceleration, 12
initial position, 12
initial speed, 12
position of the particle, 13
velocity, 11, 12
Index
N
numerical methods, 2
numerical solutions, 6
O
Order of DEs, 3
singular solutions, 28
solving a non-Exact DE, 62
swimmer model
drift, 21
trajectory of the swimmer, 20
water speed, 19
T
P
partial fractions, 217, 228
population
birth rate, 76, 78, 80, 81
death rate, 17, 76, 78, 80, 82, 322
logistic equation, 78, 80
population as a function of time, 17
threshold population, 82
properties of operators, 179
terminal speed, 90
trial solution, 121, 154
the case of duplication, 145
U
undetermined coefficients method,
158
unit step function, 203, 204, 220, 507
V
R
radius of the earth, 94
reverse motion, 92
S
solutions
variables
dependent variables, 3, 9
independent variables, 4, 9, 13, 15
W
wave function, 214
519