1. UNIVERSITY OF BUEA
FACULTY OF DEPARTMENT OF
SCIENCE MATHEMATICS
SECOND ORDER LINEAR ORDINARY DIFFERENTIAL
EQUATIONS AND THEIR APPLICATIONS TO THE STUDY OF
PROBLEMS IN PHYSICS
A PROJECT SUBMITTED TO THE FACULTY OF SCIENCE
OF THE UNIVERSITY OF BUEA IN FULFILMENT OF THE
COURSE MAT498: Research Project
BY
FOUTSE YUEHGOH
(SC11A309)
Supervisor: Dr. NKEMZI BONIFACE
July, 2014

2. UNIVERSITY OF BUEA
FACULTY OF SCIENCE
CERTIFICATION
This long essay entitle ’second order linear ordinary differential equation
and their application to the study of problems in physics’ has been
submitted to the department of Mathematics, Faculty of Science,
University of Buea, in partial fulfilment of the requirement for a B.sc in
Mathematics. The work was carried out by FOUTSE YUEHGOH.
Dr.NKEMZI BONIFACE (Supervisor)
Head of Department, Mathematics
i

3. DEDICATION
This project is dedicated to the FOUTSE FAMILY.
ii

4. ACKNOWLEDGEMENTS
Feeling gratitude and not expressing it is like
wrapping a present and not giving it.
William Arthur Ward (1921-1994)
Many people accompanied me during the endeavour of my bachelor studies
and of writing this project. I am deeply grateful of their support and thank
God for that. First of all, I would like to thank my supervisor, Dr. NKEMZI
BONIFACE, for showing me the joy of research, for shaping my path to
research by guiding me with extensive knowledge, for teaching me how to
write, for his unending encouragement, support and advises.
My thankful admiration goes to FOUTSE KHOMH, for all the support, good
advises and permanent guidance he has been giving me. Great thanks go to
all my lecturers in the University of Buea, Dr. NANA, for the continuous
advice and encouragements he has been giving us both in academics and
social live during his lectures; Mr. HENRY NGATCHU for guiding me on
how to manage a project. Great thanks also go to Mr. EKOWGE FRITZ
for the help he gave me in building up my skills in research and computer
programming during my internship, which was of great help in this project;
thanks to it I could understand latex faster.
A special thanks goes to all the maths teachers I have ever had in my life,
thanks to their good ways of teaching I could benefit from the joy one encoun-
ters when doing maths, they contributed a lot in my interest in mathematics.
Special thanks also go to FOTING GERVAIS SIKADIE, for the moral and
technical support he has been giving me since I knew him. I also thank all my
friends in the University of Buea, especially my course mates, NGUEMTO
SANDRA, MEGANG JUNILE, CHIME FRANCINE, NANA FADIMATOU,
TCHAMBA LARISSA, KEMTIM STEPHANE, KENGNE JOJO, just to
name a few, and the masters students, for encouraging me and thus con-
tributing to the success in my studies.
I am grateful to my parents, Dora and Abraham, for their love, for their
unconditional support, and for believing in me whenever and wherever. To
my brothers and sisters.
iii

5. Abstract
Here, we are concerned with the establishment of the existence and unique-
ness of solutions of second order linear differential equations, and how they
can be used to interpret physical problems. We started by introducing sec-
ond order linear equations, then established a theorem for the existence and
uniqueness of a solution to second order linear initial value problem, and
proved the theorem considering existence, uniqueness and continuous depen-
dence of the solution on the initial conditions. Furthermore, we exhibit five
problems from physics that can be modelled using second order linear dif-
ferential equations, giving details on the physical aspects with the help of
diagrams to show some practical occurrences and then we give an interpre-
tation of some results. Finally, we make a critical stability analysis of the
equilibrium solutions on one of the problems and demonstrate it graphically.

6. Contents
1 INTRODUCTION 1
1.1 Modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 SECOND ORDER LINEAR DIFFERENTIAL EQUATION 4
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 AN EXISTENCE AND UNIQUENESS THEOREM FOR SEC-
OND ORDER LINEAR EQUATIONS . . . . . . . . . . . . . 5
2.2.1 EXISTENCE . . . . . . . . . . . . . . . . . . . . . . . 6
2.2.2 UNIQUENESS . . . . . . . . . . . . . . . . . . . . . . 10
2.2.3 CONTINUOUS DEPENDENCE OF THE SOLUTION
ON THE INITIAL VALUES . . . . . . . . . . . . . . . 12
3 DESCRIPTION OF SOME PHYSICAL PROBLEMS THAT
CAN BE MODELLED BY LINEAR SECOND ORDER EQUA-
TIONS 14
3.1 Simple mass-spring system in free vibration . . . . . . . . . . 14
3.2 Simple damped mass-spring system in free vibration . . . . . . 18
3.3 Resonant vibration Analysis . . . . . . . . . . . . . . . . . . . 22
3.4 Electric circuits . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.5 An electric motor . . . . . . . . . . . . . . . . . . . . . . . . . 28
4 CRITICAL STABILITY ANALYSIS OF THE EQUILIB-
RIUM SOLUTION 30
i

7. Chapter 1
INTRODUCTION
Before embarking on a serious study of differential equations (for example,
by reading this project) one should have some idea of the possible benefits to
be gained by doing so. For some people the intrinsic interest of the subject
itself is enough motivation, but for most it is the likelihood of important
applications to other fields that makes the undertaking worthwhile. Many
of the principles, or laws, underlying the behaviour of the natural world
are statements or relations involving rates at which things happen. When
expressed in mathematical terms the resulting relations are differential equa-
tions, that is, equations containing derivatives. Therefore, to understand
and investigate problems involving the motion of fluids, the flow of current
in electric circuits, the dissipation of heat in solid objects, the propagation
and detection of seismic waves, or the increase or decrease of populations,
among many others, it is necessary to know something about differential
equations. A differential equation that describes some physical process is
often called a mathematical model of the process, and five of such models
are discussed in this project. In applying differential equations to any of the
numerous fields in which they are useful, it is necessary first to formulate the
appropriate differential equation that describes, or models, the problem be-
ing investigated. In this project we will focus more on ordinary second order
linear differential equations which can be used to model real life problems
in various aspects such as: Physics, Biology, Finances, Engineering, just to
name a few. Many interesting ordinary differential equations (ODEs) arise
from applications. One reason for understanding these applications is that
you can combine your physical intuition with your mathematical intuition in
the same problem, and the result is usually an improvement on both. Given
a differential equation
p(x)y + q(x)y + r(x)y = g(x), y(x0) = y0, y (x0) = y0, (1.1)
1

8. where p, q, r, and g are given continuous functions in some interval a < x < b
containing the points x0 and y0, y0, which are arbitrary prescribed initial
values. Any differentiable function y = ψ(x), satisfying Eq.(1.1) for all x
in some interval (a, b) is called the solution of Eq.(1.1) and our objective
is to determine whether such functions exist. Unfortunately, for arbitrary
functions p, q, r, g, there is no general method for solving the equation in
terms of elementary functions. Before we look at how to solve this prob-
lem, we need to address the following question: ”What constitute a properly
formulated problem? ”This question will be answered by the theorem we
are going to state and prove subsequently. The importance of that theorem
lies in the fact that, it defines the framework within which we can construct
solutions. These equations serves as a mathematical model, to describe or
predict a physical aspect, this shall be demonstrated in chapter three. But
before then, let us understand what modelling is all about.
1.1 Modelling
In constructing future mathematical models, we have to recognize that each
problem is different, and that successful modelling is not a skill that can be
reduced to the observation of a set of prescribed rules. Indeed, constructing
a satisfactory model may sometimes be the most difficult part of the prob-
lem. Nevertheless, it may be helpful to list some steps that are often part of
the process. The art of mathematical modelling can thus be subdivided into
three phases:
1. Formulation: After observing a physical system, we identify the appro-
priate dependent and independent variables, and develop a mathemat-
ical description of how these variables interact. Often, a differential
equation will serve as a mathematical description of the system.
2. Solution: Once the mathematical model has been formulated, we solve
it. This involves recognising the mathematical structure of the problem
and bringing the appropriate analytical and/or numerical technique to
bear.
3. Validation and interpretation: Once the problem is solved, the solution
needs to be examined carefully. Thus the following questions can be
asked: Does the solution make sense? Is it consistent with our phys-
ical intuition about what should be expected? The solution needs to
be scrutinized for its content: What does it say about physical phe-
nomenon being modelled?
2

9. The work is organised as follows. Chapter two is concerned with the
establishment of the theorem, and it’s proof. In chapter three, we shall
see how differential equations are used to model problems in physics
such as simple mass-spring system in free vibration, simple damped
mass-spring system in free vibration, resonant vibration, electric cir-
cuit and electric motor. They often serve as mathematical models that
can be used to describe and make predictions about physical systems.
In the last chapter, we will make a critical analysis on the equilibrium
solution of one of the problem in chapter three.
3

10. Chapter 2
SECOND ORDER LINEAR
DIFFERENTIAL EQUATION
2.1 Introduction
These equations are of crucial importance in the study of differential equa-
tions for two main reasons: First, linear equations have a rich theoretical
structure that establishes a number of systematic methods of solution. Fur-
ther, a substantial portion of this structure and these methods are under-
standable at a fairly elementary mathematical level. Secondly, they are vital
to any serious investigation of the classical areas of mathematical physics.
One cannot go very far in the development of fluid mechanics, heat conduc-
tion, wave motion, or electromagnetic phenomena without finding it neces-
sary to solve second order linear differential equations.
A second order linear ordinary differential equation has the form of Eq.(1.1),
where p(x) = 0. We can divide Eq.(1.1) by p(x) and obtain:
v(x) =
q(x)
p(x)
u(x) =
r(x)
p(x)
f(x) =
g(x)
p(x)
p(x) = 0
y + v(x)y + u(x)y = f(x), y(x0) = y0, y (x0) = y0. (2.1)
If the term f(x) or g(x) is equal to zero, the equation is called a homogeneous
equation, else it is called a non homogeneous equation, and thus the term
f(x) or g(x) is sometimes called the the non homogeneous term.
Here we are concerned with the non homogeneous equations. Thus the equa-
tion
L[y] := y + v(x)y + u(x)y = 0, (2.2)
4

11. where f(x) = 0 is called the homogeneous equation corresponding to the non
homogeneous equation Eq.(2.1). In discussing Eq.(2.1) and trying to solve
it, we will restrict ourselves to intervals in which u, v, and f are continuous
functions.
Observe that the initial conditions for a second order equation prescribes
not only a particular point (x0, y0) through which the graph of the solution
must pass, but also the slope y0 of the graph at that point. It is reasonable
to expect that two initial conditions are needed for a second order equation
because, roughly speaking, two integrations are required to find a solution
and each integration introduces an arbitrary constant. Hence, the two ini-
tial conditions will suffice to determine the values for these two constants.
Consequently, we should anticipate that to obtain a unique solution, it is
necessary to specify 2 conditions for example, the value of the solution y0
and its derivative y0 at the point x0, referred to as initial conditions as is
the case in Eq.(1.1) above. Thus to determine uniquely an integral curve of
a second order equation, it is necessary to specify not only a point through
which it passes, but also the slope of the curve at the point. Now, let us
consider a general second order initial value problem in the form
y = F(x, y, y ), y(x0) = y0, y (x0) = y0. (2.3)
To ensure the existence of a solution of Eq.(2.3) with the prescribed condi-
tions it is necessary to postulate certain properties for the function F. We
address this issue in the next section. For references, we used the books [2],
[4], [6], [8]
2.2 AN EXISTENCE AND UNIQUENESS
THEOREM FOR SECOND ORDER LIN-
EAR EQUATIONS
Theorem 2.2.1 Let v(x),u(x),f(x) be continuous functions on the interval
(a, b) and let x0 ∈ (a, b). Then the initial value problem
y + v(t)y + u(t)y = f(t), y(x0) = y0, y (x0) = y0. (2.4)
has a unique solution defined on the entire interval (a,b).
5

12. Notice that the theorem gives us three conclusions
1. The solution exists.
2. The solution is unique.
3. The solution exists on the entire interval (a, b).
This theorem gives us the framework within which we will work. The proof
of this theorem can be done in two ways. One can try to prove the theorem
directly by considering the second order equation or else convert the second
order equation into a system of first order equations. We will use the later
strategy.
2.2.1 EXISTENCE
Existence of a solution to an equation of type (2.4) is governed by the con-
dition that the functions v(t), u(t), f(t) are continuous. [4] Consider the
following second order linear differential equation
y + v(t)y + u(t)y = f(t), y(x0) = y0, y (x0) = y0. (2.5)
where v(t), u(t), f(t) be continuous functions on the interval (a, b) and x0 is
in (a, b). Now, let us convert Eq(2.5) in to a system of first order equations.
Let x1 = y, and x2 = y
that is,
x1 = x2, x2 = y = −v(t)x2 − u(t)x1 + f(t).
Thus; we get the system of two first order equations in the form
x1 = x2, x2 = −v(t) x2 - u(t) x1 + f(t) ,
or
˜X =
0 1
−u(t) −v(t)
x1
x2
+ f(t), (2.6)
or
X (t) + A(t)X(t) =
−→
b(t), X(t0) = y0, (2.7)
6

13. where
X (t) =
x1
x2
, A(t) =
0 −1
u(t) v(t)
, X(t) =
x1
x2
,
b(t) =
0
f(t)
, y0(t0) =
y0
y0
.
Note that, if y(t) is a solution to (2.5) then the vector-valued function
X =
y
y
is a solution to (2.7). Conversely, if the vector
X =
x1
x2
is a solution to (2.7) then x1 = x2. Hence
x1 + v(t)x1 + u(t)x1 = f(t)
which means that x1 is a solution to (2.5).
To carry out the method of integrating factor successfully to equation (2.7)
requires that we have
e
x
x0
A(s)ds
X(s) = e
x
x0
A(s)ds
X (s) + A(s)e
x
x0
A(s)ds
X(s), (2.8)
where for any square matrix A(t) = aij(t)
we define,
x
x0
A(s)ds =
x
x0
aij(s)ds
2
i,j=1
.
But Equation (2.7) is valid only if
d
dt
e
x
x0
A(s)ds
= A(t) × e
x
x0
A(s)ds
,
and this last equation is valid when
A(t)
x
x0
A(s)ds =
x
x0
A(s)dsA(t).
This equation leads to the following system of integral equations
x
x0
u(s)ds = u(x)(x − x0),
x
x0
v(s)ds = v(x)(x − x0).
7

14. The above system implies u(x), v(x) are constant functions. Thus, one can
prove the existence and uniqueness of solutions to second order linear differ-
ential equations with constant coefficients using the method of integrating
factor. The closed form of the solution is
X(t) = e−(x−x0)A
X(0) + e(x−x0)A
x
x0
e−(t−x0)
b(t)dt.
Illustration
Here, we will do the illustration using an example. Let us apply the method
of integrating factor described above to solve the initial value problem
y − y = 0, y(0) = 1, y (0) = 0.
In this problem, v(t) = 0, u(t) = −1. So
A(t) =
0 −1
u(t) v(t)
=
0 −1
−1 0
.
Hence,
−
t
0
Adt =
0 t
t 0
.
Now, one can easily see that for any non-negative odd integer n we have
−
t
0
Adt
n
=
0 tn
tn
0
,
and for non-negative even integer n
−
t
0
Adt
n
=
tn
0
0 tn .
We know that ex
= ∞
n=0
xn
n!
. Thus,
e− t
0 A(s)ds
=
∞
n=0
t2n
(2n)!
∞
n=0
t(2n+1)
(2n+1)!
∞
n=0
t(2n+1)
(2n+1)!
∞
n=0
t2n
(2n)!
=
cosh(t) sinh(t)
sinh(t) cosh(t)
,
and,
X(t) =
cosh(t) sinh(t)
sinh(t) cosh(t)
1
0
,
From this we obtain the unique solution y(t) = cosh(t).
We conclude from the above discussion that the method of integrating factor
works for first order systems with constant coefficients. Does this extension
works for any first order linear systems? We tackle this problem next.
8

15. Integrating factor method for first order linear systems
Here we show that the extended integrating factor method also works for
first order linear systems. Consider the initial value problem
y1 = a11y1 + a12y2
y2 = a21y1 + a22y2
with initial conditions
y1(t0) = y0
1,
y2(t0) = y0
2.
in matrix form we have
Y (t) = A(t)Y, Y (t0) = Y0
where
A(t) =
a11 a12
a21 a22
.
For the integrating factor method to work we need
a11 a12
a21 a22
.
t
t0
a11ds
t
t0
a12ds
t
t0
a21ds
t
t0
a22ds
=
t
t0
a11ds
t
t0
a12ds
t
t0
a21ds
t
t0
a22ds
.
a11 a12
a21 a22
This leads us to the system
a11
t
t0
a11ds + a21
t
t0
a12ds = a11
t
t0
a11ds + a12
t
t0
a21ds
a12
t
t0
a11ds + a22
t
t0
a12ds = a11
t
t0
a12ds + a12
t
t0
a22ds
a11
t
t0
a21ds + a21
t
t0
a22ds = a21
t
t0
a11ds + a22
t
t0
a21ds
a12
t
t0
a21ds + a22
t
t0
a22ds = a21
t
t0
a12ds + a22
t
t0
a22ds
The integrating factor is successful here provided that either a11,a12,a21,a22
are constants or a11=a22 and a21=a12, that is, the system matrix A is sym-
metric.
9

16. 2.2.2 UNIQUENESS
Here, we proof the existence of a unique solution to equation (2.4). To this
end, we consider the differential equation c.f. [6]
y (x) + h1(x)y (x) + h0(x)y(x) = g(x). (2.9)
Let h1(x), h0(x), and g(x) be defined on an open interval J. Assume that
for any finite closed subinterval [x1, x2] of J, there exist a constant M such
that
| h0(x) |≤ M and | h1(x) |≤ M ∀x ∈ [x1, x2] (2.10)
Suppose x0 ∈ J and u(x), v(x) are 2 different solutions of (2.9) such that
u(x0) = v(x0), and u (x0) = v (x0), that is u(x) and v(x) are functions de-
fined on J, which has two continuous derivatives and satisfies the differential
equation (2.9) we show that u(x) ≡ v(x), ∀x ∈ J
Proof:
Let w(x) = u(x) − v(x).
Then by substituting it in (2.9) we have the following:
w (x) + h1(x)w (x) + h0(x)w(x) = 0, (2.11)
Since u(x0) = v(x0) and u (x0) = v (x0), we have
w (x0) = w(x0) = 0. (2.12)
Now, choose x1, x2 ∈ J such that x1 < x0 < x2. Let M be as in (2.10), and
let z(x) = [w (x)]2
+ [w(x)]2
. Then from (2.12) we have
z(x0) = 0 (2.13)
Also, for x ∈ [x1, x2] ,
z (x) = 2w (x)w (x) + 2w(x)w (x)
= 2w (x)[−h1(x)w (x) − h0(x)w(x)] + 2w(x)w (x)
= −2h1(x)[w (x)]2
− 2h0(x)w (x)w(x) + 2w(x)w (x)
= −2h1(x)[w (x)]2
+ 2w(x)w (x)[1 − h0(x)]
(2.14)
Therefore,
|z (x)| ≤ |−2h1| [w (x)]2
+ |2w (x)w(x)| |1 − h0|
≤ 2M [w (x)]2
+ 2 |w (x)| |w(x)| (1 + M)
≤ 2M [w (x)]2
+ [w(x)]2
+ 2 (1 + M) |w (x)| |w(x)|
≤ 2M [w (x)]2
+ [w(x)]2
+ (1 + M) [w (x)]2
+ [w(x)]2
10

17. (2|w ||w|) ≤ (w )2
+ (w)2
, since (|w| − |w |) ≥ 0
|z (x)| = (1 + 3M) [w (x)]2
+ [w(x)]2
= (1 + 3M) z(x).
Thus,
−(1 + 3M)z(x) ≤ z (x) ≤ (1 + 3M)z(x).
Let k = (1 + 3M). Hence we have shown that
z (x) ≤ kz(x), ∀x ∈ [x0, x2] (2.15)
and
z (x) ≥ −kz(x), ∀x ∈ [x1, x0] (2.16)
From (2.15) we have that
z (x)e−kx
≤ kz(x)e−kx
, ∀x ∈ [x0, x2]
that is,
z (x)e−kx
− kz(x)e−kx
≤ 0, ∀x ∈ [x0, x2]
this implies,
d
dx
(z(x)e−kx
) ≤ 0, ∀x ∈ [x0, x2]
Hence, (z(x)e−kx
) is non increasing. But z(x0) = 0.
Which implies that,
(z(x)e−kx
) ≤ (z(x0)e−kx0
) = 0.
This shows that z(x) ≤ 0, ∀x ∈ [x0, x2].
But z(x) = [w (x)]2
+ [w(x)]2
≥ 0 . Therefore,
z(x) ≡ 0, ∀x ∈ [x0, x2].
Similarly, from (2.16), we have that, z (x) + kz(x) ≥ 0 , for x ∈ [x1, x0].
Multiplying through by ekx
, we get
0 ≤ z (x)ekx
+ kekx
z(x) = d
dx
(z(x)ekx
), ∀x ∈ [x1, x0]
Therefore, for x ∈ [x1, x0], we have that
(z(x)ekx
) ≤ (z(x0)ekx0
) = 0 . Hence, z(x) ≤ 0 , for x ∈ [x1, x0]. But still,
z(x) = [w (x)]2
+ [w(x)]2
≥ 0. Hence z(x) ≡ 0.
We have shown that z(x) ≡ 0. That is [w (x)]2
+ [w(x)]2
= 0. Which implies
that w(x) = 0, since for the sum of two positive numbers to be zero, each of
them must be zero. Hence u(x) − v(x) = w(x) ≡ 0, ∀x ∈ [x1, x2].
Hence u(x) − v(x) ≡ 0, ∀x ∈ [x1, x2]
That is, u(x) ≡ v(x), ∀x ∈ [x1, x2]. Since x1 and x2 are arbitrary, we have
that, u(x) ≡ v(x) for all x ∈ J. Hence the solution is unique.
11

18. 2.2.3 CONTINUOUS DEPENDENCE OF THE SO-
LUTION ON THE INITIAL VALUES
Consider the following second order differential equation transformed into a
system of first order differential equations. [2]
X (t) = A(t)X(t) +
−→
b(t), t ∈ I := (a, b) (2.17)
Where A(t) is a 2 × 2 continuous matrix, and X(t) is the vector function
on the interval I := (a, b). We will show that solutions to the equations
of the type (2.17) is defined on the hold interval I. But before, let us state
Gronwall’s inequality which we will subsequently use.
GRONWALL’S INEQUALITY 2.2.1 Let the functions µ(x), and
f(x) be continuous and non-negative on the interval I := [α, β] con-
taining the point x0. If there exist a constant C ≥ 0, such that
µ(x) ≤ C + |
x
x0
µ(s)f(s)ds|, x ∈ I
is satisfied, then also, the inequality
µ(x) ≤ Ce
| x
x0
f(s)ds|
, ∀x ∈ I
is satisfied. In particular, if C = 0, then µ(x) ≡ 0 in I.
Now we can proceed and proof the continuous dependence of the
solution. To do this, we show that for any subinterval I ⊆ I the solution in
this interval is bounded.
Proof:
Suppose X(t) = ϕ(t) is a solution of (2.17) defined on some interval
I = (β1, β2), a < β1 < β2 < b, (then it satisfies the integral equation). Let
t0 ∈ I , then ϕ(t) satisfies the integral equation
ϕ(t) = ϕ(t0) +
t
t0
A(s)ϕ(s) + b(s) ds.
Hence,
ϕ(t) ≤ ϕ(t0) +
t
t0
A(s)ϕ(s)ds +
t
t0
b(s)ds .
12

19. Let



K = ϕ(t0) + max
t∈I
b(t) (β2 − β1)
L =max
t∈I
|A(t)|
|t − t0| ≤ |β2 − β1|
Hence,
ϕ(t) ≤ K + L
t
t0
|ϕ(s)|ds .
By Gronwall’s inequality, we have
ϕ(t) ≤ Kexp (L(β2 − β1)) < ∞
Hence, ϕ(t) is bounded and so we conclude that every solution of equation
(2.17) is defined on the hold interval I := (a, b).
13

20. Chapter 3
DESCRIPTION OF SOME
PHYSICAL PROBLEMS
THAT CAN BE MODELLED
BY LINEAR SECOND
ORDER EQUATIONS
Introduction
Many problems in physics can be modelled by linear second order
differential equations. In this chapter, are given some examples. The choice
of these examples is based on the fact that they are the simplest and most
encountered physical phenomenon in our society. For references on this
chapter, we used the books [1], [3], [7], [8].
3.1 Simple mass-spring system in free
vibration
practical occurrence
Physical phenomena of solids in free vibration occurs when the vibrations
are produced by instantaneous disturbances as a result of either a force or a
deformation of the supporting spring [7]. The initial disturbance does not
exist after the vibration of the solid. This can be shown by the diagram
below.
14

21. Figure 3.1: Physical demonstration of a simple mass-spring system in free
vibration [7]
Mathematical formulation
Consider the diagram in figure 3.2. We assume that there is no air
resistance to the motion. The spring is hung freely and deflects upon
attaching a mass m on it. Then a small instantaneous push-down is applied
to the mass and released quickly. We can expect the mass to bounce down
and up passing its initial equilibrium position.
Figure 3.2: Simple mass-spring system in free vibration [7]
Forces:Weight(W), spring force(Fs), Dynamic force F(t)
15

22. Equilibrium forces acting on the mass at given time t satisfies Newton’s
first law of motion, which state’s that ”every object will remain at rest or in
uniform motion in a straight line unless coupled to change its state by the
action of an external force”
Hence,
+ ↓ [−F(t) − Fs + W] = 0,
where + ↓ means the downward direction is taken to be positive.
But dynamic and spring force equals:
F(t)=md2y(t)
dt2 , Fs=k [h + y(t)].
Hence we then have,
−m
d2
y(t)
dt2
− k [h + y(t)] + mg = 0,
but mg = kh from the static equilibrium condition.
So,
−m
d2
y(t)
dt2
− k [h + y(t)] + kh = 0.
Hence we have the following second order differential equation for the
instantaneous position y(t) of the vibrating mass,
m
d2
y(t)
dt2
+ ky(t) = 0.
Which is the same as,
d2
y(t)
dt2
+
k
m
y(t) = 0. (3.1)
We observe that, Since k = spring constant (a property of the spring) and
m = mass of the vibrating solid, then the equivalent coefficient k
m
is a
positive real number. Hence it is continuous as a constant function.
Therefore our theorem guarantee’s to us that (3.1) has a unique solution.
Solution
The solution to (3.1) can be obtained by comparing it to the second order
differential equation
d2
y(t)
dt2
+ v(t)
dy(t)
dt
+ u(t)y(t) = 0,
where v(t) = 0, u(t) = k
m
.
16

23. The common expression of the solution to (3.1) is
y(t) = c1cosω0t + c2sinω0t,
where ω0 = k
m
,
since u(t) is positive, and thus we have that v2
− 4u = −4 k
m
≤ 0
Validation and interpretation
This is an oscillating function, oscillating about the ”zero-time” axis, with
amplitude of vibration y(t), where c1 and c2 are arbitrary constants to be
determined given some initial conditions. And ω0 is the circular or angular
frequency of the mass-spring vibration of the system. It often represents the
natural frequency of the system, who’s unit is Rad/s. Corresponding to the
angular frequency ω0, the real frequency of the vibration is
f =
ω0
2π
=
1
2π
k
m
.
Combining cosine and sine function, we have the following graph:
Figure 3.3: Simple mass-spring system in free vibration [7]
17

24. 3.2 Simple damped mass-spring system in
free vibration
What makes free-vibration of a mass-spring system to stop after some time
t in reality is the damping effect. This is a more realistic phenomenon.
Some of the sources of damping in mechanical vibrations include, resistance
by the air surrounding the vibrating mass, and internal friction of the
spring, during deformation. [7]
A practical occurrence of a simple damped mass-spring in free
vibration
Damping of a simple mass-spring vibration system is induced by air
resistance to moving mass. One can use an air cylinder with adjustable air
vent to regulate the air resistance to the moving mass as shown in figure 3.4
and figure 3.5. Figure 3.5 shows an application of coil-overs in motorcycle.
Figure 3.4: air cylinder with adjustable air vent to regulate the air resistance
[7]
The damper in the physical model is characterized by a damping coefficient
c, which is usually specified by manufacturers of the damper (a dash pot)
Mathematical Formulation
The corresponding damping force is related to air resistance to the
movement of the mass. The resistance R is proportional to the velocity of
the moving mass. Thus, Mathematically, we have:
18

25. Figure 3.5: Real world application with coil-overs in motorcycle suspension
[7]
R(t) ∝
dy(t)
dt
,
where y(t) is the distance the mass has travelled from its initial equilibrium
position.
Thus the damping force R(t), is of the form
R(t) = c
dy(t)
dt
, (3.2)
Where c = damping coefficient.
The mathematical expression of this physical model can be obtained by
following similar procedure for the simple mass-spring, with the inclusion of
the additional damping force.
Newton’s first law in dynamic equilibrium gives:
+ ↓ Fy = −F(t) − R(t) − Fs + W = 0,
Fs=k [h + y(t)], F(t)=md2y(t)
dt2 .
Hence we have the following second order linear equation
m
d2
y(t)
dt2
+ c
dy(t)
dt
+ ky(t) = 0.
This is a second order homogeneous equation for the instantaneous position
of the vibrating mass.
Which is the same as:
19

26. Figure 3.6: Mathematical modelling of damped mass-spring system in free
vibration [7]
d2
y(t)
dt2
+
c
m
dy(t)
dt
+
k
m
y(t) = 0. (3.3)
We can compare it with
d2
y(t)
dt2
+ v(t)
dy(t)
dt
+ u(t)y(t) = g(t), (3.4)
where v(t) = c
m
, u(t) = k
m
and g(t) = 0 which are all constant functions,
hence continuous functions. By our theorem stated in chapter two,
equation (3.3) has a solution, it is unique and will depend continuously on
the initial conditions when given.
Solution
The solution to (3.3) depend on the signs of the discriminators, we then
analyse the different cases as follows
Case 1: c
m
2
− 4 k
m
> 0 (this is an over-damping situation) with solution:
y(t) = e( c
2m )t
AeΩt
+ BeΩt
, Ω =
√
4mk − c2
2m
.
Case 2: c
m
2
− 4 k
m
= 0 (this situation is called critical damping) with
solution:
y(t) = e( c
2m )t
(A + Bt) .
20

27. Case 3: c
m
2
− 4 k
m
< 0 (this situation is called under damping)with
solution:
y(t) = e−( c
2m )t
(AcosΩt + sinΩt) , Ω =
√
4mk − c2
2m
.
Where A and B are arbitrary constants.
Validation and interpretation
These different cases can be illustrated by the help of graphs as shown
below.
Figure 3.7: CASE 1 [7]
Figure 3.8: CASE 2 [7]
Observations
21

28. Figure 3.9: CASE 3 [7]
CASE 1: There is no oscillatory motion of the mass, an initial increase in
the displacement can occur, followed by continuous decay in the amplitudes
of vibration (which usually decays quickly in time). This is a desirable
situation in abating (mitigating) mechanical vibration.
CASE 2: No oscillatory motion of the mass by theory, amplitude reduces
with time but takes longer time to ”die down” compared to CASE 1. This
may became an unstable situation of vibration.
CASE 3: Here, there is oscillatory motion of the mass, the amplitude of
each oscillatory motion of the mass reduces continuously, but takes longer
time to die down. This is thus the least desirable situation in machine
design.
3.3 Resonant vibration Analysis
Any machine or structure subjected to Potential cyclic (intermittent)
loading is vulnerable to resonance vibration. [7]
Practical occurrences
Resonance vibration of machines or structures occurs when subjected to
cyclic loads, it is a Forced Vibration, with forces acting to the vibrating
solids at all times. Thus the consequence of resonant vibration is that, the
amplitude of the oscillatory motion of the structure continues to magnify in
short time, hence resulting in an overall structure failure [7]. The diagrams
below shows some occurrences of resonant vibration in the real world .
22

29. Figure 3.10: Occurrences of Resonance [7]
Mathematical Formulation of a resonance vibration problem
Consider the simple mass spring system subjected to an exciting force
F(t), where t=time. The mathematical formulation for this model can be
Figure 3.11: mathematical model for the occurrences of resonance [7]
derived using Newton’s first law:
+ ↓ Fy = 0 → −Fd − k [h + y(t)] + W + F(t) = 0
With Fd = md2y(t)
dt2 From Newton’s second law which say’s that:
Force = mass × acceleration.
Thus the differential equation for the instantaneous amplitude of the
vibrating mass under the influence of the force F(t) is given by
m
d2
y(t)
dt2
+ ky(t) = F(t) (3.5)
23

30. if we assume F(t) in (3.5) is of cyclic nature following a cosine nature, that
is
F(t) = F0cosωt,
F0 = Maximum magnitude of the force
And ω = circular frequency of the applied circular force. F(t) is displayed
graphically as in figure
Figure 3.12: F(t) = F0cosωt, [7]
(4) became
m
d2
y(t)
dt2
+ ky(t) = F0 cos ωt
or
d2
y(t)
dt2
+ ω2
0y(t) =
F0
m
cos ωt; ω2
0 =
k
m
(3.6)
We see that, comparing it with equation (3.4), we have that v(t) = 0,
u(t) = ω2
0 and g(t) = F0
m
cos ωt are continuous as constant function and
product of 2 continuous functions respectively. Hence the solution to (3.6)
exists and it is unique. Also, it will depend continuously on the initial
conditions.
Solution
This is a non-homogeneous second order differential equation which has as
general solution:
y(t) = yh(t) + yp(t)
where yh(t) is obtained from the homogeneous part of (5) as such:
d2
yh(t)
dt2
+
k
m
yh(t) = 0
24

31. To have:
yh(t) = c1cosω0t + c2sinωot
yp(t) is obtained from the non-homogeneous part of (3.6) as follows:
yp(t) = Acosω0t + Bsinωt
We differentiate it twice, substitute it in (3.6) and solve for A and B to get,
yp(t) =
F0
m(−ω2 + ω2
0)
cosωt.
Hence the general equation is:
y(t) = c1cosω0t + c2sinωot +
F0
m(−ω2 + ω2
0)
cosωt,
c1 and c2 are constants which can be determined by specifying the initial
conditions.
Now if ω = ω0 , then y(t) → ∞.
This means amplitude of vibration goes to infinity instantly at all times,
which is not physically possible. Hence an alternative solution needs to be
derived for such a case. This is done below as follows:
Since ω = ω0 , we can then write
d2
y(t)
dt2
+
k
m
y(t) =
F0
m
cosω0t
and,
yh(t) = c1cosω0t + c2sinω0t
yp(t) has cosω0t which is the same as that on the non-homogeneous part,
thus
yp(t) = t(Acosω0t + Bsinω0t)
hence we have,
yp(t) =
F0
2mω0
tsinω0t
Thus the general equation became
y(t) = c1cosω0t + c2sinω0t +
F0
2mω0
tsinω0t
Validation and interpretation
We now demonstrate the graphical representation of the amplitude
fluctuation of the vibrating mass.
The amplitude of vibration of the mass will increase rapidly with time;
The attached string will soon be stretched to break with elongation in a
short time at tf , as demonstrated in figure 3.13 below.
25

32. Figure 3.13: Amplitude fluctuation of the vibrating mass [7]
3.4 Electric circuits
The purpose of modelling this problem is because it is important in the
construction of electrical devices. Understanding these principles allows
electrical engineers to create stuffs like, hair dryers, toasters, and heating
devices. Also, resistor circuits are essential for comprehending voltage
dividers, which are circuit elements that split voltage between different
branches of electrical circuits. Capacitors have many practical applications,
and in order to use them, we must understand how they work in circuits.
For instance, a circuit containing a capacitor is necessary in order to change
alternating current (AC which has a sinusoidal varying voltage) voltage to a
direct current (DC, constant voltage) voltage supply. All electricity
supplied to households is AC because when the electrical grid was being
designed, AC transmission lines allowed power to be transmitted over much
greater distances [1]. Thus to improve modern circuits, we must fully
understand the relationships between the circuit components.
Mathematical Modelling
Consider the diagram in figure 3.14. [8]
1. The current I, measured in ampere is a function of time;
2. The resistance R, measured in ohms;
26

33. Figure 3.14: A simple electric circuit [8]
3. The capacitance C, measured in Farads; and
4. The inductance L, measured in henry’s
They are all positive and are assumed to be known constants. The
impressed voltage E (measured in volts) is a given function of time.
Another quantity that enters in the discussion is the total charge Q
(measured in coulombs) on the capacitor at time t.
The relationship between charge and current is given by
I =
dQ
dt
. (3.7)
The flow of current in the circuit is governed by Kirchhoff’s second law
which says that: ”In a closed circuit, the impressed voltage is equal to the
sum of the voltage drops in the rest of the circuit.”
According to the elementary law of electricity, we know that:
The voltage drop across the resistance is IR.
The voltage drop across the capacitor is Q
C
.
The voltage drop across the inductor is LdI
dt
.
Hence, by Kirchhoff Law,
L
dI
dt
+ RI +
1
C
Q = E(t). (3.8)
The units have been chosen so that: 1 volt = 1 ohm, 1 ampere = 1
coulomb/1 farad = 1 henry.1 ampere/1 second. Substituting for I from
27

34. (3.7), we obtain the differential equation
LQ + RQ +
1
C
Q = E(t), (3.9)
for the charge Q. The initial conditions are
Q(t0) = Q0, Q (t0) = I(t0) = I0. (3.10)
Thus we must know the charge on the capacitor and the current in the
circuit at some initial time t0.
Alternatively, we can obtain a differential equation for the current I by
differentiating (3.9) with respect to t, and substitute for dQ
dt
from (3.7) and
obtain the following result:
LI + RI +
1
C
I = E (t), (3.11)
with initial conditions,
I(t0) = I0, I (t0) = I0. (3.12)
From (3.8) it follows that,
I0 =
E(t0) − RI0 − (I/C)Q0
L
. (3.13)
Hence I0 is also determined by the initial charge and current, which are
physically measurable quantities.
Now, (3.11) can be written in the form
I +
R
L
I +
1
CL
I =
1
L
E (t),
and can be compared to the second order linear equation (3.4) where
v(t) = R
L
, u(t) = 1
CL
and g(t) = 1
L
E (t) which are all continuous as constant
functions. Hence the solution exist, it is unique and is continuously
dependent on the data.
3.5 An electric motor
Many industrial and consumer products rely on precisely controlling the
speed of an electric motor, for example CD players (an electronic device
that plays audio compact discs) and printers. A schematic picture of an
electric motor is shown in figure 3.15. Energy stored is stored in the
28

35. Figure 3.15: A schematic diagram of an electric motor [3]
capacitor and the inductor, and momentum is stored in the rotor. Three
state variables are needed if we are interested in the motor speed [3].
Storage can be represented by the current I through the rotor, the voltage
V across the capacitor and the angular velocity ω of the rotor. The control
signal is the voltage E applied to the motor. The output for the system is
the motor speed ω.
A momentum balance for the rotor gives
J
dω
dt
+ Dω = kI,
and Kirchoff’s laws for electricity gives
E = RI + L
dI
dt
+ V − k
dω
dt
, I = C
dV
dt
.
Introducing the state variables x1 = ω, x2 = V , x3 = I and the control
variable u = E, the equations for the motor can be written as
CL
d2
V
dt2
+
dV
dt
CR −
Ck2
J
+ V +
kDω
J
= E,
which is the same as
d2
V
dt2
+ CR −
Ck2
J
1
CL
dV
dt
+
V
CL
= E −
kDω
J
1
CL
. (3.14)
This is a second order linear differential equation with constant coefficients,
hence according to our theorem, it has a solution which is unique, and
would depend continuously on the initial conditions when given. This is so
since constant functions are continuous.
29

36. Chapter 4
CRITICAL STABILITY
ANALYSIS OF THE
EQUILIBRIUM SOLUTION
In this chapter we make a critical stability analysis of the equilibrium
solution for the equation of a driven mass-spring system, with damping.
The equation will first be converted into a system of first order equations,
and then we are going to look for the eigenvalues. The stability of the
equilibrium solution is determined by the nature of the eigenvalues as will
be shown subsequently. For references on this chapter, the books which
have been used include, [3], [5]. Consider figure 4.1. Assuming that the
Figure 4.1: A driven mass spring system, with damping [3].
damper exerts a force on the spring that is proportional to the velocity of
30

37. the system and using Hooks law (which state’s that ”the restoring force of a
spring is directly proportional to a small displacement”) to model the
spring, we have
mp + bp + kp = f(t), (4.1)
where m, is the mass, b, the coefficient of viscous friction, k, the spring
constant, f, the applied force and p the displacement. Now let us compute
the response of the system to an input of f(t) = A sin ωt. We now have
mp + bp + kp = A sin ωt.
Let x = p, and y = p . Then,
x = p = y and y = p = [A sin ωt − kx − by] 1
m
.
We now look for the critical points.
At equilibrium position, x = 0 and y = 0. Hence we have, y = 0 and
[A sin ωt − kx − by] 1
m
= 0. Solving for x and y we have that the critical
point is, (A sin ωt, 0). Now we look for the eigenvalues to determine the
nature of the stability of the critical point. Consider the following
continuous functions,
F(x, y) = y and G(x, y) = [A sin ωt − kx − by] 1
m
.
The associated linear system is given by;
Fx(x, y) = 0, Fy(x, y) = 1
Gx(x, y) = − k
m
, Gy(x, y) = − b
m
Hence at (A sin ωt, 0), we have
x
y
=
0 1
− k
m
− b
m
x
y
.
We then look for the eigenvalues as follows
−λ 1
− k
m
− b
m
− λ
= 0,
solving for λ we obtain
λ1,2 =
− b
m
+
−
b
m
− 4( k
m
)
2
.
We now analyse the different cases: Let ∆ := b
m
− 4( k
m
)
CASE 1: When ∆ < 0,
we have complex eigenvalues, hence the critical point is a spiral point, and
the system is asymptotically stable.
CASE 2: When ∆ = 0,
31

38. Figure 4.2: Asymptotically stable improper node
Figure 4.3: Asymptotically stable singular node
we have real and equal negative eigenvalues, hence the critical point is
asymptotically stable singular node.
CASE 3: When ∆ > 0,
we have distinct real eigenvalues, hence the critical point is an improper
node which is asymptotically stable if the eigenvalues are negative, else it is
unstable. Thus we have a saddle point. If both eigenvalues are negative,
the saddle is an asymptotically stable point, else it is unstable. The
straight lines directed along the eigenvalues V1, V2 are called separatrices.
These are the asymptotes of other phase trajectories that have the form of
a hyperbola. Each separatrices can be associated with a certain direction of
32

39. Figure 4.4: Saddel [5].
motion. If the separatrix is associated with a negative eigenvalue, the
movement along it occurs towards the equilibrium point X = 0. Conversely,
if the eigenvalue is positive, the movement is directed from the origin.
33

40. CONCLUSION
In this project, we have established and proved the existence and
uniqueness theorem for a solution to a second order linear equation, and
have modelled five physics problems using second order linear equations.
From this, we see that second order linear equation is an important type of
equation because it is used to model real life problems, like the ones we
encountered in the project. The most important things we can take note of
in this project are that:
• The flow of current in the circuit is described by an initial value problem
of precisely the same form as the one that describes the motion of a
spring-mass system.
• Also, understanding how resonance arises (how to minimize it) is a very
important application of second order linear equation to structural
engineering.
A poor understanding of resonance is something which, at several times in
the not-too-distant past has caused bridges to fall, air planes to crash, and
buildings to fall over. We can see from the examples that resonance arises
when an external force act on the system at the same (or very close)
frequency that the system is already oscillating at.
Of course resonance is not always bad. The general principle of applying an
external driving force at a systems natural resonance frequency is the
underlying physical idea behind the construction of many types of musical
instruments.
These examples show the unifying nature of mathematics. Once you know
how to solve second order linear equations with constant coefficients, you
can interpret the results in terms of either mechanical vibrations, electric
circuits, or any other physical situation that leads to the same problem.
34

41. Bibliography
[1] ARJUN MOORJANI, DANIEL STRAUS, JENNIFER ZELENTY: A
Model of voltage in a resistor circuit and an RC circuit, (2010).
[2] NKEMZI BONIFACE: MAT409 Notes, University of Buea, (2014).
[3] KARL J ASTROM, RICHARD M. MURRAY, An Introduction for
Scientists and Engineers, California Institute of Technology, (2006).
[4] MARCEL B FINAN,Existence and Uniqueness Proof for nth Order
Linear Differential Equations with Constant Coefficients,Department of
Mathematics, Arkansas Tech University. Russellville, Ar 72801, (2006).
[5] Math24.net, Differential equations, (2014).
[6] NAGLE, R. , SA E. ,Fundamentals of Differential Equations,
Addison-Wesley, New York, (1993).
[7] TAI-RAN HSU, ME 130 Applied Engineering Analysis. Department of
Mechanical and Aerospace Engineering, San Jose State University, San
Jose, California, USA, (2009).
[8] WILLIAM E. BOYCE, RICHARD C.DIPRIMA, Elementary
Differential Equations and boundary Value problems-7th edition, John
Wiley and Sons, Inc. New York Chichester Weinheim Brisbane Toronto
Singapore, (2001).
35