KWAME NKRUMAH UNIVERSITY OF SCIENCE
AND TECHNOLOGY, KUMASI – GHANA
INSTITUTE OF DISTANCE LEARNING
[]
MATH 251: ENGINEERING MATHEMATICS IV
(CREDITS: 4)
YAO ELIKEM AYEKPLE
1
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Course Writer
Yao Elikem Ayekple received both BSc. (Hons) and MSc. Mathematics from the Department
of Mathematics, Faculty of Science, KNUST, Kumasi.
He entered the university in 1997 for undergraduate degree and completed in 2001. After
National Service in 2002, he enrolled for his MSc. Mathematics programme. He completed
his MSc. Thesis in 2004 and defended in the same year.
He was a demonstrator in the university from 2002 to 2004. He was employed as a Teaching
Assistant from 2004 to 2005. From 2005 to date, he has been teaching at his alma mater,
Kwame Nkrumah University of Science and Technology, Kumasi at the Department of
Mathematics as a Lecturer.
4
Course Introduction
Differential Equation is among the lynchpins of modern mathematics, which along with
matrices, are essential for analyzing and solving complex problems in engineering, the
natural sciences, economics, actuarial sciences, and even business.
In the first section, we begin with general differential equation and classifying them in terms
of type, order, degree and linearity. We then narrowed to one type of differential equation,
which is ordinary differential equation.
We move on to look at first order ordinary differential equation types, put them in classes and
find an appropriate methods of solving most of them, if not all, and its applications to
practical problems.
We then consider higher order linear ordinary differential equation, that is, orders greater
than one. Initially, we concentrate on second – order homogeneous linear ordinary
differential equation with constant coefficient because their theory is typical of that of linear
differential equations of any order n (but involves much simpler formulas), so that the
transition to higher order n needs only very few new ideas. The Wronskian is used to
determine the independences of solutions to a particular equation. Elementary methods such
as undetermined coefficient, variation of parameter and linear differential operator are used to
solve second – order non – homogeneous linear ordinary differential equations with constant
coefficient.
Systems of differential equations occur in various applications. Linear systems are best
treated by the use of matrices and vectors, of which, however, only modest knowledge will be
needed here. We will deal with first – order linear systems with constant coefficient and draw
some analogies from the first order ordinary differential equation.
5
Course Objectives
On completion of the course, it is hoped that you would be able to:
Classify first order ordinary differential equations in terms of type, order,
degree and linearity.
-
Have the technique to solve most, if not all, first order differential equations
-
Be able to solve differential equation of higher order with constant coefficient
using the appropriate technique for each kind of higher order equations.
-
Apply the technique in solving the differential equation in solving differential
equation appears in actuarial science, economics and finance.
-
Reduce higher order linear differential equations to system of first order
differential equations and solve.
-
Transform system of equation to normal form and put it in matrix and vector
to solve using eigenvalues and eigenvectors.
Course Outline
Unit 1 Basic Concepts:
o Definition of Differential Equation
o Classification of Differential Equations
o Solution of a Differential Equation
o General Solution
o Initial – Value and Boundary – Value Problems
Unit 2 Ordinary Differential Equation of First Order:
o General and Particular Solution of a First – Order Differential Equation.

Method Separation of Variables

Reduction to Separable Form

Exact Equation

Integrating Factors

Linear Equations and Those Reducible to that form

Linear Equations

Reducible to Linear Forms by change of Variable

Bernoulli’s Equation
6

Riccati’s Equation
o Applications of First – Order Ordinary Differential Equation
Unit 3 Linear Differential Equation of Higher Order:
o Homogenous Equation with constant coefficient
o Theory of solution of Linear Differential Equation

Independent and Dependent function

Wronskian
o Linear Non – Homogeneous with constant coefficients

Method of Undetermined Coefficient

Variation of Parameter

Linear Differential Operator
Unit 4 LAPLACE TRNSFORM:
o Definition of Laplace Transform
o Find Laplace Inverse
o D

The Differential Operator

Reduction of n th Order Equation to a System of First – Order
Equations.

The Matrix Method

The Eigenvalue Method of Homogeneous Linear System

Non – Homogeneous Linear System.
Mode of Assessment
This would involve a combination of continuous assessment (30%) and end of semester
examination (70%). The continuous assessment would include handed in exercises and midsemester examination.
Test your understanding
Within each unit/session are exercises. These are meant to help you assess your
understanding of the unit or course. It is vital that you take time to complete or find solutions
to these exercises as they occur in the study material. Make sure you go through them. I
recommend you have a notebook/exercise book specifically for this purpose and keep it with
your study materials as record of your work.
7
References
1.
Frank Ayers, JR.: Theory and Problems of Differential Equations, McGraw –
Hill Inc.
2.
Earl D. Rainville and Philip E. Bedient: Elementary Differential Equations,
The Macmillan Company
3.
Richard Bronson and Gabriel Costa (2006):Differential Equation (Third
Edition), McGraw – Hill Inc.
4.
Erwin Kreyszig(2002): Advanced Engineering Mathematics (Eighth Edition),
John Wiley and Sons, Inc.
5.
William E. Boyce and Richard C. DiPrima: Elementary Differential Equations
and Boundary Value Problems, John Wiley and Sons, Inc.
6.
www.sosmath.com
7.
https://tutorial.math.lamar.edu/pauldawkins
8
UNIT
1
BASIC CONCEPTS
Introduction
The laws of the universe are written largely in the language of mathematics. Algebra is
sufficient to solve many static problems, but the most interesting naturally phenomena
involve change and are best described by equations that relate changing quantities.
Many important and significant problems in engineering, the physical sciences, and the social
sciences such as economics and business when formulated in mathematical terms require the
determination of a function satisfying an equation containing the derivatives of unknown
function. Such equations are called differential equation. Perhaps the most familiar example
is Newton’s law:
d 2x
m 2 F
dt
for the position
(1)
x  t  of a particle acted on by a force F . In general F will be a function of
time t , the position x , and the velocity dx / dt . To determine the motion of a particle acted
on by a given force
F it is necessary to find a function x  t  satisfying Eq. (1). If the force
is that due to gravity, then
F  mg and
d 2x
m 2  mg
dt
(2)
On integrating Eq. (2) we have:
dx
  gt  c1
dt
x  t    12 gt 2  c1t  c2
where c1 and c2 are constants. To determine
(3)
x  t  completely it is necessary to specify two
additional conditions, such as the position and velocity of the particle at some instant of time.
These conditions can be used to determine the constants c1 and c2 .
This unit introduces us to the differential equation in general and helps to differentiate
between various kinds and associated solutions.
9
Learning Objectives
After going through this section session, you would be able to:
• know what is a differential equation, and state the difference between the
independent and dependent variables.
• classify differential equations in terms of types, order, degree and linearity.
• differentiate between general and particular solution.
• derive differential equation from general solutions that is; elimination of arbitrary
constant.
• differentiate between boundary and initial conditions and be able to solve
problems with such conditions.
Session 1-1 Differential Equations
The laws of the universe are written largely in the language of mathematics. Algebra is
sufficient to solve many static problems, but the most interesting naturally phenomena
involve change and are best described by equations that relate changing quantities.
Many important and significant problems in engineering, the physical sciences, and the social
sciences such as economics and business when formulated in mathematical terms require the
determination of a function satisfying an equation containing the derivatives of unknown
function. Such equations are called differential equation. Perhaps the most familiar example
is Newton’s law:
d 2x
m 2 F
dt
(1)
d 2x
m 2  mg
dt
(2)
x  t  of a particle acted on by a force F . In general F will be a function of
time t , the position x , and the velocity dx / dt . To determine the motion of a particle acted
on by a given force F it is necessary to find a function x  t  satisfying Eq. (1). If the force
is that due to gravity, then F  mg and
for the position
On integrating Eq. (2) we have:
dx
  gt  c1
dt
x  t    12 gt 2  c1t  c2
where c1 and c2 are constants. To determine
(3)
x  t  completely it is necessary to specify two
additional conditions, such as the position and velocity of the particle at some instant of time.
These conditions can be used to determine the constants c1 and c2 .
10
Differential equations are an important part of the calculus, the fundamentals of which are
presented here. In developing the theory of differential equations in a systematic manner it is
helpful to classify different types of equations.
There are two types of derivatives which usually are (and always can be) interpreted as rates.
For example, the ordinary derivative dy / dx is the rate of change of y with respect to x
(independent variable), and the partial derivative u / x is the rate of change of u with
respect to x when all independent variables except x are given fixed values. And the study of
differential equations has two principal goals:
1.
2.
To discover differential equation that describes a physical situation;
To find the appropriate solution of that equation.
The solution of differential equations plays an important role in the study of the motions of
heavenly bodies such as planets, moons, and artificial satellites.
1-1.1 Definition of Differential Equation
A differential equation is an equation which involves one or more derivatives, or
differentials of an unknown function.
Example 1.1
The following are examples of differential equations involving their respective unknown
functions:
dy
1.1
 3x 2
dx
dR
1.2
 kR  0
dt
d2y 2
k y 0
dx2
1.3
 x2  y2  dx  2xydy  0
1.4
  2v  2v 
v

 h2 

 x2 y 2 
t


1.5
L
d 2i
dt 2
R
di 1
 i  E cos t
dt C
1.6
 2v
 2v

0
x2 y 2
1.7
11
3
 d 2w 
dw
 2   xw  w  0
 dx 
dx


1.8
d 3x
dx

x
 4 xy  0
dy
dy3
d2y m
 dy 

1  
2
dx
H
 dx 
1.9
2
1.10
3
3
2
d y  dy 
x

e 
dx3  dx 

3
1.11
d2y
g

sin y  0
dx2 l
3
1.12
 d2y 
 dy 
3  dy 
 2   3 y    y    5x
 dx 
 dx 
 dx 
7
2
1.13
When an equation involves one or more derivatives with respect to a particular variable, that
variable is called an independent variable. A variable is called dependent if a derivative of
that variable occurs.
In the equation:
d 2i
di 1
1.6

R
 i  E cos t
dt C
dt 2
i is the dependent variable, t the independent varible, and L , R , C , E , and
parameters.
The equation :
L
 2v
 2v

0
x2 y 2
 are called
1.7
has one dependent variable v , and two independent variables x and y .
Since the equation
 x2  y2  dx  2xydy  0
may be written
x2  y 2  2 xy
or
dy
0
dx
 x2  y2  dydx  2xy  0
we may consider either variable to be dependent, the other being the independent one
Oral Exercise 1
12
Identify the independent variables, the dependent variables, and the parameters in the
equations given as examples in this section.
1-1.2 Classification of Differential Equations
Recall that a differential equation is an equation (has an equal sign, but not an identity) that
involves derivatives. Just as biologists have a classification system for life, mathematicians
have a classification system for differential equations.
Differential equations are classified as to:
(a)
type (namely, ordinary or partial)
(b)
order
(c)
degree, and
(d)
linearity
(a)
Type
We can place all differential equations into two types: ordinary differential equation and
partial differential equations.
An Ordinary Differential Equation, (ODE) is one containing ordinary derivatives of one or
more unknown functions (dependent variable) with respect to a single independent variable.
Examples are equations: (1.1), (1.2), (1.3), (1.4), (1.6), (1.8), (1.9), (1.10), (1.11), (1.12) and
(1.13). So also is each of the two equations with more than one dependent variable:
dx
dy
(1.14)
a b  c
dt
dt
dx
dy
(1.15)
d e  f
dt
dt
A Partial Differential Equation is one involving partial derivatives of one or more
dependent variables with respect to one or more of the independent variables. Examples are
equation: (1.5) and (1.7).
Note:
The systematic treatment of partial differential equation lies beyond the scope of this book.
We treat only ordinary differential equation with constant coefficient in this book.
13
(b)
Order
Definition 1.1
The order of a differential equation is the order of the highest-ordered derivative appearing in
the equation. For instance,
3
d2y
 dy 
 2b    y  0
2
dx
 dx 
(1.16)
is an equation of “order two.” It is also referred at as a “second-order ordinary differential
equation.”
More generally, the equation


n
F x, y, y ', y '',..., y   0
(1.17)
is called an “ nth - order’’ ordinary differential equation. Equation (1.17) represents a
relation between the n  2 variables x, y, y,..., y
solved for y
 n
 n
which under suitable conditions can be
in terms of the other variables:

y n  f x, y, y,..., y n1

(1.18)
For the purpose of this book we shall assume that this is always possible. Otherwise, an
equation of the form of equation (1.17) may actually represent more than one equation of the
form of equation (1.18).
For example, the equation x2  y  3 y  2x  0 actually represents the two different
2
equations,
y 
3  9  8 x3
2 x2
or
y 
3  9  8 x3
2 x2
Note: Eq.(1.17) is called the general form.
Note also that Eq. (1.17) is an n th order differential equation because:
n
d y
n
y   n …………………….
dx
14
(1.19)
Example 1.2
1.
2.
3.
di
 Ri  E
dt
yy  x
L
(order 1)
(order 2)
2 z
2 z

3
0
x2
y 2
(order 2)
(c)
Degree
Definition 1.2
The degree of the differential equation is the power (exponent) or the index that its highestordered derivative is raised, if the equation is rationalized or cleared of fractions with regard
to the dependent variable and its derivatives involve in it.
From equation (1.11), squaring both sides results in:
3
2
 d 3 y   dy 3 x 
 3      e  , degree
 dx   dx 


 
2
2
5
 d3y   d2 y 
y
 e x is an ordinary differential equation,
The differential equation  3    2   2
dx
dx
x

1

 

of order three and degree two.
Oral Exercise 2
State the order and degree of equations 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 1.10, 1.11,
1.12, 1.13, 1.14, 1.15 and 1.16
(d)
Linear Differential Equation
A differential equation is linear if it can be put in the form:
an  x  y n  an1  x  y n1  ...  a2  x  y '' a1  x  y ' a0  x  y  f  x  (1.20)
where an is not identically zero and also the subscripted (or indexed) a ’s are functions of
independent variable ( x ) only.
The conditions for a linear differential equation are as follows:
(i)
The dependent variable and all its derivatives occur only in the first
degree (or to the first power).
(ii)
No product of the dependent variable, say y , and/or any of its
(iii)
derivatives present.
No transcendental function (trigonometric, logarithmic or exponential)
of the dependent variable and/or its derivatives occurs.
15
Example 1.3 (linear)
1.
dy
 y  x2
dx
It is linear, it does not matter that the independent variable x is raised to the power
dependent and derivative are not.
2.
2 , the
3x2 y  2ln  x  y  e x y  3x cos x
This is a second order linear ordinary differential equation
Example 1.4 (Non - linear)
1.
4 yy  x3 y  cos y  e2 x
This is not a linear differential equation because of the 4yy and the cos y terms.
Other examples are:
(i)
dy
 y2  0
dx
(ii)
 dy 
 dx   3 y  0
 
2
3
d 3 y  d 2 y  dy
(iii)
  2    e x .
3
dx  dx  dx
Remark: A linear differential equation is always in the first degree of the
dependent variable (variables) and the derivatives.
Oral Exercise 3
For each of the following, state whether the equations are ordinary or partial, linear or nonlinear, and give its order and degree.
1.
3.
d2y
 k 2x  0
2
dx
 2x
2

 y 2 dx  2 xydy  0
16
2.
2
2w
2 w
a
t 2
x2
4.
y ' p  x  y  q  x 
dy
 xy 2  0
dx
d4y
   x
dx4
5.
y '' 3 y ' 2 y  0
6.
7.
 2v  2v  2v


0
x2 y 2 z 2
8.
9.
d2y
 sin  x  y   sin x
dx2
10.
x2 y  yy  0
12.
x  y ''   y '  y  0
14.
dy
 1  xy  y 2
dx
16.
d 2 g
 sin  0
dt 2 l
11.
 x  y  dx  3x2  1 dy  0
2
13.
4
 d 3w 
 dw 
 3   2    xw  0
 dx 
 dx 
15.
y '' 2 y ' 8 y  x2  cos x
3
4
1-1.3 Notation
 4
The expressions y, y, y, y ,..., y
 n
are often used to represent, respectively, the first,
second, third, fourth, ..., n th derivatives of y with respect to the independent variable under
d2y
consideration. Thus, y represents
if the independent variable is x , but represents
dx 2
d2y
 4
n
if the independent variable is p . Observe that parentheses are used in y and y to
2
dp
y 4 and y n respectively. In mechanics, if the independent
variable is time, usually denoted by t , primes are often replaced by dots. Thus, y, y , and y
distinguish it from the n th power,
dy d 2 y
d3y
represent
,
, and
, respectively
dt dt 2
dt 3
Session 2-1 Solution of a Differential Equation
Unlike algebra, in which we seek the unknown numbers that satisfy an equation such as
x3  7 x2  11x  41  0 . In solving a differential equation we are challenged to find the
unknown functions, say y  f  x  , for which an identity such as f   x   2xf  x   0 – in
Leibniz notation or
dy
 2 xy  0 holds on some interval numbers.
dx
17
Ordinarily, we will want to find all solutions of the differential equation if possible. The
solution of differential equations plays an important role in the study of the motions of
heavenly bodies such as planets, moons and artificial satellites. Two questions that we will
be asking repeatedly of a differential equation in this course are:
1. Is there a solution to the differential equation?
2. Is the solution given unique?
2-1.1 Definition 1.3
A solution of a differential equation is any function, say  , satisfying the given differential
equation on a specified interval,
I.
A solution may be defined on the whole real line
 ,  or on only a part of the line often
 a, b . The n derivatives of the function must exist on the interval, say a  x  b
 n
 n 1 x for every x in a  x  b .
such that   x   f  x,  x  ,   x  ,...,
 
an interval
Thus if
interval

is a solution of some first order differential equation, say
dy 

F  x, y,   0 on an
dx 

d
I (real), then it implies F  x, ,   0 .
dx 

Thus to test whether a given function solves a particular differential equation, we substitute
the function  and its derivatives into the differential equation. If the equation reduces to
identity
 0
, then the function

solves the equation otherwise it does not.
It is also important to note that since solutions are often accompanied by intervals and these
intervals can impart some important information about the solution.
Example 1.5
Show that for
any
  c1 cos x  c2 sin x
values
of
the
arbitrary
constant
is a solution of the differential equation
c1 and c2 the function
d2y
 y0
dx2
Solution:
We will need a second derivative of the solution function,  . If

is a solution to
 d2y 
 d 2 
F  y, 2   0 , then F   , 2   0 . If  proves otherwise then it is not a solution to
 dx 
 dx 
 d2y 
d 2
F  y, 2   0 . We differentiate  twice to obtain 2 .
dx
 dx 
18
d
d 2
 c1 sin x  c2 cos x  2  c1 sin x  c2 cos x
dx
dx
d2y
d 2
 2  y  0  2   0
dx
dx
d 2
 2     c1 cos x  c2 sin x    c1 cos x  c2 sin x   0
dx
This implies we have an identity. Hence

is a solution to the given ODE and that the
d2y
formula:   c1 cos x  c2 sin x gives all possible solution of the equation
 y 0.
dx2
Since sin x and cos x are continuous in the entire real line, the solution is defined in the
entire real line  ,   for any arbitrary constant c1 and c2 .
Example 1.6
1.
1
2
is a solution of y  2 xy  0 on    1,1 but not on any
x 1
larger interval containing I .
Show that
y
2
Solution:
We will need a first derivative of the solution function.
y
1
2 x
1,1 .
and y 
2 are well-defined functions on 
2
x 1
x

1
 
2
Imitating the LHS of the differential equation y  2 xy
2
 0 , we have:
2
1
 1 
y

Thus,
is a solution of    1,1 .
y  2 xy  

2
x

0
2
2
 x2  1
2
x

1
 x  1
2x
2
Note, however, that
1
is not defined at x  1 and therefore could not be a solution on
x 1
2
any interval containing either of these two points.
2.
Show that
on I
y  ln x is a solution of xy  y  0 on I   0,  but is not a solution
  ,  .
Solution:
We will need the first and second derivative of the solution function.
y  ln x , y 
1
1
and y   2 are well-defined functions on  0,  .
x
x
19
Imitating the LHS of xy  y  0 ,we have:
y  ln x is a solution on  0,  .
Thus,
Note that
 1 1
x  2    0
 x  x
y  ln x could not be a solution on  ,   , since the logarithm is undefined for
negative numbers containing.
3.
Prove that y  e
x
d2y
 sin x is a solution of 2  y  2e  x
dx
Solution:
We will need a second derivative of solution function to do this.
y  e  x  sin x , y  e  x  cos x and y  e  x  sin x
d2y
x

y

2
e
Imitating the LHS of differential equation
dx 2
e  x  sin x  e  x  sin x  2e  x
4.

Show that the two functions y  c  x
of the equation x 
y
2

yc x

1

1
2 2
,

y   x c  x


2
1
 2
2 2 
x   c  x   x c 2  x 2


5.
Show that y  x
 32

y   c2  x2

1
2
are both solutions
dy
 0,  c  x  c.
dx
Solution:
We will need a first derivative to do this.
2

2 2
and
is a solution of


1
2 2
 12 
  0
4x2 y  12xy  3 y  0 for x  0 .
Solution:
We will need the first and second derivative of the solution function to do this.
3 5
15  7
3
y  x 2 , y   x 2 and y  x 2 are well – defined functions on x  0 or  0,  .
2
4
2
Imitating the LHS of 4x y  12xy  3 y  0 , we have:
20
 15  7 
 3 5 
4x2  x 2   12x   x 2   3x  0
4

 2 
3
Thus, y  x 2 is a solution of x  0 .
Note, however, that
3
x2
1
x3
could not be a solution on
 ,0 , since zero and any
negative real number plug into it would give an undefined number and complex number
respectively, which is not what we are looking for.

Exercise 1.1
In each of assignment 1 – 12, verify by substitution that each given function is a solution of
the given differential equation and also state the interval in which the solution exists.
Solution(s)
Differential Equation
2.
y  2e x  xe x
y 1
y  2 y  y  0
y  2 y  y  x
3.
 y  c 2  cx , c  constant
4 x  y  2 xy  y  0
4.
y  e x  e x
y'  y  2e  x
6.
y1  e 2 x , y2  xe2 x
y1  x , y2  0
y  4 y  4 y  0
y  xy  y  0
7.
y  x  x
dy 2 y 4  x4

dx
xy3
8.
y
1.
5.
9.
10.
8
4

2
1
4
1
1  x2
y'2 xy 2  0
1
 ln x
x
y1  x cosln x , y2  x sinln x
y1  x  ln x , y2 
21
x 2 y  xy  y  ln x
x 2 y  xy  2 y  0
Session 3-1 General Solution
3-1.1 Definition 1.4
A formula that gives all solutions of a differential equation is called the general solution of
the equation. A general solution to an n th order DE generally involves n independent
arbitrary constants, each admitting a range of real values.
NOTE:
From the problems of Exercise 1.1, it is clear that the number of distinct solutions depend on
the order of the differential equation, for which they are equal.
It is worth mentioning that, the differential equation
dy
 4 describes a straight line with a
dx
constant gradient of 4 . There are however infinitely constant many straight lines that take
the equation y  4x  c , where c is an arbitrary constant. Since c  R , y is a constantly
many hence the differential equation
dy
 4 has many solutions, because c can assume
dx
any real number and the differential equation would be satisfied, hence the name arbitrary
constant. The general solution is known as family of solutions.
y  4x  c
3-1.2 Determining a Differential Equation from a General Solution: The
Elimination of Arbitrary Constant
To determine a differential equation from a general solution, we need to eliminate the
arbitrary constant. Methods for the elimination of arbitrary constants vary with the way in
which the constants enter the relation. A method which is efficient for one problem may be
poor for another. One fact persists throughout. Since each differentiation yields a new
relation, number of derivatives that need be used is the same as the number of arbitrary
constants to be eliminated.
We shall in each case determine the differential equation that is:
22
(a)
(b)
(c)
Of order equal to the number of arbitrary constants in the given relation
Consistent with that relation;
Free from arbitrary constants.
Example 1.7
Find the differential equation whose general solution is
y  c1e2 x  c2e3x
Eliminating the arbitrary constants c1 and c2 from the relation:
y  c1e2 x  c2e3 x ______________ (I)
Since two constants are to be eliminated, obtain the two derivatives:
y '  2c1e2 x  3c2e3 x __________ (II)
y ''  4c1e2 x  9c2e3 x ___________ (III)
The elimination of c1 from (II) and (III) yields: y '' 2 y '  15c2e
3x
The elimination of c1 from equations (I) and (II) yields: y ' 2 y  5c2e
3x
Hence
y '' 2 y '  3 y ' 2 y  Or y '' y ' 6 y  0 .
Alternatively, (another) method for obtaining the differential equation in this example
proceeds as follows. We know from a theorem in algebra (MATH 152) that three equations
(I),(II) and (III) considered as equations in the two unknowns c1 and c2 can have solutions
only if:
e2 x
y
e3x
 y ' 2e2 x 3e3x  0 ______________ (IV)
4e2 x 9e3x
 y ''
Since e
2x
factors e
and e
2x
3x
and e
cannot be zero for any x  , equation (IV) may be rewritten, with the
3x
removed, as:
y
1 1
y ' 2 3  0
y '' 4 9
From which the differential equation:
y '' y ' 6 y  0 follows immediately.
Example 1.8
23
1.
Find the differential equation whose general solution is
y  c sin x , where c is an
arbitrary constant.
Solution:
y  c sin x _______________ (I)
There is one arbitrary constant, we then find the first derivative:
dy
 c cos x ______________ (II)
dx
From (I) and (II):
c
y
y

sin x cos x
 y sin x  y cos x
 y sin x  y cos x  0
y
Alternatively, from (I) let: c 
sin x
y sin x  y cos x
Differentiating w.r.t. x : 0 
sin2 x
 y sin x  y cos x  0
2.
Determine the differential equation whose general solution is
y  c1e x  c2e x  2 x .
Solution:
There are two arbitrary constants, so we get equation of order 2. Hence we differentiate
twice.
y  c1e x  c2e x  2 x ……………… (I)
dy
  c1e x  c2ex  2 ………………. (II)
dx
d2y
 2  c1e x  c2e  x ……………… (III)
dx
dy
Add (I) and (II): 
 y  2c1e x  2 x  2 ………. (IV)
dx
d 2 y dy

 2c1e x  2 ………… (V)
Add (II) and (III): 
2
dx
dx
Let
24
The two equations above have the same term on the RHS, hence we equate the LHS, and we
d 2 y dy dy
get:


 y  2x
dx2 dx dx
d2y
This simplifies as:
 y  2x  0
2
dx
3.
Find the differential equation whose general solution is
y  c0  c1x  c2 x 2  c3 x3 .
Solution:
The equation has four arbitrary constants
c0 , c1, c2 , c3  ; hence we need to differentiate
four times to get a differential equation of the fourth order.
y  c0  c1x  c2 x 2  c3 x3
Let
d2y
dy
d3y
 c1  2c2 x  3c3 x2 

2
c

6
c
x

 6c3
2
3
dx
dx2
dx3

4.
d4y
dx4
 0 , which is the differential equation that we need.
Eliminate the constant c from the equation
Solution:
Direct differentiation of the relation yields: 2
 x  c 2  y2  c2
 x  c   2 yy  0 from which
c  x  yy . Therefore, using the original equation, we find that:
 yy2  y2   x  yy2 , or
x
2
y 2  x 2  2 xyy , which may be written in the form:

 y 2 dx  2 xydy  0
Alternatively:
The equation
 x  c 2  y2  c2 may be put in the form:
x2  y 2  2cx  0 or
x2  y 2
 2c then differentiation of both sides leads to:
x


x  2xdx  2 ydy   x2  y 2 dx
x
2


 0 or x2  y 2 dx  2 xydy  0
25
5.
Eliminate
B and  from the relation x  B cos t    in which  is a
parameter (not to be eliminated)
Solution:
First we obtain two derivatives of x with respect to t :
dx
d 2x
  B sin t    and 2   2 B cos t   
dt
dt
2
d x
 2   2 x , since x  B cos t   
dt
d 2x
 2  2x  0
dt
Eliminate c from the equation cxy  c x  4  0
2
6.
Solution:
At once we get:
c  y  xy  c2  0 Since c  0 , c    y  xy and substitution into the
original equation leads us to the result:
2
x3  y  x2 yy  4  0
Remark:
The general solution of an n th order ordinary differential equation could be expected to have
n arbitrary constants.

Exercise 1.2
Assuming that a, b, c, c1, c2 , A,  , B are arbitrary constants, show that each function or
equation on the left satisfies the differential equation written opposite it. Remember that we
differentiate and substitute.
Solution
Differential Equation
1.
x  y c
yy  x  0
2.
y  cx  c4
y  xy   y
3.
y  c1x  c2e x
4.
x3  3 x 2 y  c
5.
y sin x  xy 2  c
2
2
4
 x  1 y  xy  y  0
 x  2 y  dx  xdy  0
y  cos x  y  dx  sin x  2xy  dy  0
26
6.
x 2 y  1  cx
7.
cy 2  x2  y
 x y  1 dx  x dy  0
2 xydx   y  2 x  dy  0
2
3
2
d 2x
 2x  0
2
dt
d 2x
 2x  0
2
dt
8.
x  Asin t    ,  : parameter
9.
x  c1 cost  c2 sin t ,  : parameter
10.
y  cx  c 2  1
11.
y  cx 
12.
y 2  4ax
2xdy  ydx  0
13.
y  c1e px  c2e px ; p R
y  p2 y  0
y  xy   y  1
2

y  xy 
c
1

y
 2u  2u

0
x 2 y 2
y
x
14.
u  tan
15.
u  log x  y
16.
y  xc  cos x 
17.
y  e x 1  x 
y  2 y  y  0
18.
y  ax2  bx  c
y  0
19.
y  c1  c2e2 x
y  2 y  0
20.
y  4  c1e2 x
y  2 y  8
21.
y  c1e x  c2e3 x
y  4 y  3 y  0
22.
y  c1e x  c2 xe x
y  2 y  2 y  0
23.
y  x2  c1e x  c2e2 x
y  y  2 y  2 1  x  x 2
24.
y  c1e2 x cos3x  c2e2 x sin3x
y  4 y  13 y  0
25.
y  c1x 2  c2e2 x

2
2

1
2
uxx  u yy  0
xy  y  x 2 sin x




x 1  x  y  2 x2  1 y  2  2 x  1 y  0
27
Session 4-1 Initial-value and Boundary-value Problems
4-1.1 Definition 1.5: Subsidiary Conditions
Subsidiary Condition(s) is/are condition(s), or set of conditions, on the differential equation
that will allow us to determine which solution that we are after.
4-1.2 Definition 1.6: Initial–Value Problem
Initial-Value Problem is the differential equation along with subsidiary conditions on the
unknown function and its derivatives, all given at the same value of the independent variable.
The subsidiary conditions are initial conditions.
4-1.3 Definition 1.7: Boundary–Value Problem
Boundary-Value Problem is the differential equation along with subsidiary conditions on
the unknown function and its derivatives, which are given at more than one value of the
independent variable. The subsidiary conditions are boundary conditions.
Theorem 1.1
An initial value problem (IVP) for an n th order DE includes a specification of the
solution’s value and n  1 derivatives at some point or a set of n algebraic conditions at a
common point:
yx0   y0 ,
dy
x0   y1,
dx
...,
d n 1 y
x0   yn 1
dxn 1
Generally in applications, an IVP has a unique solution on some interval containing the
initial value point.
Example 1.9
The problem y  2 y  e ; y    1, y    2 is an initial – value problem, because the
x
two subsidiary conditions are both given at x   .
The problem y  2 y  e ; y  0   1, y 1  1 is a boundary – value problem, because the
x
two subsidiary conditions are given at the different values x  0 and x  1 .
If an initial condition is given along with the differential equation, that is, a constraint of the
form y  y0 when x  x0 , then this information can be used to determine the particular
value of c . In this way, one particular solution (actual solution) can be selected from the
family of solutions – the one that satisfies both the differential equation and the initial
conditions.
28
The geometric interpretation of the general solution of a first order differential equation
dy 

F  x, y,   0 is an infinitely many family curves; one for each value of real number.
dx 

These curves are called integral curves, sometimes it is important to pick one particular
member of the family of integral curves. This is done by identifying a particular point
 x0 , y0  through which the graph of the solution is required to pass. This requirement is
usually written as y  x0   y0 .
This is referred to as an initial value condition and the problem of the form:
dy 

F  x, y,   0; y  x0   y0 …………………. (1.21)
dx 

is called initial value problem.
Note: If initial condition(s) is used to solve a differential equation, the solution is
then called particular solution (actual solution).
Example 1.10
1.
y  c1e x  c2e3 x is a general solution of the differential equation:
y  2 y  3 y  0
Determine the particular solution of the initial conditions y  3 when x  0 and
y  4 when x  0 .
Solution:
Since the differential equation is second order, we have a specification of the solution
y(0)  3 and one derivative at a point: y  0  4 , that is 2
conditions at a common point: x  0
y  c1e x  c2e3 x when x  0 , y  3
let
value:
 3  c1e 0  c2e3 0  c1  c2 ______(I)
y  c1e x  3c2e3x when x  0 , y  4 ,
 4  c1e 0  3c2e 0
 4  c1  3c2 _______________
Solving equation (I) and (II) we have, c1 
29
(II)
5
7
and c2 
4
4
algebraic
 the particular solution is given by:
5
7
1
y  e x  e3x or y  5e x  7e3x
4
4
4

2.

Find a solution to the boundary – value problem y  4 y  0;
y  /8  0, y  / 6  1, if the general solution to the differential equation is
y  c1 sin 2 x  c2 cos2 x .
Solution:
For x  8 ,
y  c1 sin2 8   c2 cos2 8   c1 sin 4  c2 cos 4
2
2
 c2
2
2
To satisfy the condition y  /8  0 , we have:
 y  c1
2
2
 c2
 0 ………………….(I);
2
2
For x  6 , y  c1 sin2  6   c2 cos2  6   c1 sin 3  c2 cos 3
c1
3
1
 c2
2
2
To satisfy the condition y  / 6  1 , we have:
 y  c1
c1
3
1
 c2  1 ……………………..(II)
2
2
Solving (I) and (II) simultaneously, we get:
c1 
2
2
and c2  
.
3 1
3 1
Hence a particular solution is obtained:
y
2
sin 2 x  cos2 x 
3 1
as the solution of the boundary – value problem.
3.
Find a solution to the boundary – value problem
y  4 y  0; y  0  1,
y  / 2  2 , if the general solution to the differential equation is known to be
y  c1 sin 2 x  c2 cos2 x .
30
Solution:
For x  0 , y  c1 sin0  c2 cos0  c2
To satisfy the condition
y  0  1 , we have:
c2  1 …………………........................(I).
For x   / 2 ,
y  c1 sin  2   c2 cos  2   c2
y  / 2  2 , we have:
c2  2  c2  2 ………………….(II).
To satisfy the condition
Thus, to satisfy both boundary conditions simultaneously, we must require c2
to equal both 1 and 2 , which is impossible.
Therefore, there is no solution to this problem.

Exercise 1.3
1.
Verify that the following are solution of the given differential equations.
a)
for
y  2 sin 2x
y  4 y  0
d)
y  e3 x
1
y
1 x
y  e x cos x
e)
y  2 x  x ln x
b)
c)
2.
for
y  3 y  0
for
y  y2  0
for
y  y  2 y  0
for
4x2 y  y  0
Determine c1 and c2 that y  c1 sin 2 x  c2 cos2 x  1 will satisfy the conditions
y  /8  0 and y  /8  2 .
3.
Determine c1 and c2 that
y  0  0 and y  0  1.
4.
y  c1e2 x  c2e x  2sin x will satisfy the conditions
In Problems 4 a) through e), find values c1 and c2 so that the given functions will
satisfy the given conditions. Determine whether the given conditions are initial
conditions or boundary conditions:
a)
y  c1ex  c2e x  4sin x ; y  0  1, y  0  1
b)
y  c1x  c2  x2  1;
y 1  1, y 1  2
c) y  c1 sin x  c2 cos x
y  0  1, y  2   1
d) y  c1 sin x  c2 cos x
y  4   0, y  6   1
e)
y  c1e x  c2 xe x  x2e x
y 1  1, y 1  1
31
Example 1.12
Show that the integral curves of the differential equation:
 y  x  dx   y
3
3

 x dy  0 are
given by the family y  4 xy  x  c .
4
4
Solution:
Apply implicit differentiation to the proposed family of integral curves to find y ;
y 4  4 xy  x4  c  4 y3 y '  4xy ' 4 y   4x3  0



 4 y 3 y  4xy  4 y   4x3  0  y y 3  x  y  x3  0




dy 3
y  x  y  x3  0
dx


dy
y  x3
 3
dx
y x

This last equation can then be written in terms of the differentials dx and dy
y3  xdy  y  x3 dx

 

 y  x3 dx  y 3  x dy  0
Notice that the last equation can be put in the form M x, y dx  N x, y dy  0 ,
which is simply an alternate form of the equation:
dy
M x, y 

dx
N x, y 
where
M  x, y   y  x3 and N  x, y   y3  x .

Summary
-
Definition of Differential Equation: A differential equation is an equation
which involves one or more derivatives, or differentials of an unknown
function.
-
Classification of Differential Equation:
o Types
o Order
o Degree
o Linearity
32
-
A differential equation is linear if it can be put in the form:
an  x  y n  an1  x  y n1  ...  a2  x  y '' a1  x  y ' a0  x  y  f  x 
where
an  x  is not identically zero and also the subscripted (or indexed) a ’s are
functions of independent variable ( x ) only.
The conditions for a linear differential equation are as follows:
o The dependent variable and all its derivatives occur only in the first degree
(or to the first power).
o No product of the dependent variable, say y , and/or any of its derivatives
present.
o No transcendental function (trigonometric, logarithmic or exponential) of
the dependent variable and/or its derivatives occurs.
33
UNIT 2
ORDINARY DIFFERENTIAL EQUATION OF FIRST
ORDER
Introduction
In this unit we begin our program of solving differential equations. The first – order
differential equations involve only the first derivative of the unknown function. Our usual
notation for the unknown function will be
y  x .
Learning Objectives
After going through this unit, you would be able to:
• put first order ordinary differential equation in general, standard and differential
forms
• find the difference between all first all the differential equation.
• describe the methods of solving most first order differential equations.
Session 1-2 First Order Differential Equations
First order differential equations are often used to model the behaviour of engineering
systems. For example, in a radioactive decay, mass is converted to energy by radiation. It has
been observed that the rate of change of the mass is proportional to mass itself. That is if
N t  is the mass at time t then:
dN t 
dt
 N t  
dN t 
dt
 kN t  for some
k that depends
on the element. Such an equation is called a Model in mathematics.
1-2.1 Definition 2.1
A first order differential equation is an equation containing only the first differential. In its


general form the 1st order differential equation is given by: F  x, y,
F x, y, y  0
dy 
  0 or
dx 
The above equation has x called the independent variable, the unknown function y which
depends on x and the first derivative of y (i.e. y ) with respect to x .
34
Examples of first-order differential equations
dy
 5x  0
dx
dy
 4 xy  x 3e x
2.
dx
dy
y0
3. x
dx
dy
 4 xy
4. 3
dx
1.
Standard form for a first order – differential equation in the unknown function y is:
y  f  x, y  ……………………….. (2.1)
where the derivative y appears only on the left side of (2.1). Many, but not all, first – order
differential equations can be written in standard form by algebraically solving for y and
then setting
f  x, y  equal to the right side of the resulting equation.
Example 2.1
1.
Write the differential equation
xy  y 2  0 , which is in the general form, in
standard form.
Solution:
Solving for y , we obtain:
2.
y  y 2 / x as standard form with f  x, y   y 2 / x
Write the differential equation
ex y  e2 x y  sin x general form and standard form.
Solution:
Shifting everything to the left of equation we have:
and solving for y , we obtain:
ex y  e2 x y  sin x  0 as general form
y  ex y  e x sin x as standard form, with
f  x, y   e x y  e x sin x .
3.
Write the differential equation  y  y   sin  y / x  in standard form.
5
Solution:
This equation cannot be solved algebraically for y , and cannot be written in standard form.
35
Differential form: The right side of (2.1) can always be written as a quotient of two other
functions
M  x, y  and  N  x, y  . Then (2.1) becomes: y  M  x, y  /  N  x, y  , or
dy M  x, y 
which is equivalent to the differential form:

dx  N  x, y 
M  x, y  dx  N  x, y  dy  0 ……………………….. (2.2)
where we call the functions
M  x, y  and N  x, y  the coefficients of dx and dy
respectively.
NOTE:
We will put most (if not all) first order ordinary differential equation in the form of equation
2.2 before solving.
Example 2.2
Write the differential equation
y  yy  1  x in the differential form.
Solution:
Solving for y , we have: y
with
f  x, y  
2
y  y  x  y 
 x  y  ………..........................
y2
x y
, which is the standard form
y2
(I)
There are infinitely many different differential forms associated with (I), which are as
follows:
b)
Take M  x, y   x  y, N  x, y    y .
2
Then
dy M  x, y 
x y
x y


 2 the equivalent differential form is:
2
dx  N  x, y     y 
y
 x  y  dx    y2  dy  0
c)
y2
.
x y
Take
M  x, y   1, N  x, y  
Then
dy M  x, y 
1
x y

 2
 2 the equivalent differential form is:
dx  N  x, y   y /  x  y 
y
 y2 
 1 dx  
 dy  0
x

y


36
x y
 y2
d) Take M  x, y  
.
, N  x, y  
2
2
dy M  x, y   x  y  / 2 x  y


 2 the equivalent differential form is:
Then
dx  N  x, y     y 2 / 2
y
  y2 
 x y
 2  dx   2  dy  0




x  y
y2
, N  x, y   2 .
e) Take M  x, y  
x2
x
2
dy M  x, y    x  y  / x
x y
Then
the equivalent differential form is:



dx  N  x, y 
 y 2 / x2
y2
 y2 
 x  y 
 x2  dx   x2  dy  0


 

Exercise 2.1
1.
2.
dy y
 in differential form.
dx x
2
Write the differential equation  xy  3 dx  2x  y  1 dy  0 in standard
Write the differential equation


form.
xy  y 2  0 in differential form.
3.
Write the differential equation
4.
Write the differential equation  y   5 y  6   x  y  y  2  in standard
2
form.
 y y 
 x in differential form.
5.
Write the differential equation e
6.
Write the differential equation  y   y  y  sin x in differential form.
3
37
2
1-2.2 General and Particular Solutions of a First Order Ordinary Differential
Equation
In this section, we examine methods of solving a first order differential equation. The
methods or differential equation kinds to consider are as follows;
(I)
Method of separation of variables
(II)
Reduction to Separable Form
Homogeneous equation
Transformations (coefficients linear in the two variables)
(III) Exact Differential Equations
(IV) Method of Integrating – factor used for the following:
a)
Non – exact differential equation
b)
Linear ordinary differential equation of first order and those
reducible to that form:
(i)
Bernoulli’s Equation
(ii)
Riccati’s Equation
Theorem 2.1: An Existence Theorem for Equations of Order One
Consider the equation of order one
dy
 f  x, y  ………………………………… (2.1)
dx
Let T denote the rectangular region defined by: x  x0  a and y  y0  b , where
a, bR , a region with the point  x0 , y0  at its centre. Suppose that f and f / y are
continuous functions of x and y in T .
Under the conditions imposed on
f  x, y  above, there exists an interval about x0 ,
x  x0  h , and a function y  x  which has the properties:
(a)
y  y  x  is a solution of equation (2.1) on the interval x  x0  h ;
(b)
On the interval
(c)
At x  x0 ,
x  x0  h
y  y  x0   y0 ;
,
y  x  satisfies the inequality y  x   y0  b ;
y  x  is unique on the interval x  x0  h in the sense that it is the only
(d)
function that has all of the properties (a) , (b), and (c).
The interval
x  x0  h may or may not need to be smaller than the interval x  x0  a
over which conditions were imposed upon
In other sense, the theorem states that
 x0 , y0  then the differential equation
f  x, y  .
f  x, y  is sufficiently well behaved near the point
38
dy
 f  x, y  …………………………………….(2.1)
dx
has a solution that passes through the point  x0 , y0  and that the solution is unique near
 x0 , y0  .
Session 2-2 Method of Separation of Variables
Some differential equations have a special form that allows us to separate them. Such
equations can be solved by using the method in this section.
A separable differential equation is a first-order ordinary equation that is algebraically
reducible to a standard differential form in which each of the non-zero terms contains exactly
one variable. Solutions to this kind of equation are usually quite straightforward.
2-2.1 Definition 2.2 Separable Differential Equation
A first order differential equation is said to be separable if it can be written in the form:
dy
 f  x, y   G  x  H  y  or equivalent form:
dx
M  x, y  dx  N  x, y  dy  0 ------------- (2.2)
If (2.2) is separable, then:
M  x, y   m  x  p  y  

and

N  x, y   n  x  q  y  

 2.3
Then (2.2) is called separable differential equation;
m x p  y   n  x q  y 
We then divide throughout by

dy
 0 ...……………… (2.4)
dx
p  y  n  x  , we get:
m  x  p  y  dy

0
n  x  q  y  dx
m x 
p y
dx 
dy  0 …………………………….. (2.5),
n x
q y
now it is separated in each variable x and y (hence the method’s name)

39
Integrating both sides:

m x 
p y
dx  
dy  c ……………………….
n x
q y
(2.6)
where c is arbitrary constant. (2.6) is then the general solution of the differential equation.
The evaluated term on the left-hand side of (2.6) is a function F whose total differential is
the left-hand side of (2.5)
ALGORITHM 1
Separation of Variables: To solve the initial-value problem
y  f  x, y ; y  x0   y0 ……………………………… (2.7)
(a) Separate the variables, so that all of the x -dependence is on one side of the
equation, and all of the y -dependence is on the other side of the equation.
(b) Integrate both sides of the equation.
(c) Substitute the initial condition to solve for the constant of integration.
Example 2.3
1.
Solve the following differential equations by separation of variables:
(a)
(b)
dy
 x  y 2  1
dx
dy
x2

dx y 1  x3 
 x  1
Solution:
(a) We change to differential form, separate the variables, and integrate:
 x  1 dy  x  y 2  1 dx
x
1
dx  2 dy  0
x 1
y 1
x
1
Assuming x  1 , integrating both sides: 
dx   2 dy  c we get:
x 1
y 1

x  log x  1  tan1 y  c , as our general solution.
ln
Note:
is the same as
log
40
dy
x2
x2

dx and assuming that x  1
(b) The ODE
becomes: ydy 
dx y 1  x3 
1  x3
, then:
y2 1
x2

 ln 1  x3   c
dx
3
2 3
1 x
2
2
y   ln 1  x3   2c   ln 1  x3   c1
3
3
 ydy  
Since c is arbitrary constant, any number multiplying or dividing it will not have any effect,
it will still be arbitrary constant. We replace 2c with c1 to bring a distinction but no effect in
general, so one can write the general solution as:
2. Solve the equation (IVP)
y
2
ln 1  x3   c
3
1  x2  x1  2 yy  0
y1  0
Solution:
1  x2  x1  2 yy  0

1  x2 
dy
 1  2 y   0
Divide throughout by x 
x
dx
Assuming that x  0 , the general solution is:
2


1  x dx 
x
 1  2 ydy  c
x2
 ln x 
 y  y2  c
2
To solve for the arbitrary c , we substitute x  1, y  0 ;
12
1
 ln1   0  02  c  c 
2
2
x2
1
 ln x   y  y 2  : as particular(actual)solution.
2
2
dy
x2

3. Find a particular solution of
, y  2  1
dx 1  y 2
41
Solution:
The variables are separable, since the equation can be written as:
Integrating both sides:
 x dx   1  y dy  0 , we get:
2
x2dx  1  y2  dy  0
2
x3
y3
 y   c  x3  3 y  y3  3c  c
3
3
Note the replacement of 3c with c
Solving for arbitrary constant, c : 2  3  1  c  c  4
3
Then particular solution is given by: x  3 y  y  4 .
3
3

Exercise 2.2
Solve the differential equations
Ans.
2
x
y2
c
  c  c  18

9
4
1.
9 yy  4x  0
2.
ln x
3.
x cos xdx  1  6 y2  dy  0; y    0
x sin x  cos x  1  y6  y
mydx  nxdy
xy  y  0; y 1  1
xm  cy n
4.
5.
8.
9.
10.
11.
1
2
 ln x 
2
 ln y  c
y  1/ x
cosh 1 y  sinh 1 x  c
1  x2 dy  y 2  1dx  0
6.
7.
dx x

dy y
dy
 ex y
e y  ex  c
dx
x2  1
x2
y
y
xe dy 
dx  0
 y  1 e   ln x  c
y
2
dV
V
PV  C

dP
P
dr
r  r0 exp 2t 2
 4rt ; when t  0, r  r0
dt
xy 2dx  e x dy  0 ; when x  , y  12




y  e x / 2e x  x  1
12.

d
 g ; when x  x0 ,  0
dx
42
 2 02  2g x  x0 
2-2.2 Reduction to Separable Form
Certain differential equations are not separable but can be made separable by the introduction
of a new unknown function. We illustrate the idea of this method by some typical cases, i.e.
Homogeneous forms and Non-Homogeneous forms, where the coefficients are linear in
the two variables, transform to Homogeneous form.
a)
Homogeneous Functions
Polynomials in which all terms are of the same degree, such as:
x2  3xy  4 y 2 ,
x3  y 3 ,
x 4 y  7 y5.
are called homogeneous polynomials. We wish now to extend that concept of homogeneity
so it will apply to functions other than polynomials.
A formal definition of homogeneity is: A function of two variables, f x, y , is said to be
homogeneous of degree n if there is constant n such that:
f x, y   n f x, y  …………………………………… (2.8)
for all  , x and y for which both sides are defined.
Example 2.4
1.
f x, y   x 4  x3 y
f x, y   x 4  x 3 y   4 x 4  4 x3 y

 4 x 4  x3 y

f x, y   4 f x, y 
Thus the equation of is homogeneous with degree 4.
2.
f x, y   e
y
x
 f x, y   e
e
y
y
x
y
 tan x
x
y
 tan x
y
 tan x
 f x, y   0 f x, y 
This function is homogeneous of degree zero.
43
3.
f x, y   x 2  sin x cos y
 f x, y   x 2  sin x cos y
 2 x 2  sin x cos y
1


 f x, y   2  x 2  2 sin x cos y 



 f x, y   2 f x, y 
This function is not homogenous.
4.
2x  y
f x, y  
f x, y  
x2 y2
2x  y 
x2 y 2

2x  y
2 x 22 y 2

 2x  y 
4 x 2 y 2
 2x  y 
 f x, y    3  2 2 
x y 
 f x, y   3 f x, y 
This function is homogeneous of degree  3 .
Equations with homogeneous coefficients
A differential equation of the form (differential form):
M x, y dx  N x, y dy  0 …………………….. (2.9)
is homogeneous if the functions
M  x, y  and N  x, y  are both homogeneous functions of
the same degree. Or if the equation in standard form:
y  f  x, y  ………………………….. (2.10)
depends only on the ratio
y
x
or , then it is said to be homogenous.
x
y
To solve homogeneous equations, turn them into separable ones using the substitution:
y  x …………………………. (2.11),
where  is a function of x ; also:
dy
d
  x
………………..
dx
dx
or
dy  xd  dx
44
(2.12)
Example 2.5
1.


Solve the differential equation x 2  y 2 dx  2xydy  0
Solution:
Notice that the function M x, y   x 2  y 2 and N x, y   2 xy are both homogeneous of
degree 2; therefore, the differential equation is homogeneous.
The substitution y  x implies that dy  xd  dx , and given equation is transformed
into:
 x2   xv 2  dx  2 x  x  xd  vdx   0


After little algebra, this becomes the separable equation:
1
2
dx 
d
x
1  2
Integrating both sides gives: log x   log 1   2  log c , in which we wrote the arbitrary
constant of integration as log c because it makes the next step neater:
log x  log
c
1  2
Finally, to write the solution in terms of the original variables, x and y , we replace  by
y x:
x
2.
c
1   y / x
2
 x 2  y 2  cx
Solve the differential equation
xdy  ydx  x 2  y 2 dx  0
Solution:
The differential equation is clearly homogeneous degree 1. Using the transformation y  x
,  dy  dx  xd and dividing by
x , we have:
dx  xd  dx  1   2 dx  0
d
 xd  1   2 dx  0 
Integrating both sides:  
d
1
2

1  2

dx
0
x
dx
x
1
We get: sin   ln x  ln c
And returning to the original variables, using 
45
 y/x
We get:
3.
sin 1 x  ln xc  xc  exp sin 1  y / x 
y
Solve the differential equation xy   y  x   y .
3
Solution:
3
y
 y
2 y
Express the differential equation in form of : y  x   1 
x
x  x
y
Substitute   and y    x by equation (2.11) and (2.12) on page 34, we get:
x
dv
  x  x2   1    x  x 2   1
dx
1
x2

dv   xdx  log   1   c
 1
2
y
Returning back to the original variables:   , we get:
x
 x2

y
  1  exp   c 
x
 2


 x2

y  x 1  exp   c 
2



Exercise 2.3
Solve the following differential equations (IVP)
1.
xy
dy
 x 2  y 2 ; y 1  2 ;
dx
Ans.
y 2  x2 ln x2  4x2 or  x 2 ln x 2  4 x 2
Note the negative square root is taken to be consistent with the initial condition
xy  y  x
2.
y  x ln xc

3.
xy  y  x 2 sec x ;
y1  
4.
xy  y  3x 4 cos2  y / x;
y1  0
5.
xyy  2 y 2  4x 2 ;
y2  4
y
46
A Transformation (Equations in which
M  x, y  and N  x, y  are linear but not
homogeneous)
Another type of equation that can be reduced to a more basic type by means of a suitable
transformation is an equation of the form:
a1x  b1 y  c1 dx  a2 x  b2 y  c2 dy  0 ……….. (2.13)
where a1 , b1 , c1 , a2 , b2 , c2 are constants where
M x, y   a1x  b1 y  c1 and N  x, y   a2 x  b2 y  c2
written as M x, y   N x, y y  0 , which are not homogeneous.
However the differential equation can be made homogeneous and solved accordingly
provided;
Case I
a2 b2
 , then the transformation: x  X  h and y  Y  k , where h, k  is
a1 b1
the solution of the system a1 x  b1 y  c1  0 and a2 x  b2 y  c2  0 , that is
x  h and y  k , which reduces (2.13) to a homogeneous equation:
If
 a X  b Y d X   a X  b Y dy  0 …..(2.14)
1
1
2
2
in the variables x and Y .
Case II
a2 b2
  k , then the transformation z  a1x  b1 y reduces the equation
a1 b1
(2.13) to a separable equation in the variables x and z :
If
dz
dy
 a1  b1 …………………………
dx
dx
(2.15)
Example 2.6
1.
Solve the differential equation:
x  2 y  1dx  4 x  3 y  5dy  0 …………
(I)
Solution:
Here a1
 1, b1  2, a2  4, b2  3 and so
case I transformation is applied.
47
a2
4
a1
and
b2 3 a2 b2
 
 ,
b1 2
a1 b1
Hence, we then solve:
x  2 y  1  0, 4x  3y  5  0 simultaneously to obtain:
x  3, y  2 .
This implies the solution of the system is
h  3, k  2 , and so the transformation is:
x  X  3 and y  Y  2
This reduces the (I) to the homogeneous equation:
 X  2Y dX   4X  3Y dy  0 …………… (I)
dy 1  2 


d x 3   4
y
x
y
x
Let Y  X to obtain:
d 1  2
3  4 d  d x ….. (II)


d x 3  4
1  2  3 2 x
 3  4 d  d x
Integrating both side:  
1  2  3 2  x
 x

3  4 d  15 d  1 d  d x
 4 1   
x
1  3 1     4 1  3 
5
1
  ln 1  3  ln 1    ln x  ln c1
4
4
 5ln 1  3  ln 1    4ln x  4lnc1

1
4 4
5  x c1
1  3 
1    c 1  3  x4 , where c  c14
5
These are the solutions of the separable equation (II).
y
Now, substituting  
, we obtain the solutions of the homogeneous equation (I) in the
x

form: x  y  c x  3y
Finally, replacing X by

5
x  3 and y by y  2 , from the original transformation, we obtain
the solution of the differential equation in the form:
x  3   y  2  cx  3  3 y  25  x  y  1  cx  3 y  95
2.
If
x  2 y  3dx  2 x  4 y  1dy  0 , determine the general solution.
48
Solution:
Here a1
 1 , b1  2 , a2  2 , and b2  4 
a2 b2
  2,
a1 b1
therefore it follows case (II)
So, let z  x  2 y ,  z  1  2 y
The transformations give us:
z  3dx  2 z  1 dz  dx  0
2
 7dx  2 z  1dz  0
This is separable.

7dx 
Now, replacing

2 z  1dz  c  7 x  z 2z  c
z by x  2 y , we obtain the solutions as: 7 x  x  2 y 2  x  2 y   c

Exercise 2.4
Solve each differential equation by making a suitable transformation.
1.
2.
3.
4.
5x  2 y  1dx  2 x  y  1dy  0
3x  y  1dx  6 x  2 y  3dy  0
x  2 y  3dx  2 x  y  1dy  0
10x  4 y  12dx  x  5 y  3dy  0
Solve the following initial-value problems
5.
6.
7.
8.
6x  4 y  1dx  4x  2 y  2dy  0; y12   3
3x  y  5dx  x  y  2dy  0; y2  2
2 x  3 y  1dx  4 x  6 y  1dy  0; y 2  2
4 x  3 y  1dx  x  y  1dy  0; y3  4
49
2-2.3 Exact Equation
Given a function
f  x, y  , its total differentials, df , is defined as:
f
f
dx  dy ---------------------(٠)
x
y
This shows that the family of curves (or general solution) f x, y   c satisfies the
differential equation df  0 .
f
f
dx  dy  0 ----------------------(٠٠)
x
y
So if there exists a function f x, y  such that:
df 
M  x, y  
f
f
and N x, y  
y
x
then M x, y dx  N x, y dy is called an exact differential, and the equation:
M x, y dx  N x, y dy  0 ……………………….(2.2)
is said to be an exact equation, whose solution is the family f x, y   c .
How can you tell when a differential equation is exact? Recall that in partial derivatives if
f x, y  continuous second partial derivatives, then:
2 f 2 f
(that is, the order of

xy yx
differentiation does not matter).
So if
f
M N
f
2 f 2 f

 N , then:
 M and
(which is
)

x
y
x
y
xy yx
Not only is this condition necessary for exactness, it also determines exactness.
Example 2.6
1.


Solve the differential equation 1  2 xy dx  4 y 3  x 2 dy  0 .
Solution:
With M x, y   1  2 xy and N x, y   4 y 3  x 2 , we notice that:
M
N
 2 x 
y
x
So the given functions M and N pass the test for exactness. All that is left to do is to find
the function f x, y  such that:
f
f
 N.
 M and
x
y
50
The way to accomplish this is to integrate M with respect to x , integrate N with respect to
y , and then merge the results:
 M x, y dx   1  2xydx  x  x y  (a function of y alone)
3
2
4
2
 N x, y dy   4 y  x dy  y  x y  (a function of x alone)
These calculations imply that a function f x, y  which satisfies both:
2
f
f
 N  4 y 3  x 2 is: f x, y   x  x 2 y  y 4
 M  1 2 xy and
x
y
Therefore, the exact equation we were given is satisfied by the family of curves:
x  x2 y  y 4  c

 
Solve the differentiation equation y cos x  2 xe y  sin x  x 2e y  2
2.
Solution:
0
 dy
dx
M  x, y   y cos x  2 xe , N  x, y   sin x  x e  2
y
M
y
 cos x  2 xe ,
N
2 y
 cos x  2 xe
y
y
x
Therefore, the differentiation equation is exact.
We know that
f
f
 N x, y  . Thus there is an f  x, y  , such that:
 M  x, y  ;
x
y
f
x
y
 y cos x  2 xe , and
f
y
Integrating the first of these equations:  df
 sin x  x e  2 ….(I)
2 y
  y cos x  2xe y  dx , keeping y constant
gives: f  x, y   y sin x  x e    y  …………….(II)
2 y
and hence differentiating the new function w.r.t
f
y
 sin x  x e
2 y
d
dy
y
keeping
x
constant:
 sin x  x e  2 ,
2 y
i.e. the differential is equal to the second equation in (I).
d
 2   d   2dy and   2y the constant of integration can be omitted; we do
dy
not require the most general one.
Thus
Substituting for   y  in (II) gives: f  x, y   y sin x  x e  2 y
2 y
51
Before f x, y   c ,constant. Hence
3.
c  y sin x  x 2e y  2 y
Solve the differential equation
3x2  2xy x  y2 dy
0
dx
Solution:
Here
M x  2 x , N x  1 ; since M y  N x , the given equation is not exact. To see that it
cannot be solved by the procedure above, let us suppose that there is a function
f  x, y 
such that:
f x  3x 2  2 xy , f y  x  y 2 ………….(I)
Integrating the first of (I) gives:
f ( x, y)  x3  x 2 y    y ………………(II)
where   y  is an arbitrary function of y only. To try to satisfy the second of (I) we
compute f y from (II), obtaining:
f y  x2 
d
 x  y2
dy
d
 x  y 2  x 2 ……………(III)
dy
Since the right – hand side of (III) depends on x as well as y , it is impossible to solve (III)
for   y  . Thus there is no f  x, y  satisfying both of (I)
or
So in the next sub – topic we try to find a factor to make the equation exact and have a
function satisfying both (I) – the factor is called integrating factor.
Session 3-2 Integrating factors
3-2.1 Non- Exact Equations
If M x, y dx  N x, y dy  0 .................. (2.2) is not exact, sometimes we can turn it into an
exact differential equation by multiplying the whole equation by an appropriate factor, called
an integrating factor, let’s say    x, y  , which is a function of x, y . For which we
obtain:
M x, y dx  N x, y dy  0 ---------------- (2.13)
and expect it to be exact. Then for (2.13) to be exact then:
52
M  N 

y
x
Applying product rule we get:

M

N

M

N
y
y
x
x
 N M 


 ------------- (2.14)
N
  

y
x

x

y


M y  N x   N x  M y
M

Simply put:

 N x  M  y    M y  N x 
Remark: Finding the integrating factor for a given equation can be very difficult.
Some of the important rules and procedures follow.
Theorem 2.1
a)

d  M y  Nx
  y  0 ,then:  x 

, which is a
y
dx
N
function of x alone, call this function  x  .
Assume
   x  
 x  e   x dx is an integrating factor of (2.2).

d  M y  Nx
 0 , then:  y 
Assume     y  
, which is a

x
dy
 M 
function of y alone, call this function   y  .
Then
b)
Then   y   e 
c)
If
  y dy
is an integrating factor of (2.2).
1
M N

 0 , then:   x, y  
, is the integrating factor of
M N
x y

x y
(2.2).
d)
If M x, y dx  N x, y dy  0 can be written in the form
yg  x, y  dx  xh  x, y  dy  0 , where f  x, y   g  x, y  , then
  x, y  
1
xM  yN
53
The following observations are often helpful in finding integrating factors:
(1) If a first-order differential equation contains the combination
1
xdx  ydy  d  x2  y 2  try some function x2  y 2 as a multiplier.
2
(2) If a first-order differential equation contains the combination
xdy  ydx  d  xy  , try
some function xy as a multiplier.
(3) If a first-order differential equation contains the combination
xdy  ydx , try
1
1
or
x2 y 2
1
1
or 2
or some function of these
xy x  y 2
xdy  ydx
1  y 
expressions, as an integrating factor, remember that d tan   
and
xy
 x
 y  xdy  ydx
.
tan 1    2
x  y2
 x
as a multiplier. If neither of these works, try
Sometimes an integrating factor may be found by inspection, after grouping the terms in the
equation, and recognizing a certain group as being a part of an exact differential. The table
below has some of the forms already listed above:
54
Table 2.1
Group of Terms Integrating factor
ydx  xdy

ydx  xdy
Exact differential
xdy  ydx
 y
d 
2
x
 x
1
x2
x
ydx  xdy
d 
2
y
 y
xdy  ydx
 y
 d  ln 
xy
 x
xdy  ydx
y


d
arctan

x2  y 2
x 

ydx  xdy
 d  ln xy 
xy
1
y2
ydx  xdy
ydx  xdy
ydx  xdy

1
xy

1
x  y2
2
1
xy
1
,n 1
 xy n
ydx  xdy
ydy  xdx


ydx  xdy
1

d

n1 
 xy n
  n  1 xy  
ydy  xdx
1

 d  ln  x2  y 2 
2
2
x y
2

1
x  y2
2
ydy  xdx
x
aydx  bxdy
 a, b constant 
1
2
df  x, y 
y

2 n
,n 1
xa1 yb1
ydy  xdx
x
2
 y2 
n


1


d
 2  n  1  x 2  y 2 n1 


xa1 yb1  aydx  bxdy   d  xa yb 
NOTE: If a non – exact equation has a solution, then an integrating factor is
guaranteed to exist, but that does not mean it is easy to find.
Example 2.7
1.
Solve the differential equation
Solution:
3xy  y2  x2  xy dy
 0.
dx
M ( x, y)  3xy  y 2 , N x, y   x 2  xy
55
M y  3x  2 y
and N x
 2x  y .
M N

, therefore the differential equation is not exact.
y
x
Involve (2.14): M y  N x   N x  M y
Clearly,


Let  be a function of x alone, that is,
Applying theorem 2.1a:
 x  
M y  Nx
N
   x  ,
3x  2 y  2 x  y  1

x
x 2  xy

 x  e   x dx  e  x
1 dx
Multiplying
 eln x  x
 x   x through the differential equation we obtain:
dy
3x 2 y  xy2  x3  x 2 y
0
dx

 

which is a new equation in same form i.e.: M x, y   N x, y y  0


M x, y   3x 2 y  xy 2 , N ( x, y)  x3  x 2 y
M y  3x 2  2 xy , N x  3x 2  2 xy ,which is exact.
f x  3x 2 y  xy 2
Let
Integrating:
f x, y  

3x 2 y  xy 2 dx  x3 y  12 x 2 y 2    y 

Find the derivative w.r.t
y and equating with N  x, y  we obtain:
 f y  x3  x 2 y    y   x3  x 2 y
   y   0 ,
x 3 y  12 x 2 y 2  c .
Finally
2.
Solve
Solution:
 y  c
y
2
 y  dx  xdy  0 .
M  x, y   y 2  y , N  x, y   x
M y  2 y 1
Nx  1
56
Hence the differential equation is not exact, and no integrating factor is immediately
apparent. Note, however, that if terms are strategically regrouped, the differential equation
can be rewritten as:
  ydx  xdy   ydx  0
(I)
The group of terms in parenthesis has many integrating factors in table (2.1). Trying each
integrating factor separately, we find that the only one that makes the entire equation exact is:
  x, y  
1
.
y2
Using this integrating factor we can rewrite (I) as:

ydx  xdy
 1dx  0
y2
(II)
Since (II) is exact, it can now be solved. Alternatively, we note from table 2.1 that (II) can be
rewritten as
x
d  x / y   1dx  0 , or as d    1dx .
 y
Integrating, we obtain the solution:
x
x
 x  c or y 
xc
y

Exercise 2.5
Solve the following differential equation
Ans.
1.
2xy  1  x  y  0
x y y c
2.
y  xy  0
y  cx
3.
 x  sin y    x cos y  2 y  y  0
4.
y  xy  0
1 2
x  x sin y  y 2  c
2
xy  c
5.
2
2
3x2 y2   2x3 y  4 y3  y  0
6.
x
7.
ydx  1  x  dy  0
8.
2xy  y 2dy  0
y 2  2  c  x2 
9.
1  2 xyy  0
y 2  ln cx
10.
x
1
14  1 
y  x2   
3
3  x
2
2
 y  y2  dx  xdy  0
x3 y 2  y 4  c
y  x tan  x  c 
cy  x 1
 y  dx  xdy  0; y 1  5
57
3-2.2 Linear Equations and those Reducible to that form
A first – order linear equation is defined as a differential of the form:
A( x) y  Bx y  Cx , ……………………….. (2.15)
where Ax   0 such that the derivative of the dependent variable exist otherwise it is not a
differential equation.
Then (2.14) in the standard form is written as
y  px y  qx  ……………………………… (2.16)
p x  
To divide by
B x 
C x 
and qx  
Ax 
Ax 
Ax  would violate the fundamental commandment of mathematics, if
Ax   0 , which prohibits division by zero.
We recognise (2.16) as an inexact differential equation which requires an integrating factor
 x  , that is:
d x 
0
dy
 x  to obtain the following:
 xyx    pxyx  qx  0 ……………… (2.17)
M x, y    x  px y  qx  and N x, y    x
Multiplying through (2.16) by
M N

.
y
x

py  q    x  d x
y
x
dx
d
 p 
…………. (2.18)
dx
If (2.17) is exact, then
Thus,
Separating variables:

d

  pdx  ln    pdx
  e  pdx …………………….. (2.19)
Without loss of operating the constant of integration of (2.19) is unity.
Consider (2.16) and multiply through by:
 x  e  px dx , then we have the exact equation:
y  py  q
y   y  q by (2.18) that is p   
d
p   p
dx
58
Integrating both sides:
y   qdx  c
 y   1  qdx   1c
 p x dx  p x dx
 p x dx
 ye 
qxdx  ce 
…… (2.20)
e
The solution is in two parts in the R.H.S. The term
ce   p x dx   1c , called the
complementary function is the solution of the reduced equation: y  px  y  0 obtained
by setting qx   0 in (2.16).
The differential equation which has qx   0 is called Homogeneous differential equation.
The other term on the R.H.S of the solution is dependent on qx  , that  1  qdx is called
the particular integral and it satisfies (with the complementary function) the differential
equation: y  px  y  qx  , called the Non – Homogeneous differential equation. Notice
that the particular integral contains no constant.
ALGORITHM
The solution process for a first order linear differential equation is as follows:
1.
Put the differential equation in the correct initial form (2.16).
  x  , using (2.19).
2.
Find the integrating factor,
3.
Multiply everything in the differential equation by
becomes the product rule
4.
5.
  x  and verify that the left side
  y  and write it as such.
Integrate both sides; make sure you properly deal with the constant of integration.
Solve for the solution y .
Let’s work a couple of examples.
Example 2.8
1.
Find the solution of
dy
3y
 5x 
such that y1  2
dx
x
Solution:
First, we rearrange the equation to put tin the standard form (2.16) of a linear equation:
dy 3
 y  5x
dx x
Multiply both sides of this by the integrating factor:
 x  e  px dx  e 
To give the equivalent equation: x 3
3x dx  e3 log x  x3
dy
 3x 2 y  5 x 4
dx
59
 
d 3
x y , so by integrating both sides, we get:
dx
Notice that the left – hand side is
 
x 3 y   5x 4 dx  x 5  c  y  x5  cx 3
Now, to satisfy the initial condition, y1  2 , the value of c must be 1. Therefore, the
solution to the IVP is: y  x 5  x 3 .
2.
Solve the differential equation
xy  2 y  4 x 2
Solution:
y  2x y  4 x …………………………..(*)
Integrating factor,
 x   e  x
2 dx
2
Multiply (*) through by x :
2
 e 2 ln x  eln x  x 2 .
 

x 2 y  2 xy  4 x3  x 2 y  4x3  x 2 y 

4 x3dx  x 4  c
3-2.3 Equations Reducible to Linear Form
Other first order equations in first degree which are not linear may be reduced to the linear
form by means of appropriate transformations. No general rule can be stated; in each
instance, the proper transformation is suggested by the form of the equation. Examples of
such equations are Bernoulli’s equation, Ricatti’s equation and the one considered below.
Consider
xs y   t  y y    y  ……………………………. (2.21)
in which the coefficient of y  is a linear function of x and No other x appears anywhere
again.
Thus (2.21) can be expressed as a linear differential equation with x as the dependent
variable and y the independent variable. To be able to do this, multiply (30) through by
to obtain:   y 
dx
dy
dx
dx
 xs y   t  y  or   y   s y x  t  y 
dy
dy
Thus this is a linear differential equation of x as a function of the independent variable y .
60
Example 2.9


Solve the differential equation 3x  4 y 3 y  y  0 .
Solution:
Clearly the coefficient of y  is linear in x and further more x appears nowhere again.
dx
to obtain:
dy
dx
dx
3x  4 y 3  y  0  y  3 x  4 y 3
dy
dy
dx 3
 x  4y2
Whose standard form is :
dy y
Therefore we multiply through by


Integrating factor:   y   e
dy
3 y
 d xy
3
3
 e ln y  y 3  y 3
dx
 3y 2 x  4 y5
dy
   4 y5  xy3  4 / 6y 6  c
 3xy 3  2 y 6  c
BERNOULLI’S EQUATION
A Bernoulli equation is a first order differential of the form:
y  pxy  qxy n …………… (2.22)
in which n is any real number. You would notice that the equation differs from a
standardized first order linear differential equation in first degree at the right side of the
equation. Here there is a multiplication of y n , n  0, n  1 , at the right hand side(RHS)
Then divide (2.22) through by y n and substitute
z  y1 n .
yy n  pxy1 n  qx
Let
z  y1 n  z  1  ny  n y 
The equation becomes:
z
 y n y
1 n
z  1  n  px z  1  n qx 
The solution is obtained using first order linear equation in first degree solution techniques.
Example 2.10
Solve the differential equation
x 2 y  x3
dy
 y 4 cos x .
dx
61
Solution:
Write it in the form y  pxy  qxy n
y 
i.e.
Divide through by
Let
z  y 3
We get:
y
y
cos x
  3 y4
x
x
3
:
y
4
(I)
y 3
cos x
y 
 3
x
x
z  3 y 4 y , and substitute into (I)
3
3 cos x
, which is linear first order in first degree.
z  z 
x
x3
and
  x   e  p x dx
Therefore,
But
….
zy
z  x3 
3

e
3 cos x 3
x dx 
x3
 3x dx

 e3 ln x  x3
3 cos xdx  zx3  sin x  c
x3
 3  3 sin x  c
y
x3
y 
3 sin x  c
3
Hence,

Exercise 2.6
Solve the following differential equation.
1.
2.
3.
4.
5.
xy  3 y  6 x3
Ans
y  x3  x3c
1
1
y  ce x  sin x  cos x
y  y  sin x
2
2
dQ
2
4t 2  40t  1304

Q  4; Q  2   100 Q  t  
t  2
dt 10  2t
2t  10
1
y x
y  y  y 2
ce  1
2
1
31 8
y  y   x9 y 5 ; y  1  2


x  2 x10
4
x
y
16
62
RICCATI’S EQUATION
The equation
dy
 Px   Qx  y  Rx  y 2 ………… (2.23) is non – linear, and needs a
dx
particular solution, before one can have the chance to solve it.
Consider the new function z defined by: z 
1
, where y1 is the particular solution.
y  y1
dz
 Qx   2 y1Rx z  Rx  …… (2.24)
dx
Which is a linear equation satisfied by new function z . Once it is solved, we go back to via
1
the relation: y  y1  .
z
Then easy calculations give:
Example 2.11
Solve the equation
dy
 2  y  y 2 where y1  2 is a particular solution.
dx
Solution:
First we verify that y1  2 is a solution, otherwise our calculation is fruitless.
1
1
 y  2
z
y2
z
y   2
z
Hence, from the equation satisfied by y , we get:
Let
z
z 3 1
1 
1

 2  2   2     2     2   2
z z
z 
z

z
z
z
2
Hence z   3z  1 .
This is a linear equation. The general solution is given by:
1
 1

z  e 3x   e3x  c     ce3x
3
 3

1
Therefore, we have: y  2 
 13  ce  3x
63
Note:
If one remembers the equation satisfied by z , then the linear equation satisfied by the
dz
 Qx   2 y1Rx z  Rx    1  4z  1  3z  1 , since
dz
Px   2, Qx   1, and Rx   1.
functions z is:

Exercise 2.7
dy 2 cos2 x  sin 2 x  y 2
and solve the

dx
2 cosx
differential equation using the initial condition y0  1
1. Verify that y1  sin x is a solution to
2. Solve the following;
i)
y  1  x y 2  2 x  1y  x ; given solution y  1
ii)
y   y 2  xy  1; given solution y  x
iii)
y  8xy 2  4 x4 x  1 y  8x3  4 x 2  1 ; given soln. y  x

64

Session 4-2 Applications of First-Order Differential Equation
(Modeling)
We now move into one of the main applications of differential equations. Modeling is the
process of writing a differential equation to describe a physical situation. Almost all of the
differential equations that you will use in your job (as an engineer) are there because
somebody, at some time, modeled a situation to come up with the differential equation that
you are using.
This session is not intended to teach you how to go about modeling physical situations. A
whole course could be devoted to the subject of modeling and still not cover everything! This
session is designed to introduce you to the process of modeling and show you what is
involved in modeling.
In all of these situations we will be forced to make assumptions that do not accurately
depict reality in most cases, but without them the problems would be very difficult and
beyond the scope of this discussion (and the course in most cases to be honest).
We will look at different situations in this session: Law of exponential change – that deals
with growth and decay problems – Temperature problems, Falling Bodies, Mixing
Problems and Electrical Circuits. But we will part particular attention on electrical
problems.
4-2.7 Electrical Circuits
Fig. 2.2
The diagram in Figure 2.2 represents an electrical circuit whose total resistance is a constant
R ohms and whose self-inductance, show as a coil, is L henries, also a constant. There is a
switch whose terminals at a and b can be closed to connect a constant electrical source (may
be emf) of E (or V ) volts.
Ohm’s law, E  IR or V  IR  , has to be modified for such a circuit. The modified form is:
L
dI
dt
 RI  E …………………..(2.38)
65
where I is the intensity of the current in amperes and t is the time in seconds. By solving
this equation, we can predict how the current will flow after the switch is closed.
For an RC circuit consisting of a resistance, a capacitance C (in farads), an emf and no
inductance (Fig 2.2) the equation governing the amount of electrical charge q (in coulombs)
on the capacitor is:
dq
1

dt RC
The relationship between q and
I
dq
dt
q
I
E
R
is:
………………..(2.39)
…………………………(2.40)
Example
1.
The switch in the RL circuit in (Fig 2.2) is closed at time t  0 . How will the
current flow as a function of time?
Solution:
Equation (2.38) is a liner first-order differential equation for
standard form is:
dI

R
I
I
as a function of t . Its
E
……………………(2.41)
dt L
L
and the corresponding solution, given that I  0 when t  0 , is:
E E  R/ Lt
I  e
……………………(2.42)
R R
Since
R and L are positive,   R / L  is negative and e R / Lt  0 as t  .
Thus, lim I
t 
E
E E
 E E
 lim   e R/ Lt     0 
t   R
R
R
 R R
At any given time, the current is theoretically less than E / R , but as time passes, the
current approaches the steady-state value E / R . According to the equation:
dI
 RI  E ,
dt
I  E / R is the current that will flow in the circuit if either L  0 (no inductance) or
dI
 0 (steady current, I  constant).
dt
L
Equation (2.42) expresses the solution of equation (2.41) as the sum of two terms: a
 R / Lt
steady-state solution V / R and a transient solution  V / R e
that tends to

zero as t  .
66

An RL circuit has emf of 5 volts, a resistance of 50 ohms, an inductance of 1
henry, and no initial current. Find the current in the circuit at any time t
2.
Solution:
Here
E  5, R  50 , and L  1, hence (2.41) becomes:
The corresponding general solution is: I  ce
50t
dI
dt
 50 I  5
 101
Using the initial condition I  0 when t  0 , we get
I   101 e50t  101 ………………….(I)
as a particular solution.
1
e
The quantity  10
50t
in (I) is called the transient current, since this quantity goes to
zero (“dies out”) as t  . The quantity
As t  , the current
I
1
10
in (I) is called the steady – state current.
approaches the value of the steady – state current.
The RL circuit has an emf given (in volts) by 3sint , a resistance of 10 ohms, an
inductance of 0.5 henry, and an initial current of 6 amperes. Find the current in
the circuit at any time t
3.
Solution:
Here, E  3sin 2t , R  10 , and L  0.5 ; hence (2.41) becomes:
General solution is: I  ce
At t  0 , I  6 ;
20t
dt
 20I  6sin 2t
30
3
 101
sin 2t  101
cos2t
30
3
6  ce200  101
sin2 0  101
cos2  0  c 
The current at any time t is: I 
The steady state current:
4.
dI
609
101
609
101
30
3
e20t  101
sin 2t  101
cos2t
30
3
I  101
sin 2t  101
cos2t
The RC circuit has an emf given (in volts) by 400sint , a resistance of 100
2
ohms, an capacitance of 10 farad. Initially there is no charge on the capacitor.
Find the current in the circuit at any time t
Solution:
We first fine the charge q and then use (2.40) to obtain the current. Here,
E  400cos 2t , R  100 , and C  102 ; hence (2.39) becomes:
67
dq
 q  4cos 2t
dt
This equation is linear, and its solution is (two integrations by parts are required)
q  cet  85 sin 2t  54 cos2t .
When there is no charge, that means, q  0 , so at t  0, q  0 ; hence
0  ce
 0
 85 sin 2  0  54 cos  0 
 c   54
Thus, q   54 e
t
 85 sin 2t  54 cos 2t and using (2.40), we obtain:
I

dq
dt
t
 54 e  165 sin 2t  85 cos 2t
Summary
-
General first – order differential equation: F x, y, y  0
-
Standard form:
-
Differential form:
-
First – order differential equation is said to be separable, if the differential
form have
y  f  x, y 
M  x, y  dx  N  x, y  dy  0
M  x, y   m  x  p  y  , which the product of a function x and
a function y and also
N  x, y   n  x  q  y  , also a function of x and a
function, then we can write the differential form as:
m x  p  y  dx  n  x  q  y  dy  0
We then divide throughout by
p  y  n  x  , we get:
m x 
p y
dx 
dy  0
n x
q y
now it is separated in each variable x and y .

Integrating both sides:

m x 
p y
dx  
dy  c
n x
q y
where
-
c
is arbitrary constant.
Separation of Variables. To solve the initial – value problem
y  f  x, y ; y  x0   y0
68
(a)
Separate the variables, so that all of the x -dependence is on one side of the
equation, and all of the y -dependence is on the other side of the equation.
(b)
(c)
Integrate both sides of the equation.
Substitute the initial condition to solve for the constant of integration
-
Reduction to Separable
o A formal definition of homogeneity is: A function of two variables, f x, y 
, is said to be homogeneous of degree n if there is constant n such that:
f x, y   n f x, y 
o A differential equation of the form (differential form):
M x, y dx  N x, y dy  0 is homogeneous if the functions M  x, y 
and
N  x, y  are both homogeneous functions of the same degree.
Or
If the equation in standard form:
y  f  x, y  depends only on the ratio
y
x
or , then it is said to be homogenous.
x
y
To solve homogeneous equations, turn them into separable ones using
the substitution: y  x , where  is a function of x ; also:
dy
d
  x
or dy  xd  dx
dx
dx
o A Transformation (Equations in which
N  x, y  are linear but not homogeneous)
M  x, y  and
Another type of equations that can be reduced to a more basic type by means
of a suitable transformation is an equation of the form:
a1x  b1 y  c1 dx  a2 x  b2 y  c2 dy  0 where
a1 , b1, c1 , a2 , b2 , c2 are constants where
M x, y   a1 x  b1 y  c1 and N  x, y   a2 x  b2 y  c2
written as M x, y   N x, y y  0 , which are not homogeneous.
However the differential equation can be made homogeneous and solved
accordingly provided;
69
Case I
a2 b2
 , then the transformation:
a1 b1
x  X  h and y  Y  k , where h, k  is the solution of the
system a1 x  b1 y  c1  0 and a2 x  b2 y  c2  0 , that is x  h and
y  k , which reduces to a homogeneous equation:
a1X  b1YdX  a2X  b2Ydy  0
in the variables x and Y .
If
Case II
a2 b2
  k , then the transformation z  a1x  b1 y reduces the
a1 b1
equation to a separable equation in the variables x and z :
If
dz
dy
 a1  b1
dx
dx
-
Exact Equations
Given a function
f  x, y  , its total differentials, df , is defined as:
df 
f
f
dx  dy
x
y
This shows that the family of curves (or general solution) f x, y   c satisfies
the differential equation df  0 .
f
f
dx  dy  0
x
y
So if there exists a function f x, y  such that:
M  x, y  
f
f
and N x, y  
y
x
then M x, y dx  N x, y dy is called an exact differential, and the equation:
M x, y dx  N x, y dy  0
is said to be an exact equation, whose solution is the family f x, y   c .
-
Integrating Factors – Non-exact
If M x, y dx  N x, y dy  0 is not exact, sometimes we can turn it into an
exact differential equation by multiplying the whole equation by an
appropriate factor, called an integrating factor, let say    x, y  , which is
a function of x, y .
70
For which we obtain:
M x, y dx  N x, y dy  0 and expect it to be exact.
M  N 
For it to be exact then:

y
Applying product rule we get: 
x
M

N

M

N
y
y
x
x
 N M 



N
  

y
x

x

y


M y  N x   N x  M y
M
Simply put:


 N x  M  y    M y  N x 
o Linear First – Order Equations
A first – order linear in the standard form is written as
y  px y  qx  ………………………………(2.16)
We recognise (2.16) as an inexact differential equation which requires
an integrating factor
 x  , that is:
d x 
0
dy
 x  to obtain the following:
 xyx    pxyx  qx  0 ……………………..(2.17)
M x, y    x  px y  qx  and N x, y    x
Multiplying through (2.16) by
M N

.
y
x

py  q    x  d x
Thus,
y
x
dx
d
 p 
………….(2.18)
dx
d
  pdx  ln    pdx
Separating variables: 
If (2.17) is exact, then

  e  pdx ……………………..(2.19)
Without loose of operating the constant of integration of (2.19) is
unity.
Consider (2.16) and multiply through by:  x  e 
then we have the exact equation:
y  py  q
y   y  q by (2.18) that is p   
71
p x dx
,
d
p   p
dx
Integrating both sides:
y   qdx  c
 y   1  qdx   1c
 p x dx  p x dx
 p x dx
 ye 
qxdx  ce 
……(2.20)
e
The solution is in two parts in the R.H.S. The term
ce   p x dx 
 1c , called the complementary function is the solution of the
reduced equation:
y   p x  y  0
obtained by setting qx   0 in (2.16). The differential equation which
has qx   0 is called Homogeneous differential equation.
The other term on the R.H.S of the solution is dependent on qx  , that
 1  qdx is called the particular integral and it satisfies (with the
complementary function) the differential equation:
y   p x  y  q x  ,
called the Non – Homogeneous differential equation.
Notice that the particular integral contains no constant.
o Bernoulli’s Equation
A Bernoulli’s equation is a first order differential of the form:
y  pxy  qxy n ……………(2.22)
in which n is any real number. Here there is a multiply of y n , n  0, n  1 ,
at the right hand side(RHS)
n
Then divide (2.22) through by y and substitute
yy n  pxy1 n  qx
Let
z  y1 n
 z  1  ny  n y
z

 y n y
1 n
The equation becomes:
z  1  n  px z  1  n qx 
72
z  y1 n .
The solution is obtained using first order linear equation in first degree
solution techniques.
o Riccati’s Equation
dy
 Px   Qx  y  Rx  y 2
dx
Consider the new function z defined by:
1
z
y  y1
The equation
where y1 is the particular solution.
dz
 Qx   2 y1Rx z  Rx 
dx
Which is a linear equation satisfied by new function z . Once it is solved, we
1
go back to via the relation: y  y1  .
z
Then easy calculations give:
73
UNIT 3
LINEAR DIFFERENTIAL EQUATION OF HIGHER
ORDER
Introduction
So far, all equations we have studied have been first order, which means that they contained
only the first derivative, say
y . Now we will turn to higher – order equation. We will start
with one of the most common (and, fortunately, easily solved) types of higher – order
equation – the linear equations with constant coefficients then we proceed with other forms.
Learning Objectives
After going through this unit, you would be able to:
• solve most linear differential equation of higher order with constant coefficient.
• explain the method of attack to a particular linear differential equation and which
method is most efficient
• recognise that solutions of higher order differential equation are linearly
independent and can use the Wronskian to determine it.
Session 1-3 Linear Differential Equation of Higher Order
A general n th – order linear differential equation has the form:
dny
d n1 y
d2y
dy

a
x

...

a
x
 a1  x   a0  x  y  R  x  ….. (3.1)
n1  
2 
n
n1
2
dx
dx
dx
dx
where an  x   0 and the coefficients ai  x i  0,1,...n  1, n  ,and f  x  are given functions of x or
an  x 
are constants.
This equation (3.1) is homogeneous if f x   0 and non-homogeneous if f x   0 .
The solution of the homogeneous differential equation is called the complimentary function
(conventionally yc ) and the solution of the non-homogeneous part is known as the particular
integral (conventionally y p ).
Then the general solution = complimentary function + particular integral (that is y  yc  y p
).
74
1-3.1 Homogeneous Equation with Constant Coefficients
A general homogeneous differential equation with constant coefficients is:
an y n  an1 y n1  ...  a1 y  a0 y  0 …..(3.2)
We start problem solving with second order differential equation of the form:
2
a2
d y
dx
2
 a1
dy
 a0 y  0 ……………………...
dx
(3.3)
We consider a trial solution y  e , exponential function, suitably chosen r to be determined
rx
for the constant coefficient linear differential equation. Then (3.3) becomes:
a r
2
clearly
2

 a1r  a0 e  0 …………………………….(3.4)
erx  0 , hence;
rx
a2 r  a1r  a0  0 ……………………….(3.5)
2
or for convenience, ar  br  c  0 ……………………………(3.5)
2
Hence if r is root of this quadratic equation (3.5), often called the auxiliary or characteristic
rx
equations, e is a solution of (3.3).
b  b  4ac
2
The roots of (3.5) are: r1 
2a
There are three cases to consider:

b  b  4ac
2
and r2 
2a

Case 1: If the roots are real and distinct b  4ac  0 , that is, r  r1 ,
2
r  r2 and r1  r2
then the general solution of the differential equation is:
y  c1e r1 x  c2er2 x


Case 2: If the roots are real and identical b  4ac  0 , that is, r  r1  r2 , if we denote the
2
double root by r , then the general solution of the differential equation is:
y  c1e  c2 xe
rx
rx
Case 3: If the roots are not real. In this case, the roots are complex conjugates
b
2

 4ac  0 , that is r    i , then the general solution of the differential
x
equation is y  e
 c cos  x  c sin  x  where  
1
2
Example 3.1
1.
Solve the differential equation y  2 y  y  0
75
b
2a
b  4ac
2
; 
2a
Solution:
Let y  e , y  re , y  r e :
rx
rx
2 rx
r e  2re  e  0 r  2r  1  0 : r  1
x
x
y  c1e  c2 xe
2 rx
2.
rx
rx
2
Solve the differential equation y  2 y  5 y  0 .
Solution:
Aux: r  2r  5  0 r  1  2i
2
ye
3.
x
 c cos 2x  c sin 2x 
1
2
d 2 y dy
  6y  0
2
dx
dx
Solve
Solution:
Aux: r  r  6  0  r1  2, r2  3
2
y  c1e  c2e
2x
3 x
1-3.2 Homogeneous with constant coefficient of order 3 and higher
Theorem I: Distinct Real Roots
In case 1; for an n th order linear differential equation with constant coefficients, if r1 , r2 ,...rn
are distinct roots of the characteristic equation resulting from the differential equation, then
the general solution is of the form: y  c1e 1  c2e 2  ....  cne n
r
r
r
Theorem II: Repeated Roots
In case 2; if the characteristic equation has a repeated root r of multiplicity k , then the part
of a general solution of the differential equation of the form:

y  c1  c2 x  c3 x  ...  ck x
2
k 1
e
rx
Example 3.2
Find a general solution of y
1.
 3
 y  6 y  0
Solution:
Aux: r  r  6r  0  r  r  3 r  2  0 ; r  0, r  3, r  2
3
2
Hence, y  c1  c2e  c3e
3x
2.
2 x
Find a general solution of y
 4
 3y
 3
76
 3 y  y  0
Solution:
Aux:
r  3r  3r  r  0
4
3
2
r1  0, and it has also the triple  k  3 root r2  1


y  c1  c2  c3 x  c4 x e
2
x

Exercise 3.1
Solve the following linear differential equation.
1.
y  4 y  0
Ans.
y  c1 cos 2 x  c2 sin 2 x
2.
4 y  4 y  3 y  0
y  c1e 2  c2e
3.
2 y  9 y  0
y  c1  c2e 2
4.
9 y  12 y  4 y  0
y  c1  c2e 2 x / 3  c3 xe 2 x / 3
5.
y  y  y  y  0
6.
y  4 y  4 y  0; y  0  1, y  0  1
y  c1e  c2e
7.
y  2.2 y  1.17 y  0; y  0   2, y  0   2.6
y  0.3e
8.
y  6 y  25 y  0; y  0  3, y  0  1
ye
9.
y 4  4 y  0
10.
y  10 y  25 y  0; y  0  3, y 0  4, y 0  5
x
 23 x
9x
x
y  1  3x  e
3x
 x /4
x
 c3 xe
x
2 x
 0.5e
x /2
3cos 4 x  2sin 4 x 
y  c1  c2 x  c3 cos 2 x  c4 sin 2 x
1-3.3 Theory of Solutions of Linear Differential Equations
Consider a Homogeneous second – order linear differential equation
y  px y  qx y  0 ……………………………(3.6)
Principle of Superposition
Let y1 and y2 be two solutions of the homogeneous linear equation (3.6) on the interval I .
If c1 and c2 are constants, then the linear combination
y  c1 y1  c2 y2 ………………………(3.7)
is also a solution of (3.6) on I . Prove as exercise.
77
Definition:
Linear Independence of Two Functions and the Wronskian
Two functions defined on an open interval are called linearly independent provided that
neither is a constant multiple of the other. Two functions are said to be linearly dependent if
they are not linearly independent; that is, one is a constant multiple of the other.


In general, a set of functions (solutions to differential equation) y1  x  , y2  x  ,..., yn  x  is
linearly dependent on a  x  b if there exist constants, c1 , c2 ,..., cn , not all zero, such that
c1 y1  c2 y2  ...  cn yn  0 ………. (3.8)
on a  x  b .
Example 3.2
y1  cosx and y2  sin x are two solutions of the equation y  y  0 . By principle of
superposition y  3 y1  2 y2  3cosx  2 sin x is also a solution.
We can always determine whether two given functions f and g are linearly dependent on an
interval I by noting at a glance whether either of the two quotients f / g and g / f as a
constant on I .
The following pairs are linearly independent on the entire real line:
sin x,cos x , e x , e  2 x , e x , xe x , x  1, x 2 , and x, x .
Given two functions f and g , the Wronskian of f and g is the determinant :
W f , g 
f g
 fg   f g
f  g
For example, W cosx, sin x 


ex
while W e , xe  x
e
x
x
cosx sin x
 cos2 x  sin 2 x  1  0
 sin x cosx
xe x
 e 2 x  0x
x
x
e  xe
Given two functions f x  and g  x  that are differentiable on some interval I :
1.
2.
If W  f , g x   0 for some x0 in I , then f x  and g  x  are linearly
independent on the interval F .
If f x  and g  x  are linearly dependent on I then W  f , g x   0 for
all x is the interval.
78
Note:
It is possible for two linearly independent functions to have a zero Wronskian.
Suppose that y1 and y2 are two solutions of homogeneous second order linear equation
y  px y  qx y  0 on open interval I on which px  and qx  are continuous. If
W  y1, y2   0 , then the two solutions are called a fundamental set of solutions and the
general solution to (3.6) is y  c1 y1  c2 y2 .
Theorem 3.1
The
nth - order linear homogeneous differential equation (3.2) always has n linearly
independent solutions. If y1  x  , y2  x  ,..., yn  x  represent these solution, then the general
solution (3.2) is
y  x   c1 y1  x   c2 y2  x  
 cn yn  x  …… (3.9)
where c1, c2 ,..., cn denote arbitrary constants.

Exercise 3.2
Determine if the following sets of functions are linearly dependent or linearly dependent
a)
f x   9 cos 2 x , g x  2 cos2 x  2 sin x
b)
f x   2x 2 , g x   x 4
1-3.4 Reduction of Order
Consider the non – constant coefficient, second order differential equations of the form
y  px y  qx y  0 ………………(3.6)
In general, finding solutions to these kinds of differential equations can be much more
difficult than finding solutions to constant coefficient differential equation.
We discuss here the method of reduction of order, which enables us to use one known
solution, say, y1 to find a second linearly independent solution y2 .
Solution of Reduction Order:
y  px y  qx y  0 ……………… (3.6)
Consider
Let
y2 x  vxy1x………………………. (3.10)
79
We begin by substituting the expression in (3.10) into (3.6), using the derivatives:
y2  vy1  vy1  y2  vy1  2vy1  vy1
We get: vy1  2vy1  vy1   pvy1  vy1   qvy1  0 and rearrangement gives
vy1  py1  qy1   vy1  2vy1  pvy1  0
but y1  py1  qy1  0 is y1 is a solution of (3.6).
This leaves the equation y1v  2 y1  py1 v  0 …………. (3.11)
The key to the success of this method is that (3.11) is linear in v  . Thus the substitution of
(3.10) has reduced the second order linear equation in (3.6) to the first order (in v  ) in linear
equation in (3.11)
If we write u  v and assume that y1 never vanishes on I , then (3.10) yields:
 2 y

u    1  p( x) u  0 ……………………………… (3.12)
 y1

An integrating factor for (3.9) is:

  y
 

  exp 2 1  px  dx  exp2 ln y1 
 

  y1
  y12 e  p
Thus
 x dx
2
We now integrate the equation in (3.6) to obtain: uy1 e 
So
v  u 
p ( x) dx


p( x)dx

c
  p x dx
e
2
c
y1
 p x dx
y2
e 
vc
dx  k
Another integration now gives:
y1
y12
With the particular choices c  1 and k  0 , we get:
 p x dx
e 
y2  y1
dx ………………(3.13)
y12


Example 3.3
1.
1
2
Find the general solution to 2 x y  xy  3 y  0 given that y1 ( x)  x is a
solution.
Solution:
80
2x 2 y  xy  3 y  0  y 
y2  y1

 p x dx
e 
y12
1
1
3
y  2 y  0 where p( x) 
.
2x
2x
2x
dx  y2  x
1

 1 dx
e  2x
x 
1
x 2
dx  x  2 dx
2
x
1
1
3
3
 5

 y2  x 1  x 2 dx  x 1 52 x 2  c  let c  0  y2  2 x 2
5


3
y  c1 x 1  c2 x 2
Thus a general solution is:
Given the solution y1  x , find a general solution of
4.
x
2

2
 1 y  2 xy  2 y  0 (x  1)
Solution:
2
First divide each term by x  1 to obtain: y  


With px  2x / x 2  1 .

 p x dx
 exp
Hence e 


2x
x 1
2
y 
2
x 1
2
y0



dx  exp ln x 2  1  1  x 2 since x 2  1 .
x 1 
2x
2
Therefore, the formula in (37) yields: y2  x
Thus a general solution is:


1 x2
x
y  c1 x  c2 1  x 2
2
dx  x


x  2  1dx  1  x 2


Exercise 3.3
Given one solution y1 for each question, find a second linearly independent solution y2 of
the following differential equation:
1.
y  4 y  4 y  0;
2.
x 2 y  3xy  y  0; y1  1x
3.
x 2 y  4xy  x 2  6 y  0 ; y1  x 2 sin x
4.
x 2 y  xy  9 y  0; y1  x 3
5.
x  1y  x  2y  y  0;

y1  e 2 x

x
6.
4 y  4 y  y  0 ;
y1  e 2
7.
xy  3 y  0 ;
y1  1
y1  x 3
81
1-3.5 Linear Non – Homogeneous Equation with Constant Coefficients
A general linear non – homogeneous constant coefficient equation is:
dny
d n 1 y
dy
 a0 y  f x ………………… (3.14)
dx
dxn
dxn 1
f x  is a given function of x , and a0 , a1,...,an 1, an also given constants.
an
 an 1
 ...  a1
The general solution of (3.11) is given as:
General solution = Complimentary Function yc x + Particular Integral y p  x 
There are several different methods for finding solutions of non-homogeneous linear
equations. Here we study three methods to find our Particular Integral, since the
complimentary is from the homogeneous part, which we have done already;
(I)
Method of Undetermined Coefficients
(II)
Method of Variation of Parameter
(III) Method of the Linear Differential Operator
1-3.6 Method of Undetermined Coefficients
The main focus of this section is the structure of solutions of non-homogeneous linear
differential equations rather than solution methods. Knowledge of the structure of solutions
makes it easier to find solutions.
The method of undetermined coefficients is a method for finding particular solutions for
some differential equations of the form (3.14). The method consists of three steps:
a.
We assume a trial solution function y p  x  , a family of functions that is
guaranteed to include a correct particular solution;
Find the image of the trial solution whose form depends on f x  of (3.14).
b.
y p  x  is then substituted into (3.14) and the constants take values which satisfy
c.
(3.14) completely
(a)
bx
For f x   ae a, b are constants, if the auxiliary equation of (3.14) has r  b as
a root of multiplicity k , then the trial function is of the form y p x   Ax k ebx ,
where A constant to be determined. If r  b is Not a root, take k  0 .
Example 3.4
Find a particular integral of y  3 y  2 y  e 2 x .
Solution:
The auxiliary equation is: r  3r  2  0 ,   r  2 r  1  0  r  2,1
2
82
Hence f x   e 2 x that is b  2, a root.
The trial function is: y p x   Axe 2 x
Substituting this into the differential equation:
y p x   Axe 2 x , yp x   Ae 2 x  2 xAe2 x
yp x   2 Ae 2 x  2 Ae 2 x  4 Axe2 x
 4 Ae2 x  4 Axe2 x  3Ae2 x  6 Axe2 x  2 Axe2 x  e 2 x
 Ae2 x  e 2 x  A  1
 y p x   xe 2 x
General solution: y  yc  y p  c1e
(b)
2x
 c2e x  xe2 x
For f x   a sin bx or cos bx or both, a, b constants, if
r 2  b2 is a factor of the
auxiliary equation of multiplicity k , then: y p x   x k  A sin bx  B cos bx  , where
A, B are constant to be determined. If r  b is Not a factor, then k  0 .
2
2
Example 3.4
Solve the differential equation y  4 y  3sin 2 x .
Solution:
r  2  0, b  2, k 1
y p  Ax sin 2 x  Bx cos 2 x --------------- (I)
2
2
yp  Asin 2 x  2 x cos 2 x   Bcos 2 x  2 x sin 2 x  ------ (II)
yp  A2 cos2 x  2cos2 x  2 x sin 2 x  
B 2 sin 2 x  2sin 2 x  2 x cos2 x 
Substitute back into the equation gives:
--------------- (III)
 A2 cos 2 x  2 cos 2 x  4 x sin 2 x 
B 2 sin 2 x  2 sin 2 x  4 x cos 2 x 
4Ax sin 2 x  Bx cos 2 x  3sin 2 x
 4 A cos 2 x  4B sin 2 x  3sin 2 x
Comparing the coefficients:
A  0, B  
3
4
3
 y p  x    x cos2 x
4
83
General solution: yx   c1 cos 2 x  c2 sin 2 x 
(c)
3
x cos 2 x .
4
For f x   axs , a, s (integer) constant, if the auxiliary equation has

r  0 as a root
of multiplicity k , then trial function: y p x   x k Ax s  Bx s 1      Px  Q

where A, B,  , P, Q are constants to be determined.
If r  0 is Not a root, then k  0
Example 3.5
Find a particular solution of y  4 y  3x3 .
Solution:
y p  Ax3  Bx 2  Cx  D
yp  3 Ax 2  2Bx  C
yp  6 Ax  2B
Substituting in the differential equation, we get:

y  4 y p  6 Ax  2B  4 Ax3  Bx 2  Cx  D

 4 Ax3  4Bx2  6 A  4C x  2B  D  3x 3 .
We equate coefficients of like powers of x to get the equations:
4 A  3,
4B  0
6 A  4C  0, 2B  D  0
with solution A  34 , B  0, C   98 , and D  0 . Hence a particular solution of the
3
differential equation is: y p  3 x  9 x .
4
8
If f x  is the sum of any two or all these special functions, the particular integral
(d)
is the appropriate sum of individual particular integral. Hence f x  takes the trial
solution:




y p  x k A0  A1 x  ...  An x n e mx cos bx  B0  B1 x  ...  Bn x n e mx sin bx
Example 3.6
Determine the appropriate form of for a particular solution of
y  6 y  13 y  e 3x cos2x
Solution:
The characteristic equation r  6r  13  0 has roots: r  3  2i ,
2
3 x
So the complementary function is: yc  e  A cos 2 x  B sin 2 x 
84

This is the same form as a first attempt e 3x  A cos2 x  B sin 2 x at a particular
solution, so we must multiply by x to eliminate duplication.
Hence we would take y p  e 3x  Ax cos 2 x  Bx sin 2 x 

Exercise 3.4
Solve the following linear differential equation.
1.
y  y  2 y  4x2
y  c1e x  c2e2 x  2x2  2x  3
2.
y  y  2 y  e3x
y  c1e x  c2e2 x  14 e3 x
3.
y  y  2 y  sin 2 x
4.
y  4 y  2e2 x
y  c1e2 x  c2e2 x  12 xe2 x
6.
y  6 y  10 y  20  e 2 x
y  4 y  3x cos 2 x
7.
y  6 y  13 y  xe3x sin 2x
8.
y  3 y  2 y  x e  x  e  2 x
9.
y  2 y  3 y  4e3x  5
5.
10.

d3y
3

d2y
2
y  c1e x  c2e2 x  203 sin 2 x  201 cos2 x

 3e x  4x 2
11.
dx
dx
y  6 y  8 y  cosh x given y0  0
12.
y  y  3  2e  x
85
y0  1
1-3.7 Method of Variation of Parameters
In method (1-3.6), we presented the method of undetermined coefficients which, when it is
applicable, is usually the simplest method of finding a particular solution of a nonhomogeneous linear differential equation with constant coefficients. This method (1-3.6)
cannot succeed with an equation such as y  y  tan x , because the function f x   tan x
has infinitely many linearly independent derivatives.
Consider a non-homogeneous linear second order differential equation of the form:
y  px y  qx y  f x  ……………….. (3.15)
Suppose y1 and y2 are the two solutions of the homogeneous equation
y  px y  qx y  0 …………………… (3.16)
Then the complimentary function is:
yx  c1 y1x  c2 y2 x ……………………. (3.17)
where c1 and c2 are arbitrary constant.
To find the particular integral, this method requires the trial function
y p x   u1 x  y1 x   u 2 x  y2 x  …………………….(3.18)
where we replace the constants, or parameters c1,c2 in the complementary function in
(3.17) by variables: functions u1 and u2 of x . (3.18) is substituted into (3.15) to determine
u1x and u2 ( x) .
To determine u1 x and u2 x the following conditions should be fulfilled:
(I)
(II)
(3.18) must satisfy (3.15)
A condition imposed arbitrarily to facilitate the calculation of the function
u1x and u2 x .
y p x   u1 y1  u 2 y2 ……………………………………..(3.19)
y p x   u1 y1  u1 y1  u 2 y2  u 2 y2
Let u1 y1  u2 y2  0 ……………………….(3.19) : condition imposed
 y p x   u1 y1  u 2 y2 …………………….. (3.20)
 y p x   u1 y1  u1 y1  u 2 y2  u 2 y2 …. (3.21)
Substituting (3.21), (3.20) and (3.18) into (3.15), we have:
 u1 y1  u1 y1  u2 y2  u2 y2  pu1 y1  pu2 y2  qu1 y1  qu2 y2  f x
 u1 y1  py1  qy1   u2  y2  py2  qy2   u1 y1  u2 y2  f ( x)
86
But both y1 and y2 satisfy the homogeneous equation (3.16) associated with the nonhomogeneous equation (3.15).
So y1  py1  qy1  0 and y2  py2  qy2  0
Equation (3.15) finally becomes:
u1 y1  u2 y2  f x ………………….. (3.22)
Using (3.19) and (3.22) u1 and u2 can be determined and thus we have a system:
u1 y1  u2 y2  0
…………………. (3.23)
u1 y1  u2 y2  f x 
The Cramer’s Rule for the solution of a linear system of equation yields:
0
f x 
u1 
y1
y1
 u1 
 y2 f x 
W  y1, y2 
Wronskian.
u1  

y2
y2
,
y2
y2
y1
0
y  f x 
u 2  1
y1 y2
y1 y2
,
 u2 
y2 f x 
dx ,
W  y1 , y2 
u2 
Then (3.18) becomes:

y p   y1 x 
y1 f x 
; since the denominator is the
W  y1 , y2 

y1 f x 
dx
W  y1 , y2 
y2 x  f x 
y x  f x 
dx  y2 x  1
dx ……..(3.24)
W  y1 , y2 
W  y1 , y2 

Example 3.7
Find the general solution of the differential equation
y  y  sec x
0  x  2
Solution:
The two solutions of the homogeneous equation are:
y1  cosx and y2  sin x .
Then y p  u1 x cos x  u 2 x sin x
 y p  u1 cos x  u1 sin x  u2 sin x  u 2 cos x
Imposition: u1 cosx  u2 sin x  0 …………….(I)
yp  u1 x sin x  u2 x cos x
y p  u1 sin x  u1 cos x  u 2 cos x  u 2 sin x
87
Then differential equation becomes:
 u1 sin x  u1 cosx  u2 cosx  u2 sin x  u1 cosx  u2 sin x  sec x
 u1 sin x  u2 cosx  sec x ………………………….(II)
Using (I) and II, we have:
u1 cos x  u2 sin x  0
 u1 sin x  u2 cos x  sec x
The Cramer’s Rule for the solution of a linear system of equation yields:
0
sec x
u1 
cos x
 sin x
sin x
cos x
,
sin x
cos x
cos x
0
 sin x sec x
u2 
cos x sin x
 sin x cos x
 u1  tan x ,
 u2  1
 u1  ln cosx ,
 u2  x
Therefore, y p  ln cos x cos x  x sin x .
Hence the General solution:
yx  c1 cosx  c2 sin x  ln cosxcosx  x sin x

Exercise 3.5
Determine a particular solution of the following using the method of variation of parameter.
1.
y  5 y  6 y  e 2 x
9.
y  3 y  2 y  e x  e 2 x
2.
y  y  2 y  2e  x
10.
y  y  cos x
3.
y  2 y  y  3e  x
11.
y  y  cot x
4.
y  y  tan x
12.
y  5 y  6 y  e 2 x  e3x
5.
y  9 y  9 sec2 3x 0  x  6
13.
y  y  sin x
6.
y  4 y  3 cos ec2x 0  x  2
7.
y  3 y  2 y  e x
8.
y  2 y  y  e x
0  x  2
88
Session 2-3 Linear Differential Operator Method (D-Operator)
A differential operator is a rule D that uses derivatives to assign a function D  y  to any
sufficiently differentiable function y . The function D  y  is called the image of y under D .
This method is very efficient for the solution of a general linear equation of n th order and
with constant coefficients. It is most efficient because it can be manipulated algebraically.
2
Let D denote differentiation with respect to x , D differentiation twice with respect to x ,
k
Then D( x 2 )  2x
d y
d
, that is D 
k
dx
dx
D2 ( x3  4x 2  1)  6x  8
k
and so on; that is, for positive integral k , D y 
D2 (e3x )  9e3x
The expression f  D   a0 D  a1D
n1
d2
D  2 etc
dx
2
 an1D  an is called a differential operator
of order n . It may be defined as that operator which, when applied to any function y , yields
n

the result
f  D  y  a0
n
d y
 a1
d
n1
y

 an1
dy
 an y
n
n1
dx
dx
dx
where the function y is assumed to possess as many derivatives as may be encountered in
whatever operation take place. Let f  D be a polynomial in D .
2-3.1 Fundamental Properties of the D–operator
The following properties are satisfied by the D-operator.
D n (u  v)  D n u  D n v (Distributive property)
Dn (ku)  kDnu , k is a constant
D m (D n u)  D mn (u)
2-3.2 Factorability
Any constant coefficient n th order linear differential equation (LDE) can be written as:
f ( D)y  R( x) , where f (D) is a polynomial function of the operator D .
Example: y  3y  2 y  D2 y  3Dy  2 y  e 2 x
 ( D 2  3D  2) y  e 2 x  f (D)  D2  3D  2
This polynomial can be factorized if required. In this case
f (D)  D2  3D  2  (D  2)(D  1)
89
We recognize that (D  2)(D  1)  (D  1)(D  2) which is valid ONLY for constant
coefficient linear equations.
2-3.3 Linear Equations of First Order
Recall the familiar first-order linear equation with constant coefficient:
dy
 ay  R(x) , a is
dx
a constant, which is equivalent to:
(D  a) y  R( x)  y 
1
R( x )       
Da
The integrating factor is I  e   e ax . Thus multiplying both sides by the integrating
factor we obtain
dy
e ax
 aeax y  e ax R(x)
dx
 adx


d
ye ax  e ax R( x)
dx
 ye ax   e ax R( x)dx

 y  eax  e ax R( x)dx        (1.1)
1
1
R( x) 
R( x)  e ax  e ax R( x)dx
Da
Da
1
e2x
Therefore if we have ( D  2) y  e 2 x  y 
D2
From (  )
y
2x
2x
2 x
2x
dx  e2 x x  c  xe2 x  ce2 x
then y  e  e  e dx  y  e

2-3.4 Linear Equation of Higher Order


Here we discuss second order linear equations, aD2  bD  c y  f ( x)


Example 1: Solve D 2  3D  2 y  e 2 x
Solution
D
2

 3D  2 y  D  2D  1y  e 2 x
Let D 1y  z , then D  2z  e 2 x
1 2x
e  e 2 x  e  2 x e 2 x dx  e 2 x x  c1 
D2
2x
Thus D  1y  e x  c1 
1
y 
e 2 x x  c1   e 2 x  e  x e 2 x ( x  c1 ) dx
D 1
y  xe2 x  Ae 2 x  Be x , A and B are arbitrary constants.
z



90

Alternatively;
D 1D  2y  e2x  y 
1
e2x
D  1D  2
Using partial fraction:
1  2x
1 2x
1 2x
 1
y 

e 
e
e 
D2
D 1
 D  2 D 1
 y  e 2 x  e 2 x e 2 x dx  e x  e  x e 2 x dx  xe 2 x  c1e 2 x  e 2 x  c2 e x
 y  xe 2 x  (c1  1)e 2 x  c2 e x  xe 2 x  Ae 2 x  B2 e x
The efficiency of the operator method becomes very clear where the auxiliary equation has
repeated roots.


Example 2: Solve D2  2D  1 y  2xex
Solution:
Let z  D  1
D  12 y  2xex


1
2 xe x  e x  e  x 2 xe x dx  e x x 2  c1
D 1
D  1y  e x ( x 2  c1 )
1
y
e x ( x 2  c1 )  e x  e  x  e x x 2  c1 dx
D 1
Then D  1z  2xex  z 

 ye
x
 (x
2



 c1 )dx  e x 13 x3


x 3e x
 c1x  c2 
 c1xe x  c2e x
3

Exercise 3.6
1.
2.
Demonstrate Commutation of the factor in the operator D  2D  3 applied
onto:
i)
ii)
e2x  x3
ln x  x 3
Use equation 1.1 to solve the following;
i)
ii)
D  2y  e x
D 1y  e x
iii)
v)
vii)
D  2y  x 2  e2x
2D  3y  sin x
3D  2y  cos 23 x
iv)
vi)
viii)
91
D  2y  x 2  e2x
2D  3y  cos x
3D  2y  sin 23 x
Solve
D
D
i)
2

 9 y  e3x

 5D  6 y  e
y  7 y  12y  x
iii)
v)
2
D  32 y  e 3x
D  32 y  e3x
D  22 y  xe3x
vii)
ix)
xi)
ii)
3x
iv)
vi)
viii)
x)
xii)
D
D
2

 9 y  e 3x

 5D  6 y  e 2 x
y  9 y  2 y  x
2
D  32 y  e3x
D  32 y  e 3x
D  32 y  xe3x
2-3.5 f  D  Polynomials with complex roots
Consider a differential equation
f  D y  R  x  for which the auxiliary equation
f  m  0 has real coefficients. From elementary algebra we know that if the auxiliary
equation has any imaginary roots, those roots must occur in conjugate pairs. Thus, if
m1    i is a root of the homogeneous equation f  m  0 , with a and b real and
b  0 , then m2    i is also a root of f  m  0 .
For
 aD
2

 bD  c y  R  x  if f  m  0 have complex roots then we have:
 D    i  D    i  y  R  x 
Let z   D    i  y then  D    i  z  R  x  .


Example 3: Solve D 2  4 y  3
Solution
D  2i D  2i y  3
Let D  2i y  z
 zD  2i   3


1
3
3  e  2ix  3e 2ix dx  e  2ix e 2ix  c
D  2i
2i
3 3
3
3
 z   ce  2ix   c1e  2ix , where c1  c
2i 2i
2i
2i
1 3
 2ix 
2ix
 2ix  3
 2ix 
y 

c
e

e
e
  c1e
dx
1



D  2i  2i

 2i

z
92
c
3
 3 c
 y  e 2ix  e  2ix  1 e  4ix  c2    1 e  2ix  c2e 2ix
4i
4
 4 4i
c
3
 y   c3e 2ix  c2 e 2ix where c3  1
4
4i
y
3
 A cos 2 x  B sin 2 x
4

Exercise 3.7
1.
3.
5.
7.
9.
D( D  1) y  5
D
D
D
D

2
 D 1 y  2
4.
2
 4 y  e2x
6.
2
2

 9y  sin 2x
 2D  1y  e
D  4y  0
D  4y  cos x
D  1y  e
D  2D  1y  e sin x
2
2.
8.
x
2
2
2x
2
x
cos x
A general n th – order linear differential equation has the form:
dny
d n1 y
d2y
dy
an  x  n  an1  x  n1  ...  a2  x  2  a1  x   a0  x  y  f  x  ….. (3.1)
dx
dx
dx
dx
where an  x   0 and the coefficients ai  x i  0,1,...n  1, n  ,and f  x  are given functions
of x or are constants.
This equation (3.1) is homogeneous if f x   0 and non – homogeneous if
f x   0 .
The solution of the homogeneous differential equation is called the complimentary
function (conventionally yc ) and the solution of the non – homogeneous part is
known as the particular integral (conventionally y p ).
Then the general solution = complimentary function + particular integral (that is
y  yc  y p ).
-
Homogeneous Equation of Second – Order with constant coefficient
d2y
dy
a2 2  a1  a0 y  0
dx
dx
rx
A trial solution y  e substituted gives:
a2r 2  a1r  a0 erx  0
93
clearly
erx  0 , hence; a2 r 2  a1r  a0  0
or for convenience,
ar 2  br  c  0 , where a  a2 ; b  a1; c  a0
whose roots:
b  b2  4ac
b  b2  4ac
r1 
and r2 
2a
2a
There are three cases to consider:
Case 1:
If the roots are real and distinct
b
2

 4ac  0 , that is,
r  r1 , r  r2 and r1  r2
then the general solution of the differential equation is:
y  c1e r1 x  c2er2 x
Case 2:
If the roots are real and identical
b
2

 4ac  0 , that is,
r  r1  r2 , if we denote the double root by r ,
then the general solution of the differential equation is:
y  c1e rx  c2 xe rx
Case 3:
conjugates
b
If the roots are not real. In this case, the roots are complex
2

 4ac  0 , that is r    i , then the general solution of
the differential equation is:
y  ex c1 cos x  c2 sin x 
where  
-
b2  4ac
b


;
2a
2a
Wronskian
Given two functions f and g , the Wronskian of f and g is the
determinant :
W f , g 
f g
 fg   f g
f  g
Given two functions f x  and g  x  that are differentiable on some interval
I:
i. If W  f , g x   0 for some x0 in I , then f x  and g  x  are
linearly independent on the interval F .
ii. If f x  and g  x  are linearly dependent on I then W  f , g x   0
for all x is the interval.
94
-
Reduction Order
In reduction of order, it enables us to use one known solution , say, y1 to
find a second linearly independent solution y2 .
Solution of Reduction Order:
y  px y  qx y  0
Consider
Then y2  y1
-

 p x dx
e 
y12
dx
Second – Order Non – Homogeneous Equation
A general linear non – homogeneous constant coefficient equation is:
an
dny
n
dx
 an 1
d n 1 y
n 1
dx
 ...  a1
dy
 a0 y  f x
dx
We have three Methods of attack. Each depend on the external function,
f  x  . The methods are:
(IV)
(V)
(VI)
Method of Undetermined Coefficients
Method of Variation of Parameter
Method of the Linear Differential Operator
Method of Undetermined Coefficients
(e)
For f x   aebx a, b are constants, if the auxiliary equation of has
r  b as a root of multiplicity k , then the trial function is of the form
y p x   Ax k ebx , where A constant to be determined. If r  b is
Not is a root, take k  0 .
(b)
For f x   a sin bx or cos bx or both, a, b constants, if
r 2  b2 is a
factor of the auxiliary equation of multiplicity k , then:
y p x   x k  A sin bx  B cos bx 
where A, B are constant to be determined.
If
(c)
r 2  b2 is Not a factor, then k  0 .
For f x   axs , a, s (integer) constant, if the auxiliary equation has
r  0 as a root of multiplicity k , then trial function:

y p x   x k Ax s  Bx s 1      Px  Q

where A, B,  , P, Q are constants to be determined.
If
r  0 is Not a root, then k  0
95
(d)
If f x  is the sum of any two or all these special functions, the
particular integral is the appropriate sum of individual particular
integral. Hence f x  takes the trial solution:




y p  x k A0  A1 x  ...  An x n e mx cos bx  B0  B1 x  ...  Bn x n e mx sin bx
Method of Variation of Parameter
The method of undetermined coefficient does not succeed if the external
function,
f  x  is not analytic everywhere. The parameters of the
homogeneous solution are then varied (that is change to variables).
py  qy  f  x  ,
The homogeneous part, y  py  qy  0 , gives the solution
yc  c1 y1  c2 y2 . If the parameters c1 and c2 is varied to u1 ( x) and u2  x 
Given a differential equation: y 
respectively, we have:
y p  u1 y1  u2 y2
We then determined u1 and u2 to get:

y p   y1 x 
y2 x  f x 
y x  f x 
dx  y2 x  1
dx
W  y1 , y2 
W  y1 , y2 

Method Linear Differential Operator
dy
 ay  R(x) , a is a constant,
dx
which is equivalent to: (D  a) y  R( x)
1
y
R( x )       
Da
 adx
The integrating factor I  e   e ax
Let
y  eax  eax R( x)dx
Then
from

1
R( x)
Da
1

R( x)  e ax  e ax R( x)dx
Da
y
96

UNIT FOUR
LAPLACE TRANSFORM
4.0
INTRODUCTION
In this section we will be looking at how to use Laplace transforms to solve
differential equations. Laplace transforms reduce a differential equation to an
algebra problem.
We use Laplace transforms when the force function becomes complicated, that is,
discontinuous – for instance, when the voltage supplied to an electrical circuit is
periodically turned off and on.
Learning Objectives
After going through this section session, you would be able to:
•
Know the basic definition of Laplace Transform.
•
Solve improper integral with an infinite limit.
•
Transform continuous functions like polynomials, exponentials,
trigonometric (only cosine and sine ones) to Laplace forms
•
Find the inverse a Laplace transform back to polynomial or Laplace
Transform.
4.1
DEFINITION: LAPLACE TRANSFORM
Given a function f t  defined for all t  0 , the Laplace transform of f is
the function F of s defined as follows:

L f t    e  st f t dt  F s  ……………….. (1.1)
0
for all values of s for which the improper integral converges.
The improper integral over an infinite interval is defined as a limit of
integrals over finite intervals, that is:

m
f t dt ………………………..(1.2)
a f t dt  mlim

 a
97
If the limit in (1.2) exists, then we say that improper integral converges.
Otherwise, it diverges or fails to exist. Note that the integrand of the improper
integral in (1.1) contains the parameter s in addition to the variable of
integration t . Therefore, when the integral (1.1) converges, it converges not
merely to a number, but to a function F of s .
Example 4.1
Compute the Laplace transforms of the function
(i)
1,
(ii)
t
cost , for t  0 .
eat , (iv)
(iii)
,
Solution:
m
(i)

1
 1

L1   1 e  st dt  lim  e  st   for every s  0
0
m   s
0 s
(ii)

 1
 t

Lt    t  e  st dt  lim  e  st      e  st dt
0
0  s
m   s
0
m
m
1
 1

 lim  2 e  st   2 for every s  0
m   s
0 s
 

 e  s  a t 
 at  st
  s  a t
e e dt  e
dt  

0
0
 s  a  0
(iii)
L e at  

(iv)
L cos at    e st cos atdt

1
, sa
sa

s   st
e sin atdt
a 0

0
m
1
1

 lim  e st sin at     s  e st sin atdt
m a

0 0 a
m
1
1

 lim  e st sin at     s  e st sin atdt
m a

0 0 a

s2
  st

e
a2 a2 0
s

cosatdt 
s
a2

s2
a2
Lcosat
s s2
s
 L cos at   2  2 L cos at   2 2
a a
s a
  

Find the following: L t 2 , L t 3 , L t 4 , and L t n
98
4.2
ELEMENTARY PROPERTIES OF LAPLACE TRANSFORM
Theorem 4.1: (Linear Property)
Let f and g be functions whose Laplace transforms exist, and let c1,c2 be
any two real numbers, then:
Lc1 f t   c2 gt   c1L f t   c2 Lgt 
Prove as exercise.
Example 4.2
Find the Laplace transforms of the given functions:
f t   6e 5t  e3t  5t 2  9
g t   4 cos 4t  9 sin 4t  cos10t
Solution:
1.
F s   6
1
1
3!
1

 5 31  9
s   5 s  3
s
s
 F s  
2.
Gs   4
6
1
30 9

 4
s 5 s 3 s
s
s
s 4
 Gs  
2
2
9
4s
s  16
2

4
s 4
2
36
s  16
2
2
2

s
s  10 2
2
2s
s  100
2
Example 4.3
1.
Let f t   cosh at . Find L f t .
Solution:


   
Lcosh at   L 12 e at  e  at  12 L e at  12 L e  at 
 Lcosh at  
s
s2  a2
, sa0
99
1 1
1 1

2 sa 2 sa
2.
Let f t   eiwt . Find L f t .
Solution:
 
L eiwt 
1
s  iw
s  iw
s
iw

 2


s  iw s  iws  iw s  w2 s 2  w2 s 2  w2
By Euler formula:
 
eiwt  cos wt  i sin wt
L eiwt  Lcoswt  i sin wt  Lcoswt  iLsin wt
Equating the real and imaginary parts of these two equations, we obtain the
transforms of cosine and sine.
Lcos wt  
s
s w
2
2
, Lsin wt  
w
s  w2
2
.
Theorem 4.2: (First Shifting Theorem)
Replacement of s by s  a in the function.
If f t  has the transformation F s  (where s  k ), then e at f t  has the
transform F s  a  (where s  a  k ). In formula:


L e at f t   F s  a
or, if we take the inverse on both sides
e at f t   L1F s  a.
Example 4.4




1.
L e at coswt 
2.
L e at sin wt 
sa
s  a2  w2
w
s  a2  w2
100
Theorem 4.3: (Multiplying by t and t n )
If L f t   F s  then Ltf t    F s  . Because:


 de st 
L tf  t    tf  t e st dt   f t   
 dt
0
0
ds



For example; Lsin 2t  
Therefore,
Since
dn
ds
n
d 
f  t e st dt   F   s 

0
ds
2
s 4
2
L t sin 2t   
d 2 
4s
.



2
ds  s  4  s 2  4 2

e st   1n t ne st , we have:
e f t dt  1 
ds
  1 Lt f t 
F n s  


0
dn
n   st n
e t f
0
 st
n
n
t dt
n
n
n d
n
n


 F s    1 F  n  s 
Hence L t f  t    1
n   
ds
Theorem 4.4:
Let f be a piecewise – continuous function on 0,   exponential order.
Define a function g by:
0t a
0
g t   
 f t  a a  t
where a positive number,
then Lg   e as L f t 
101
Example 4.5
1.
0 0  t  a
1 a  t
Let g t   
Then Lg   e  as
1
s
2.
0  t 
0
cost      t
Let g t   
 s
Then e
s
s 1
2
Theorem 4.5:
If
L  f t   F  s  and if lim
t 0
t 0
f t 
exists, then
t
1
 
L  f  t    F  x dx
t
 s
Theorem 4.6:
If
L  f t   F  s  , then
t
1
L   f  x  dx   F  s 
 0
 s
Theorem 4.7:
If
f  t  is periodic with period  , that is, f t    f t  , then
L  f  t  

  st
0
e
f  t  dt
1  es
Example 4.5 Further examples
1.
et
Find the Laplace transform of f  t   
3
102
t2
t2
Solution:

2

0
0
2
L  f  t    e st f  t dt   e st et dt   e st  3dt
1  e2 s1 3 2s
 L  f t  
 e  s  0
s 1
s
2.
Find the Laplace transform of the function graph below:
1
2
3
4
5
6
7
8
Solution:
The function is
1 x  4
f t   
1 x4
Therefore,

4

0
0
4
L  f  t    e st f  t  dt   e st  1dt   e st 1dt
2e4 s 1
 L  f  t  
  s  0
s
s
3. Find
 sin3t 
.
L
 t 
Using Theorem 1.5, let
can be written as:
f t   sin3t , then F  s  
F  x 
3
x 9
2
103
3
, which also
s 9
2
Therefore,

 sin3t   3
s
L
  3 dx   tan 1  

2
 t  s x 9
 3
4. Find
t
L   sinh 2 xdx 
 0

Solution:
Taking
f  x   sinh 2x , we have f t   sinh 2t
F s 

2
, and then, from Theorem 1.6 that
s2  4

t
1 2 
2
L   sinh 2 xdx    2


 0
 s  s  4  s s 2  4


5. Find the Laplace for the square wave shown below
Solution:
Note that
f  t  is periodic with   2 , and in the interval
0  t  2 it can be defined analytically by:
 1 0  t 1
f t   
1 0  t  2
From Theorem 1.7, we have:
L  f t  
104

2  st
0
e
f t  dt
1  e2s
Since

2  st
0
e
f  t dt   e st 1 dt   e st  1dt
1
2
0

1

 

2
1 2 s
1
e  2e s  1  e s  1
s
s
It follows that,

  
  1 e
F s 
s 1  e  s 1  e 1  e  s 1  e 
e s  1
2
1  e s
2 s
s
 e s / 2   1  e s
 es / 2  
s

  s 1  e


2
s
s
s

e s / 2  e s / 2
1
s


tanh
s/2
s / 2
s
2
 s e  e


Exercise4.1
Find the Laplace Transforms of the following functions
1.
(i)
f t   sin t
F s 

s  2
1
s 1
2
(ii)
f t   sinh t
F s 
(iii)
f  t   2 x 2  3x  4
F  s   s43  s32  4s
(iv) f t  
2

tk
 k! .
k 1
F  s 
s 3
 s  32  36
2.
f  t   e3t cos6t
3.
f t   3sinh 2t  3sin 2t
4.
f t   e3t  cos6t  e3t cos6t
5.
f t   t sin 2t
12s 2  16
F s 
 s  43
6.
f t   t 2
F s 
f t   tg t 
F  s   G  s   sG  s 
7.
2
3
105
3 
5
4s 2
8.
f t   sin 3t cos 3t
9.
f t   1  t 3
10.
1 0  t  1
f t   
0 1  t
11.
t 0  t  1
f t   
0 1  t
12. Find the Laplace transform of the period function
0t 
 t
f t   
2  t   t  2
13.
Find
1
s
tanh
2
s
2
t1


L e4tt  e4 x sin3xdx 
0 x



2  s  4
2

1
3
s
tan 1   
2
 3   s  4 s 2  9
 s  4

106

4.3
INVERSE LAPLACE
The inverse Laplace is given by:
L1F s   f t  .
Linearity of inverse Laplace transforms:
L1aFs   bGs   aL1F s   bL1Gs  for any constant a, b .
For example:
 
1/ s  t , L


L1  1 / s , Lt   1/ s 2 , L e at  1/s  a , and Lcosat  s / s 2  a 2 .
Hence L11/ s  1 , L1
2
1
1/s  a  eat .
Example 4.6
1.
2 3 
1
1 
L1    2   2L1    3L1  2   2  3t
s
s s 
s 
2.
s 
 5
1  1 
1  s 
L1 
 2

5
L

L
 5e3t  cos 2t





2
 s  3
 s  3 s  4
 s  4
Example 4.7
Find the Laplace inverse of the function
F s  
2s  3
s  5s  4
Solution:
By partial fraction:
2s  3
13
5


s  5s  4 9s  5 9s  4
 2s  3 
 13
5 
L1 
 L1 



 s  5s  4
 9s  5 9s  4

13 1  1  5 1  1  13 5t 5  4t
L 
  L  s  4  9 e  9 e
9


 s  5 9
107
Exercise 4.2
Find the Laplace inverse of the following;
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
s
s 6
cos 6t
2
s
5s
2

1
5
t sin t
2
2
s 1
s2  9
1
cosh3t  sinh3t
3
s4
s  4s  8
e2t cos2t  e2t sin2t
1
 s  1 s 2  1
1 t 1
1
e  cos t  sin t
2
2
2
2


s 2  3s  1
s  2s 2  16
e as
s
s5
s 2  25

1  2s
e
 e s
s
e s

s s2  4


s2
s  23
108
4.4
CONVOLUTION
Definition: Let
f  t  and g  t  be piecewise – continuous functions on
0,  of exponential order. The function called the convolution of f t 
and
g  t  , denoted by f  g , is defined by:
 f  g t   0 f xg t  xdx
t
Note: f  g  g  f
Theorem: Let f and g be piecewise continuous on 0,   of exponential
order. Then:
L f  g   L f Lg  or, equivalently, L1L f Lg   f  g
Example 4.8
t
1 

1.
t
1 
1
1
L 
 1 e  3t   1 e  3t  x dx  e  3t  x   1  e  3t

0
3
 ss  3
0 3
2.
Using convolution, find the inverse f t  of;
F s  
1
s 2  12

1


1
s2 1 s2 1
Solution:
f t   L1F s   sin t  sin t   sin  sin t   d
t
0

3.
Let F s  
1 t
1 t
1
1



cos
td


cos
2


t
d



t
cos
t

sin t
2 0
2 0
2
2
1
s s  a 
2
. Find f t 
Solution:
1
L1  2   t
s 
 1   at
L1 
e
 s  a 
f t   t  e at   xe at  x dx  e at  xe  ax dx 
t
t
0
0
109
at

e
 at  1
a2
1
4.5
SOLUTION OF LINEAR DIFFERENTIAL EQUATION
The use of Laplace Transforms enables us to reduce the problem of finding a
solution y to a linear differential equation with constant coefficients to the
problems of solving a linear algebraic equation for L y  and then determining
y.





L f t    e  st f t dt  e  st f t  0  s  e  st f t dt
0
0
 L f t   sL f t   f 0
L f t   s 2 L f   sf 0  f 0
In general;
 
L f n  s n L f   s n 1 f 0  s n  2 f 0    f n 1 0
Example 4.9
1.
Let f t   t 2 . Derive L f  for L1 .
Solution:
Since f 0  0 , f 0  0 . f t   2 , and L2  2L1  2 / s ,
we obtain

L f   L2  2 / s  s 2 L f 
Hence L t 2  2 / s 3
2.
Let f t   sin 2 t . Find L f .
Solution:
We have f 0  0 ,
Lsin 2t  

f  t   2sin t cos t  sin 2t
2
s2  4
 sL f  or
 ss 2 4
L sin 2 t 
2
110
Example 4.10
1. Solve the differential equation (IVP)
y0  2
y  3 y  1
Solution:
Ly  3 y  1  Ly  3Ly  L1
But
Ly  sLy  y0  sLy  2
Hence sL y   2  3L y  
 s  3L y   2 
 L y  
1
s
1
s
2
1

s  3 ss  3
 2
1 
 1  1  1 
 y  L1 

 2L1 
L 


 s  3
 s  3 ss  3
 ss  3
1 1
5
1
 y  2e  3t   e  3t  e  3t 
3 3
3
3
2. Solve the differential equation (IVP)
y  3 y  2 y  e t
y0  3, y0  4
Solution:
 
Ly  3Ly  2Ly  L e t ………..(I)
but
Ly  s 2 Ly  sy0  y0
 Ly  s 2 Ly  3s  4 …………….(II)
and
L y  sLy  y0  sLy  3 ………(III)
Put (III) and (II) into (I)
s 2 Ly  3s  4 3sLy  3  2Ly  s 1 1


 s 2  3s  2 L y   3s  5 
 L y  
1
s 1
3s  5
1

s  2s  1 s  2s  1s  1
111
 3s  5

1
 y  L1 


 s  2s  1 s  2s  1s  1
 3s  5  1 

1
 y  L1 
L 


 s  2s  1
 s  2s  1s  1
by partial fraction, we have:
 3s  5  1 1  1  1 1  1  1 1  1 
y  L1 
  L  S  2   2 L  s  1  6 L  s  1






 s  2s  1 3
 y  e2t  2et  13 e2t  12 et  16 et
 y  43 e2t  32 et  16 et
Exercise 4.3
Work out the following (on convolution)
2. 1 sin t
1. 11
4. sin t  cost
3. t  et
5. eit  e it
Inverse Transforms by Convolution. Find f t  by “the convolution theorem”
1.
6
ss  3
2.
1
s 2 s  1
3.

1
s s2  4

4.
s2
s 2   2 2
5.
e as
ss  2
Solve the following differential equations using Laplace Transforms
y0  0 , y0  0
1.
y  2 y  5 y  5
2.
1
y  5 y  6 y  2 cos2t  et y0  0 , y0  10
3.
y  y  F t 
4.
y   2 y  Asin t y0  1 , y0  0
5.
y  y  F t 
1 0  t  1
y0  0 , y0  0 ; where F t   
0 1  t
 4 0t 2
y0  0 , y0  0 where F t   
t 2
t  2
112
4.6
DIRAC’S DELTA FUNCTION


Definition:
Also
f t  t  adt  f a and

 t  adt  1 .
m
m
n f t  t  adt  f a and n  t  adt ,
provided in each case, that n  a  m .
In the Laplace transform form:
L f t  t  a  f ae as and L t  a   e as
Note:  t  a  is not a function in the ordinary sense.
Example 4.11
1.
2 2t
2.
0 sin 5t   t  2 dt  f 2   sin 52   1
3.
0 2t
5
2

 
 3  t  2dt  f 2  2 22  3  11


3

 2t 2  5t  6  t  1dt  7
5  2t
e 
2
t  3dt  f 3  e 23  e6
4.

5.
L t  3  e 3s
6.
L2 t  3  f 3e 3s  2e 3s
7.
L 2t 2  t  1  t  2  f 2e  2s  2 22  2  1 e  2s  7e  2s
8.
L sin 3t   t  2  sin 32  e
9.
Lcosh2t   t   cosh20  e0  cosh0 1  11  1



  


s
2
113
 

s
 e 2
Exercise 4.4
Evaluate the following;
1.
2.
1 3  2t  4t
5
2
 t  3dt

0 sinh 2t  t dt
5  2t
e 
0
t  3dt
3.

4.
0 cos 5t   t  2 dt

Determine the following;
5.
L2   t  1
6.
L e 3t   t  2
7.
Lsin 3t   t   


Solve the following differential equation;
8.
y  y   t      t  2 
y0  0, y0  1
9.
y  4 y  5 y   t  1
y0  0, y0  3
10.
y  2 y  3 y  8et   t  12
 
y0  3, y0  5
114
UNIT 5
SYSTEM OF LINEAR DIFFERENTIAL EQUATIONS
Introduction
In the preceding course (MATH 251) we have discussed methods for solving Ordinary
Differential Equations with constant coefficient that involves only one dependent variable.
Many applications, however, require the use of two or more dependent variables, hence in
this unit, we shall deal with more than one dependent variable and more than one equation.
Learning Objectives
After going through this unit, you would be able to:
•
put systems of differential equations in normal form.
•
reduce a higher order linear differential equation into first order system of
differential equations.
•
put systems of equations in matrix form and solve.
Session 1-1 Introduction and Motivation
In real life situations quantities and their rate of change depend on more than one variable.
For example, the rabbit population, though it may be represented by a single number,
depends on the size of predator populations and the availability of food. In order to represent
and study such complicated problems we need to use more than one dependent variable and
more than one equation.
Such problems lead naturally to a system of simultaneous Ordinary Differential Equations.
Systems of differential equations are the tools to use. The differential equations are very
much helpful in many areas of science. But most of interesting real life problems involves
more than one unknown function. Therefore, the use of system of differential equations is
very useful.
115
As a motivation let us consider an island with two types of species: Rabbits and Fox. Clearly
one plays the role of predator while the other one the role of a prey. If we are interested to
model the populations’ growths of both species, then we have to keep in mind that if, for
example, the population of the Fox increases, then the Rabbit population will be affected. So
the rate of change of the population of one type will depend on the actual population of the
other type. For example, in the absence of the Rabbit population, the Fox population will
decrease (and fast) to face a certain extinction, something that most of us would like to avoid.
A model for this Predator-Prey problem was developed by Lotka (in 1925) and Volterra (in
1926) and is known as the Lotka-Volterra system
 dR
 aR   RF

 dt

 dy  bF   RF

 dt
where R t  measures the Rabbit population, F t  measures the Fox population, and all the
involved constant  a, b, ,   are positive numbers. Note that a and b are the growth rate of
the prey, and the death rate of the predator.  and  are measures of the effect of the
interaction between the Rabbits and The Fox. Can we check Iraq population with system of
differential equation?
Note that in the Lotka-Volterra system, the variable t is missing. This kind of system is called
autonomous system and is written
 dy
 dt  f  x, y 

 dy  g  x, y 
 dt
We will usually denote the independent variable by t and the dependent variables (the
unknown functions of t ) by x1 , x2 , x3 ,, or x, y, z,.
For instance, a system of two first order differential equations in the dependent variables x
and y has the general form:
116

dx dy 

f  t , x, y, ,   0

dt dt 

        ( )
dx dy 

g  t , x, y, ,   0 

dt dt 


where the functions f and g are given.
A solution of this system is a pair ( x(t ), y(t ) of functions of t that satisfy both equations
identically over some real interval of values of t .
1-1.1 Types of Linear Systems
Definition 1:
The general linear systems of two first-order differential equations in two unknown functions
x and y are of the form:
a1 (t ) x  a2 (t ) y   a3 (t ) x  a4 (t ) y  f1 (t )
b1 (t ) x  b2 (t ) y   b3 (t ) x  b4 (t ) y  f 2 (t )
............. (4.1)
where the primes denote derivatives with respect to the independent variable t .
i.e.
x 
dx
dt
y 
dy
dt
Note: We shall be concerned with systems of this type that have constant
coefficients and the number of equations
Example 1.1
1.
3.
2
x  3 y  2 x  y  t
t
x  2 y  3x  5 y  e
2 x  y  2 x  y  e
2.
t
4.
x  3x  y  cos t
117
2
x  2 y  x  y  t  1
x  y  x  t  2
( D  2) x  ( D  1) y  1
( D  1) x  ( D  2) y  t  1
We shall say that a solution of system (1.1) is an ordered pair real functions ( f , g ) such that
x  f (t ) , y  g(t ) simultaneously satisfy both equations of the system (4.1) on some real
interval a  t  b .
Definition 2:
The general linear systems of three first-order differential equations in three unknown
functions x, y and z is of the form:
a1 (t ) x  a2 (t ) y   a3 (t ) z   a4 (t ) x  a5 (t ) y  a6 (t ) z  f1 (t )
b1 (t ) x  b2 (t ) y   b3 (t ) z   b4 (t ) x  b5 (t ) y  b6 (t ) z  f 2 (t ) ..... (4.2)
c1 (t ) x  c2 (t ) y   c3 (t ) z   c4 (t ) x  c5 (t ) y  c6 (t ) z  f 3 (t )
Example 1.2
x  y   2 z   2 x  3 y  z  t
1. 2 x  y   3z   x  4 y  5z  sin t
x  2 y   z   3x  2 y  z  cost
3.
2.
x  z   2 x  5 y  z  1  t
y   3x  5 y  z  tan t
x  y   z   2 z  e t
D  1x1  x2  x3  0
 2 x1  D  3x2  4 x3  0
4.
 4 x1  x2  ( D  4) x3  0
dx1
 3x1  2 x2  2 x3
dt
dx2
 x1  4 x2  x3
dt
dx3
 2 x1  4 x2  2 x3
dt
We shall say that a solution of system (4.2) is an ordered triple of real functions  f , g , h such
that: x  f (t ), y  g (t ), z  h(t ) simultaneously satisfy all three equations of the system (4.2)
on some interval a  t  b .
Systems of the form (4.1) and (4.2) contained only first derivative, and we now consider the
basic linear system involving higher derivatives. This is the general linear system of two
second-order differential equations in two unknown functions x and y , and is a system of
the form:
a1 (t ) x  a2 (t ) y   a3 (t ) x  a4 (t ) y   a5 (t ) x  a6 (t ) y  f1 (t )
...... (4.3)
b1 (t ) x  b2 (t ) y   b3 (t ) x  b4 (t ) y   b5 (t ) x  b6 (t ) y  f 2 (t )
An example is provided below with constant coefficient:
2 x  5 y   7 x  3 y   2 y  3t  1
3x  2 y   2 y   4 x  y  0
118
For given fixed positive integers m and n , we could proceed in like manner, to exhibit other
th
general linear systems of n m - order differential equations in n unknown functions and
give examples each of such type of system (I hope you can do that on your own). Instead we
proceed to introduce the standard type of linear system.
We now introduce standard type as a special case of the system (4.1) of two first-order
differential equations in two unknown functions x and y . We consider the special type of
linear system (4.1), which is of the form:
x  a11 (t ) x  a12 (t ) y  f1 (t )
.................. (4.4)
y   a21 (t ) x  a22 (t )  f 2 (t )
This is called the normal form in the case of two linear differential equations in two
unknown functions. The characteristic feature of such a system is apparent from the manner
in which the derivatives appear in it. An example of such a system with variable coefficients
is
x  t 2 x  (t  1) y  t 3
y   tet x  t 3 y  e t
,
while one with constant coefficients is
x  5x  7 y  t 2
y   2 x  3 y  2t
Note that example (4) of equation (4.2) is in the normal form.
The normal form in the case of a linear system of three differential equations in three
unknown functions x, y and z is:
x  a11 (t ) x  a12 (t ) y  a13 (t ) z  f1 (t )
y   a21 (t ) x  a22 (t ) y  a23 (t ) z  f 2 (t )
z   a31 (t ) x  a32 (t ) y  a33 (t ) z  f 3 (t )
An example of such a system is the constant coefficient system:
119
x   3x  3 y  z  t
y   2 x  4 y  5z  t 2
z   4 x  y  3z  2t  1
The normal form in the general case of a linear system of n differential equations in n
unknown functions x1 , x2 ,, xn is:
x1  a11 (t ) x1  a12 (t ) x2    a1n (t ) xn  f1 (t )
x2  a21 (t ) x1  a22 (t ) x2    a2n (t ) xn  f 2 (t )
............. (4.5)

xn  an1 (t ) x1  an 2 (t ) x2    ann (t ) xn  f n (t )
An important fundamental property of a normal linear system (4.5) is its relationship to a
single n th – order linear differential equation in one unknown function. Specifically,
consider the so-called normalized (meaning, the coefficient of the highest derivative is one)
n th – order linear differential equation
x ( n)  a1 (t ) x ( n1)    an1 (t ) x  an (t ) x  f (t ) ______ (4.6)
in the one unknown function x .
Let
x1  x,
x2  x
x3  x

......................... (4.7)
xn1  x ( n2)
xn  x ( n1)
From (4.7), we get:
x  x1 , x  x2 ,, x ( n1)  xn 1 , x ( n)  xn .......... (4.8)
Then using both (4.7) and (4.8), the single n th – order equation (4.6) can be transformed
into:
120
x1  x2
x2  x3

xn 1  xn
xn  an (t ) x1  an1 (t ) x2    a1 (t ) xn  f (t )
.......... (4.9)
which is a special case of the normal linear system (4.5) of n equations in n unknown
functions. Thus we see that a single n th – order linear differential equation of the form (4.6)
in one unknown function is indeed intimately related to a normal linear system (4.5) of n
first – order differential equation in n unknown functions.
Example 1.3
Reduced the following to normal form:
(a)
y(t )  a1 y(t )  a2 y(t )  a3 y(t )  f (t )
(b)
d 2 x t dx
e
 (t  1) x  0
dt
dt 2
(c)
d 2x
 sinh x  3x
dt 2
Solution (a):
y(t )  a1 y(t )  a2 y(t )  a3 y(t )  f (t ) ................ (α)
Let
y(t )  x0 (t )
y(t )  x0 (t )  x1 (t )
y(t )  x0(t )  x1 (t )  x2 (t )
y(t )  x0(t )  x1(t )  x2 (t )
then from (α), we have
y(t )  f (t )  a1 y(t )  a2 y(t )  a3 y(t ) ................. (β)
from above
x0 (t )  x1 (t ) ............................................................. (γ)
x1 (t )  x2 (t ) ............................................................. (δ)
x2 (t )  f (t )  a1x2 (t )  a2 x1 (t )  a3 x0 (t ) ................ (ε)
121
system
so equation (γ), (δ) and (ε) are the 1st – order linear differential equation equivalent to (α).
These three (3) equations can be solved to obtain the solution to (α).
Solution (b):
x  x0 
Let
dx
dt

dx0
dt
2
 x1 
d x
dt
2
2

d x0
dt
2

dx1
dt
 e
t
dx
dt
 t  1 x
 The normal form is:
dx0
dt
dx1
dt
 x1
 e x1  (t  1) x0
t

Exercise 1.2
In each of exercises 1 – 4, transform the single linear differential equation of the form (1.6)
into a system of first – order differential equations of the form (1.9).
1.
dx 2
dx
 3  2x  t 2
2
dt
dt
2.
d 3x
d 2 x dx

2
  2 x  e 3t
3
2
dt
dt
dt
3.
d 3x
d 2x
dx

t
 2t 3
 5t 4  0
3
2
dt
dt
dt
4.
d 4 x 2 d 2 x dx
t
  2tx  cost
dt 4
dt 2 dt
2.
dx
dy
 2  2x  y
dt
dt
Example 1.5
Reduce the following to normal form;
1.
2
dx dy
  2 x  5 y  2t
dt dt
dx
dy
 2  x  3t  0
dt
dt
2
Solution:
2
dx dy

 2 x  5 y  2t _________(1)
dt dt
122
dx dy
  x  y
dt dt
dx
dy
 2   x  3t ____________(2)
dt
dt
Solving for
dy
dx
by the method of elimination (i.e. eliminating
first) then (1)  2  (2)
dt
dt
5
Also solving for
dx
dx
1
 3x  5 y  t 
 3x  y  t
dt
dt
5
dy
then (1)  (2)  2
dt
5
dx
dy 4
8
 4 x  5 y  8t 
 x y t
dt
dt 5
5
Therefore we have the normal linear system as:
dx 3
1
 x y t
dt 5
5
dy 4
8
 x y t
dt 5
5

Exercise 1.3
Find the first order system of the following differential equations
1.
d2y
 4y  0
dt 2
2.
d3y
d2y
dy
9 3  12 2  4  cost
dt
dt
dt
3.
d 2x
dx
m 2  c  kx  f (t ) , where f (t ) is an external force.
dt
dt
4.
d 3x
d 2x
dx
3 3  2 2  12  8x  e t sin t
dt
dt
dt
5.
d 4x
d 2x
 5 2  4 x  e t  te 2t
4
dt
dt
d5x
d 3x
d 2x

2

2
 3t 2  1
6.
5
3
2
dt
dt
dt
123
Reduce each of these to the normal form.
1.
dy
dx
 2  et
dt
dt
dy dx

 x  y  sin t
dt dt
2.
dx dy
2   x  et
dt dt
dx
dy
 3  2 x  4 y  sinh t
dt
dt
3.
dy dx

 x  2 y  sin t
dt dt
dy dx

 4e t  x
dt dt
4.
d 2x
dx
dy
6 3  0
2
dt
dt
dt
2
d y dx
  x  y  4t
dt 2 dt
1-1.2 Solution to a Linear System of Differential Equations
The solution to the system:
dx1

 f1 t , x1 , x2 , x3 , xn  
dt

dx2
 f 2 t , x1 , x2 , x3 , xn 
dt
 ....................... (1.10)



dxn
 f n t , x1 , x2 , x3 , xn 
dt

 x1 (t ) 
 x (t )
2
 on open real interval a  t  b for which the system is
is the parametric vector x  




 xn (t )
true and the vector x must be continuous on the open real interval.
Session 2-1 Methods of Solving SLDE
The most elementary approach to Systems of Linear Differential Equations (SLDE) with
constant coefficients involves the elimination of dependent variables by appropriately
combining pairs of equations.
The main methods we will consider in solving such systems are:
A)
The differential operator method,
B)
The matrix method.
124
The differential operator method for linear differential system is quite similar to the solution
of linear algebraic systems by elimination of variables until only one remains. It is most
convenient in the case of manageably small systems: those consisting of no more than four
equations. For large systems of differential equations, the matrix method is preferable.
2-1.1 The Differential Operator Method
Recall that D 
d
d2
dn
 D 2  2 and also D n  n
dt
dt
dt
System of 1st Order Equations: The differential operator method for linear differential
system is quite similar to the solution of linear algebraic systems by elimination of variables
until only one remains. It is most convenient in the case of manageably small systems: those
consisting of no more than four equations. For large systems of differential equations, the
matrix method is preferable.
Example 1.6
Solve the system
dx

 x y 
dx
dy

dt


dy
x  y 4x  y
 4 x  y 
dt

Note: this system is called Autonomous system since time is not explicitly expressed
in the equation.
Solution:
We use matrix/determinant techniques. We rewrite the problem as follows:
D  1x  y  0
 4 x  D  1 y  0
 D  1 1  x   0 
 y    0 

4
D

1

   

Using Cramer’s Rule:
125
D  1 1
0 1
x
4 D  1
0 D 1
D  1
2



 4 x  0  D 2  2D  3 x  0  x  0 D  3,1
(Just like finding the roots of a characteristic equation in linear differential equation
we did last semester)
 x  Ae3t  Be t
Then go back to the system to pick one equation to determine y .
y


dx
 x  3 Ae3t  Bet  Ae3t  Bet  2 Ae3t  2Bet
dt
Then the solution is written as:
x(t )  Ae 3t  Be t
y(t )  2 Ae 3t  2Be t
Example 1.7
1.
Solve the system
dx
dy

, when x(0)  1 , y(0)  2
3x  y x  y
Solution:
dx
 3x  y  D  3x  y  0
1  x   0 
D 3
dt
    
 
 1 D  1 y   0 
dy

 x  y   x  D  1y  0
dt
Using Cramer’s Rule:
D3
1
0
1
  D  3 D  1  1 x  0
x
1 D  1
0 D 1

 D

 D2  D  3D  4 x  0
2



 4D  4 x  0  D  22 x  0 but x  0
 x   At  Be 2t _________(*)
Solving for y , we have: y  3x 
dx
dt
 y  3e 2t At  B  A  2At  Be 2t
126
 y  3e 2t At  B  2 At  2B  Ae 2t
 y  Ate 2t  B  Ae 2t _____ (**)
with the initial conditions:
(*)
 B 1
(**)
 2  B  A  A  1

x  1 t e 2t
y  te2t  2e 2t
2.
Solve the system:
 D  1 x  Dy  2t  1
 2D  1 x  2Dy  t
Solution:
 D  1 x  Dy  2t  1........()
 2D  1 x  2Dy  t...........()
Subtracting twice (I) from (II), we have: 3x  3t  2 .
Substituting x  t  2 / 3 in (I), we obtain:
Dy  2t  1   D  1 x  t  43
y  12 t 2  43 t  c
The complete solution:
x  t  23 , y  12 t 2  43 t  c .

Exercise 4.4
1.
dx
y
dt
dy
x
dt
4.
dx
 2 x  y
dt
dy
 x y
dt
2.
dx
 y
dt
dy
x
dt
5.
dx
 3x  2 y  t
dt
non-homogeneous
dy
 2x  3 y  3
dt
127
3.
dx
 6x  8 y
dt
dy
 6x  2 y
dt
6.
dx
 6x  8 y
dt
dy
 6 x  2 y  e t
dt
8.
dx1
 3x1  x2
dt
dx2
 3x1  2 x2
dt
dx3
 x1  x2  x3
dt
10.
7.
dx
 3x  2 y  2 z
dt
dy
 x  4y  z
dt
dz
 2 x  4 y  z
dt
9.
dx1
 x1  x2  x3  e t
dt
dx2
 x1  x2  t
dt
dx3
 x1  x3  1
dt
Find the solution of the system of differential equation Ex. 1 such that x0  1
, y0  2 .
11.
Find the solution of the system of differential equation is Ex.5 such that
x0  3 , y0  1 .
12.
Find the solution of the system of differential equations in Ex.7 such that
x0  2 , y0  1 , z0  0 .
13.
Find the solution of the system of differential equations in Ex.9 such that
x1 0  1 , x2 0  1 , x3 0  0 .
128
2-1.2 Reduction of nth – Order Equation to a System of 1st Order Equations
The number of integration constants in an n th order differential equation, in the same vein, is
the same as the number of integration constant in a system of n 1st order differential
equation.
Example 1.8
1.
Convert y  3y  2 y  et into a system of equations and solve completely.
Solution:
y(t )  x0 (t )
Let
y(t )  x0 (t )  x1 (t )
y(t )  x1 (t )
The system becomes:
x0 (t )  x1 (t )

 non – homogeneous example
x1 (t )  et  3x1 (t )  2 x0 (t )
Then the solution of the system:
 D  1  x0   0 

    t 
2
D

3

 x1   e 
Using Cramer’s Rule:
D 1
0
x1  t
2 D3
e
1
D3
 DD  3  2x1  et


 D2  3D  2 x1  et
 D  2D  1x1  et
Let D  2x1  z
D  1z  et  z 
 z  et
e
t
1 t
e
D 1
 z  tet  k1et
e t dt
Then
D  2x1  tet  k1et


 x1  e 2t  tet  k1e t e 2t dt
129



 x1  e 2t  te t  k1e t dt
 x1  e 2t  tet  e t  k1e t  k2

 x1  et 1  t   k1e t  k2 e 2t
Since we are solving for y(t ) , then we have
y  Ae t  Be 2t  et 1  t 

Exercise 1.4
y  7 y  12y  t
1.
D
D
3.
5.
7.
D
2
2.
D  32 y  te3t
DD  1y  5
2
 5D  6 y  e 2t

4.
2
 2D y  e  x cos x
6.


 1 y  tan t 2 ,   2  t   2 8.
D
D
2
2

 Dy  2
D y x
t
2-1.3 The Matrix Method
Although the simple differential operator method suffice for the solution of small linear
systems containing only two or three equations with constant coefficients, the general
properties of linear systems – as well as solution methods suitable for large systems – are
most easily and concisely described in matrix notation.
We discuss here the general system of n first order linear equations being the form: (note
equation 4.5)
dx1
 a11 (t ) x1  a12 (t ) x2   a1n (t ) xn  f1 (t )
dt
dx2
 a21 (t ) x1  a22 (t ) x2   a2 n (t ) xn  f 2 (t )
______ (4.5)
dt
dxn
 an1 (t ) x1  an 2 (t ) x2 
dt
 ann (t ) xn  f n (t )
where f1 t , f 2 t , f 3 t ,, f n t  are continuous on a real interval a  t  b .
130
The Matrix Notation form of (4.5) is given as:
 dx1 


 dt   a11 a12
 dx2  
 dt   a21 a22
 dx3    a
a32

  31

 dt   
   a
 dxn   n1 an 2


 dt 
This can simply be written as:
a13
a23
a33

an3
a1n   f1 t  
 

a2n   f 2 t 
a3n    f 3 t 
 

    
ann   f n t 





dx
 Ax  F(t ) …………............... .(I)
dt
which is a NON-HOMOGENEOUS FORM.
where F(t ) is the non-homogeneous term or the force term.
If F(t )  0 , that is f1 (t )  f 2 (t )  f 3 (t )    f n (t )  0 for the real interval a  t  b then
we have
dx
 Ax ………………............... (II)
dt
Equation (II) is known as the HOMOGENEOUS FORM.
Examples of non – homogeneous systems of linear differential equation
1.
dx
 x  y  e t
dt
dy
 x  y  6sin t
dx
The matrix notation form is given as:
 dx 
 dx 
 dt  1 1 x   et 
1 1
 x
dx  dt 
 , x    and
   , A  

 
 where


dt  dy 
 dy  1 1 y   6sin t 
1  1
 y
 
 
dt


 dt 
 e t 

F(t )  
 6 sin t 
131
2.
dx1
 3x1  x2  2 x3  t 2
dt
dx2
 4 x1  2 x2  x3  e t ,
dt
dx3

 2 x 2  5 x3  sin t
dt
The matrix notation form is given as:
 dx1 
 dx1 




2
dt 
1  2  x1   t 
1  2
3
 x1 
 dt   3


  


 
dx  dx2 
t
 dx2    4  2
1 x2     e  where

, A  4  2
1 , x   x2 

 dt 
dt  dt 
0  2
x 
5  x3    sin t 
 dx3   0  2
 dx3 
5 

 3




 dt 
 dt 
 t2 


and F(t )    e t 
  sin t 


Note that
dx
 Ax  F(t ) can be simply be written as x  Ax  Ft  .
dt
Examples of Homogeneous systems of linear differential equation
1.
dx
 5x  3 y
dt
dy
 2 x  4 y
dt
 dx 
 dx 
 
 
dx  dt 
dt    5  3 x 


The matrix notation form is given as :
where
,
 dy    2
4  y 
dt  dy 
 
 
 dt 
 dt 
 5  3
 x
 and x   
A  
4
 2
 y
2.
dx1
 x1  x2  x3
dt
dx2
 2 x1
 4 x3
dt
dx3

 2 x1  3x3
dt
132
The matrix notation form is given as:
 dx1 
 dx1 




1  1 x1 
1  1
1
 x1 
 dt   1
 dt 
 


 
dx  dx2 
 dx2    2
0  4  x2  where

, A  2
0  4  and x   x2 

 dt 
dt  dt 
0  2
x 
3 x3 
 dx3   0  2
 dx3 
3

 3




 dt 
 dt 
Note that
dx
 Ax can simply be written as x  Ax
dt
Definition
Let A and F be defined on an open interval I. A vector function u is a solution of equation
(I) on I if:
u(t )  Au(t )  F(t )
Example 1.10
Consider the differential equation
 5  3
x -------------- (1i)
x  
4 
 2
 3e 7t 
 is a solution of equation (1i) on
The vector function: u(t )  
7t 
  2e 
 21e 7t 
 5  3 3e 7t   21e 7t 





(, ) since u(t )  
and
7t 
4   2e 7t    14e 7t 
 2
  14e 
 e 2t 
 2e 2t 
Similarly, v(t )   2t  is a solution of equation (1) on (, ) since: v(t )   2t  and
e 
 2e 
 5  3 e 2t   2e 2t 

 2t    2t 

2
4

 e   2e 
133
Example 1.11
Consider the equation:
 1 1  1


x   2 3  4 x ----------- (1ii)
 4 1  4


 e 3t 
 e 2t 
 3t 


The vector functions: u(t )   7e  and v(t )   2e 2t  are solutions on (, ) of equation
11e 3t 
 e 2t 




  3e 3t 
 2e 2t 




(1ii) since: u(t )    21e 3t  v(t )   4e 2t  and
  33e 3t 
 2e 2t 




3t
3t
 1 1  1 e    3e 


3t
3t
 2 3  4  7e     21e 
 4 1  4 11e 3t    33e 3t 


 

2t
2t
 1 1  1 e   2e 

 2t
2t
 2 3  4  2e    4e 
 4 1  4  e 2t   2e 2t 


 

 et 
 
It is now left for you to verify that : w(t)   e t  is also a solution on (, ) of equation (1ii)
 et 
 
Theorem 1.1
If u and v are solutions on an open interval I of x  Ax and c1 ,c2 are any two constants,
then w(t )  c1u(t )  c2 v(t ) is also a solution on I .
Example 1.12
 3e 7t 
 e 2t 


In example 1.10 we found that u(t )  
and v(t )   2t  are solutions
7t 

2
e


e 
 5  3
x
of x  

2
4


134
 3e 7t 
 e 2t   3c1e 7t  c2 e 2t 
 is also a solution for any choice

  

c
By Theorem 1.1, c1 
2  2t   
7t
2t 
7t 
  2e 
 e    2c1e  c2 e 
of the constants c1 and c 2 .
The vector – valued functions x1 , x 2 ,x n are linearly dependent on the interval I
provided that there exist constants c1 , c2 ,, cn not all zero such that:
c1x1 (t )  c2 x 2 (t )   cn x n (t )  0 for all t in I . Otherwise they are linearly independent.
Equivalently, they are linearly independent provided that no one of them is a linear
combination of the others.
135

Exercise 1.5
a. Write the given system in the form: x(t )  A  F(t )
1.
x   3x  2 y
y  2x  y
2.
x  x  2 y  cost
y   x  y  sin t
6.
3.
x  2 x  4 y  3e t
y   5x  y  t 2
x   3x  4 y  z  t
4.
y   x  3z  t
2
z  6 y  7z  t 3
x  4 x  y  z
5. y   2 x  y  z
z  x  3y  z
State whether homogeneous or non – homogeneous.
b. Verify that the given vectors are solutions to the given system, and then use the
Wronskian to show that they are linearly independent. Finally write the general
solution of the system.
1.
 e 3t 
 2e 2t 
  3 2
x ; x1   3t  , x 2   t 
x  
  3 4
 3e 
 e 
2.
 3  1
1
 1
x ; x1  e 2t   , x 2  e 2t  
x  
 5  3
1
 5
3.
 4 1
 1
 2
x ; x1  e 3t   , x 2  e 2t  
x  
  2 1
  1
  1
4.
 0 1 1
1
 1
 0


 
 
 
2t
t
t
x   1 0 1x ; x1  e 1 , x 2  e  0  , x 3  e  1
 1 1 0
1
  1
  1


 
 
 
5.
0
 3 2
 2
  2
 2


 
 
 
t
3t
5t
x    1
3  2 x ; x 1  e  2  , x 2  e  0  , x 3  e   2 
 0 1
 1
 1
 1
3

 
 
 
136
Theorem 1.2: Principle of Superposition
Let x1 , x 2 ,x n be n solutions of homogeneous linear equation (II) i.e.:
dx
 Ax on the
dt
open interval (, ) . If c1 , c2 ,, cn are constants, then the linear combination,
x  c1x1 (t )  c2 x 2 (t )    cn x n (t ) is also a solution of (II). Prove as exercise.
2-1.4 The Eigenvalue Method of Homogeneous Linear Systems
In this section we will present a method for finding a general solution of the differential
equation:
dx
 Ax ………. (III)
dt
where A is an n n matrix.


Recall that ordinary differential equation: Dy  ay or D 2  aD  b y  0 , has solutions of
the form: y(t )  ce rt , where c is an arbitrary constant, r is the root(s) of the characteristic.
We will show that equation (III) has analogous function for solutions in system of
differential equation. That is, we will show that equation (III) has solutions of the form:
x(t )  vet where  is a constant (possibly a complex number) and v is a constant.
  
 
  
To begin, we note that if: v    then:

 
 
 
 e t1   e t1 
 1 
 t   t 
 
d t
d  e  2   e  2 
t  2 
e v  

 e    e t v



dt
dt      
 
 
 e t   e t 
 n
n
n


 
Thus x(t )  vet is a solution of x  Ax if and only if e t v  et Av or, equivalently, if and
only if Av  v
2-1.5 Definition: Eigenvalue and Eigenvector
The number  (either zero or non – zero) is called an eigenvalue of n n matrix provided:
det(A  I)  A  I  0
137
An eigenvector associated with the eigenvalue  is a non-zero vector v such that Av  v ,
so that: A  v   0
Lemma: If x(t )  v k e k t is a solution to (III) for the k th eigenvalue of A of order n n
then: x(t )  c1 v1e 1t  c2 v 2 e 2t    cn v n e nt is also a solution, where ci is a constant for
i  1,2,, n
In outline, the eigenvalue method for solving system x  Ax goes as follows:
(I)
Solve the characteristic equation det(A  I)  A  I  0 for the eigenvalues
1 , 2 ,, n of matrix A .
(II)
We attempt to find n linearly independent eigenvector v1 , v 2 ,, v n (so that
v k  0 for each k ) associated with these eigenvalues. That is
A  Iv k
 0 for each k . This will not always be possible, but when it is,
we get n linearly independent solutions:
x1 (t )  v1e t , x 2 (t )  v 2 e t ,, x n (t )  v n e t
(III)
The general solution is then a linear combination of these n solutions :
x(t )  c1 v1e t  c2 v 2 e t    cn v n e t
We will discuss the various cases that can occur separately, depending upon whether the
eigenvalues are distinct or repeated, real or complex.
Case 1: Distinct Real Eigenvalues
If the eigenvalues 1 , 2 ,, n are real and distinct, then we substitute each of them in turn in
A  I   0 and solve for the associated eigenvectors
Example 4.13
Find a general solution of the system
x1  4 x1  2 x2
x2  3x1  x2
138
v1 , v 2 ,, v n .
Solution:
 4 2
x
The matrix form of the system is: x  
 3  1
The characteristic equation of the coefficient matrix is:
4
2
0
3
1 
 4    1     6  0  2  3  10  0    2  5  0
So we have the distinct real eigenvalues: 1  2 and 2  5
For the coefficient matrix A the eigenvector equation A  I   0 takes the form:
2  1   0 
4  

     ………..(δ)
 1    2   0 
 3
When we substitute the first eigenvalue 1  2 , we find that the two scalar equations that
follow are: 61  2 2  0 and 31   2  0 .
These equations are equivalent i.e. they are redundant, so only one is needed; we can choose
1 arbitrary (but non-zero) and solve for  2 . The simplest choice is 1  1 , which yields
 1
 2  3 , and thus: v1    is an eigenvector associated with 1  2 (as is any zero
  3
constant multiple of 1 ).
Remark: If, instead of the “simplest” choice 1  1 ,  2  3 , we had made another
choice, 1  c ,  2  3c , we would have obtained the eigenvector
 c   1
  c 
v1  
  3c    3 
Because this is a constant multiple of our previous result, any choice we make leads to the
 1
same solution: x1 (t )   e 2t
  3
139
Next we substitute in (δ) the second eigenvalue 2  5 and get the equivalent scalar
equations  1  2 2  0 and 31  6 2  0 .
 2
With  2  1 we obtain 1  2 , so: v 2    is an eigenvector associated with 2  5 . A
1
different choice 1  2c ,  2  c would merely give a (constant) multiple of v 2 .
 1
These two eigenvalues and associated eigenvectors yield the two solutions x1   e 2t
  3
 2
and x 2   e 5t .
 1
e 2t
They are linearly independent because their Wronskian
 3e 2t
2e 5t
 7e 3t  0 , since
5t
e
e 3t  0 is nonzero. Hence a general solution of the system is:
 1
 2
x  c1x1  c2 x 2  c1  e 2t  c2  e 5t ;
  3
1
in scalar form: x1  c1e 2t  2c2 e5t ,
x2  3c1e 2t  c2 e5t .
Case 2: Repeated Real Eigenvalues
Suppose that the n n matrix A has m  n distinct eigenvalues 1 , 2 ,, n . Then at least
one of these eigenvalues is a repeated root of the characteristic equation: A  I  0 .
An eigenvalue is of multiplicity k if it is a k  fold root.
Let 1 and 2 be eigenvalues of the system matrix such that 1  2   then the system has
two linearly independent solutions of the form: x  c1x1  c2 x 2 , where x1  ve t and
x 2  vtet  wet . And we use the equation A  I w  v for the second part that is x 2 .
Example 1.15
Find the general solution of the system
140
x1  3x1  x2
x2   x1  x2
Solution:
The characteristic equation is:
3
1
   22  0    1  2  2
1
1 
1  1   0 
 3  
    
 1   2   0 
 1
The eigenvector will be solution of: 
For
     0
1
1
 2t
  2,  1 2
  e  2t
then v1     x1  v1e
1
1
 1  2  0
With this eigenvector v1 and eigenvalue
  2
we use the equation: A  I w  v1
  1 1 1  1
      1  2  1
 
  1 1 2  1
Let 1  0 then
2  1
Therefore x 2  v1te
 2t
1
 0
 we  2t   te 2t   e  2t
1
 1
1
Hence x  c1x1  c2 x 2  c1  e
1
 
 2t
 t   2t
e .
 c2 
t

1


Use WRONSKIAN to determine whether their linearly independent.
Case 3: Complex Eigenvalues
To find the general solution of a two – dimensional system x  Ax , where the eigenvalues of
A are nonreal (i.e. complex conjugates),     i :
I.
Find a complex eigenvector v  v1  iv 2 corresponding to an eigenvalue  .
II.
Construct a complex – valued solution of the system as:
t
xe
cos  t  i sin t  v ,     i .
III.
Separate the solution into a real part x1 and an imaginary x2 .
IV.
The general solution of the system is x  c1x1  c2x2
141
Example 1.14
1.
Find a general solution of the system
x1   x1  4 x2
x2  2 x1  3x2
Solution:
The characteristic equation of the coefficient matrix is given as:
1 
4
  1   3     8  2  2  5  0
2
3
  1  2i , Note: the complex roots are conjugates.
Next we choose   1  2i and find an eigenvector.
Thus,
The eigenvectors will be solutions of:
4
  1  1  2i 
 v1   0 

    



2
3

1

2
i

 v2   0 
4  v1   0 
  2  2i

    
2  2i  v2   0 
 2
 2  2i v1  4v2  0

 2v1  2  2i v2  0
As before, the equation corresponding to the tow row are redundant, so only one is needed.
Using the top row, we obtain the equation:  2  2i v1  4v2  0 or v2  1  i  / 2v1 we
may avoid fractions by choosing v1  2 ; then v2 1  i . Hence, we have the eigenvector:
 2 

v  
1  i 
t
A complex – valued solution is constructed from the formula x  e v and Euler’s formula:
 2  12i t
e
1  i 
x
2cos t  2i sin 2t

 2  t
t
e  cos 2t  i sin 2t   e 


1  i 
 1  i  cos 2t   1  i  sin 2t 

The real part of this solution is: x1  e
t
 2cos 2t 
 cos 2t  sin 2t  and the imaginary part is:


2sin 2t 
t
x2  e 

 cos 2t  sin 2t 
142
Both of these real-value functions are solutions of the system, and with them we get a general
solution: x  x1  x 2
2cos 2t 
 2sin 2t 
t  
x  e c1 
 c2 


 cos 2t  sin 2t 
  cos 2t  sin 2t 
Case 4: Systems with Zero as an Eigenvalue
We discussed the case of system with two distinct real eigenvalues, repeated (nonzero)
eigenvalue, and complex eigenvalues. But we did not discuss the case when one of the
eigenvalues is zero.
Example 1.15:
Find the general solution to
x1  2 x1  x2
x2  2 x1  x2
Solution:
The procedure for solving a system with zero eigenvalue is the same as system with distinct
real eigenvalue.
2   1
0
2 1  


The characteristic polynomial of this system is:   2  1   0  0 , which reduces to
2
 2  3  0 . The eigenvalues are 1  0 and 2  3 .
Let us find the associated eigenvectors;


For 1  0 , A  I1 v1  0
2  0  1   v1  0
  2 1  0 v   0

 2   


2v1  v2  0
 2v1  v2  0
The equation A  I1 v1  0 translates into: 
1

 2
The two equations are the same. So we have 2v1  v2 . Hence an eigenvector is: v1  
143
 v1 
For 2  3 , set: v 2   
 v2 


2  3  1   v1  0
 v   0 or

2
1

3

 2   
The equation A  3I v 2  0 translates into: 
 v1  v2  0
 2v1  v2  3v1


 2v1  v2  3v2  2v1  2v2  0
The two equations are the same (as  v1  v2  0 ). So we have v2  v1 . Hence an
1
eigenvector is: v 2    .
  1
0t 1 
3t 
1
1 
3t 
1
Therefore the general solution is: x  c1e    c2 e    c1    c2 e  
 2
  1
 2
  1

Exercise 1.6
a. Apply the eigenvalue method of this section to find a general solution of the given
system. If initial values are given, find the corresponding particular solution.
1.
x1  x1  2 x2
x2  2 x1  x2
4.
x1  2 x1  3x2
x2  2 x1  x2
2.
x1  3x1  4 x2
; x (0)  x2 (0)  1
x2  3x1  2 x2 1
5.
x1  4 x1  x2
x2  6 x1  x2
3.
x1  6 x1  7 x2
x2  x1  2 x2
6.
x1  9 x1  5 x2
;
x2  6 x1  2 x2
b. Find general solutions of the systems in these exercises.
1.
x1  5x1
 6 x3
x2  2 x1  x2  2 x3
x3  4 x1  2 x2  4 x3
x1  3x1  2 x2  2 x3
3. x2   5x1  4 x2  2 x3
x3  5x1  5x2  3x3
144
2.
x1  3x1  x2  x3
x2   5x1  3x2  x3
x3  5x1  5x2  3x3
4.
x1  2 x1  x2  x3
x2   4 x1  3x2  x3
x3  4 x1  4 x2  2 x3
x1  2 x1  3x2  3x3
 x2  3x3 has the
c. Show that the coefficient matrix of the system x2 
x3 
2 x3
eigenvalue   2 of multiplicity 2. Find two linearly independent eigenvectors
associated with   2 , and then find a general solution of the system.
d. i) Show that the coefficient matrix of the system
x1  3x1  x2
x2   x1
 x3
x3  x1  2 x2  3x 3
has the single eigenvalue   2 of multiplicity 3 , but with only one
linearly independent eigenvector u associated with it. Then one
solution is x1 (t )  ue 2t .
ii) Find a second solution of the form x 2 (t )  ute2t  ve 2t where v is a nontrivial
solution of A  2I v  u .
iii) Find a third solution of the form x 3 (t )  12 ut 2 e 2t  vte 2t  we 2t where w is a
nontrivial solution of A  2I w  v .
e. Solve for the system
x  y   2 x  4 y
2 x  3 y   2 x  6 y
Hint: the system above is not in the normal form, so reduce it to normal form first
before solving it.
145
2-1.6 Non–Homogeneous Linear Systems
Here we discuss two ways (methods) of solving non – homogeneous linear systems. The
methods are;
I.
Undetermined coefficients
II.
Variation of parameters
These are analogous to finding a particular solution of a single non-homogeneous n th order
linear differential equation.
Given the non-homogeneous first order linear system:
x(t )  A(t )x  F(t ) …………… (1)
The general solution of (1) has the form:
x(t )  x c (t )  x p (t ) …………….(2)
where x c denotes a general solution of the associated homogeneous system: x(t )  A(t )x
and x p is a single particular solution of the equation (1)
I.
Undetermined Coefficient
The method is essentially the same for a system as for single differential equations – we make
an intelligent guess as to the general form of a particular solution x p and then attempt to
determine the coefficients in x p by substitution in x(t )  Ax  F(t ) .
Example 1.16
Find a general solution of the system
x  4 x  2 y  8t
………...... (3)
y   3x  y  2t  3
Solution:
We found the general solution
x 
 1
 2
x c   c   c1  e 2t  c2  e 5t ……….... (4)
  3
 1
 yc 
of the associated homogeneous system. Because there is no duplication between the terms
x c and the non – homogeneous terms, we assume a trial solution of the form:
146
 xp 
 a  b 
x p     at  b   1 t   1  ……....... (5)
 a2   b2 
 yp 
Upon substitution of (5) into (3) we get:
 a1   4 2  a1t  b1    8t 
   
  


 a2   3  1 a2 t  b2   2t  3
 4a  2a 2 8   4b1  2b2 
t  

  1
 3a1  a2  2   3b1  b2  3
When we equate the coefficients of t and constant terms, we get the equations:
4a1  2a2  8,
3a1  a2  2,
4b1  2b2  a1 ,
3b1  b2  a2  3
hence a1  52 , a2  165 , b1 
2
25
and b2 
1
25
.
 52 t  252 
 and the general solution of the system is given
Thus our particular solution is: x p   16
1 
t

25 
5
as:
x  c1e 2t  2c2 e 5t  52 t  252
y  3c1e 2t  c2 e 5t  165 t  251
Example 1.17
Find a general solution of the system
x   4 x  2 y,
y   3x  y  e  2t
…………….(6)
Solution:
The complimentary solution is given as:
x 
 1
 2
x c   c   c1  e 2t  c2  e 5t …….. (7)
  3
 1
 yc 
Because of duplication of the term e 2t in non-homogeneous system, we assume a trial
 xp 
 a1t  b1  2t
e ……(8)
solution of the form: x p     ate 2t  be 2t  
y
a
t

b
p
2
2


 
(rather than ate2t alone). Then xp  (a  2b)e 2t  2ate 2t ……. (9)
147
Substitution of (8) and (9) into (6), and cancellation of e 2t throughout, yields:
 a1  2b1  2a1t   4 2  a1t  b1   0 

  
   

 a2  2b2  2a2 t   3  1 a2 t  b2   1 
If we equate the coefficients of t and the constant terms, we get the equations:
6a1  2a2  0,
3a1  a2  0,
6b1  2b2  a1 ,
3b1  b2  a2  1
The first two equations imply that a2  3a1 . In order for the last two equations to be
consistent, we must have a1  2(a2  1)  2(3a1  1)  6a1  2 .
It follows that a1   72 , and so a2  76 and b2  3b1  17
Hence the general solution of the system is given by:
x  c1e 2t  2c2 e 5t  72 te 2t
y  3c1e 2t  c2 e 5t  76 te 2t  17 e 2t
Fundamental Matrices
Suppose that x1 (t ), x 2 (t ),, x n (t ) are n linearly independent solutions on some open
interval of the homogeneous system x  Ax of n linear equations. Then the n n matrix
 x11 (t ) x12 (t )  x1n (t ) 


 x21 (t ) x22 (t )  x2 n (t ) 
Φ(t )  


 


 x (t ) x (t )  x (t ) 
n2
nn
 n1

having as column vectors the solution vectors x1 , x 2 ,, x n is called a fundamental matrix
for the system. Because its column vectors are linearly independent, it follows that the matrix
Φ(t ) is non-singular and therefore has an inverse matrix Φ1 (t ) . In terms of the fundamental
matrix Φ(t ) , the general solution: x  c1x1 (t )  c2 x2 (t ) 
form: x(t )  Φ(t )c ………….(α)

where c  c1 , c2 ,..., cn


148
 cn xn (t ) can be written in the
In order that the solution x(t ) satisfy the initial condition: x(t 0 )  b where b1 , b2 ,bn are
given, it therefore will suffice for the coefficient vector c in x(t )  Φ(t )c to satisfy
1
Φ(t0 )c  b ; that is: c  Φ (t 0 )b …………….(β)
When we combine (α) and (β), it becomes clear that the solution of the initial value problem
x  Ax , x(t 0 )  b is given by: x(t )  Φ(t )Φ 1 (t 0 )b ……………(γ)
Example 1.18
Find the fundamental matrix for the system:
x  4 x  2 y
and use it to find the solution of the
y   3x  y
system which satisfies the initial conditions x(0)  1, y(0)  1 .
Solution:
 e 2t 
 2e 5t 

 5t  yield the fundamental
x
(
t
)

The linearly independent solutions: x1 (t )  
and
 2t 
  3e 
e 
 e 2t
matrix: Φ(t )  
 2t
  3e
2e 5t 

e 5t 
 1 2
 and the inverse matrix is given as: Φ 1 (0) 
Then Φ(0)  

3
1


1
7
1  2


3 1 
Hence the formula (γ) gives the solution:
 e 2t
x(t )  
 2t
  3e
2e 5t  1  1  2  1
7
  
1  1
e 5t   2
1
7
 3e 2t  4e 5t 


 2t
5t 

9
e

2
e


Thus the solution of the original initial value problem is given by
x  73 e2t  74 e5t
y   97 e2t  72 e5t
149
II.
Variation of Parameters
Theorem
If Φ(t ) is the fundamental matrix of
dx
dx
 Ax . Let x p be a solution to
 Ax  F(t ) , then
dt
dt
we have the following; x p (t )  Φ(t ) Φ 1 (t )F(t )dt
x(t )  Φ(t )c  Φ(t ) Φ 1 (t )F(t )dt ………….. (ζ)
Then also with definite integration we have x p  Φ(t ) Φ(s)F(s)ds
t
t0
If we add this particular solution and the homogeneous solution, we get the solution:
x(t )  Φ(t )Φ 1 (t 0 )b  Φ(t ) Φ 1 (s)F(s)ds ……….(ε)
t
t0
of the initial value problem x(t )  A(t )x  F(t ), x(t0 )  b .
Example 1.19
 4 2  15 2t
 1
x   te , x(0)   
Solve the initial problem x  
 3  1  4 
  1
Solution:
If we were to use the method of undetermined coefficients, our trial solution would be of the
form x p (t )  at 2 e 2t  bte 2t  ce 2t , and we would have six scalar coefficients to determine.
Here it will be simpler to use the method of variation of parameters.
 e 2t
We already know the fundamental matrix: Φ(t )  
 2t
  3e
2e 5t 
 of the associated
e 5t 
homogeneous system. The determinant , Φ(t )  7e3t , so the inverse matrix:
 e 2t
Φ 1 (t )  17 e 3t 
 2t
  3e
2e 5t 

e 5t 
From the theorem gives the particular solution:
x p  Φ(t ) e
3 s
 Φ (t ) e
3 s
t
1
0 7
t
1
0 7
 e 5s
 2 s
 3e
 2e 5s   15se2 s 

ds
e 2 s   4se2 s 
  7se3s 

ds
4 s 
  49se 
150
t
s 
ds
 Φ (t ) 
0  7 se 7 s 


s t
  12 s 2

 Φ (t ) 7 s 1 7 s 
 se  7 e  s 0
 e 2t
 
 2t
  3e

 12 t 2
2e 5t 



e 5t  te 7t  17 e 7t  17 
Therefore:
 4  28t  7t 2  1 5t  2 
  7 e  
x P (t )  141 e 2t 
2
2

14
t

21
t
1


This is a particular solution such that x p (0)  0 .
 3e 2t  4e 5t 
 of the
We already know (from previous examples) the solution: x c (t )  17 
 2t
5t 
  9e  2e 
 1
associated homogeneous systems such that: x c (0)   
  1
Hence the solution of the initial value problem is given by:


 16  14t  7t e
x  141 10  28t  7t 2 e 2t  72 e 5t
y  141
2
 2t
 17 e 5t

Summary: The Method of Solving Non-Homogeneous systems
dx
 Ax .
dt
1.
Solve for the homogeneous part
2.
Form your fundamental matrix Φ(t )
3.
Find Φ1 (s)
4.
Evaluate
5.
Find x p  Φ(t ) Φ(s)F(s)ds .
6.
Your final step x(t )  Φ(t )Φ 1 (t 0 )b  Φ(t ) Φ 1 (s)F(s)ds .
Φ
t
t0
1
(s)F(s)ds
t
t0
t
t0
151

Exercise 1.7
a. Apply the method of undetermined coefficients to find a particular solution of each of
the system below. If initial conditions are given, find the particular solution that
satisfies these conditions.
1.
2.
3.
x  x  2 y  3
y  2x  y  2
x   3x  4 y  3
y   3x  2 y  t 2
7.
x(0)  y(0)  0
8.
x  6 x  7 y  10
9.
y   x  2 y  2e t
4.
x  3x  4 y  sin t
x(0)  1, y(0)  0
y  6x  5 y
10.
5.
x  x  5 y  cos 2t
y  x  y
11.
6.
x  2 x  4 y  2
x(0)  1, y(0)  1
y  x  2 y  3
12.
x  2 x  3 y  5
y   2 x  y  2t
x  4 x  y  e t
y  6x  y  et
x(0)  y(0)  1
x  9 x  y  2e t
y   8x  2 y  tet
x  x  5 y  2 sin t
y   x  y  3 cost
x  x  2 y
y   2 x  y  e t sin t
x  x  y  2t
y   x  y  2t
b. Find the fundamental matrix of each of the systems below and then find a solution
satisfying the given initial conditions.
1.
 2 1
 3
x, x(0)   
x  
 1 2
  2
2.
 2  1
 2
x, x(0)   
x  
  4 2
  1
3.
 3  1
 1
x, x(0)   
x  
 1 1
 0
4.
  3  2
 1
x, x(0)   
x  
3
 9
  1
152
5.
 7  5
 2
x, x(0)   
x  
3
4
 0
6.
0  6
5
 2


 
x   2  1  2 x, x(0)   1
 4  2  4
 0


 
7.
2
2
 3
 1


 
x    5  4  2 x, x(0)   0 
 5
  1
5
3

 
c. In each of the problem below apply the method of variation of parameters to find a
particular solution of the given system.
7.
2   3e 2t 
1

x  
x  
2

2
5
t

 

2.
 4  1  e t 
x   t 
x  
 5  2   2e 
8.
 3  1  t12 
x   1 
x  
 9  3  t 3 
3.
 2  4   cos3t 
x  

x  
 5  2   sin 3t 
9.
 3  5   cos 4t 
x  

x  
 5  3   sin 4t 
4.
 2  1  sect 
x  

x  
 5  2   tan t 
10.
 2  1  e 2t tan t 

x  
x  
 1 2  0 
5.
0
 3  2 

x   3t

x  
3  e cos 2t 
2
11.
 2  4   ln t 
x   
x  
 1  2  t 
6.
Find the particular solution for each of the following above such that
1.
 6  7   2t 
x   
x  
 1  2  3 
x1 (1)  x2 (1)  0 .

Summary
-
Normal Form
The normal form in the general case of a linear system of n differential equations in
n unknown functions x1 , x2 ,, xn is:
153
x1  a11 (t ) x1  a12 (t ) x2    a1n (t ) xn  f1 (t )
x2  a21 (t ) x1  a22 (t ) x2    a2n (t ) xn  f 2 (t )

xn  an1 (t ) x1  an 2 (t ) x2    ann (t ) xn  f n (t )
real
-
Solution To A Linear System Of Differential Equations
The solution to the system:
dx1

 f1 t , x1 , x2 , x3 , xn  
dt

dx2
 f 2 t , x1 , x2 , x3 , xn 
dt




dxn
 f n t , x1 , x2 , x3 , xn 
dt

is the parametric vector
 x1 (t ) 
 x (t )
x   2  on open real interval a  t  b for which the




 xn (t )
system is true and the vector x must be continuous on the open
-
Matrix Form
The Matrix Notation form of the normal is given as:
 dx1 
 dt 

 a
dx
 2   11
 dt   a21
 dx    a
 3   31
 dt  

 

  an1
 dxn 


 dt 
a12
a22
a32
a1n   f1  t  


a2n   f 2  t  
a3n    f3  t  

 




ann   f n  t  
a13
a23
a33
an 2 an3
can simply be written as:
dx
 Ax  F(t ) ………….(I)
dt
154
which is a NON – HOMOGENEOUS FORM.
where F(t ) is the non – homogeneous term or the force term.
If F(t )  0 , that is f1 (t )  f 2 (t )  f 3 (t )    f n (t )  0 for the real interval
a  t  b then we have
dx
 Ax ………………(II)
dt
Equation (II) is known as the HOMOGENEOUS FORM.
-
Solution of Homogeneous Equation
In outline, the eigenvalue method for solving system x  Ax goes as follows:
(IV)
Solve the characteristic equation
det(A  I)  A  I  0
for the eigenvalues 1 , 2 ,, n of matrix A .
(V)
We attempt to find n linearly independent eigenvector v1 , v 2 ,, v n (so that
v k  0 for each k ) associated with these eigenvalues.
That is A  Iv k  0 for each k . This will not always be possible, but when
it is, we get n linearly independent solutions:
x1 (t )  v1e t , x 2 (t )  v 2 e t ,, x n (t )  v n e t
(VI)
The general solution is then a linear combination of these n solutions :
x(t )  c1 v1e t  c2 v 2 e t    cn v n e t
-
Non – Homogenous
The Method of Solving Non – Homogeneous systems
7.
Solve for the homogeneous part
dx
 Ax .
dt
8.
Form your fundamental matrix Φ(t )
9.
Find Φ1 (s)
155
Φ
t
1
(s)F(s)ds
10.
Evaluate
11.
Find x p  Φ(t ) Φ(s)F(s)ds .
t0
t
t0
Your final step x(t )  Φ(t )Φ 1 (t 0 )b  Φ(t ) Φ 1 (s)F(s)ds .
t
t0
156
UNIT SIX
SOLUTION SERIES
2.0
INTRODUCTION
We have seen that a linear differential equation with constant coefficients can
be reduced to the algebraic problem of finding the roots of its characteristic
equation. There is no similar procedure for solving linear differential equations
with variable coefficients, (at least not routinely and in finitely many steps)
with the exception of special types, and the occasional equation that can be
solved by inspections (perhaps followed by reduction of order).
For linear ordinary differential equations with variable coefficients and of
order greater than one, probably the most generally effective method of
attack is that based upon the use of power series.
To simplify our work and statements, the equations treated (that is the ones
to be solved completely) here will by restricted to those with polynomial
coefficients. The difficulties to be encountered, the methods of attack, and
the results accomplished all remain essentially unchanged when the
coefficients are permitted to be functions that have power series expansion
valid about some point (such function are called analytic functions).
2.1
SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATION
There are two standard methods of solving linear differential equation in
series form:
1.
Power Series Method
2.
Frobenius Method (which is an extension of Power
Series Method).
Recall the theory of power series:
We first remember that a power series (in powers of x  x0 ) is an infinite
series of the form:
157

 an x  x0 n  a0  a1 x  x0   a2 x  x0 2    ak x  x0 k   (2.1
n 0
where a0 , a1 , a2 , are constants, called the coefficients of the series, x0 is a
constant, called the centre of the series, and x is a variable.
If in particular x0  0 , we obtain a power series in powers of x as:

 an x n  a0  a1 x  a2 x 2    ak x k   _____ (2.2)
n 0
we assume that all variables and constant are real.
Recall typical Maclaurin’s series:
ex  1

x x 2 x3
xk
xn
…………………………. (I)

  
  
1 2! 3!
k!
n0 n
sin x  x 
 1k x 2k 1      1n x 2n1 ….. (II)
x3 x5 x7
 
 

2k  1!
3! 5! 7!
n0 2n  1!
cos x  1 
 1k x 2k      1n x 2n ………. (III)
x2 x4 x6


 
 2n!
2k !
2! 4! 6!
n0
cosh x  1 

x2 x4 x6
x 2k
x 2n
……………… (IV)


 
  
2k !


2! 4! 6!
2
n
!
n0

x3 x5 x7
x 2k 1
x 2n1
……… (V)
sinh x  x  

 
  
2k  1!
3! 5! 7!
n0 2n  1!



x 2 x3
 1k 1 x k
 1n1 x n
………….. (VI)
ln1  x  x 
 

2
3
k!
n!
n0

1
 1  x  x 2  x 3   x k     x n ………………………….. (VII)
1 x
n 0
Power series such as those listed above are often derived as Taylor Series.
The Taylor series with centre x  x0 of the function f is the power series:


n0

f n  x0 
x  x0 n  f x0   f x0 x  x0   f x0  x  x0 2  
n!
2!
158
in powers of x  x0 , under the hypothesis that f is infinitely differentiable at
x0 (so that the coefficients of the power series are all defined). If the Taylor
series of f converges to f x  for all x in some open interval containing x0 ,
then we say that the function is analytic at x  x0 or we say converges to
f x  in some neighbourhood of x0 .
x
Polynomials, sin x , cos x and e are analytic everywhere; so too are sums,
difference and products of these functions. Quotients of any two of these
functions are analytic all powers where the denominator is not zero.
Note: 0! 1
2.2
SHIFT OF INDEX OF SUMMATION OF INFINITE SERIES
1.
The index of summation of an infinite series as a dummy parameter
that is:



2n x n
2k x k
2p xp



n0 n!
k 0 k!
p0 p!
2.
Indices used to maintain a particular generic term in an infinite series.
Example 2.1

 a n x n  a 2 x 2  a3 x 3  a 4 x 4  
n2

  ar 2 x
r 0
(ii)
Consider
r 2
 a0 2 x 0 2  a1 2 x1 2  a22 x 2 2  

  an 2 x n 2
n 0

 n  2n  1an x  x0 n2
and maintain the generic
n2
term x  x0  in the infinite series:
n

 n  2n  1a x  x 
n2
n2
n
0
 43a2 x  x0  
0
54a3 x  x0   65a4 x  x0 2  
 4  03  0a20 x  x0 0  4  13  1a21 x  x0  
4  23  2a22 x  x0 2  
159

  4  n3  na2n x  x0 n
n 0
(iii)

x 2  n  r an x nr 1 and maintain the generic term x nr .
n 0


n 0
n 0
x 2  n  r an x nr 1   n  r x nr 1
 0  r a0 x 0r 1  1  r a1 x1r 1 
2  r a2 x 2r 1  

  n  r  1an1 x nr
n1
(iiii)
Solve the expression


n 1
n 0
 nan x n1   an x n
Solution:
we shift index of summation of the first series, then we get:


n 0
n 0
 n  1an1 x n   an x n .

  n  1an1  an x n  0 , which is true for all x .
n 0
Then n  1an1  an  0 n  1,2,3,
 an1 
an
n 1
Hence for a0  0
a1 
a 0 a0

1
1!
a2 
a1 a0 a0


2 1 2 2!
a3 
a2 a0 a0


3 2!3 3!
160

Thus  allows the determination of:
an 
If asked to find
a0
n!
n .

 an x n , with
n0
a n so determined we have:


a0 n
xn
 n! x  a0  n!  a0e x .
n0
n0
Exercise 2.1
With (iv) determine a n so that


n1
n 0
 nan x n1  2 an x n  0 2.
1.


n1
n 0
 nan x n1  2 an x n  0
Theorem 2.1: IDENTITY PRINCIPLE
If


n 0
n 0
 an x  x0 n   bn x  x0 n
for every point x in some open interval I ,
Then an  bn for all n  0 .
2.3
THE POWER SERIES METHOD
Consider the general inhomogeneous linear differential equation of n th order as:
bn ( x)
dny
d n1 y
d n 2 y
dy

b
(
x
)

b
(
x
)
   b1 ( x)  b0 ( x) y  f ( x)
n1
n 2
n
n1
n2
dx
dx
dx
dx
Then the second – order inhomogeneous equation would be:
b2 ( x)
d2y
dy
 b1 ( x)  b0 ( x) y  f ( x) ………….. (2.3)
2
dx
dx
or b2 ( x) y  b1 ( x) y  b0 ( x) y  f ( x) ………………… (2.3)
The idea of the power series method for solving differential equations is simple
natural.
For (2.3) b2 ( x) y  b1 ( x) y  b0 ( x) y  f ( x)
161
 y  
b ( x)
b1 ( x)
f ( x)
y  0
y
b2 ( x)
b2 ( x)
b2 ( x)
This can be simplified as:
y  p( x) y  q( x) y  g( x) ………..…… (2.4)
where p( x) 
b ( x)
b1 ( x)
f ( x)
and g( x) 
.
, q( x)  0
b2 ( x)
b2 ( x)
b2 ( x)
For a given homogeneous linear equation of the second – order of (2.4)
That is y  p( x) y  q( x) y  0 …………………….. (2.5)
we first represents px  and qx  by power series in powers of x
(or
x  x0 , if solutions in power of x  x0 is what we needed).
Let

y  a0  a1 x  a2 x 2  a3 x 3    ak x k     an x n ………………………..(a)
n 0

 y   a1  2a2 x  3a3 x 2    kak x k 1     nan x n1 ……………........(b)
n1

 y   2a2  3  2a3 x  4  3a4 x 2    k k  1ak x k 2     nn  1an x n2 (c)
n1
Then we substitute (a), (b) and (c) into (2.5).

The coefficient a0 , a1 , a2 , of an analytic solution yx    an x n (i.e. (a)) of
n 0
equation (2.5) can be computed as in the following steps;
Step 1:

Insert yx    an x n into the differential equation [ or (a), (b), (c) into
n 0
(2.5)].
Step 2:
Rearrange the left – hand side of the equation so that the equation has
the form

 cn x n  0 where cn is an equation in terms of
n 0
Step 3:
a j terms.
By theorem (2.1) cn  0 for every n . Solve this equation for the a j
term having the largest subscript. The resulting equation is known as
the Recurrence formula for the given differential equation.
162
Step 4:
Use the Recurrence formula to sequentially determine a j  j  2,3,
in terms of, a0 and a1 if it is second order differential equation, and a0
if it is first – order.
Example 2.2
1.
Solve the linear differential equation y  y  0 .
Solution:

Let y  a0  a1 x  a2 x 2  a3 x 3    ak x k     an x n .
n1

 y   a1  2a2 x  3a3 x 2    kak x k 1     nan x n1
n1
Substituting these expressions into the equation, we get:


n1
n 0
 nan x n1   an x n  0
we shift index in the summation of first sum, i.e. replacing n  1 by n in
the summation to maintain the generic term x n , we have


 n  1an1 x   an x n  0
n
n 0
n 0

  n  1an1  an x n  0
n 0
by Identity principle:
n  1an1  an  0 .
 an 
an
,
n 1
which is a recursion formula that gives as the following:
a1 
a0
a
a
a
a
a
a
, a2  1  0  0 , a3  2  0  0 ,…,
1
2 1 2 2!
3 3  2! 3!
Continuing the processes, then:
Therefore,


a0 n
x
n0 n!
y   an x n  
n 0
an 
163
a0
n!

xn
n0 n!
 y  a0 
The solution is given as:
y  a0 e x , where a0 is an arbitrary constant.
Alternatively,
y  y  0
a
1
 

 2a2 x  3a3 x 2    a0  a1 x  a2 x 2    0
Then we collect like powers of x , finding
a1  a0   2a0  a1 x  3a3  a2 x 2    0
Equating the coefficient of each power of x to zero, we have:
a1  a0  0, 2a0  a1  0, 3a3  a2  0
Solving these equations, we may express a1 , a2 , a3 , in terms of a0 ,
which remains arbitrary:
a1  a0 , a2 
a
a1 a0
a
 , a3  2  0 ,…
2
2!
3
3!
with these coefficients the power series:
y  a0  a1 x  a2 x 2    ak x k  
becomes:
y  a0  a0 x 
a0 2 a0 3
a
x  x  0 xk  
2!
3!
k!
and we see that we have obtained the familiar general solution:

xn
 a0 e x , where a0 is arbitrary.
n
!
n0
y  a0 
2.
Solve y  2xy  0 .
Solution:
y  2xy


 a1  2a1 x  3a3 x 2    2 x a0  a1 x  a2 x 2  
 a1  2a2 x  3a3 x 2    2a0 x  2a1 x 2  2a2 x 3  
equating coefficients, we see that:
164
a1  0, 2a2  2a0 , 3a3  2a1 , 4a4  2a2 , 5a5  2a3 , …
hence coefficients with odd subscript, a3  0 , a5  0 ,… and for the
coefficients with even subscripts:
a2  a0 , a4 
a
a2 a0
a
, a6  4  0 ,

2
2!
3
3!
a0 remains arbitrary. With these coefficients the power series gives the
following solution, which you should conform by the method of
separating variables:


2
x4 x6
2

y  a0 1  x 

   a0 e x
2! 3!


Alternatively,
Solve y  2xy  0
Solution:


n 0
n 1
Let y   a n x n , then y'   nan x n 1
By substitution:


n 1
n 0
 nan x n1  2 x an x n  0


n 1
n 0
  nan x n1  2 an x n1  0
For the same generic term, we shift the left index by  2 , to maintain
n  1:



n  1
n 0
 n  2an2 x n1  2 an x n1  0
Since the two summations do not start from the same point (i.e either
n  0 or n  1). We let them start from the same point. We then find
the term n  1(that is the left summation, which the start with n  0 ):


n 0
n 0
  1  2a1 2  x 11   n  2an 2 x n1  2 an x n1  0
165
Group like terms:

 1a1 x 0   n  2an 2  2an x n1  0
n 0
Comparing coefficients, we get:
a1  0 , and n  2an2  2an  0
 an2 
2a n
n  2
Which is the recursion formula for n  0,1,2,
a1  0
a5 
2a 3
20

0
3  2 5
a2 
2a 0
2a
 0  a0
0  2 2
a6 
2a
a
2a4
 0  0
4  2 2!6 3!
a3 
2a1
20

0
1  2 3
a7 
2a5
20

0
5  2 5
a4 
2a
a
2a 2
 0  0
2  2 4 2!
a8 
2a 6
2a
a
 0  0
6  2 3!8 4!
then we have:
if n is odd
0
a
an   0 if n is even
  n2 !
a2  a0 , a4 
or a2n 
a0
n!
a
a2 a0
a
, a6  4  0 ,

2
2!
3
3!
Hence by substitution:
2


x4 x6
y  a0 1  x 2      a0 e x
2! 3!


3.
Solve y  y  0 .
Solution:
Let

y   an x n
n 0


n2
n 0
Then we get:  nn  1an x n2   an x n  0
166
Shifting index in the first summation, and maintaining the
generic term xn .


n 0
n 0
 n  2n  1an2 x n   an x n  0

  n  2 n  1 an2  an  xn  0
n 0
Now, by theorem 2.1:  n  2  n  1 an 2  an  0
 an2  
an
 n  2 n  1
Which is the recursion formula for n  0,1,2,
a2 
a0
2!
a3 
a1 a1

3  2 3!
a4 
a2   a0  a0


4  3 4  3  2! 4!
a5 
a3   a1  a1


5  4 5  4  3! 5!
a6 
a0
a
a4

 0
6  5. 6  5  4! 6!
a7 
a5
a1
a

 1
7  6 7  6  5! 7!
Since, y  a0  a1 x  a2 x 2    ak x k  
Thus,
 a 
a 
 a 
a 
y  a0  a1x   0  x 2   1  x3   0  x 4   1  x5 
 3! 
 5! 
 2! 
 4! 
 x 2 x 4 x6
y  a0 1    
 2! 4! 6!


x3 x5 x 7
  a1  x    
3! 5! 7!


n
 1 n x 2 n1
1 x2n

 
y  a0 
 a1 
n0  2n !
n0  2n  1!

y  a0 cos x  a1 sin x
167



Exercise 2.2
In each of 1-16, find a power series solution of the given differential equation.
Determine the radius of convergence of the resulting series.
1.
y  2 y  0
9.
2  x  1 y  y
2.
y  y
10.
 x  1 y  2 y  0
3.
2 y  3 y  0
11.
2  x  1 y  3 y
4.
y  2xy  0
12.
 x  3 y   2 y  0
5.
y  x2 y
13.
xy  3 y  0
6.
 x  2  y  y
14.
 x  2  y  2 xy
7.
y  4 y
15.
8.
 2 x  1 y  2 y  0
16.
1  x  y  2xy
xy   x  1 y
2
2
Exercise 2.3
Fine two linearly independent power series solutions of to given differential equation.
Determine the radius of convergence of each series, and identify the general
solution in terms of familiar elementary functions.
1.
y  y
7.
y  4 y
2.
y  9 y  0
8.
y  y  x
3.
y  4 y  0
4.
y  4 xy  4 x 2  2 y  0
5.
y  3 y  2 y  0
6.
1  x  y  2xy  2 y  0 .


2
168
2.4
SOLUTIONS ABOUT NEIGHBOURHOOD POINTS
The power series method is used to solve linear differential equations such as
y  p  x  y  q  x  y  g ( x) ………………………. (2.4)
where p  x  and q  x  are given function as if  x  x0  solutions obtained are power
series solution in the neighbourhood of x  x0 .
To obtain this solution p  x  and q  x  need to satisfy the following;
(1)
x  x0 is an ordinary point if both of the following limits exist and
lim p  x  , lim q  x 
are finite:
x x0
i.e.:
(2)
xx0
lim p  x    , lim q  x    .
x  x0
x x0
x  x0 is singular point if either
lim p  x    , lim q  x   
x x0
x x0
i.e. they are not finite (infinite)
a)
x  x0 is a regular point if it is a singular point and if both
of the following limits exist and are finite:
lim  x  x0  p  x  , lim  x  x0  q  x 
2
x x0
xx0
i.e. lim  x  x0  p  x    , lim  x  x0  q  x   
2
x x0
xx0
b) x  x0 is a irregular singular point if either (or both) of limits
do not exist.
lim  x  x0  p  x    , lim  x  x0  q  x   
2
x  x0
xx0
i.e. they are not finite (infinite).
NOTE:
169
The solutions we have obtained from the examples of Power series is about the
point x  0 (i.e x0  0 ) p  0  and q  0  are ordinary point.
Example 2.3
1.
☞
The differential equation
1  x  y  6xy  4 y  0
2
has x  1 and x  1 as its only singular points in the finite plane.
☞
The equation
y  2xy  y  0
has no singular points in the finite plane.
☞
The equation
xy  y  xy  0
has the origin x  0 as the only singular point in the finite plane.
2.
Determine whether x  0 is an ordinary point of the differential equation
y  xy  2 y  0 .
Here p  x    x and q  x   2 are both polynomials; hence they are analytic
everywhere. Therefore, every value of x , in particular x  0 , is an ordinary
point.
3.
Determine whether x  0 is an ordinary point of the differential equation
y  y  0 .
Here p  x   0 and q  x   1 are both constants; hence they are analytic
everywhere. Therefore, every value of x , in particular x  0 , is an ordinary
point.
4.
Determine whether x  0 is an ordinary point of the differential equation
x2 y  2 y  xy  0 .
170
Here p  x  
2
1
and q  x   . Neither of these functions is analytic at x  0 ,
2
x
x
so x  0 is not an ordinary point, rather, a singular point.
5.
Determine whether x  0 or x  1 is an ordinary point of the differential
equation (called Legendre’s equation)
1  x  y  2xy  n(n 1) y  0
2
for any positive integer n .
Hence p  x  
n  n  1
2 x
and q  x  
;
2
1  x2
1 x
If x  0 , both are analytic there and x  0 is an ordinary point.
If x  1 , neither function is analytic there. Consequently, x  1 is a singular
point.
6.
Classify the singular points, in the finite plane, of the equation
x 1  x   x  2 y  x2 y  ( x3  2x 1) y  0
2
Hence p  x  
x
 x 12  x  2
and q  x  


 x3  2 x  1
x  x 1  x  2
2
;
The singular points in the finite plane are x  0,1, 2 .
Consider x  0 :
The factor x is absent from the denominator of p  x  and it appears to the
first power in the denominator of q  x  . Hence x  0 is a regular singular
point.
Consider x  1 :
The factor  x  1 appears to the second power in the denominator of p  x  .
That is a higher power than is permitted in the definition of a regular singular
171
point. Hence it does not matter how  x  1 appears in q  x  ; the point x  1 is
an irregular singular point.
Consider x  2
The factor  x  2  appears to the first power in the denominator of p  x  , just
as high as is permitted, and to the first power also in the denominator of
q  x  , so x  2 is a regular singular point.
In summary, the equation has in finite plane the following singular points:
7.
~
Regular singular points at x  0, x  2 .
~
Irregular singular points at x  1 .
Classify the singular points in the finite plane for the equation


x4 x2 1  x 1 y  4x3  x 1 y  ( x  1) y  0
Hence p  x  
q  x 

2
4

x x  1  x 1
2

4
and
x  x  i  x  i  x 1
x 1
x  x  i  x  i  x 1
4
2
Therefore the desired classification is;
Regular Singular Point (R.S.P) at x  i, i,1 and
Irregular Singular Point (I.S.P) at x  0
8.
In the equation
sin 2x  y  x  x 1 y  x2 y  0
there are infinite number of singular points, namely,
x


,  , 3 ,
2
2
Note that x  0 is not a singular point,
172
since lim
x0
 x  x  1 1
x2
   and lim
0
x0 sin 2 x
sin 2 x
2
Exercise 2.4
For each equation, list all of the singular points in the finite plane.
1.
x
2.
x  3  x  y   3  x  y  4 xy  0 12.
x2 x2  9 y  3xy  y  0
3.
4 y  3xy  2 y  0
x2
4.
x  x 1 y  3xy   x 1 y  0 14.
5.
x2 y  xy  1  x2 y  0
15.
 2 x  1 x  3 y  y   2 x  1 y  0
6.
x4  y  0
16.
x3 x 2  4
17.
x x2  1 y  y  0
18.
x
19.
 4 x  1 y  3xy  y  0
2

 4 y  6 xy  3 y  0
11.
13.
2


8.
1  x  y  2xy  6 y  0
 x  4x  3 y  x y  4 y  0
9.
x2 1  x  y  1  2x  y  0
10.
6 xy  1  x2 y  2 y  0
7.
2
2
2
3

4 y  y  0
 
1  4x  y  4xy  y  0
2
4xy  y  0


2

2


y  2 x 2  4 y  xy  0


 6 x  8 y  3 y  0

Exercise 2.5
For each equation, locate and classify all its singular points in the finite plane.
1.
x3  x 1 y   x 1 y  4xy  0 10.
2.
x2 x2  4 y  2 x3 y  3 y  0
3.
y  xy  0
12.
 x 1 x  2 y  5  x  2 y  x2 y  0
4.
x2 y  y  0
13.
 x 12  x  42 y   x  4 y  7 y  0
5.
x4 y  y  0
14.
1 4x  y  6x 1 4x  y  9 y  0


11.
2
173
x2  x  2 y   x  2 y  4 y  0
x  x  3 y  y  y  0
2
6.
 x 1  x  4
7.
y   x  4 y  y  0
15.
x3 y  4 y  0
x2  x  2 y  3 x  2 y  y  0
16.
1  4x  y  6xy  9 y  0
8.
x2  x  4 y  3xy   x  4 y  0
17.
x4 y  2x3 y  4 y  0
9.
1  4x 
 2x 14 y   2x 1 y  8 y  0
3
2
2
2
2 2
y  6 xy  9 y  0
18.
2
Exercise 2.6
Determine and classify the singular points (if any) of each of the following equation.
3.
x
x
x
4.
sin 2xy   x  3 y  e x y  0
5.
3x2  x  2 y  2x  x 1 y  x2  2 y  xe x
1.
2.
6.
7.
8.

3
 7 x  8 y   x  3 y  xy  0
4
 16 y   x  2 y   x  2 y  0
4
 4 x2  4 y  xy  0


sin 3x  y  x  x  2 y  5xe y  2e
9.
2

2
x
2x

 x  x 1 y  sin 2x  y  4x y   sin 2x
1  cos x  y  3xy  4 y  0 10. x sinh x y  3sin x y 14e y  0
e 1 y 14 y 12e cos x  y  0
2
2
1
2
x
2
x
x
Examples of Solution about a neighbourhood point x  x0
1.
Find the general solution near x  0 (i.e. x0  0 ) of y  xy  2 y  0 .
Solution:
Here p  x    x and q  x   2 .
x  0 is an ordinary point since p( x) and q  x  are polynomials that
means it is analytic everywhere.
Let

y   an x n
n 0

 y   nan x n1
n1
Therefore,
174

 y   n  n  1 an xn2
n2
2a2  6a3 x  12a4 x2 

 n  n  1 an x n2   n  1 nan1x n1 
 n  2 n  1 an2 xn 
   x a1  2a2 x  3a3 x 2  4a4 x3   nan x n1 


n
n1
 n  1 an1x   n  2 an2 x    2 a0  a1x  a2 x2  a3 x3  
an xn  an1xn1    0
Combining terms that contain like powers of x , we have:
 2a2  2a0   x  6a3  a1   x2 12a4   x3  20a5  a3  
xn  n  2 n  1 an2  nan  2an    0  0x  0x2 

 0xk 
The last equation holds if and only if each coefficient in the left – hand
sides is zero. Thus:
2a2  a0  0,6a3  a1  0,12a4  0,20a5  a3  0
In general:  n  2  n  1 an2   n  2  an  0 , or
an2 
 n  2 a
 n  2 n  1 n
which is the recurrence formula for this problem.
Calculate
a 2  a 0
a4  0
1
a3   a1
6
1
1  1 
1
a5  a3    a1   
a1
20
20  6 
120
1 1 
1
a7   
a1
a1  
14  120 
1680
2
1
a4  0
30
15
4
1
a8  a6  0  0 a9  .....
56
14
a6 
Note that since a4  0 , it follows from the recurrence formula that all
the even coefficients beyond a4 are also zero.
Substituting we have
175
1
1
1
y  a0  a1x  a0 x2  a1x3  0 x 4 
a1x5  0 x6 
a1x7 
6
120
1680


1
1 5
1 7

 y  a0 1  x 2  a1  x  x3 
x 
x 
6
120
1680

2.



Solve the equation y   x 1 y  4  x 1 y  0 about the ordinary point
2
x  1.
Solution:
To solve an equation “about the point x  x0 ” means to obtain solution
valid in a region surrounding the point, solutions expressed in powers
of x  x0 .
We first translate the axes, putting x 1  v .
The equation then becomes
d 2 y 2 dy
v
 4vy  0
dv
dv2
always in a pure translation, x  x0  v ,
we have
dy dy
 , and so on.
dx dv
As usual we put :

y   anv n
n 0
and we obtain:



n2
n1
n 0
 n  n 1anvn2   nanvn1  4 anvn1  0
collecting like terms yields:


n2
n 0
 n  n 1anvn2    n  4 anvn1  0
which, with a shift of index from n to  n  3 in the second series, gives:


n 0
n 3
 n  n 1anvn2    n  7  an3vn2  0
Therefore a0 and a1 are arbitrary and for the remainder we have
n  2: 2a2  0,
n  3: n  n  1   n  7  an3  0
176
an  
n7
an3
n  n  1
4
a0
3 2
1
a6  
a3
65
2
a9  
a6
9 8
3
a1
43
0
a7  
a4  0
76
3
a10  
a7  0
10  9
a3  
a4  
2
a2  0
5 4
a7    a5  0
a5  
a11  0
a3n2  0, n  1
a3n1  0, n  2
3n  7
a3n  
a3n3
3n  3n  1
with the usual multiplication scheme, the first column yields:
n  1:
 1n  4   1  2   3n  7  a0
a3n 
3  6  9    3n  2  5  8   3n  1
   1n  4 1  2   3n  7  v3n 


  a1 v  14 v4
y  a0 1  
 n1 3  6  9   3n  2  5  8   3n 1 




since v  x 1 , the solution appears as:
   1n  4 1  2   3n  7   x 13n 


  a1  x 1  14  x 14
y  a0 1  
 n1 3  6  9   3n  2  5  8   3n 1 



but 3  6  9 
 3n  3n 1 2  3 

n  3n n!
Furthermore, all but the first two factors inside the square bracket in the
numerator also appear in the denominator, and from n  2 , they cancelled
out, it can be shown that :
4  1  x  1
n

y  a0 
3n
4
 a1  x  1  14  x  1 


n0 3  3n  1 3n  4  n !
n
177
Exercise 2.7
1.


Find the general solution in powers of x of x2  4 y  3xy  y  0 .
Then find the particular solution with y  0   4, y  0   1
2.
Determine the general series solution of the equation
 x 1 y  xy  4 y  0 about the point x  0 .
2
3.
Find the general solution near x  2 of y   x  2  y  2 y  0 .
4.
y  2  x  3 y  3 y  0 . Solve about x  3 .
5.
y   x  2  y  0 . Solve about x  2
6.
Find the general solution of the Legendre’s equation
1 x  y  2xy  n  n 1 y  0
2
2.5
THE FROBENIUS METHOD
The standard form of linear homogeneous differential equation of second order of
equation (2.5) is:
y  p( x) y  q( x) y  0 ………….(2.5)
We shall confine our discussion to a solution of (2.5) in the neighbourhood of x  0 .
The Frobenius method applicable when x  0 is ordinary or regular point of (2.5).
Any differential equation of the form:
y 
p  x
q  x
y  2 y  0 ………(2.6)
x
x
where the functions p  x  and q  x  are analytic at x  0 , we use a more general
method to assume a trial solution of the form:

y  xr a0  a1x  a2 x2  a3 x3 

  x a x
r
n 0
n
where a0 is the first coefficient that is not zero (i.e. a0  0 ).
178
n
…(2.7)
Indicial Equation
Assume a series solution of the form:

y  xr a0  a1x  a2 x2  a3 x3 

  x a x
r
n 0
n
n
……..(2.7) a0  0
of the differential equation
x2 y  xp( x) y  q( x) y  0 …………………………(2.6a)
we first expand p  x  and q  x  in power series:
q  x   q0  q1x  q2 x2  q3 x3 
p( x)  p0  p1x  p2 x 2  p3 x3 
Termwise differentiation of trial solution in (2.7)
y  x   xr 1 ra0   r  1 a1x 

    n  r  an x nr 1

n 0
 y  x   xr 2 r  r  1 a0  r  r  1 a1x 

    n  r  n  r  1 an x nr 2

n 0
by inserting all these series into (2.6a) we readily obtain:
xr r  r 1 a0     q0  q1x   xr ra0   r  1 a1   
..(2.8)
 q0  q1x   xr  a0  a1x    0
we now equate the sum of the coefficients of each power xr , xr 1, xr 2 ,
to
zero. This yields a system of equations involving these unknown coefficients
an . The equation corresponding to the power xr is
r  r  1  p0r  q0  a0  0 …….(2.9)
Since by assumption a0  0 , the expression in the brackets must by zero. This
gives
r  r  1  p0 r  q0  0 ……………..(2.10)
This important quadratic equation is called indicial equation of the
differential equation (2.7).
This is the coefficient of the lowest power of x gives the indicial equation
from which values of r are obtained, r  r1 and r  r2 , the roots of the
equation.
179
Three Cases Arises
Case 1: Distinct roots not differing by an Integer.
A basis is

y  x  x c
y1  x   xr1 a0  a1x  a2 x2 
and
2
r2
0
 c1x  c2 x2 
 …………(I)
 …………(II)
with coefficients obtained successively from (2.8) with r  r1 and r  r2 ,
respectively.
Case 2: Double root r  r1  r2 .
A basis is

y1  x   xr a0  a1x  a2 x 2 
 ……….(III)
 r  12 1  p0 
and

 …..(IV)
y2  x   y1  x  ln x  xr c1x  c2 x 2 
x0
Case 3: Roots differing by an integer.
A basis is

 ………..(V)
c  c x  c x   …….(VI)
y1  x   xr1 a0  a1x  a2 x2 
and
y2  x   ky1  x  ln x  xr2
2
0
1
2
where the roots are so denoted that r1  r2  0 and k may turn out to
be zero.
180
Example 2.5
Solve the equation 4 x
d2y
dy
2  y  0.
2
dx
dx
Solution:
y 
1
1
y 
y0
2x
4x
clearly p  x  
1
1
, q  x 
.
2x
4x
Therefore the problem makes the point x  0 a regular singular point.

Let y   an x nr
n 0

 y    n  r an x nr 1
n 0

 y    n  r  n  r  1 an x nr 2
n 0
substituting the above into the differential equation we get:



n 0
n 0
n 0
4  n  r  n  r  1 an x nr 1  2  n  r  an x nr 1   an x nr  0


n 0
n 0
2  n  r  2n  2r  1 an x nr 1   an x nr  0


n1
n 0
2r  2r  1 a0 xr 1  2  n  r  2n  2r  1 an x nr 1   an x nr  0


n 0
n 0
2r  2r  1 a0 xr 1  2  n  r  1 2n  2r  1 an x nr   an x nr  0

2r  2r  1 a0 xr 1   2  n  r  1 2n  2r  1 an1  an  x nr  0
n 0
This last expression is true for all x , therefore the coefficients of all powers of
x must vanish.
The coefficient of the lowest power of x ,
2r  2r  1 a0
is called the indicial term and it determines the values of the index r .
181
In this case i.e. Case 1, assuming a0  0 then r  0, 12 .
The other coefficients of x n  r for n  0,1,2,3
lead to the recurrence
relationship:
2  n  r  1 2n  2r  1 an1  an  0
an1 
an
2  n  r  1 2n  2r  1
2r  2r  1 a0  0 …….. indicial equation
If r  0 , a0  0 , then the recurrence relation of coefficient becomes:
an1 
an
2  n  1 2n  1
if n  0 , a1 
a0
a
 0
2 11 2!
if n  1 , a2 
a
a1
a
 1  0
2  23 4 3 4!
if n  2 , a3 
a
a2
a
 2  0
2  35 6 5 6!
if n  3 , a4 
a3
a
a
 3  0
2  4 7  8  7  8!
therefore
n
1 a0

an 
 2n !

Recall that (I), y1  x r1  an x n .
n 0

 x0 a0  a1x  a2 x 2 
 ak x k 
 x x2 x3
 a0 1    
 2! 4! 6!


 a0 1 

k
1 xk



 2k !
 x   x   x
2
2!
4
4!

6
6!






 1k 
 y  a0 cos x , a0 is arbitrary constant
182
2!
x

2k

 

The other value obtained for r  12 .
Recall the recurrence relation:
cn1 
if n  0 , c1 
if n  1 , c2 
if n  2 , c3 
Therefore,
cn
c is arbitrary,
2  n    2n  2  0
3
2
c0
c
 0
3 2 3!
2
c1
5
2
  4
c2
2
7
2
  6

c1 c0

5  4  5!

c2 c0

7  6  7!
n
1 c0

cn 
 2n  1!

Recall that (II), y2  xr2  cn x n
n 0
1

 x 2 c0  c1x  c2 x2 
 x x 2 x3
 x c0 1    
 3! 5! 7!

1
2

 c0  x 


 ck xk 
k
1 xk



 2k 1!
 x   x   x
3
3!
y2  c0 sin x ,
Hence for y  y1  y2
The general solution is given as:
y  a0 cos x  c0 sin x or.
183
5
5!
7
7!

c0 is arbitrary





 1k 
k!
x

k





Example 2.6
Find the exponents in the possible Frobenius series solutions of the equation


2x2 1  x  y  3x 1  x  y  1  x2 y  0
3
Solution:
We divide each term by 2 x2 1  x  to recast the differential equation in the
form
y 
 32  1 2x  x2 
x
and thus see that p0 
3
2
y 
 12  1 x  y  0
x2
and q0   12
Hence the indicial equation is:
r  r  1  32 r  12  r 2  12 r  12   r  1  r  12   0
with roots r1 
1
2
and r2  1.
The two possible Frobenius series solutions are then of the forms


n 0
n 0
y1  x 2  an x n and y2  x1  cn x n .
1
Example 2.7
Find a Frobenius series solution of Bessel’s equation of order zero:
x2 y  xy  x2 y  0
Solution:
y 
1
x2
y  2 y  0
x
x
Hence x  0 is a regular singular point with p  x   1 and q  x   x2 , so our
series will converge for all x  0 .
Because p0  1 and q0  0 , the indicial equation is
r  r 1  r  r 2  0
Thus we obtain only the single exponent r  0 and so there is only one
Frobenius series solution
184

y  x0  an x n
n 0
substitute in the differential equation; the result is:



n 0
n 0
n 0
 n  n 1 an xn   nan xn   an xn2  0
we combine the first two sums and shift the index of summation in the third
by 2 to obtain:


n 0
n2
 n2an xn   an2 xn  0
Therefore, the recursion formula is:
an  
an2
n2
for n  2
a2  
a0
,
22
a4  
a
a
a2
a
 2 0 2 and a6   42   2 02 2
2
4
24
6
246
n
n
1 a0
1 a0


a2n 

2
2
22  42   2n 
22n  n!
The choice a0  1 gives us one of the most important special functions in
mathematics, the Bessel function of order zero of the first kind,
denoted by J0 ( x) .
Thus
n
1 x2n

x2 x4
x6
J 0 ( x)  

1




2
2n
4 64 2304
n0 2  n!

In this example we have not been able to find a second linearly independent
solution of Bessel’s equation of order zero.
185
Exercise 2.9
In each exercise, determine whether x  0 is an ordinary point, a regular singular
point, or an irregular singular point.
If it is a regular point, find the exponent of the differential equation at x  0 .
1.
xy  x  x3 y   sin x  y  0


1a.
xy  x2 y  e x  1 y  0
2.
x2 y   cos x  y  xy  0
2a.
x2 y   6sin x  y  6 y  0
3.
3x3 y  2 x2 y  1  x2 y  0
4.
Apply the method of Frobenius of Bessel’s equation of order





1
2

x2 y  xy  x2  14 y  0 , to derive its general solution for x  0 ,
y  a0
5.
cos x
sin x
 a1
x
x
Show that Bessel’s equation of order 1,


x2 y  xy  x2  1 y  0
has exponents r1  1 and r2  1 at x  0 , and that the Frobenius series
2n
x   1 x
solution corresponding to r1  1 is J1  x   
2 n0 n! n  1!22n
n
6.
Obtain the solution
  x  
y  y  0
by Frobenius method.
7.
Find the series solutions in powers of Airy’s equation
  x  
y  xy
8.
Find the general series solution of the following
a) 2 xy   3  2 x  y  4 y  0

b)

c) x2 y  xy  x2  1 y  0
186


2 x  x2 y  1  9 x  y  3 y  0
UNIT SEVEN
FOURIER SERIES
Definition: Periodic Function
If the value of each ordinate
f  t  repeats itself at equal intervals in abscissa, t , the
f  t  is said to be periodic. Thus if f t   f t  T  f t  2T  ... , then T is
called the least period (or period) of the function
f t  .
Example 3.1
1:
The function sin x has periods 2 ,4 ,6 since sinx  2  , sinx  4  ,
sinx  6  all equal sin x . So the least period is 2 .
2:
The period of sin nx or cosnx , where n a positive integer, is 2 / n .
3:
The period of
4:
A constant has any positive number as period.
tan x is  .
Other examples of periodic functions are shown in graphs below:
period
period
Piecewise Smooth Functions
Definition: A function f  x  is piecewise continuous on the open interval
a  x  b if:
1.
f  x  is continuous everywhere in a  x  b with the possible
exception of at most a finite number of points x1, x2 ,..., xn and
2.
at these points of discontinuity, the right- and left – hand limits of
f  x  , respectively lim f  x  and lim f  x  , exist ( j  1,2,..., n )
x x j
x x j
x x j
x x j
187
(Note that a continuous function is piecewise continuous).
Definition: A function f  x  is piecewise continuous on the closed interval
a  x  b if:
1.
it is piecewise continuous on the open interval a  x  b ,
2.
the right – hand limit of
3.
the left – hand limit of
f  x  exists at x  a , and
f  x  exists at x  b .
Definition: A function f  x  is piecewise smooth on  a, b if both f  x  and
f   x  are piecewise continuous on  a, b .
Examples 3.2
1.
Determine whether
 x2  1 x  0
f  x  
is piecewise continuous on -1,1
1/
x
x

0

Solution:
1,1 except at x  0 .
Therefore, if the right- and left- hand limits exist at x  0 , f  x  will be
piecewise continuous on  1,1 . We have:
The given function is continuous everywhere on


lim f  x   lim x2  1  1
x0
x0
x0
x0
1
lim f  x   lim  
x0
x0 x
x 0
x 0
Since the left – hand limit does not exist,
on
2.
f  x  is not piecewise continuous
1,1 .
x 1
sin  x
 0
0  x 1

Is f  x    x
piecewise continuous on  -2,5 ?
e

1

x

0

 x3
x  1
188
Solution:
The given function is continuous on
2,5 except at the two points
x1  0 and x2  1.(Note that f  x  continuous at x  1 ). At the two points
of discontinuity, we find that:
lim f  x   lim0  0
lim f  x   lim e x  1
lim f  x   lim e x  e1
lim f  x   lim x3  1
x0
x 0
and
x1
x 1
x1
since all require limits exist,
3.
x0
x 0
x0
x1
x 1
x0
x1
f  x  is piecewise continuous on 2,5
 x2  1
x<0

Is f  x    1
0  x  1 piecewise smooth on -2,2?
2 x  1
x 1

Solution:
The function is continuous everywhere on
Since the require limits exist at
Differentiating
2,2 except at
x1  1.
x1, f  x  is piecewise continuous.
f  x  , we obtain
x<0
2 x

f  x    0 0  x  1
2
x 1

The derivative does not exist at x1  1 but is continuous at all other points in
2,2 . At
x1 the required limits exist; hence f   x  is piecewise continuous.
It follows that
4.
f  x  is piecewise smooth on  2,2 .
1
x<0

Is f  x    x 0  x  1 piecewise smooth on -1,3?
 x3
x 1

189
Solution:
The function is continuous everywhere on
require limits exist at
1,3 except at
x1  0 . Since the
x1, f  x  is piecewise continuous. Differentiating f  x 
, we obtain:
 0
x<0
 1

f  x   
0  x 1
2
x

x 1
 3x 2
which is continuous everywhere on
1,3 except at the two points
x1  0
and x2  1where the derivative does not exist.
At x1  0 ,
lim f   x   lim
xx1
x x1
x0
x0
1
2 x

Hence , one of the required limits does not exist. If follows that
piecewise continuous, and therefore that
1,3 .
190
f   x  is not
f  x  is not piecewise smooth on
Definition: Fourier Series
f  x  (which is integrable in   ,  ) can be
A non – sinusoidal period function
expressed in a converging series of the form:


n1
n1
f  x   a0  an cos nx  bn sin nx ………..(3.1)
called the Fourier Series, where a0 , a1, a2 ,..., b1, b2 , b3 ,... are
constants that can easily be determined, and

  f xdx
a0 
1
2
an 
f x cos nxdx,
 
bn 
1
1





  f xsin nxdx

n  1,2,...
……… (3.2)
n  1,2,...
where a0 , an , and bn are called Fourier coefficients.
The function cos x and sin x are called fundamentals for n  1 an integral cosnx
and sin nx are called harmonics, which are multiples of the fundamental
frequencies.
Orthogonality Conditions For The Sine And Cosine Functions
The following integrals are very useful in the Fourier expansions:
For
m, n integers m  n :
2
2
1.
 sin nxdx  0
2.
0
2
3.
 sin
2
nxdx  
0
2
4.
0
 sin nx sin mxdx  0
6.
nxdx  
 cos nx sin mxdx  0
0
2
 cos nx cos mxdx  0
2
2
0
7.
 cos
0
2
5.
 cos nxdx  0
2
8.
0
 sin nx cos mxdx  0
0
191
The iterated integration by parts is also very useful:
 uvdx  uv  uv
1
2
 uv3  uv4  ...
du
d 2u
d 3u
, u  2 , u  3 ,...
where u 
dx
dx
dx
v1   vdx, v2   v1dx, v3   v2dx, v4   v3dx, …
Also for
N which is an integer n  N ,
sin n  0 , cos n   1
n
Determination of Fourier Coefficients


n1
n1
f  x   a0  an cos nx  bn sin nx with period 2 .
Let
a)
Then to find a0 : Integrate both sides of (3.1) w.r.t. x for 0  x  2

2
0
2
2 
2 
0
0
0
f  x dx   a0dx  

2
0
an cos nxdx  
n1
b sin nxdx
n1
n
2
f  x dx  a0  dx  a0 2 , other term vanishes by orthogonal
0
properties.
 a0 
b)
1
2
2

f  x dx ……….(I)
0
To find an : Multiply each side of equation (3.1) by cosnx and integrate
throughout with respect to x , 0  x  2

2
0
2
2 
0
0
f  x  cos nxdx   a0 cos nxdx  

2
0
 an 
a cos
2
n1
n
2
nxdx  
2 
0
f  x  cos nxdx  an  cos nx cos nxdx  an
0
1
2
 0
f  x  cos nxdx ……….(II)
192
b cos nx sin nxdx
n1
n
c)
To find bn : Multiply each side of equation (3.1) by sin nx and integrate
throughout with respect to x , 0  x  2

2
0
2
2 
2 
0
0
0
f  x  cos nxdx   a0 sin nxdx  

2
an cos nx sin xdx  
n1
b sin
n1
n
2
nxdx
2
f  x  sin nxdx  bn  sin nx sin nxdx  bn
0
0
bn 
1
2
 0
f  x  sin nxdx ………….(III)
Example and Solution
If the following function are defined over the interval   x   and
f  x  2  ,
state whether or not each function can be represented by a Fourier series.
1.
f  x   x3 - Yes
2.
f  x   4x  5 - Yes
3.
f  x 
2
- No, infinite discontinuity at x  0
x
4.
f  x 
1
- Yes
x4
5.
f  x   tan x - No: infinite discontinuity at x   / 2 .
f  x   y , where x2  y 2  9 - No; two valued
Dirichlet Conditions
Suppose that;
1.
f x  is defined except possibly at a finite number of points in  L, L  .
2.
f x  is periodic outside  L, L  with period 2L .
3.
f x  and f x  are piecewise continuous in  L, L  .
Then the series with Fourier coefficients converges to
a)
f x  if x is a point of continuity
b)
if x is point of discontinuity.
193
Here f x  0 and f x  0 are the right – and left – hand limits of f x 
at x represents
lim f  x    and lim f  x    respectively.
 0
 0
The conditions 1, 2 and 3 imposed on
f  x  are sufficient but not
necessary, and are generally satisfied in practice.
Example 3.1:
f  x   x and 0  x  2
Find the Fourier series representing the periodic function
and sketch its graph from x  4 to x  4 .
Solution:
Clearly
f  x   x is periodic then,


n1
n1
f  x   a0  an cos nx  bn sin nx
Recall:
a0 
1
2

2
0
f  x dx , an 
1
2
 0
f  x  cos nxdx , bn 
1


2
0
f  x  sin nxdx
Thus;
1
a0 
2

2
0
an 
1
2
 0
bn 
1
2
 0
2
1  x2 
xdx 
 ,
2  2  0
2

1 x
1 2
x cos nxdx   sin nx   sin nxdx   0
  n
n 0

0
2

1  x
1 2
2
x cos nxdx   cos nx   cos nxdx   
  n
n 0
n

0
Put the values into:


n1
n1
f  x   a0  an cos nx  bn sin nx
1
1
1


 f  x     2 sin x  sin 2x  sin3x  ...  sin kx  ...
2
3
k


194
Sketch
Function Defines in two or more sub – interval
Example
1.
Find the Fourier coefficients corresponding to the function
0 5  x  0
f  x  
3 0  x  5
period  10
2.
Write the corresponding Fourier series
3.
How should
f  x  be defined at x  5 , Fourier series will converge to
f  x  for 5  x  5 ?
Solution:
1.
Period 2 L  10 and L  5 .
Choose the interval c to c  2 L as 5 to 5 , so that c  5 .
5
1 L
1  0
f x dx 
0dx   3dx



0
2L  L
2(5)  5
a0 
 a0  0 
Then
an 
1  5 3
3x 
25  0 2
1 c2 L
n x 1 5
n x
f
x
cos

f
x
cos
dx






L c
L
5 5
5
5
1 0
n x
n x  3 5
n x
 5  0 cos
dx  0 3 cos
dx
  0 cos
5
5
5  5
5
3 5
n x 
  sin
  0 if n  0
5  n
5 0
5
bn 
1 c2 L
n x 1 5
n x
f
x
sin

f
x
sin
dx




L c
L
5 5
5
195
5
1 0
n x
n x  3 5 n x
 5  0 sin
dx  0 3 sin
sin
dx

5
5
5  5 0
5
31  cos n 
3 5
n x 
   cos


5  n
5 0
n
5
2.
The corresponding Fourier series is:

n
n

f x  a0   an cos x  bn sin
L
L
n1 

x

3  31  cos n  nx
 
sin
2 n1
n
5
3 6   x 1 3 x 1 5 x

   sin
 sin
 sin
 ... 
2 
5 3
5
5
5

3.
f  x  satisfying the Dirichlet conditions, we can say that the series
Since
converges to
f  x  at all points of continuity and to t at points of
discontinuity. At x  5,0 and 5 , which are points of discontinuity, the
series converges to 3  0 / 2  3 / 2 .
If we redefine f x  as follows:
3 / 2
0

f x   3 / 2
0

3 / 2
x  5
5  x  0
x0
0 x5
x5
Then the series will converges to f x  for  5  x  5 .
Example 3.5
Find the coefficients of the period function
k
f  x  
 k
if    x  0
if 0  x  
and
Solution:
a0 
1
2



f x dx 

1  0

k
dx

0 kdx  0
2 
196
f  x  2   f  x 
an 


f  x  cos nxdx     k  cos nx    k  cos nx 



0


  
1
1
0
0

1  sin nx
sin nx 
 k
k
0
 
n 
n 0 
because sin nx  0 at
bn 
 ,0, and  for all n  1,2,...


f  x  sin nxdx     k  sin nx    k  sin nx 



0


  
1
1
0
0

1  cos nx
cos nx 
 k
 k

 
n 
n 0 
since
cos     cos ;cos0  1, this yields.
bn 
Now,
k
2k
cos0  cos  n   cos n  cos0  1  cos n 
n
n
cos  1,cos2  1,cos3  1, etc;
1 for odd n
cos n  
and thus
 1 for even n
In general
2 for odd n
1  cos n  
0 for even n
Hence the Fourier Coefficients bn of our function are
b1 
4k

, b2  0, b3 
4k

, b4  0, b5 
Since the an are zero, the Fourier series of
4k 
1
1
 sin x  sin3x  sin5x 
 
3
5
197



4k
5
f  x  is:
ODD AND EVEN FUNCTION
Even Functions
Definition: A function
f  x  is said to be even (or symmetric ) if f   x   f  x 
The graph of a symmetric function is symmetrical about the y  axis
The area under such a curve from  to
is



Thus
 is also twice the area from 0
to

f  x dx  2 f  x dx .
0
x4 ,2x6  4x2  5,cos x, e x  e x are even function
Odd functions
Definition: A function is called odd (or skew symmetric) if
f  x   f  x .
Some graphs of odd functions:
The area under the curve from  to
Thus

is zero; i.e.
  f  x dx  0 .
x3, x5  3x3  2x,sin x,tan3x are odd functions.
Products of Odd and Even Functions
 even    even    even 
 odd    odd    even 
 odd    even    odd 
198



that
Expansion of the Even Function,



0
f  x
a0 
1
2
1
  f  x dx    f  x dx
an 
1

 
f  x  cos nxdx 
bn 
1

f  x  sin nxdx  0
 

2
 0
f  x  cos nxdx
Therefore the series of the even function contains only cosine terms
Expansion of the Odd Function,
f  x

a0 
1
2
  f  x dx  0
an 
1

 
f  x  cos nxdx  0
bn 
1

f  x  sin nxdx 

 
2

 0
f  x  sin nxdx
Therefore the series of the odd function contains only sine terms.
Remark: Functions which are neither even nor odd contains both sine and cosine
terms.
Example 3.5
Consider the function
6   x  0
f  x  
 6 0 x 
f  x  2   f  x 
Before we do any evaluation, we can see that this is an odd function and therefore
sine terms only; i.e. a0  0
an  0 bn 
1

 
f  x  sin nxdx
f  x  sin nx is a product of odd functions and is therefore even.
199

12  cos nx 
bn   f  x  sin nxdx   6sin nxdx  
 0
 0
 
n 0
 0 n even
12

 1  cos n    24
n
 n n odd
2

2

The Fourier series is then:
f  x 
24 
1
1
sin x  sin3x  sin5x 
 
3
5



Example 3.6
Obtain a Fourier expansion for f  x   x for   x   .
3
Solution:
f  x   x3 is an odd function, therefore a0  an  0 n
for
bn 
2


0
6
n  
x3 sin nxdx  2  1 
 3
 n n 
   2 6 

  2 6 
  2 6 
x3  2   
 3  sin x  
 3  sin 2 x  
 3  sin x  ...
2 
3 
 2
 3
  1 1 

200
Half – Range Series, Periodic for
 0, 
 0,  . To get the series of cosine
f  x  is an even function in the interval   ,  .
Let the given function be defined in the interval
only, the function
Then,
an 

2
 0
f  x  cos nxdx and bn  0 n
f  x  as series of sine only, extend the interval to   ,  and treat
To expand
f  x  as odd function.
Then,
bn 

2
 0
f  x  sin nxdx , an  0 n
Example 3.7
Find the Fourier sine series for the function f  x   e
ax
for 0  x  
Solution:

Let
eax  bn sin nx
n1
bn 
2

 0
eax sin nxdx 

2n
1   1n ea 

a 2  n2  

Thus,
b1 
eax 

2 1  ea
a
2

,b
 12 
2 

2  2 1  ea
a
2


 22 

 1  ea 
2  1  ea 
sin
x

2
 2 2 
 2 2  sin 2 x  ...
  a  1 
a 2 

201
Example 3.8
Expand
f  x   x,0  x  2 in a half range
a.
sine series
b.
cosine series
Solution:
a)
an  0
bn 
2 L
n x
2 2
n x
f
x
sin
dx

x sin
dx




L 0
L
2 0
2
2
  2
n x    4
n x  
4
  x   cos
cos 
  1  2 2 sin
  
n

2
2
n

n





 
 0


Then
4
n x
cos n sin
2
n 1 n
f  x  
4   x 1 2 x 1 3 x

  sin
 sin
 sin
 ... 

2 2
2
3
2

b)
Thus bn  0 ,
an 
2 L
n x
2 2
n x
f
x
cos
dx

x
cos
dx


L 0
2
2 0
2
2
n x 
n x  
  2
 4
  x   sin

1
cos



 2 2

2 
2  0
n 
  n

If
4
n
2 2
 cos n  1
2
n  0, a0   xdx  2
0

Then
1
if n  0 .
f  x  1  
n 1 n
4

2
cos n  1 cos
2
8  x 1
3 x 1
5 x

cos

cos

cos
 ... 

2
2
2
2 3
2
2
 
5

202
n x
2
Change of Integral and Functions of Arbitrary Period
For some problems the period of the function is not 2 but t or
2L . To conform
to the general theory, this period must be converted to its equivalent of 2 , and
such will equally affect the argument of the function proportionally.
Let
f  x  be defined in the interval   L, L  , to convert this to 2 ;
2L is the interval for variable x
1 is the interval for the variable
x
2L
and therefore 2 is the interval for the variable
2 x  x

2L
L
let
z
x
L
, then
x
zL

zL
f  x  of period 2L is transformed to the function f   of
 
Thus the function
period 2 .Thus,


 zL 
f    a0   an cos nz  bn sin nz
 
n1
n1
a0 
where
but
an 
1

x
0
1
2
 0
zL

1 2L
 zL 
x 
f  dz 
f  x d   o

2 0
 
 L 
and so
a0 
1 2L
f  x dx
2L 0
1 2L
n x   x 
 zL 
f   cos nzdz   f  x  cos
d
L 0
L  L 
 
2
an 
bn 
1 2
2 0
1 2L
n x
f
x
cos
dx


L 0
L
1 2L
n x
 zL 
f   sin nzdz   f  x  sin
dx
L 0
L
 
n x 
n x
 f  x   a0   an cos
 bn sin
L
L
n1
n1

203
In general the Fourier series with period   2L is stated as:

n
n 

f x   a0    an cos
x  bn sin
x  …………….(3.3)
L
L 
n 1 
with the Fourier coefficients of f x  given by the Euler formulas:
a0 
1 L
f x dx
2L  L
an 
1 L
f x cos nxdx ,
L  L
n  1,2,...
bn 
1 L
f x sin nxdx
L  L
n  1,2,...
Corollary: For half range series in the interval

f  x   a0  an cos
Cosine series:
n1
a0 
where
f  x   bn sin
n1
where
bn 
n x
L
1 L
2 L
n x
,
f
x
dx
a

f
x
cos
dx




n
L 0
L 0
L

Sine series:
 0, L  ;
n x
L
2 L
n x
f
x
sin
dx


L 0
L
Example 3.9
1.
Given
f  x  
Expand
0
1
0 xc
c  x  2c
f  x  in the Fourier series of period 2c
Solution:

f  x   a0   an cos
n1
n x 
n x
 bn sin
L
L
n1
Recall that:
a0 
1 2c
1 c
1 2c
1
f
x
dx

0
dx

1dx 






2c 0
2c 0
2c c
2
204
an 
2c
1 2L
n x
1 c
n x
n x 
f
x
cos
dx

0
cos
dx

1
cos
dx   0






c
L 0
L
c 0
c
c

bn 
2 L
n x
f
x
sin
dx


L 0
L
 bn 
2c
1 2c
n x
1 c
n x
n x 
f
x
sin
dx

0
sin
dx

1
sin
dx 






c
c 0
L
c 0
c
c

 2
 bn   n
0
n is odd
n is even
1 2   x 1 3 x

 f  x    sin
 sin
 ...
2 
c 3
c

2.
0 2  x  1

1  x  1
Find the Fourier series of the function f  x   k
0 1  x  2

Solution: We determine the Fourier series, but the period 2c  4
c2
a0 
1 c
1 2
f
x
dx

f  x dx


2c c
2  2 2
1
2
1 1
k
 a0     0 dx   kdx    0 dx  
 2
1
1
4  2
an 
1 2L
n x
1 c
n x
f
x
cos
dx

f
x
cos
dx




L 0
L
c c
c
 an 
bn 
1 2
n x
1 1
n x
2k
n
f
x
cos
dx

k
cos
dx

sin


2 2
2
2 1
2
n
2
1 2L
n x
1 c
n x
f
x
sin
dx

f
x
sin
dx




L 0
L
c c
c
 an 
1 2
n x
1 1
n x
f
x
sin
dx

k sin
dx  0




2 2
2
2 1
2
f  x 
k 2k 1 
 x 1 3
3

  sin cos
 sin cos x  ...
2  1 2
2 3
2
2

then
205
f  x 
k 2k   x 1
3

 cos
 cos x  ...
2  
2 3
2

Fourier Series from the Sketches of the Wave Form
Example 1: Derive the Fourier series for the sequence wave form up to the fire four
terms. The figure is shown below:
Solution:
Let
d 0  x  
y  f  x  defined as f  x   
 0   x  2
Then by Fourier series expansion, with period 2
n x 
n x
f  x   a0   an cos
 bn sin
L
L
n1
n1

where L  
a0 
an 
1
2

2
0
f  x dx 

1  
d
d
dx

0
dx




0
 2
2  0
2
1 2L
n x
1 
f
x
cos
dx

d
cos
xdx

0  cos xdx   0










0
0

L
L

2
 2nd
1 

bn 
 d  sin xdx    0 sin xdx   
  0
0
n is odd
n is even
Then
y  f  x 
d 2d 1
1
1


sin
x

sin3
x

sin5
x

...

2  1
3
5
206
Example 2: Find the first three terms of the Fourier series for the complete square
waveform shown in the following figure:
Solution:
Let y  f
 d / 2 0  t  
d / 2  t  2
t   
Notice that the square wave form is skew – symmetric. Then the Fourier
series is to comprise of only sine terms. Thus,
a0  0 , an  0 n
bn 
1
2
 0
 bn 
f t  sin t  d t 
2  d 
1  d 

sin

t
d

t


sin

t
d

t














  2 
  0  2 

 2d
d 
n

bn 
1   1    n

n 
 0
y  f t  
odd
even
2d 1
1
1

sin

t

sin3

t

sin5

t

...

 1
3
5
207
Problems
1. Determine whether the following functions are piecewise continuous on
1,5 :
Ans
a)
 x2
x0

f  x   4 0  x  2
x
x0

yes
b)
1/  x  2 2
f  x  
2
 5x  1
no
c)
1
f  x 
x2
d)
f  x 
x2
x2


 lim f  x    
2
 xx

 2



no  lim f  x    
2
 xx

 2

1
x2
yes,
2. Which of the following functions are piecewise smooth on
a)
 x3
x0

f  x    sin  x 0  x  1
 x2  5x
x 1

yes
b)
x

e
f  x  

 x
yes
x 1
x 1
c)
f  x   ln x
d)
2

  x  1
f  x  
1/ 3

 x  1
2,3 ?


no  limln x   
0
 xx

 0

x 1
x 1
208


1



2/3
0
 xx

3
x

1

 0 

no  lim
3. State whether each of the following products is odd, even or neither
(a) x sin x
2
(f).
 2x  3 sin 4x
(b) x cos x
3
(g).
sin3x cos3x
(c) cos2 x cos3x
(h).
x3e x
(d) x sin x
(i).
x
(e) 3sin x cos4 x
(j).
1
cosh x
x2
4

 4 sin 2 x
4. Find the smallest positive period of
(a)
cos x,sin x,cos2 x,sin 2 x,cos  x,sin  x,cos2 x,sin 2 x
(b)
cos nx,sin nx,cos 2k x ,sin 2k x ,cos 2knx ,sin 2knx
5. Find the Fourier series of the function
f  x  , which is a assumed to have the
period 2 , and plot a accurate graphs of the first three partial sum.
1 for    x  0
f  x  
 1 for 0  x  
(a)
f  x   x;    x    (d)
(b)
f  x   x2 ;   x   
(e)
 / 4 for    x  0
f  x  
and also f     f  0  f    0
  / 4 for 0  x  
(c)
f  x   x3;    x    (f)
 1 for    x  0
f  x  
1 for 0  x  
1 for   x   / 2

(g) f  x    0 for  / 2  x   / 2
 1 for
 /2  x 

209
6. Given that f  x   x  x for   x   , find the Fourier expression for
2
f  x  . Deduce that
2
6
1
1 1 1
   ...
22 32 42
7. State whether the given function is even or odd. Find the Fourier series. Sketch
the function and some partial sums.
8.
a)
k if   / 2  x   / 2
f  x  
 0 if  / 2  x  3 / 2
b)
x if   / 2  x   / 2

f  x  
  x if  / 2  x  3 / 2
c)
f  x   x2 / 2
   x   
Find the Fourier cosines as well as the Fourier sine series.
(a) Find a Fourier sine series for
Ans:

f  x   1 on  0,1 .
 1   1  sin n x
 n
2
1
n
n1
(b) Find a Fourier sine series for
Ans: 
6



n1
f  x   x on  0,3 .
 1n sin n x
n
3
(c) Find a Fourier cosine series for f  x   x on
2
 0,  .

1 2
 1 cos nx
Ans:   4
2
3
n1 n
n
0 x  2
on  0,3
2 x  2
(d) Find a Fourier cosine series for f  x   
2 4  1 2n
n x
Ans:   sin
cos
3  n1 n
3
3
(e) Find a Fourier cosine series for
Ans:
f  x   1 on  0,7  .
1
210
x x  1
on  0,2 
2 x  1
(f) Find a Fourier sine series for f  x   

Ans:
 n 
n1
9.
4
2
2
sin
n 2
n 4
 n x
 cos
 cos n  sin
2 n
2 n
3

Find the first four terms of the Fourier series for the saw – tooth wave
form as shown below:
Ans:
10.
y  f t  
2L 1
1
1
1

sin

t

sin
2

t

sin3

t

sin
4

t

...

 1
2
3
4
Determine the first four terms the Fourier series for the triangular wave
form in the following figure;
211
 2d
0  t   / 2
  t

 2d
Hint: f  x   
t  2d  / 2  t  3 / 2


 2d
3 / 2  t  2
  t  4d
Ans:
y  f t  
8d  1
1
1
1

sin t  2 sin3t  2 sin5t  2 sin7t  ...
2  2
 1
3
5
7

212
UNIT EIGHT
TOTAL DIFFERENTIAL EQUATIONS
4.0
Recall that a function f x, y , its total differentials, df , is defined as:
df 
f
f
dx  dy ---------------------(٠)
x
y
This shows that the family of curves f x, y   c satisfies the differential
equation df  0 .
f
f
dx  dy  0 ----------------------(٠٠)
x
y
So if there exists a function f x, y  such that:
M  x, y  
f
y
and N x, y  
x
y
then M x, y dx  N x, y dy is called an exact differential, and the equation:
M x, y dx  N x, y dy  0
is said to be an exact equation, whose solution is the family f x, y   c .
The general form of Total Differential Equation is given as:
Px1, x2 ,..., xn dx1  Qx1, x2 ,...xn dx2  ...  S x1, x2 ,..xn dxn  0 ..(4.1)
Examples:
1.
3x
 

2.
3xz  2 y dx  xdy  x 2 dz  0
3.
ydx  dy  dz  0


y  e x z dx  2 x 3 y  sin z dy  y cos z  e x dz  0
2 2
It may be verified readily that example 1 is the exact differential of:
f x, y, z   x 3 y 2  e x z  y sin z  c , c is an arbitrary constant. Such
an equation is called exact.
Example 2 is not exact, but the use of x as an integrating factor yields:
3x z  2xydx  x dy  x dz  0
2
2
3
213
which is the exact differential of x z  x y  c . Equations 1 and 2 are called
3
2
integrable.
Example 3 is not integrable; that is, no primitive (or general solution).
4.1
CONDITION OF INTEGRABILITY AND EXACTNESS
The condition of integrability of the total differential equation
Px, y, z dx  Qx, y, z dy  Rx, y, z dz  0 …(4.2)
is
 Q R 
 P Q 
 R P 
  0 , identically….(4.3)
P
   Q
   R


z

y

x

z

y

x






And the conditions for exactness are:
P Q Q R R P



,
,
…………(4.4)
y Dx z y x z
Example 4.1
1.
The equation 3xz  2 y dx  xdy  x dz  0
2
P  3xz  2 y ,
R  x2 ,
P
Q
Q
P
 3x ; Q  x ,
1,
 0 ; and
 2,
z
x
z
y
R
R
 2x ,
 0 .Then from (4.3),we have:
x
y
3xz  2 y 0  0  x2 x  3x  x 2 2  1  0  x 2  x 2  0
then the
equation is integrable.
2.
The equation
P  y,
ydx  dy  dz  0
Q Q
P
P
R R

 0 ; R  1,

 0.
 1,  0 ; Q  1,
x z
x y
y
z
Then from (4.3),we have:
y0  0  10  0  11  0  1  0 , the equation is not integrable
214
3.
The equation
3x
 



y  e x z dx  2 x 3 y  sin z dy  y cos z  e x dz  0
2 2
P  3x 2 y 2  e x z ,
P
P
 e x
 6x2 y ,
z
y
Q  2 x 3 y  sin z ,
Q
Q
 6x 2 y ,
 cos z
x
z
R  y cos z  e x ,
R
R
 e x ,
 cos z
x
z
Hence the equation is exact.
4.2
AN APPROACH TO SOLVE AN INTEGRABLE TOTAL DIFFERENTIAL
EQUATION (in three variables)
I)
If the equation is exact, the solution is evident after, at most, a
regrouping of terms.
II)
If the equation not exact, it may be possible to find an integrating
factor (note: integrating factor would be given to help you solve).
III)
If the equation is homogeneous, one variable, say
z , can be separated
from the others by the transformation x  uz , y  vz .
IV)
If no integrating factor can be found, consider one of the variables, say
z , as a constant. Integrate the resulting equation, denoting the
arbitrary constant of integration by  z , a function of z . Take the
total differential of the integral just obtained and compare the
coefficients of its differentials with those of the given differential
equation, thus determining
  z .
Example 4.2
1.
Solve x  y dx  xdy  zdz  0
Solution:
Here
P  x  y , Q  x , R  z
215
P
Q Q
P
R R
 1 
0
0
,
,
, the equation is
x
x z
z
y x
Then
exact.
Upon regrouping thus xdx  xdy  ydx  zdz  0
 xdx   xdy  ydx   zdz  k
 12 x 2  xy  12 z 2  k
 x 2  2 xy  z 2  c
2.
Solve y dx  zdy  ydz  0 .
2
Solution:
Here
P  y2 ,
R  y,
Q
Q
P
P
 0,
 1,
 2 y,  0 ; Q   z ,
x
z
y
z
R
R
 0,  1 :
x
y
Then
 Q R 
 R Q 
 R P 
 
P
   Q    R 
 x z 
 z y 
 y X 
y 2  1  1  z0  0  y2 y  0  0
and the equation is integrable. The integrable factor
equation to
dx 
x
ydz   zdy
 0 , whose solution is:
y2
z
 c.
y
216
1
reduce the
y2
3.
Solve the homogeneous equation
2 y  z dx  x  z dy  2 y  x  z dz  0 .
Solution:
The equation is integrable since:
2 y  z  1  2  x  z  1  2  2 y  x  z 2  1  0 The
transformation x  uz, y  vz reduces the given equation to:
2zv  1udz  zdu  zu  1vdz  zdv  z2v  u  1dz  0
Dividing by z and rearranging, we have:
2zv  1du  zu  1dv  uv  u  v  1dz  0 or, dividing by
zuv  u  v  1  zu  1v  1 ,

2du
dv dz

 0
u 1 v 1 z
Then
2 lnu  1  lnv  1  ln z  ln k
 zu  12  k v  1
 x  z 2  k  y  z 
 y  z  cx  z 2 .
Exercise 4.1 (Solve the following)
2.
2x  1dx  x dy  x tan zdz  0 I.F. 1/ x
2x  z zdx  2x yzdy  xz  xdz  0 I.F. 1/ x z
3.
yzdx  z 2 dy  xydz  0
4.
2 y  z dx  2x  z dy  x  2 y dz  0
1.
3
3
4
2
2
2
2
217
UNIT NINE
PARTIAL DIFFERENTIAL EQUATION
5.0 NOTATION AND TERMINOLOGY
Let u denote a function of several independent variables; say,
u  u  x, y, z, t  . At
 x, y, z, t  , the partial derivative of u with respect to x is defined by
u  x  h, y, z, t   u  x, y, z, t 
u
 lim
……..(5.0)
x
h0
h
provided the limits exists. We will use the following subscript notation:
u
u
 ux
 uy
x
y
5.1
 2u
 uxx
x2
 2u
 u yy
y 2
 2u
 2u
 u yx or
 uxy …
xy
yx
Definition
A partial differential equation (abbreviated PDE) is an equation involving one or
more partial derivatives of an unknown function of several independent variables.
The order of a PDE is the order of the highest derivative that appears in the
equation.
The partial differential equation
be linear if the function
F  x, y, z, t;u, ux , u y , uz , ut , uxx , uxy ,...  0 is said to
F is algebraically linear in each of the variables u, ux , u y ,...,
and if the coefficients of u and its derivatives are functions only of the independent
variables. An equation that is not linear is said to be nonlinear; a nonlinear equation
is quasilinear if it is linear in the highest – order derivatives.
Example 5.1
z
z
 y  z is of order one.
x
y
1.
x
2.
2 z
2 z 2 z
3

 0 is of order two.
xy y 2
x 2
3.
 2u 2  2u
c
t 2
x2
One – dimensional wave equation
218
4.
 2u 2  2u
c
t
x2
One – dimensional heat equation
5.
2u 2u

0
x2 y 2
Two – dimensional Laplace equation
6.
 2u  2u

 f  x, y 
x2 y 2
Two – dimensional Poisson equation
7.
 2u 2   2u  2u 
c  2  2 
t 2
y 
 x
Two – dimensional wave equation
8.
 2u  2u  2u


0
x2 y 2 z 2
Three – dimensional Laplace equation
Their solution is in the form:
5.2
u  f  x, y, z, t ,...
SOLUTION TO A FIRST ORDER PARTIAL DIFFERENTIAL EQUATION
A linear partial differential equation of order one, involving a dependent variable
z and two independent variables x and y , is of the form:
P
where
z
z
 Q  R ……………………(5.1)
x
y
P, Q, R are functions of x, y, z .
If P  0 , or
Q  0 , equation (5.1) may be solved easily. Thus, the equation
z
 2 x  3 y has as solution z  x 2  3xy    y  , where  is an arbitrary
x
function.
Lagrange reduced the problem of finding the general solution of 5.1 to that of
solving an auxiliary system (called the Lagrange system) of ordinary differential
equations:
dx dy dz
  ……………….(5.2)
P Q R
219
by showing that:
 u, v  0
, arbitrary
is the general solution of equation (5.1) provided u  ux, y, z   a and
v  vx, y, z   b are two independent solutions of equation (5.2). Here, a
………(5.3)
and
b are arbitrary constants and at least one of u, v must contain z .
Example 5.1
Find the general solution of
x
z
z
 y  3z .
x
y
Solution:
The auxiliary system is:
dx dy dz
  .
x
y
z
From
v
dx dz
dx dz
z
 , we obtain u  3  a ; and from
 , we obtain
x 3z
x
y
x
y
b.
x
Thus, the general solution is
Of course, from
 
z y
,   0 , where  is arbitrary.
 x3 x 
dy dz
z
 , we obtain 3  c , and we may write
y 3z
y
 z z 
 z y
  0 or   ,   0
,
3
3
 y3 x 
x y 


 
where  and
 are arbitrary. However, these are all equivalent and
we shall call any one of them the general solution
220
COMPLETE SOLUTIONS
If u  a and v  b are two independent solutions of equations (5.2) and if

,
are arbitrary constants,
u  v  
is called a complete solution of (5.1).
Thus, for the equation is the last example above ()
z / x 3    y / x    is a complete solution.
Example 5.2
Find the general solution of
2
z
z
 3 1
x
y
Solution:
The auxiliary system is :
From
dx dy dz
 
2
3 1
dx dy
dx dz
 , we have
 , we have x  2 z  a ; and from
2
3
2 1
3x  2 y  b . Thus, the general solution is:
 x  2 y,3x  2 y   0 .
The complete solution x  2 z   3x  2 y    is a two – parameter
of family of planes.
5.3
TYPES OF SECOND – ORDER EQUATIONS
In the linear PDE of order two in two variables,
auxx  2buxy  cu yy  dux  eu y  fu  g ………………..(5.3)
if uxx is formally replaced by
 2 , uxy
by
 , u yy by  2 , u x
then associated with (5.3) is a polynomial of degree two in
P  ,    a 2  2b  c 2  d  e  f
221
by

 ,and u y by  ,
and
:
The mathematical properties of the solutions of (5.3) are largely determined by the
algebraic properties of the polynomial
P  ,   . P  ,   - and along with it, the
PDE (5.3) – is classified as hyperbolic, parabolic, pr elliptic according as its
discriminant,
b2  ac is positive, zero, or negative. Note that the type of (5.3) is
determined solely by its principal part (the terms involving the highest – order
derivatives of u ) and that the type will generally change with position in the xy 
plane unless
a, b, and c are constants.
Example 5.2
1.
The PDE
3uxx  2uxy  5uyy  xuy  0 is elliptic, since
b2  ac  12  3 5  14  0
2.
The Tricomi equation for transonic flow,
uxx  yuyy  0 , has
b2  ac  02  1 y   y
Thus, the equation is elliptic for
hyperbolic for
y  0 , parabolic for y  0 , and
y  0.
The general linear PDE of order two in n variables has the form:
n
a u
i , j 1
If
ij xi x j
n
  biuxi  cu  d ……………………….(5.4)
i 1
uxi x j  ux j xi , then the principal part of (5.4) can always be arranged so that
aij  a ji ; thus, the n  n matrix A  aij  can be assumed symmetric. In linear
algebra it is shown that every real, symmetric n  n matrix has n real eigenvalues.
These eigenvalues are the (possibly repeated) zeros of an n th – degree polynomial
in
 , det  A  I   0 , where I is the n  n identity matrix. Let P denote the
number of positive eigenvalues, and
Z the number of zero eigenvalues (i.e., the
multiplicity of the eigenvalues zero), of the matrix A . Then (5.4) is:
hyperbolic
if Z  0 and P  1 or Z  0 and P  n  1
222
parabolic
if Z  0 (equivalently, if det A  0 )
elliptic
if Z  0 and P  n or Z  0 and P  0
ultrahyperbolic
if Z  0 and 1  P  n  1
if any of the
aij is nonconstant, the type of (5.4) can vary with position.
Example 5.3
For the PDE
3ux1x1  ux2 x2  4ux2 x3  4ux3x3  0 ,
0
0 
3 0 0
3  
A  0 1 2 and det  0 1  
2    3        5  0
0 2 4
 0
2
4   
Because
 0
is an eigenvalues, the PDE is parabolic (throughout x1x2 x3  space).
Exercise 5.1
1. Classify according to type:
a)
uxx  2 yuxy  xu yy  ux  u  0
b)
2xyuxy  xu y  yux  0
c)
uxx  uxy  5uyx  uyy  2uyz  uzz  0
2. Describe the regions where the equation is hyperbolic, parabolic, elliptic:
a)
uxx  uxy  2u yy  0
b)
uxx  2uxy  u yy  0
c)
2uxx  4uxy  3uyy  5u  0
d)
uxx  2 xuxy  u yy   cos xy  ux  u
e)
yuxx  2uxy  exuyy  u  3
f)
exyuxx  sinh x  uyy  u  0
g)
xuxx  2xyuxy  yu yy  0
h)
xuxx  2xyuxy  yu yy  0
223
SOLUTIONS TO SECOND ORDER LINEAR PARTIAL
DIFFERENTIAL EQUATION.
Example 5.3
u
 2u
2
 sin t .
x

0
u

cos2
t

12
x
t

1
Solve
given
that
,
and


x
x2
Solution
u
 12 x2  t  1 dx  4 x3  t  1    t 
x
u  x4 t  1  x t    t 
Using initial conditions
   t   sin t
hence
x  0;
u
 sin t at u  cos2t
x
 t   cos2t
u  x4 t  1  x sin t  cos2t .
Exercise
 2u
 sin  x  y  , given that
Solve the equation
xy
y  0,
u
2
 1; x  0, u   y  1
x
Initial Conditions and Boundary Conditions
As with any differential equation, the arbitrary constant or arbitrary functions in any
particular case are determine from initial conditions or, more generally, the boundary
conditions since they do not always refer to zero values of independent variables.
224
5.4
SEPARATION OF VARIABLES
Solving second – order PDE with two independent variables we assume a trial
solution of the form
u  x, t     x  t  (or simply u   )where;
  x  is a function of x only.
t  is a function of t only.
Therefore
u
 2u
   2  ;
x
x
u
 2u
   2  
t
t
 2u 2  2u
 1 

c
Hence; the Wave equation
can
be
written
as
.

 c2 
t 2
x2
the Heat equation
 2u 2  2u
 1 
c
can be written as
.

2
t
 c2 
x

Y
2u 2u
the Laplace equation
can
be
written
as
.




0

Y
x2 y 2
5.5
SOLUTION OF THE WAVE EQUATION
u  f  x, t 
P
u  x, t 
0
l
x
 2u 1  2u

The wave equation
has a solution u  x, t  .
x2 c2 t 2
Boundary conditions:
a) The string is fixed at both ends, i.e. at x  0 and x  l for all values of
time t . Therefore
u  x, t  becomes
u  0, t   0
 for all values of t  0
u l, t   0 
225
Initial conditions:
b) If the initial deflection of
P at t  0 is denoted by f  x  , then
u  x,0  f  x  .
c) Let the initial velocity of
 u 
P be g  x  , then    g  x 
 t t 0
 2u 1  2u
 1 


The wave equation
becomes
.

 c2 
x2 c2 t 2
Let

1 
 k also 2  k

c 
  k   0 and   c2k  0
For oscillatory motion let k   p i.e. k  0 .
2
  p 2   0
   Acos px  B sin px
and also   c p   0
  C cos cpt  Dsin cpt
2 2
so our suggested solution u   now becomes
u  x, t    Acos px  B sin px C cos cpt  D sin cpt 
and, if we put
cp    p   / c

 

 u  x, t    Acos x  B sin x  C cos t  D sin t 
c
c 

where
A, B, C, D are arbitrary constants.
Now, using the boundary conditions
u  0 and x  0  0  AC cos t  D sin t  t  A  0
Therefore,
u0

u  x, t   B sin x  C cos t  D sin t 
c
x  l x  0  B sin
Now, B  0 or
l
c
C cos t  D sin t 
u  x, t  would be identically zero.
226
sin
l
c

0
l
c
 n
 
nc
for n  1,2,3,...
l
As we can see, there is an infinite set of values of
value gives a particular solution for
 and each separate
u  x, t  . The values of  are called
the eigenvalues and each corresponding solution the eigenfunction.
Eigenvalues Eigenfunctions
nc
l
u  x, t   B sin
x
C cos t  D sin t
n

1
1 
c
l
u1  sin
2
2 
2c
l
u2  sin
2 x 
2c t
2c t 
 D2 sin
C2 cos

l 
l
l 
3
3 
3c
l
u3  sin
3 x 
3c t
3c t 
 D3 sin
C3 cos

l 
l
l 
r
r 
rc
l
ur  sin
r x 
rc t
rc t 
 Dr sin
Cr cos

l 
l
l 
c
x
c t
c t 
 D1 sin
C1 cos

l 
l
l 
Note BC  Cn and BD  Dn , where C1, C2 , C3 ,.... and D1, D2 , D3 ,.... are
arbitrary constants.
If u1, u2 , u3 ,.... are solutions, then the linear combination is also a solution, which
is a more general solution, i.e. u  u1  u2  u3 
So


rc t
rc t 
 r x 
u  x, t    ur   sin
 Dr sin
 Cr cos

l 
l
l 
r 1
r 1 
To solve for the arbitraries Cr , Dr we use the initial conditions which we have
not yet taken into account.
227
Example5.2
A stretched string of length 20cm and is set oscillating by displacing its midpoint a
distance 1cm from its rest position and releasing it with zero initial velocity. Solve
 2u 1  2u
the wave equation
 2 2 where c2  1 to determine the resulting motion,
2
x
c t
u  x, t  .
Solution:
2
1
20
10
u
x
10
u 2
x
10
From data given the boundary conditions are
u  0, t   0;
u  20, t   0 fixed endpoints.
x
0  x  10
10
u  x,0  f  x   
 20  x 10  x  20
 10
 u 
 t  (zero initial velocity)
t 0
Since c  1 , then we have   
Which implies   p   0 and
2
  p2  0
The respective solutions are
  Acos px  B sin px and   C cos pt  Dsin pt
u  x, t      Acos px  B sin px C cos pt  D sin pt 
 u  x, t    Acos  x  B sin  x C cos t  D sin t  where   cp but c  1
228
u  0, t   0, x  0 0  AC cos t  D sin t   A  0
 x, t   B sin  x C cos t  D sin t 
u  20, t   0, x  20 0  B sin 20 C cos t  D sin t 
Therefore,
B  0 or u would be identically zero sin 20  0  20  n
Hence,
u  x, t   B sin
 
n
20
n 
n
n 
x  C cos t  D sin t 
20 
20
20 
Follow the table below:
Eigenvalues Eigenfunctions
n
20
u  x, t   B sin  xC cos t  D sin t
n

1
1 
2
2 
2
20
u2  sin
2 x 
2 t
2 t 
 D2 sin
C2 cos

20 
20
20 
3
3 
3
20
u3  sin
3 x 
3 t
3 t 
 D3 sin
C3 cos

20 
20
20 
r
r 
r
20
ur  sin
r x 
r t
r t 
 Dr sin
Cr cos

20 
20
20 

u1  sin
20
x
t
t 
C1 cos  D1 sin 
20 
20
20 
Note BC  Cn and BD  Dn , where C1, C2 , C3 ,.... and D1, D2 , D3 ,.... are
arbitrary constants.
u  u1  u2  u3 


r t
r t 
 r x 
u  x, t    ur   sin
 Dr sin
 Cr cos

20 
20
20 
r 1
r 1 
229
To find Cr and Dr we use the rest of the conditions i.e.;
x
0  x  10
10
u  x,0  f  x   
 20  x 10  x  20
 10

u  x,0    Cr sin
r 1
Then
Cr  2  mean value of f  x  sin
Cr 
10
0
10Cr  
20 20  x
x
r x
r x
sin
dx  
sin
dx
10 10
10
20
20
20
r
40
r 20
r
40
cos  2 2 sin  cos  2 2 sin r
r
2 r
2 r
2 r
4
For r=1,2,3,... Cr 

u  x, t    sin
r 1
and for the condition

r x
between x  0 and x  20
20
2 20
r x
f  x  sin
dx

20 0
20
10Cr  
Then
r x
20
r
2 2
sin
r
2
r x  4
r
r t
r t 
 Dr sin
 2 2 sin cos

20  r 
2
20
20 
u
 u 
t

0;
0
(zero
initial
velocity),
which
is
 t 

t
t 0
u  x, t  
r x  4
r  r
r t 
r
r t 
  sin
sin

sin

D
cos



r
t
20  r 2 2
2  20
20 
20
20 
r 1
Therefore, at t  0 ,

0   sin
r 1
So finally we have
u  x, t  
r x r
D
20 r 20
4

2

1
 r 2 sin
r 1
 Dr  0
r x r
r t
sin cos
20
2
20
230
5.5
HEAT AND LAPLACE EQUATION
The Heat and Laplace equations look slightly different from the wave equation, but
the method of solution is very much along the same lines.
1.
For Heat equations;
c2uxx  ut

 1 

 c2 
   p2  0 given  = A cos px  B sin px
   p2c2  0 given   Ce p c t
2 2
u  x, t      Acos px  B sin px  Ce p c t
2 2
pc  
2.

  2

u  x, t    AC cos x  BC sin x  e t
c
c 

For Laplace’s equation:
uxx  u yy  0


Y


Y
   p2  0 given  = A cos px  B sin px
 Y  p2Y  0 given Y  C cosh py  D sinh py, which we can
express as: E sinh p  y   
u  x, y    Acos px  B sin px  E sinh p  y   
Exercise 5.2
1.
A bar of length 2m is fully insulated along its sides. It is initially at a
uniform temperature of 10C and at t  0 the ends are plunged into ice
and maintained at a temperature of 0C . Determine an expression for
the temperature at a point P a distance x from one end at any
subsequent time t seconds after t  0 .
231
2.
 2u  2u
Determine a solution u  x, y  of the Laplace equation

0
x2  2 y
subject to the following boundary conditions:
u  0 when x  0; u  0 when x   ;
u  0 when y   ; u  3 when y  0 .
SOLUTIONS TO SECOND ORDER LINEAR PARTIAL
DIFFERENTIAL EQUATION.
Example 5.3
u
 2u
 sin t .
 12 x2  t  1 given that x  0 , u  cos2t and
Solve
2
x
x
Solution
u
 12 x2  t  1 dx  4 x3  t  1    t 
x
u  x4 t  1  x t    t 
Using initial conditions
x  0;
u
 sin t at u  cos2t
x
232
   t   sin t
hence
 t   cos2t
u  x4 t  1  x sin t  cos2t .
Exercise
Solve the equation
y  0,
 2u
 sin  x  y  , given that
xy
u
2
 1; x  0, u   y  1
x
Initial Conditions and Boundary Conditions
As with any differential equation, the arbitrary constant or arbitrary functions in any
particular case are determine from initial conditions or, more generally, the boundary
conditions since they do not always refer to zero values of independent variables.
5.4
SEPARATION OF VARIABLES
Solving second – order PDE with two independent variables we assume a trial
solution of the form
u  x, t     x  t  (or simply u   )where;
  x  is a function of x only.
t  is a function of t only.
u
 2u

    2  ;
Therefore
x
x
Hence; the Wave equation
the Heat equation
u
 2u

   2  
t
t
 2u 2  2u
 1 

c
can
be
written
as
.

 c2 
t 2
x2
 2u 2  2u
 1 
c
can
be
written
as
.

t
 c2 
x2

Y
2u 2u
 .
the Laplace equation
 2  0 can be written as
2

Y
x y
5.5
SOLUTION OF THE WAVE EQUATION
233
u  f  x, t 
P
u  x, t 
0
l
x
 2u 1  2u

The wave equation
has a solution u  x, t  .
x2 c2 t 2
Boundary conditions:
d) The string is fixed at both ends, i.e. at x  0 and x  l for all values of
time t . Therefore
u  x, t  becomes
u  0, t   0
 for all values of t  0
u l, t   0 
Initial conditions:
e) If the initial deflection of
P at t  0 is denoted by f  x  , then
u  x,0  f  x  .
f) Let the initial velocity of
 u 
P be g  x  , then    g  x 
 t t 0
 2u 1  2u
 1 


The wave equation
becomes
.

 c2 
x2 c2 t 2
Let

1 
 k also 2  k

c 
  k   0 and   c2k  0
For oscillatory motion let k   p i.e. k  0 .
2
  p 2   0
and also   c p   0
2 2
   Acos px  B sin px
  C cos cpt  Dsin cpt
234
so our suggested solution u   now becomes
u  x, t    Acos px  B sin px C cos cpt  D sin cpt 
and, if we put
cp    p   / c

 

 u  x, t    Acos x  B sin x  C cos t  D sin t 
c
c 

where
A, B, C, D are arbitrary constants.
Now, using the boundary conditions
u  0 and x  0  0  AC cos t  D sin t  t  A  0

u  x, t   B sin x  C cos t  D sin t 
c
Therefore,
x  l x  0  B sin
u0
Now, B  0 or
sin
l
c
l
c
C cos t  D sin t 
u  x, t  would be identically zero.

0
l
c
 n
 
nc
for n  1,2,3,...
l
As we can see, there is an infinite set of values of
value gives a particular solution for
 and each separate
u  x, t  . The values of  are called
the eigenvalues and each corresponding solution the eigenfunction.
Eigenvalues Eigenfunctions
nc
l
u  x, t   B sin
x
C cos t  D sin t
n

1
1 
c
l
u1  sin
2
2 
2c
l
u2  sin
2 x 
2c t
2c t 
 D2 sin
C2 cos

l 
l
l 
3
3 
3c
l
u3  sin
3 x 
3c t
3c t 
 D3 sin
C3 cos

l 
l
l 
c
x
c t
c t 
 D1 sin
C1 cos

l 
l
l 
235
r
r 
rc
l
ur  sin
r x 
rc t
rc t 
 Dr sin
Cr cos

l 
l
l 
Note BC  Cn and BD  Dn , where C1, C2 , C3 ,.... and D1, D2 , D3 ,.... are
arbitrary constants.
If u1, u2 , u3 ,.... are solutions, then the linear combination is also a solution, which
is a more general solution, i.e. u  u1  u2  u3 

So

rc t
rc t 
 r x 
u  x, t    ur   sin
 Dr sin
 Cr cos

l 
l
l 
r 1
r 1 
To solve for the arbitraries Cr , Dr we use the initial conditions which we have
not yet taken into account.
Example5.2
A stretched string of length 20cm and is set oscillating by displacing its midpoint a
distance 1cm from its rest position and releasing it with zero initial velocity. Solve
 2u 1  2u
 2 2 where c2  1 to determine the resulting motion,
the wave equation
2
x
c t
u  x, t  .
Solution:
236
2
1
20
10
u
x
10
u 2
x
10
From data given the boundary conditions are
u  0, t   0;
u  20, t   0 fixed endpoints.
x
0  x  10
10
u  x,0  f  x   
 20  x 10  x  20
 10
 u 
 t  (zero initial velocity)
t 0
Since c  1 , then we have   
Which implies   p   0 and
2
  p2  0
The respective solutions are
  Acos px  B sin px and   C cos pt  Dsin pt
u  x, t      Acos px  B sin px C cos pt  D sin pt 
 u  x, t    Acos  x  B sin  x C cos t  D sin t  where   cp but c  1
u  0, t   0, x  0 0  AC cos t  D sin t   A  0
 x, t   B sin  x C cos t  D sin t 
u  20, t   0, x  20 0  B sin 20 C cos t  D sin t 
Therefore,
B  0 or u would be identically zero sin 20  0  20  n
Hence,
u  x, t   B sin
n 
n
n 
x  C cos t  D sin t 
20 
20
20 
237
 
n
20
Follow the table below:
Eigenvalues Eigenfunctions
n
20
u  x, t   B sin  xC cos t  D sin t
n

1
1 
2
2 
2
20
u2  sin
2 x 
2 t
2 t 
 D2 sin
C2 cos

20 
20
20 
3
3 
3
20
u3  sin
3 x 
3 t
3 t 
 D3 sin
C3 cos

20 
20
20 
r
r 
r
20
ur  sin
r x 
r t
r t 
 Dr sin
Cr cos

20 
20
20 

20
u1  sin
x
t
t 
C1 cos  D1 sin 
20 
20
20 
Note BC  Cn and BD  Dn , where C1, C2 , C3 ,.... and D1, D2 , D3 ,.... are
arbitrary constants.
u  u1  u2  u3 


r t
r t 
 r x 
u  x, t    ur   sin
 Dr sin
 Cr cos

20 
20
20 
r 1
r 1 
To find Cr and Dr we use the rest of the conditions i.e.;
238
x
0  x  10
10
u  x,0  f  x   
 20  x 10  x  20
 10

u  x,0    Cr sin
r 1
Then
Cr  2  mean value of f  x  sin
Cr 
r x
between x  0 and x  20
20
2 20
r x
f
x
sin
dx


20 0
20
10Cr  
10
0
Then
r x
20
10Cr  
20 20  x
x
r x
r x
sin
dx  
sin
dx
10 10
10
20
20
20
r
40
r 20
r
40
cos  2 2 sin  cos  2 2 sin r
r
2 r
2 r
2 r
For r=1,2,3,... Cr 

u  x, t    sin
r 1
and for the condition
4
r 2
sin
2
r
2
r x  4
r
r t
r t 
 Dr sin
 2 2 sin cos

20  r 
2
20
20 
u
 u 
 t  (zero initial velocity), which is t  0; t  0
t 0
u  x, t  
r x  4
r  r
r t 
r
r t 

  sin
 2 2 sin   sin
  Dr cos

t
20  r 
2  20
20 
20
20 
r 1
Therefore, at t  0 ,

0   sin
r 1
So finally we have
u  x, t  
4
r x r
D
20 r 20

1
 sin
 2 r2
r 1
 Dr  0
r x r
r t
sin cos
20
2
20
239
5.5
HEAT AND LAPLACE EQUATION
The Heat and Laplace equations look slightly different from the wave equation, but
the method of solution is very much along the same lines.
3.
For Heat equations;
c2uxx  ut

 1 

 c2 
   p2  0 given  = A cos px  B sin px
   p2c2  0 given   Ce p c t
2 2
u  x, t      Acos px  B sin px  Ce p c t
2 2
pc  
4.

  2

u  x, t    AC cos x  BC sin x  e t
c
c 

For Laplace’s equation:
uxx  u yy  0


Y


Y
   p2  0 given  = A cos px  B sin px
 Y  p2Y  0 given Y  C cosh py  D sinh py, which we can
express as: E sinh p  y   
u  x, y    Acos px  B sin px  E sinh p  y   
Exercise 5.2
3.
A bar of length 2m is fully insulated along its sides. It is initially at a
uniform temperature of 10C and at t  0 the ends are plunged into ice
and maintained at a temperature of 0C . Determine an expression for
the temperature at a point P a distance x from one end at any
subsequent time t seconds after t  0 .
240
4.
 2u  2u
Determine a solution u  x, y  of the Laplace equation

0
x2  2 y
subject to the following boundary conditions:
u  0 when x  0; u  0 when x   ;
u  0 when y   ; u  3 when y  0 .
241