UECM2023 Ordinary Differential Equations
Name of Instructor : Dr. Yap Hong Keat
E-mail address : yaphk@utar.edu.my
Lessons:
Tuesday 4.00pm-5.00pm (Lecture)
5.00pm-6.00pm (Tutorial)
Wednesday 1.00pm-3.00pm (Lecture)
Course Assessment :
Assessment Component Marks Allocation
Test 1
20%
Test 2
20%
Final Exam
60%
Total
100%
Tentative Teaching Plan:
Week
Date
Activities/Holiday
1
11 February 2025
Thaipusam
2
3
4
5
6
18 March 2025
Nuzul Al-Quran
7
25 March 2025 (Tuesday) tentatively
Test 1
8
01 April 2025
Hari Raya Puasa
9
10
11
12
29 April 2025 (Tuesday) tentatively
Test 2
13
14
January 2022 Denis CK Wong
Topic 1: First Order Equations
1.1 Definitions and Terminology
Definition. An equation containing the derivatives of one or more dependent variables, with respect to one or more
independent variables, is said to be a differential equation (DE).
Classification by Type.
An equation contains only ordinary derivatives of one
or more dependent variables with respect to a single
independent variable is called an ordinary differential
equation (ODE).
An equation involving partial derivatives of one or
more dependent variables of two or more independent
variables is called a partial differential equation
(PDE).
Suppose ċ is a function in term of Ċ, where Ċ is independent variable and ċ is dependent variables.
čċ
Ā2ć Ā2ć
+ 5ċ = Ď ý
+
=0
čĊ
ĀĊ 2 Āċ 2
2
č ċ čċ
Ā2ć Ā2ć
Āć
2
+ 6ċ = 0
=
22
čĊ 2 čĊ
ĀĊ 2 ĀĆ 2
ĀĆ
čĊ čċ
Āć
Āć
+
= 2Ċ + ċ
=2
čĆ čĆ
Āċ
ĀĊ
Leibniz Notation or Prime notation
or
ċ′ + 5ċ = Ď ý
ċ′′ 2 ċ′ + 6ċ = 0
ċ (1) + 5ċ = Ď ý
ċ (2) 2 ċ (1) + 6ċ = 0
Classification by Order.
The order of a differential equation (either ODE or PDE) is the order of the highest derivative in the equation.
ODE
čċ 3
č ċ
+ ( ) 2 4ċ = Ď ý
čĊ
čĊ 2
2
(ċ 2 Ċ)čĊ + 4Ċčċ = 0
Order
Second-order ordinary differential equation
First-order ordinary differential equation
First-order differential equation is often written in differential form ā(Ċ, ċ)čĊ + Ă(Ċ, ċ)čċ = 0.
ċ′ =
čċ Ċ 2 ċ
=
čĊ
4Ċ
4Ċċ ′ + ċ = Ċ
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January 2022 Denis CK Wong
In symbols, we can express an n th-order ordinary differential equation in one dependent variable by the general form
Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0, where Ă is a real-valued function of Ā + 2 variables: Ċ, ċ, ċ ′ , & , ċ (Ą) . It is possible to solve
an ordinary differential equation in the form Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0 uniquely for the highest derivative ċ (Ą) in terms
of the remaining Ā + 1 variables. The equation
čĄ ċ
= ď(Ċ, ċ, ċ ′ , & , ċ (Ą21) )
čĊ Ą
where ď is a real-valued continuous function, is referred to as the normal form of Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0. Thus,
č2ċ
= ď(Ċ, ċ, ċ ′ )
čĊ 2
čċ
= ď(Ċ, ċ),
čĊ
to represent general first- and second- order ordinary differential equations.
Example.
(a) The normal form of the first-order equation 4Ċċ ′ + ċ = Ċ is ċ ′ =
′′
′
ý2þ
4ý
.
′′
(b) The normal form of the second-order equation ċ 2 ċ + 6ċ = 0 is ċ = ċ ′ 2 6ċ.
Classification by Linearity.
An Ā 2th order ordinary differential equation Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0 is said to be linear if Ă is linear in ċ, ċ ′ , & , ċ (Ą) ,
that is,
or
Example.
ĊĄ (Ċ)ċ (Ą) + ĊĄ21 (Ċ)ċ (Ą21) + ⋯ + Ċ1 (Ċ)ċ ′ + Ċ0 (Ċ)ċ 2 Đ(Ċ) = 0
ĊĄ (Ċ)
Ā =1
linear first-order
čċ
Ċ1 (Ċ)
+ Ċ0 (Ċ)ċ = Đ(Ċ)
čĊ
č Ą21 ċ
čċ
čĄ ċ
(Ċ)
+
Ċ
+ ⋯ + Ċ1 (Ċ)
+ Ċ0 (Ċ)ċ = Đ(Ċ)
Ą21
Ą21
Ą
čĊ
čĊ
čĊ
Ā =2
linear second-order
č2ċ
čċ
Ċ2 (Ċ) 2 + Ċ1 (Ċ)
+ Ċ0 (Ċ)ċ = Đ(Ċ)
čĊ
čĊ
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January 2022 Denis CK Wong
Two properties of a linear ODE are as follows:
1.
2.
The dependent variable ċ and all its derivatives ċ9, ċ99, & , ċ (Ą) are of the first degree, that is, the power of
each term involving ċ is 1.
The coefficients Ċ0 , Ċ1 , & , ĊĄ of ċ, ċ9 & , ċ (Ą) depend at most on the independent variable Ċ.
ODE
(ċ 2 Ċ)čĊ + 4Ċčċ = 0
ċ ′′ 2 2ċ ′ + ċ = 0
č3ċ
čċ
+Ċ
2 5ċ = Ď ý
čĊ 3
čĊ
(1 2 ċ)ċ ′ + 2ċ = Ď ý
č2ċ
+ sin ċ = 0
čĊ 2
4
č ċ
+ ċ2 = 0
čĊ 4
Degree
First-order
Linearity
Linear
Second-order
Linear
Third-order
Linear
First-order
Second-order
Nonlinear 3 coefficient depends on ċ
Fourth-order
Nonlinear-Power of y is not 1
Nonlinear - Nonlinear function of ċ
Solution of an ODE. Any function �㔙, defined on an interval �㔼 and possessing at least Ā derivatives that are continuous
on �㔼, which when substituted into an Ā th-order ordinary differential equation reduces the equation to an identity, is
said to be a solution of the equation on the interval �㔼. In other words, a solution of Āth-order differential equation
Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0 is a function �㔙 that possesses at least Ā derivatives and for which
Ă(Ċ, �㔙(Ċ), �㔙′(Ċ), & , �㔙 (Ą) (Ċ)) = 0 for all Ċ in �㔼. We say that �㔙 satisfies the differential equation on �㔼. We should
assume that a solution �㔙 is a real-valued function. The interval �㔼 is called the interval of definition, the interval of
existence, the interval of validity, or the domain of the solution.
Example. Verify that the indicated function is a solution of the given differential equation on the interval (2 ∞, ∞).
Ăþ
1
(a)
= Ċ √ċ; ċ = Ċ 4
Ăý
16
′′
(b) ċ 2 2ċ ′ + ċ = 0; ċ = ĊĎ ý
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January 2022 Denis CK Wong
Solution Curve. The graph of a solution �㔙 of an ODE is called a solution curve. Since �㔙 is a differentiable function,
it is continuous on its interval �㔼 of definition. Thus, there may be a difference between the graph of the function �㔙 and
the graph of the solution �㔙. In other word, the domain of the function �㔙 need not be the same as the interval �㔼 of
definition (or domain) of the solution �㔙.
Example.
The domain of ċ = 1/Ċ, considered as a function, is the set of all real numbers Ċ except 0. The rational function ċ =
1/Ċ is discontinuous at 0. The function is not differentiable at Ċ = 0, since the ċ 2axis is a vertical asymptote of the
graph.
Now ċ = 1/Ċ is also a solution of the linear first-order differential equation Ċċ9 + ċ = 0. But when we say that ċ =
1/Ċ is a solution of this DE, we mean that it is a function defined on an interval �㔼 on which it is differentiable and
satisfies the equation. In other words, ċ = 1/Ċ is a solution of the DE on any interval that does not contain 0, such
as (23, 21), (1/2, 10), (2 ∞, 0) or (0, ∞). It is make sense to take the interval �㔼 to be as large as possible. Thus, we
take �㔼 to be (2 ∞, 0) or (0, ∞). The solution curve on (0, ∞) is shown above.
Explicit Solutions. A solution in which the dependent variable is expressed solely in terms of the independent variable
and constants is said to be an explicit solution. An explicit solution as an explicit formula ċ = �㔙(Ċ) that we can
manipulate, evaluate, and differentiate using the standard rules.
ODE
čċ
= Ċ √ċ
čĊ
ċ ′′ 2 2ċ ′ + ċ = 0
Ċċ9 + ċ = 0
ċ = 0 is an explicit solution, which is trivial.
5
Explicit
solution
1 4
ċ=
Ċ
16
ċ = ĊĎ ý
ċ=
1
Ċ
January 2022 Denis CK Wong
Definition. Implicit Solution of an ODE. A relation ă(Ċ, ċ) = 0 is said to be an implicit solution of an ordinary
differential equation Ă(Ċ, ċ, ċ ′ , & , ċ (Ą) ) = 0 on an interval �㔼, provided that there exists at least one function �㔙 that
satisfies the relation as well as the differential equation on I.
Example. The relation Ċ 2 + ċ 2 = 25 is an implicit solution of the differential equation čċ/čĊ = 2Ċ/ċ on the open
interval (25,5).
Families of Solutions. A solution �㔙 is sometimes called as an integral of the equation, and its graph is called an
integral curve. When solve a first-order differential equation Ă(Ċ, ċ, ċ9) = 0, we usually obtain a solution containing
a single arbitrary constant or parameter Č.
A solution containing an arbitrary constant represents a set ă(Ċ, ċ, Č) = 0 of solutions called a one-parameter family
of solutions. When solving an Āth-order differential equation Ă(Ċ, ċ, ċ9, & , ċ (Ą) ) = 0, we seek an Ā 2parameter
family of solutions ă(Ċ, ċ, Č1 , Č2 , & , ČĄ ) = 0. This means that a single differential equation can possess an infinite
number of solutions corresponding to the unlimited number of choices for the parameters. A solution of a differential
equation that is free of arbitrary parameters is called a particular solution.
Example. ċ = ČĊ 2 Ċ Čāą Ċ is an explicit solution of the linear first-order equation Ċċ9 2 ċ = Ċ 2 sin Ċ on the interval
(2∞, ∞).
On the interval (2∞, ∞), ċ = Č1 Ď ý + Č2 ĊĎ ý is two-parameter family solutions of the linear second-order equation
ċ99 2 2ċ9 + ċ = 0. Can you list down some of the particular solutions?
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January 2022 Denis CK Wong
Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation 3
that is, a solution that cannot be obtained by specializing any of the parameters in the family of solutions. Such an
extra solution is called a singular solution.
For example ċ =
1
16
Ċ 4 and ċ = 0 are solutions of
Ăþ
Ăý
= Ċ √ċ on (2∞, ∞). Later we will demonstrate that
possesses the one-parameter family of solutions ċ = (
4
1
4ý 2
2
Ăþ
Ăý
= Ċ √ċ
+ Č) . When Č = 0, the resulting particular solution is
ċ = 1/16Ċ . But the trivial solution ċ = 0 is a singular solution, since it is not a member of the family ċ =
(
1
4ý 2
2
+ Č) ; there is no way of assigning a value to Č to obtain ċ = 0.
Example. The functions Ċ = Č1 cos 4Ć and Ċ = Č2 sin 4Ć , where Č1 and Č2 are arbitrary constants or parameters,
are both solutions of the linear differential equation Ċ ′′ + 16Ċ = 0.
Example. Verify that the one-parameter family ċ = ČĊ 4 is a one-parameter family of solutions of the differential
equation Ċċ9 2 4ċ = 0 on the interval (2∞, ∞). The piecewise-defined differentiable function
ċ={
2Ċ 4 ,
Ċ 4,
Ċ<0
Ċg0
is a particular solution of the equation but cannot be obtained from the family ċ = ČĊ 4 by a single choice of Č; the
solution is constructed from the family by choosing Č = 21 for Ċ < 0 and Č = 1 for Ċ g 0.
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January 2022 Denis CK Wong
Systems of Differential Equations. A system of ODEs is two or more equations involving the derivatives of two or
more unknown functions of a single independent variable. For example, if Ċ and ċ denote dependent variables and Ć
denotes the independent variable, then a system of two first-order differential equations is given by
čĊ
= ď(Ć, Ċ, ċ)
čĆ
čċ
= Đ(Ć, Ċ, ċ)
čĆ
A solution of a system above is a pair of differentiable function Ċ = �㔙1 (Ć) , ċ = �㔙2 (Ć), defined on a common interval
�㔼, that satisfy each equation of the system on this interval.
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January 2022 Denis CK Wong
1.2 Initial-Value Problems
On some interval �㔼 containing Ċ0 the problem
Solve:
�㖅�㖏 þ
�㖅ý�㖏
= �㖇(ý, þ, þ′ , & , þ(�㖏2Ā) )
Subject to: þ(ýÿ ) = þÿ , þ′ (ýÿ ) = þĀ , & , þ(�㖏2Ā) (ýÿ ) = þ�㖏2Ā ,
where ċ0 , ċ1 , & , ċĄ21 are arbitrary specified real constant, is called an initial-value problem (IVP). The values of
ċ(Ċ) and its first Ā 2 1 derivatives at a single point Ċ0 , ċ(Ċ0 ) = ċ0 , ċ9(Ċ0 ) = ċ1 , & , ċ (Ą21) (Ċ0 ) = ċĄ21 , are called
initial conditions.
First-Order IVP
Ăþ
Solve:
= ď(Ċ, ċ)
Ăý
Subject to: ċ(Ċ0 ) = ċ0
Second-Order IVP
Ă2þ
Solve: 2 = ď(Ċ, ċ, ċ ′ )
Ăý
Subject to: ċ(Ċ0 ) = ċ0 , ċ ′ (Ċ0 ) = ċ1
Example. ċ = ČĎ ý is a one-parameter family of solutions of the simple first-order equation ċ9 = ċ and all the solutions
in this family are defined on the interval (2∞, ∞).
(a) Impose an initial condition ċ(0) = 3 and find a solution of the IVP.
(b) Impose an initial condition ċ(1) = 22 and find a solution of the IVP.
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January 2022 Denis CK Wong
Example. The solutions of the first-order differential equation ċ9 + 2Ċċ 2 = 0 is ċ =
condition ċ(0) = 21.
1
ý 2 +ā
. Impose the initial
Example. Ċ = Č1 cos 4Ć + Č2 sin 4Ć is a two-parameter family of solutions of Ċ99 + 16Ċ = 0. Find a solution of the
�㔋
�㔋
initial-value problem Ċ99 + 16Ċ = 0, Ċ ( ) = 22, Ċ9 ( ) = 1.
2
2
Two fundamental questions arise in considering an IVP:
(a) Does a solution of the problem exist? (Existence)
(b) If a solution exists, is it unique? (Uniqueness)
For first-order initial-value problem, we ask
Existence
Ăþ
Does the differential equation
= ď(Ċ, ċ) possess
Ăý
solutions?
Do any of the solution curves pass through the point
(Ċ0 , ċ0 )?
Example. Each of the functions ċ = 0 and ċ =
condition ċ(0) = 0, so the initial-value problem
1
16
Ăþ
Ăý
Uniqueness
When can we be certain that there is precisely one
solution curve passing through the point (Ċ0 , ċ0 )?
Ċ 4 satisfies the differential equation
1
Ăþ
Ăý
= Ċċ 2 , ċ(0) = 0 has at least two solutions.
10
1
= Ċċ 2 and the initial
January 2022 Denis CK Wong
Theorem. Existence of a Unique Solution. Let ā be a rectangular region in the Ċċ 2plane defined by Ċ f Ċ f
ċ, Č f ċ f č that contains the point (Ċ0 , ċ0 ) in its interior. If ď(Ċ, ċ) and
ĀĄ
Āþ
are continuous on ā, then there exists
some interval �㔼0 : (Ċ0 3 /, Ċ0 + /), / > 0, contained in [Ċ, ċ], and a unique function ċ(Ċ), defined on �㔼0 , that is a
solution of the initial-value problem:
Ăþ
Solve:
= ď(Ċ, ċ)
Ăý
Subject to: ċ(Ċ0 ) = ċ0
Example. The differential equation
Ăþ
Ăý
1
= Ċċ 2 possesses at least two solutions whose graphs pass through (0,0).
1
Inspection of the functions ď(Ċ, ċ) = Ċċ 2 and
ĀĄ
Āþ
=
ý
2√þ
shows that they are continuous in the upper half-plane defined
by ċ > 0. Hence, Theorem above enables us to conclude that through any point (Ċ0 , ċ0 ), ċ0 > 0 in the upper halfplane there is some interval centered at Ċ0 on which the given differential equation has unique solution. Thus, for
example, even without solving it, we know that there exists some interval centered at 2 on which the initial-value
problem
Ăþ
Ăý
1
= Ċċ 2 , ċ(2) = 1 has unique solution.
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January 2022 Denis CK Wong
Interval of Existence/Uniqueness. Suppose ċ(Ċ) represents a solution of the initial-value problem
Solve:
Ăþ
Ăý
= ď(Ċ, ċ)
Subject to: ċ(Ċ0 ) = ċ0
The following three sets on the real Ċ 2axis may not be the same: the domain of the function ċ(Ċ), the interval �㔼 over
which the solution ċ(Ċ) is defined or exists, and the interval �㔼0 of existence and uniqueness.
Suppose (Ċ0 , ċ0 ) is a point in the interior of the rectangular region ā in the theorem. It turns out that the continuity of
the function ď(Ċ, ċ) on ā by itself is sufficient to guarantee the existence of at least one solution of čċ/čĊ = ď(Ċ, ċ),
ċ(Ċ0 ) = ċ0 , defined on some interval �㔼. The interval �㔼 of definition for this initial-value problem is usually taken to
be the largest interval containing Ċ0 over which the solution ċ(Ċ) is defined and differentiable. The interval �㔼 depends
on both ď(Ċ, ċ) and the initial condition ċ(Ċ0 ) = ċ0 .
The extra condition of continuity of the first partial derivative
ĀĄ
Āþ
on ā enables us to say that not only does a solution
exist on some interval �㔼0 containing Ċ0 , but it is the only solution satisfying ċ(Ċ0 ) = ċ0 .
However, Theorem above does not give any indication of the sizes of intervals �㔼 and �㔼0 ; the interval �㔼 of definition
need not be as wide as the region ā, and the interval �㔼0 of existence and uniqueness may not be as large as �㔼. The
number / > 0 that defined the interval �㔼0 : (Ċ0 2 /, Ċ0 + /) could be very small, so it is best to think that the solution
ċ(Ċ) is unique in a local sense-that is, a solution defined near the point (Ċ0 , ċ0 ).
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January 2022 Denis CK Wong
1.3 Differential Equations as Mathematical Models.
Differential Equations as Mathematical Models. The mathematical description of a system of phenomenon is called
a mathematical model and is constructed with certain goals in mind. Construction of a mathematical model of a system
starts with
(i) identification of the variables that are responsible for changing the system. We may choose not to incorporate all
these variables into the model at first. In this step we are specifying the level of resolution of the model.
Next,
(ii) make a set of reasonable assumptions, or hypotheses, about the system we are trying to describe. These
assumptions will also include any empirical laws that may be applicable to the system.
POPULATION DYNAMICS. One of the earlies attempts to model human population growth by means of
mathematics was by the English economist Thomas Malthus in 1978. Basically, the idea behind the Malthusian model
is the assumption that the rate at which the population of a country grows at a certain time is proportional to the total
population of the country at that time. In other words, the more people there are at time Ć, the more there are going to
be in the future. In mathematical terms, if ÿ(Ć) denoted the total population at time Ć, then this assumption can be
expressed as
Ăÿ
where ý is a constant of proportionality.
ĂĊ
∝ ÿ or
Ăÿ
ĂĊ
= ýÿ, (1)
This simple model, which fails to take into account many factors that can influence human populations to either grow
or decline, nevertheless turned out to be fairly accurate in predicting the population of the united states during the
years 1790-1860. Populations that grow at a rate described by (1) are rare; nevertheless, (1) is still used to model
growth of small populations over short intervals of time (bacteria growing in a petri dish, for example).
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January 2022 Denis CK Wong
RADIOACTIVE DECAY. The nucleus of an atom consists of combinations of protons and neutrons. Many of these
combinations of protons and neutrons are unstable 4 that is, the atoms decay or transmute into atoms of another
substance. Such nuclei are said to be radioactive. For example, over time the highly radioactive radium, Ra-226,
transmutes into the radioactive gas radon, Rn-222. To model the phenomenon of radioactive decay, it is assumed that
Ă�㔴
the rate
at which the nuclei of substance decay is proportional to the amount (more precisely, the number of nuclei)
ĂĊ
ý(Ć) of the substance remaining at time Ć:
Ă�㔴
ĂĊ
∝ ý or
Ă�㔴
ĂĊ
= ýý, (2)
Of course, equations (1) and (2) are exactly the same; the difference is only in the interpretation of the symbols and
the constants of proportionality. For growth, as we expect in (1), ý > 0, and for decay, as in (2), ý < 0.
ĂĂ
= ĄĂ, which describes the growth of capital Ă when an
The model (1) for growth can also be seen as the equation
ĂĊ
annual rate of interest Ą is compounded continuously. The model (2) for decay also occurs in biological applications
such as determining the half-life of a drug 4the time that it takes for 50% of a drug to be eliminated from a body by
excretion or metabolism. In chemistry the decay model (2) appears in the mathematical description of a ûrst-order
chemical reaction. The point is this:
A single differential equation can serve as a mathematical model for many different phenomena.
Mathematical models are often accompanied by certain side conditions. For example, in (1) and (2) we would expect
to know, in turn, the initial population ÿ0 the initial amount of radioactive substance ý0 on hand. If the initial point in
time is taken to be Ć0 , then we know that ÿ(0) = ÿ0 and ý(0) = ý0 . In other words, a mathematical model can
consist of either an initial-value problem or a boundary-value problem.
SPREAD OF A DISEASE. A contagious disease 4for example, a üu virus 4is spread throughout a community by
people coming into contact with other people. Let Ċ(Ć) denote the number of people who have contracted the disease
Ăý
and ċ(Ć) denote the number of people who have not yet been exposed. It seems reasonable to assume that the rate
ĂĊ
at which the disease spreads is proportional to the number of encounters, or interactions, between these two groups
of people. If we assume that the number of interactions is jointly proportional to Ċ(Ć) and ċ(Ć) 4that is, proportional
to the product Ċċ4then
Ăý
= ýĊċ, (4)
ĂĊ
where ý is the usual constant of proportionality. Suppose a small community has a ûxed population of Ā people. If
one infected person is introduced into this community, then it could be argued that Ċ(Ć) and ċ(Ć) are related by Ċ +
ċ = Ā + 1. Using this last equation to eliminate ċ in (4) gives us the model
Ăý
= ýĊ(Ā + 1 2 Ċ). (5)
ĂĊ
An obvious initial condition accompanying equation (5) is Ċ(0) = 1.
CHEMICAL REACTIONS. The disintegration of a radioactive substance, governed by the differential equation (1),
is said to be a first-order reaction. In chemistry a few reactions follow this same empirical law: If the molecules of
substance ý decompose into smaller molecules, it is a natural assumption that the rate at which this decomposition
takes place is proportional to the amount of the ûrst substance that has not undergoes conversion; that is, if ć(Ć) is the
Ă�㕋
amount of substance ý remaining at any time, then
= ýć, where ý is negative constant since ć is decreasing.
ĂĊ
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January 2022 Denis CK Wong
SERIES CIRCUITS. Consider the single-loop series circuit shown in Figure below, containing an inductor, resistor,
and capacitor. The current in a circuit after a switch is closed is denoted by ÿ(Ć); the charge on a capacitor at time Ć is
denoted by ă(Ć). The letters Ā, ā, ÿ are known as inductance, resistance, and capacitance, respectively, and are
generally constants.
Now according to Kirchhoff’s second law, the impressed voltage ā(Ć) on a closed loop must equal the sum of the
voltage drops in the loop. Figure 1.3.3(b) shows the symbols and the formulas for the respective voltage drops across
an inductor, a capacitor, and a resistor.
Since current ÿ(Ć) is related to charge ă(Ć) on the capacitor by ÿ =
Inductor
čÿ
č2ă
Ā =Ā 2
čĆ
čĆ
Resistor
čă
ÿā = ā
čĆ
Ăć
ĂĊ
, adding the three voltages
Capacitor
1
ă
ÿ
and equating the sum to the impressed voltage yields a second-order differential equation
Ā
Ă2ć
ĂĊ 2
+ā
Ăć
ĂĊ
1
+ ă = ā(Ć). (10)
�㔶
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January 2022 Denis CK Wong
1.4 Solution Curves Without a Solution
Direction Fields. A derivative
graph.
Ăþ
Ăý
of a differentiable function ċ = ċ(Ċ) gives slopes of tangent lines at points on its
Slope. Because a solution ċ = ċ(Ċ) of a first-order differential equation
Ăþ
Ăý
= ď(Ċ, ċ) (1)
is necessarily a differentiable function on its interval �㔼 of definition, it must also be continuous on �㔼. Thus, the
corresponding solution curve on �㔼 must have no breaks and must possess a tangent line at each point (Ċ, ċ(Ċ)). The
function ď in the normal form (1) is called the slope function or rate function.
The slope of the tangent line at (Ċ, ċ(Ċ)) on a solution curve is the value of the first derivative
Ăþ
Ăý
at this point, and we
know from (1) that this is the value of the slope function ď(Ċ, ċ(Ċ)). Now suppose that (Ċ, ċ) represents any point in
a region of the Ċċ 2plane over which the function ď is defined. The value ď(Ċ, ċ) that the function ď assigns to the
point represents the slope of a line, or, as we shall envision it, a line segment called a lineal element.
Consider the equatin
Ăþ
Ăý
= 0.2Ċċ, where ď(Ċ, ċ) = 0.2Ċċ. At, say, the point (2,3) the slope of a lineal element is
ď(2,3) = 0.2(2)(3) = 1.2. (a) shows a line segment with slope 1.2 passing though (2,3). For (b), if a solution
curve also passes through the point (2,3), it does so tangent to this line segment; in other words, the lineal element is
a miniature tangent line at that point.
Direction Field. If we systematically evaluate ď over a rectangular grid of points in the Ċċ 2plane and draw a line
element at each point (Ċ, ċ) of the grid with slope ď(Ċ, ċ), then the collection of all these line elements is called a
direction field or a slope filed of the differential equation
Ăþ
Ăý
= ď(Ċ, ċ).
Visually, the direction field suggests the appearance or shape of a family of solution curves of the differential equation,
and consequently, it may be possible to see at a glance certain qualitative aspects of the solutions 3 regions in the
plane, for example, in which a solution exhibits an unusual behaviour.
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January 2022 Denis CK Wong
A single solution curve that passes through a direction field must follow the flow pattern of the field; it is tangent to a
line element when it intersects a point in the grid.
Direction field of
Ăþ
Ăý
= sin(Ċ + ċ) over a region of the Ċċ 2plane.
Example. The direction field for the differential equation
Ăþ
Ăý
= 0.2Ċċ is shown below, was obtained in a 5 × 5 grid
of points (ÿ/, Ā/), ÿ and Ā integers, was defined by letting 25 f ÿ f 5, 2 5 f Ā f 5, and / = 1.
Any point along the Ċ 2axis (ċ = 0) and the ċ 2axis (Ċ = 0), the slopes are ď(Ċ, 0) = 0 and ď(0, ċ) = 0,
respectively, so the lineal elements are horizontal.
In the first quadrant, for a fixed value of Ċ, the values of ď(Ċ, ċ) = 0.2Ċċ increase as y increases; similarly, for a fixed
ċ the values of ď(Ċ, ċ) = 0.2Ċċ increase as Ċ increases. This means that as both Ċ and ċ increase, the lineal elements
almost become vertical and have positive slope (ď(Ċ, ċ) > 0 for Ċ > 0 , ċ > 0).
In the second quadrant, |ď(Ċ, ċ)| increases as |Ċ| and ċ increase, so the lineal elements again become almost vertical
but this time have negative slope (ď(Ċ, ċ) < 0 for Ċ < 0, ċ > 0).
Reading from left to right, imagine a solution curve that starts at a point in the second quadrant, moves steeply
downward, becomes üat as it passes through the ċ 2axis, and then, as it enters the ûrst quadrant, moves steeply upward
4in other words, its shape would be concave upward and similar to a horseshoe. It could be surmised that ċ → ∞ as
Ċ → ±∞.
In the third and fourth quadrants, since ď(Ċ, ċ) > 0 and ď(Ċ, ċ) < 0, respectively, the situation is reversed.
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January 2022 Denis CK Wong
Example. Use a direction field to sketch an approximate solution curve for the initial-value problem
sin ċ , ċ(0) = 23/2.
Interpretation of the derivative
Ăþ
Ăý
Ăþ
Ăý
=
as a function that gives slope plays the key role in the construction of a direction
field. Another telling property of the first derivative will be used nest, namely, if
Ăþ
Ăý
interval �㔼, then a differentiable function ċ = ċ(Ċ) is increasing (or decreasing) in �㔼.
> 0 (or
Ăþ
Ăý
< 0) for all Ċ in an
Autonomous First-Order DEs. An ODE in which the independent variable does not appear explicitly is said to be
autonomous. If Ċ is the independent variable, then an autonomous first-order differential equation can be written as
ď(ċ, ċ9) = 0 or in normal form as
Ăþ
Ăý
= ď(ċ). (2)
The function ď in (2) and its derivatives ď9 are continuous functions of ċ on some interval �㔼.
Autonomous
čċ
= 1 + ċ2
čĊ
Nonautonomous
čċ
= 0.2Ċċ
čĊ
The zeros of the function ď in (2) are of special importance. We say that a real number Č is a critical point of the
autonomous differential equation (2) if it is a zero of ď 3 that is, ď(Č) = 0. A critical point is also called an
equilibrium point or stationary point. If Č is a critical point of (2), then ċ(Ċ) = Č is a constant solution of the
autonomous differential equation.
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January 2022 Denis CK Wong
A constant solution ċ(Ċ) = Č of (2) is called an equilibrium solution; equilibria are the only constant solutions of
(2).
As was already mentioned, we can tell when a nonconstant solution ċ = ċ(Ċ) of (2) is increasing or decreasing by
Ăþ
determining the algebraic sign of the derivative ; in the case of (2) we do this by identifying intervals on the ċ 2axis
Ăý
over which the function ď(ċ) is positive or negative.
Example. Consider
Ăÿ
ĂĊ
= ÿ(Ċ 2 ċÿ), where Ċ and ċ are positive constants.
Solution Curves. Without solving an autonomous differential equation, we can usually say a great deal about its
solution curves.
Since the function ď in
Ăþ
Ăý
= ď(ċ) is independent of Ċ, we may consider ď defined for 2∞ < Ċ < ∞ or for 0 f Ċ <
∞. Also since ď and its derivative ď9 are continuous functions of y on some interval �㔼 of the ċ 2axis, the fundamental
results of theorem hold in some horizontal strip or region ā in the Ċċ 2plane corresponding to �㔼, and so through any
point (Ċ0 , ċ0 ) in ā there passes only one solution curve of
Suppose that
Ăþ
Ăý
Ăþ
Ăý
= ď(ċ). See figure (a).
= ď(ċ) possesses exactly two critical points Č1 and Č2 and that Č1 < Č2 . The graphs of the
equilibrium solutions ċ(Ċ) = Č1 and ċ(Ċ) = Č2 are horizontal lines, and these lines partition the region ā into three
subregions ā1 , ā2 and ā3 , as shown in Figure (b).
1.
If (Ċ0 , ċ0 ) is in a subregion āÿ , ÿ = 1,2,3, and ċ(Ċ) is a solution whose graph passes through this point, then
ċ(Ċ) remains in the subregion āÿ for all Ċ. AS in figure (b), the solution ċ(Ċ) in ā2 is bounded below by Č1
and above by Č2 , that is, Č1 < ċ(Ċ) < Č2 for all Ċ. The solution curve stays within ā2 for all Ċ because the
graph of a nonconstant solution of
2.
3.
Ăþ
Ăý
= ď(ċ) cannot cross the graph of either equilibrium solution ċ(Ċ) =
Č1 or ċ(Ċ) = Č2 .
By continuity of ď we must then have either ď(ċ) > 0 or ď(ċ) < 0 for all Ċ in a subregion āÿ , ÿ = 1,2,3. In
other words, ď(ċ) cannot change signs in a subregion.
Since
Ăþ
Ăý
= ď(ċ(Ċ)) is either positive or negative in a subregion āÿ , ÿ = 1,2,3, a solution ċ(Ċ) is strictly
monotonic- that is, ċ(Ċ) is either increasing or decreasing in the subregion āÿ . Therefore, ċ(Ċ) cannot be
oscillatory, nor can it have a relative extremum (maximum or minimum).
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January 2022 Denis CK Wong
4.
If ċ(Ċ) is bounded above by a critical point Č1 (as in subregion ā_1 where ċ(Ċ) < Č1 for all Ċ), then the
graph of ċ(Ċ) must approach the graph of the equilibrium solution ċ(Ċ) = Ċ1 either as Ċ → ∞ or as Ċ →
2∞. If ċ(Ċ) is bounded 3 that is, bounded above and below by two consecutive critical points (as in subregion
ā2 , where Č1 < ċ(Ċ) < Č2 for all Ċ)-then the graph ċ(Ċ) must approach the graphs of the equilibrium
solutions ċ(Ċ) = Č1 and ċ(Ċ) = Č2 , one as Ċ → ∞ and the other as Ċ → 2∞. If ċ(Ċ) is bunded below by
a critical point (as in subregion ā3 where Č2 < ċ(Ċ) for all Ċ), then the graph of ċ(Ċ) must approach the
graph of the equilibrium solution ċ(Ċ) = Č2 either as Ċ → ∞ or as Ċ → 2∞.
Example. Revisited
Ăÿ
ĂĊ
= ÿ(Ċ 2 ċÿ), where Ċ and ċ are positive constants.
The three intervals determined on the ÿ 2axis or phase line by the critical points ÿ = 0 and ÿ =
in the Ćÿ-plane to three subregions defined by:
ā1 : 2∞ < ÿ < 0, ā2 : 0 < ÿ <
ÿ
Ā
ÿ
Ā
now correspond
ÿ
ĊĀč ā3 : < ÿ < ∞, where : 2∞ < Ć < ∞.
Ā
The phase portrait is shown above tells us that ÿ(Ć) is decreasing in ā1 , increasing in ā2 , and decreasing in ā3 . If
ÿ(0) = ÿ0 is an initial value, then in ā1 , ā2 and ā3 we have. respectively, the following:
(i)
(ii)
(iii)
For ÿ0 < 0, ÿ(Ć) is bounded above. Since ÿ(Ć) is decreasing, ÿ(Ć) decreases without bound for
increasing Ć, and so ÿ(Ć) → 0 as Ć → 2∞. This means that the negative Ć 2axis, the graph of the
equilibrium solution ÿ(Ć) = 0, is a horizontal asymptote for a solution curve.
ÿ
For 0 < ÿ0 < Ċ/ċ , ÿ(Ć) is bounded. Since ÿ(Ć) is increasing, ÿ(Ć) → as Ć → ∞ and ÿ(Ć) → 0 as
Ā
ÿ
Ć → 2∞. The graphs of the two equilibrium solutions, ÿ(Ć) = 0 and ÿ(Ć) = , are horizontal lines that
Ā
are horizontal asymptotes for any solution curve starting in this subregion.
ÿ
ÿ
For ÿ0 > , ÿ(Ć) is bounded below. Since ÿ(Ć) is decreasing, ÿ(Ć) → as Ć → ∞. The graph of the
Ā
Ā
ÿ
equilibrium solution ÿ(Ć) = is a horizontal asymptote for a solution curve.
Ā
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January 2022 Denis CK Wong
1.5 Separable Variables
Definition. A first-order differential equation of the form
variables.
Ăþ
Ăý
= Đ(Ċ)/(ċ) is said to be separable or to have separable
For convenience, the equation can always be written in the form Ă(ċ)
Example 1. Solve (1 + Ċ)čċ 3 ċ čĊ = 0.
Example 2. Solve the initial-value problem
Example 3. Solve
Ăþ
Ăý
= ċ 2 3 4.
Example 4. Solve (Ď 2 ċ 3 ċ)Čāą Ċ
Ăþ
Ăþ
Ăý
Ăþ
Ăý
= Đ(Ċ).
ý
= 2 , ċ(4) = 23.
þ
= Ď þ ąÿĀ 2Ċ, ċ(0) = 0.
Ăý
Solutions Defined by Integrals. If Đ is a function continuous on an open interval �㔼 containing Ċ, then for every Ċ in
�㔼,
Ă ý
∫ Đ(Ć)čĆ
Ăý ÿ
ý
= Đ(Ċ). ∫ÿ Đ(Ć)čĆ is an antiderivative of the function Đ. If Đ is continuous on an interval �㔼 containing
Ċ0 and Ċ, then a solution of the simple initial-value problem
ý
ċ(Ċ) = ċ0 + ∫ý Đ(Ć)čĆ .
0
Example. Solve
Ăþ
Example. Solve
Ăý
Example. Solve
Ăþ
Ăý
ĂĊ
ĂĊ
Ăþ
Ăý
= Đ(Ċ), ċ(Ċ0 ) = ċ0 , that is defined on �㔼 is given by
2
= Ď 2ý , ċ(3) = 5.
�㔋
= 4(Ċ 2 + 1), Ċ ( ) = 1.
4
5
+ 2ċ = 1, ċ(0) = .
2
1.6 Linear Equation
Definition. A first-order differential equation of the form Ċ1 (Ċ)
Ăþ
Ăý
+ Ċ0 (Ċ) ċ = Đ(Ċ) is said to be a linear
equation in the dependent variable y. When Đ(Ċ) = 0, the linear equation is said to be homogeneous; otherwise, it
is nonhomogeneous.
By diving both sides by the lead coefficient Ċ1 (Ċ), we obtain the standard form, of a linear equation :
Ăþ
Ăý
+ ÿ(Ċ)ċ =
ď(Ċ).We seek a solution of this equation on an interval �㔼 for which both coefficient functions ÿ and ď are continuous.
The differential equation
Ăþ
Ăý
+ ÿ(Ċ)ċ = ď(Ċ) has its solution as the sum of two solutions ċ = ċā + ċĆ , where ċā is
a solution of the associated homogenous equation
nonhomogeneous equation
Ăþ
Ăý
+ ÿ(Ċ)ċ = ď(Ċ).
Ăþ
Ăý
21
+ ÿ(Ċ)ċ = 0 and ċĆ is a particular solution of the
January 2022 Denis CK Wong
From the standard form
identify ÿ(ć) and then find the
integrating factor Ď ∫ ÿ(ý)Ăý
Put a linear equation of the
form (1) into the standard form
(2)
Example. Solve
Ăþ
Example. Solve
Ăþ
Example. Solve
Ăþ
2 3ċ = 6.
Ăý
2 4ċ = Ċ 6 Ď ý .
Ăý
Ăþ
Example. Solve
Ăý
+ ċ = Ċ, ċ(0) = 4.
Ăþ
Ăý
Integrate both sides of this last
equation
3 3ċ = 0.
Ăý
Example. Find the general solution of (Ċ 2 2 0)
Example. Solve
Multiply the standard form of
the equation by the integrating
factor. The Left-hand-side of
the resulting equation is the
derivative of the integrating
factor and y:
č ∫ Ć ý Ăý
Ď
ċ
čĊ
= Ď ∫ ÿ ý Ăý ď(Ċ)
Ăþ
Ăý
+ Ċċ = 0.
1, 0 f Ċ f 1
+ ċ = ď(ċ), ċ(0) = 0 where ď(Ċ) = {
0, Ċ g 1
Functions Defined by Integrals
Error function
Complementary error function
∞
2
From the know result ∫0 Ď 2Ċ čĆ =
ĎĄď(0) = 0.
erf(Ċ) =
2
ý
2
∫ Ď 2Ċ čĆ
√�㔋 0
2 ∞ 2Ċ 2
erf ą(Ċ) =
∫ Ď čĆ
√�㔋 ý
2 ∞ 2Ċ 2
√�㔋
, so
∫ Ď čĆ = 1. Thus, ĎĄď(Ċ) + ĎĄďČ(Ċ) = 1. Note that obviously
2
√�㔋 0
Example. Solve the initial-value problem
Ăþ
Ăý
2 2Ċċ = 2, ċ(0) = 1.
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January 2022 Denis CK Wong
1.7 Exact Equations
Although the simple first-order equation ċ čĊ + Ċ čċ = 0 is separable, we can solve the equation in an alternative
manner by recognizing that the expression on the LHS of the equality is the differential of the function ď(Ċ, ċ) =
Ċċ; that is, č(Ċċ) = ċ čĊ + Ċ čċ.
Here we examine first-order equations in differential form ā(Ċ, ċ) čĊ + Ă(Ċ, ċ) čċ = 0. By applying a simple test
to ā and Ă, we can determine whether ā(Ċ, ċ) čĊ + Ă(Ċ, ċ) čċ is a differential of ď(Ċ, ċ). If the answer is yes, we
can construct ď by partial integration. If Č = ď(Ċ, ċ) is a function of two variables with continuous first partial
derivatives in a region ā of the Ċċ 2plane, then its differential is čČ =
ď(Ċ, ċ) = Č, where Č is a constant, then
ĀĄ
Āý
čĊ +
ĀĄ
Āþ
čċ = 0.
ĀĄ
Āý
čĊ +
ĀĄ
Āþ
čċ. In the special case when
Definition. A differential expression ý(ý, þ)�㖅ý + þ(ý, þ) �㖅þ is an exact differential in a region ā of the
Ċċ 2plane if it corresponds to the differential of some function ď(Ċ, ċ) defined in ā. A first-order differential equation
of the form ā(Ċ, ċ) čĊ + Ă(Ċ, ċ)čċ = 0 is said to be an exact equation if the expression on the left-hand side is
an exact differential.
Theorem. Criterion for exact differential. Let ā(Ċ, ċ) and Ă(Ċ, ċ) be continuous and have continuous first partial
derivatives in a rectangular region ā defined by Ċ < Ċ < ċ, Č < ċ < č. Then a necessary and sufficient condition
that ā(Ċ, ċ)čĊ + Ă(Ċ, ċ) čċ be an exact differential is
�㕏ý
�㕏þ
=
�㕏þ
�㕏ý
.
Example. Solve 2Ċċ čĊ + (Ċ 2 2 1)čċ = 0
Example. Solve (Ď 2 ċ 2 ċ Čāą Ċċ) čĊ + (2ĊĎ 2 ċ 3 Ċ Čāą Ċċ + 2ċ) čċ = 0
Example. Solve
Ăþ
Ăý
=
ýþ 2 2cos ý sin ý
þ(12ý 2 )
, ċ(0) = 2.
Integrating Factors. For non-exact differential equation ā(Ċ, ċ)čĊ + Ă(Ċ, ċ)čċ = 0, sometime is possible to find
an integrating factor Ā(Ċ, ċ) so that after multiplying, the LHS of Ā(Ċ, ċ)ā(Ċ, ċ)čĊ + Ā(Ċ, ċ)Ă(Ċ, ċ)čċ = 0 is an
exact differential. In an attempt to find Ā, we turn to the criterion for exactness.
Ā(Ċ, ċ)ā(Ċ, ċ)čĊ + Ā(Ċ, ċ)Ă(Ċ, ċ)čċ = 0 is exact iff (Āā)þ = (ĀĂ)ý
iff Āāþ + Āþ ā = ĀĂý + Āý Ă or Āý Ă 2 Āþ ā = (āþ 2 Ăý )Ā. (*)
ā, Ă, āþ , Ăý are known functions of Ċ and ċ, the difficulty here in determining the unknown Ā(Ċ, ċ) from (*) is that
we must solve a partial differential equation.
Suppose Ā is a function of one variable; say that Ā depends only on Ċ. Then Āý =
as
ĂĀ
Ăý
=
ýþ 2þý
þ
23
Ā. (**)
ĂĀ
Ăý
and Āþ = 0, so (*) can be written
January 2022 Denis CK Wong
If
ýþ 2þý
þ
only depend on Ċ, then (**) is a first-order ODE. We can finally determine Ā because (**) is separable as
well as linear. Thus, Ā(Ċ) = Ď ∫
ýþ 2þý
Ăý
þ
.
Similarly, if Ā depends only on ċ, then
If
þý 2ýþ
ý
ĂĀ
Ăþ
=
þý 2ýþ
ý
Ā. (***)
is a function of ċ only, then we can solve (***) for Ā. Thus, Ā(Ċ) = Ď ∫
þý 2ýþ
Ăþ
ý
.
Example. Solve Ċċ čĊ + (2Ċ 2 + 3ċ 2 2 20)čċ = 0.
Example. Solve (2ċ 2 + 3Ċ)čĊ + 2Ċċ čċ = 0.
Example. Solve ċ(Ċ + ċ + 1)čĊ + (Ċ + 2ċ)čċ = 0.
1.8 Solutions by Substitutions
Suppose we wish to transform the first-order differential equation čċ/čĊ = ď(Ċ, ċ) by the substitution ċ = Đ(Ċ, ć),
where ć is regarded as a function of the variable Ċ.
If Đ possesses first-partial derivatives, then the chain rule
If we replace
Ăþ
Ăý
Ăþ
Ăý
=
Āą Ăý
Āý Ăý
+
Āą Ăċ
Āċ Ăý
gives
Ăþ
Ăý
= Đý (Ċ, ć) + Đċ (Ċ, ć)
by the foregoing derivative and replace ċ in ď(Ċ, ċ) by Đ(Ċ, ć). then the DE
Đý (Ċ, ć) + Đċ (Ċ, ć)
Ăċ
Ăý
= ď(Ċ, Đ(Ċ, ć)), which, solved for
Ăċ
Ăý
, has the form
Ăċ
Ăý
Ăþ
Ăý
Ăċ
Ăý
.
= ď(Ċ, ċ) becomes
= Ă(Ċ, ć). If we can determine a
solution ć = �㔙(Ċ) of this last equation, then a solution of the original equation is ċ = Đ(Ċ, �㔙(Ċ)).
Homogenous Equations. If a function ď possesses the property ď(ĆĊ, Ćċ) = Ć ÿ ď(Ċ, ċ) for some real number ÿ, then
ď is said to be a homogeneous function of degree ÿ.
A first-order DE in differential form ā(Ċ, ċ) čĊ + Ă(Ċ, ċ) čċ = 0 is said to be homogeneous if both coefficient
functions ā and Ă are homogeneous equation of the same degree. In other word, if ā(ĆĊ, Ćċ) = Ć ÿ ā(Ċ, ċ) and if
Ă(ĆĊ, Ćċ) = Ć ÿ Ă(Ċ, ċ). In addition, if ā and Ă are homogenous functions of degree ÿ, we can also write ā(Ċ, ċ) =
þ
Ċ ÿ ā(1, ć) and Ă(Ċ, ċ) = Ċ ÿ Ă(1, ć), where ć = , and ā(Ċ, ċ) = ċ ÿ ā(Ĉ, 1) and Ă(Ċ, ċ) = ċ ÿ Ă(Ĉ, 1),
ý
ý
where Ĉ = . This suggests the substitution that can be used to solve a homogeneous differential equation. That is,
þ
either of the substitutions ċ = ćĊ or Ċ = Ĉċ, where ć and Ĉ are new dependent variables, will reduce a
homogeneous equation to a separable first-order DE.
ā(Ċ, ċ)čĊ + Ă(Ċ, ċ)čċ = 0
Ċ ÿ ā(1, ć)čĊ + Ċ ÿ Ă(1, ć)čċ = 0
ā(1, ć)čĊ + Ă(1, ć)čċ = 0
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January 2022 Denis CK Wong
where ć = ċ/Ċ or ċ = ćĊ.
By substituting čċ = ć čĊ + Ċ čć , we obtain a separable DE in variables ć and Ċ:
ā(1, ć)čĊ + Ă(1, ć)[ć čĊ + Ċ čć] = 0
[ā(1, ć) + ćĂ(1, ć)]čĊ + ĊĂ(1, ć)čć = 0
čĊ
Ă(1, ć)čć
+
=0
Ċ
ā(1, ć) + ćĂ(1, ć)
Example. Solve (Ċ 2 + ċ 2 ) čĊ + (Ċ 2 2 Ċċ) čċ = 0.
The differential equation
Ăþ
Ăý
+ ÿ(Ċ)ċ = ď(Ċ)ċ Ą , where Ā is any real number, is called Bernoulli’s equation. Note
that for Ā = 0 and Ā = 1, the equation is linear. For Ā b 0 and Ā b 1, the substitution ć = ċ 12Ą reduces any
equation of the of Bernoulli to a linear equation.
Example: Solve Ċ
Ăþ
Ăý
+ ċ = Ċ 2 ċ2.
Reduction to Separation of Variables. A differential equation of the form
Ăþ
Ăý
= ď(ýĊ + ċċ + ÿ) can always be
reduced to an equation with separable variables by means of the substitution ć = ýĊ + þċ + ÿ, þ b 0.
Example.
Ăþ
Ăý
= (22Ċ + ċ)2 3 7, ċ(0) = 0.
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January 2022 Denis CK Wong
1.9 A Numerical Method
Using Tangent Line. Let us assume that the ûrst-order initial-value problem
ċ ′ = ď(Ċ, ċ), ċ(Ċ0 ) = ċ0 (*)
possesses a solution. One way of approximating this solution is to use tangent lines.
For example, let ċ(Ċ) denote the unknown solution of the first-order initial-value problem ċ9 = 0.1 √ċ + 0.4 Ċ 2 ,
ċ(2) = 4. The nonlinear differential equation in this IVP cannot be solved directly using previously discussed
methods, but, we can find an approximate numerical values of the unknown ċ(Ċ). Specifically, suppose we wish to
know the value of ċ(2.5).
The IVP has a solution, and as the flow of the direction field of the DE suggests, a solution curve must have a shape
similar to the curve shown in blue. The direction field was generated with lineal elements passing through points in a
grid with integer coordinates. As the solution curve passes through the initial point (2,4), the lineal element at this
point is a tangent line with slope given by ď(2.4) = 1.8. The <zoom in= figure shown that when Ċ is close to 2, the
points on the solution curve are close to the points on the tangent line (The lineal element). Using the point (2,4), the
slope ď(2,4) = 1.8, and the point-slope form of a line, we find that an equation of the tangent line is ċ = Ā(Ċ) =
1.8Ċ + 0.4. This equation is called a linearization of ċ(Ċ) at Ċ = 2, can be used to approximate values of ċ(Ċ)
within a small neighborhood of Ċ = 2. If ċ1 = Ā(Ċ1 ) denotes the ċ 2coordinate on the tangent line and ċ(Ċ1 ) is the
ċ 2coordinate on the solution curve corresponding to an Ċ 2coordinate Ċ1 that is close to Ċ = 2, then ċ(Ċ1 ) j ċ1 .
If we choose, say, Ċ1 = 2.1, then ċ1 = Ā(2.1) = 4.18, so ċ(2.1) j 4.18.
Euler’s method. We use the linearization of the unknown solution ċ(Ċ) of (*) at Ċ = Ċ0 :
Ā(Ċ) = ċ0 + ď(Ċ0 , ċ0 )(Ċ 2 Ċ0 ). (**)
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January 2022 Denis CK Wong
The graph of this linearization is a straight-line tangent to the graph ċ = ċ(Ċ) at the point (Ċ0 , ċ0 ). We now let h be
a positive increment of the Ċ 2axis. Then by replacing Ċ by Ċ1 = Ċ0 + / in (2), we get
Ā(Ċ1 ) = ċ0 + ď(Ċ0 , ċ0 )(Ċ0 + / 2 Ċ0 ) or ċ1 = ċ0 + / ď(Ċ1 , ċ1 ), where ċ1 = Ā(Ċ1 ).
The point (Ċ1 , ċ1 ) on the tangent line is an approximation to the point (Ċ1 , ċ(Ċ1 )) on the solution curve.
The accuracy of the approximation Ā(Ċ1 ) j ċ(Ċ1 ) āĄ ċ1 j ċ(Ċ1 ) depends heavily on the size of the increment /.
Usually, we must choose this step size to be <reasonably small.= We now repeat the process using a second <tangent
line= at (Ċ1 , ċ1 ). By identifying the new starting point as (Ċ1 , ċ1 ) with (Ċ0 , ċ0 ), we obtain an approximation ċ2 j
ċ(Ċ2 ) corresponding to two steps of length / from Ċ0 , that is,
Ċ2 = Ċ1 + / = Ċ0 + 2/, and
ċ(Ċ2 ) = ċ(Ċ0 + 2/) = ċ(Ċ1 + /) j ċ2 = ċ1 + /ď(Ċ1 , ċ1 ).
Continuing in this manner, we see that ċ1 , ċ2 , ċ3 , & , can be defined recursively by the general formula
ċĄ+1 = ċĄ + / ď(ĊĄ , ċĄ ), (***)
where ĊĄ = Ċ0 + Ā/ , Ā = 0,1,2, &. This procedure of using successive tangent lines is called Euler’s method.
Example. Consider the initial-value problem ċ9 = 0.1 √ċ + 0.4 Ċ 2 , ċ(2) = 4. Use Euler’s method to obtain an
approximation of ċ(2.5) using first / = 0.1 and then / = 0.05.
Example. Consider the initial-value problem ċ9 = 0.2Ċċ, ċ(1) = 1. Use Euler’s method to obtain an approximation
of ċ(1.5) using first / = 0.1 and then / = 0.05.
The absolute error is defined to be |actual value 3 approximation|.
ÿĀĉąĂċĊă ăĈĈąĈ
ÿĀĉąĂċĊă ăĈĈąĈ
ĊĀč
× 100.
The relative error and percentage relative error are, in turn,
|ÿāĊċÿĂ ČÿĂċă|
27
|ÿāĊċÿĂ ČÿĂċă|
January 2022 Denis CK Wong
1.10 Linear Models
Growth and Decay. The initial-value problem
Ăý
ĂĊ
= ýĊ, Ċ(Ć0 ) = Ċ0 (*), where k is a constant of proportionality,
serve as a model for diverse phenomena involving either growth or decay.
Knowing the proportional at some arbitrary initial time Ć0 , we can then use the solution of (*) to predict the population
in the future 3 that is, at times Ć > Ć0 . The constant of proportionality ý in (1) can be determined from the solution
of the initial-value problem, using a subsequent measurement of Ċ at a time Ć1 > Ć0 .
In physics and chemistry (*) is seen in the form of a first-order reaction 3 that is, a reaction whose rate, or velocity,
dx/dt is directly proportional to the amount x of substance that is unconverted or remaining at time Ć.
Example. Bacterial Growth. A cultural initially has ÿ0 number of bacteria. At Ć = 1 h the number of bacterial is
3
measured to be ÿ0 .
2
If the rate of growth is proportional to the number of bacteria ÿ(Ć) present at time Ć, determine the time necessary for
the number of bacteria to triple.
Half-life. In physics the half-life is a measure of the stability of a radioactive substance. The half-life is simply the
time it takes for one-half of the atoms in an initial amount ý0 to disintegrate, or transmute, into the atoms of another
element. The longer the half-life of a substance, the more stable it is.
Example. Half-life of Plutonium. A breeder reactor converts relatively stable uranium 238 into the isotope plutonium
239. After 15 years it is determined that 0.043% of the initial amount ý0 of plutonium has disintegrated. Find the halflife of this isotope if the rate of disintegration is proportional to the amount remaining.
Carbon Dating. About 1950 the chemist Willard Libby devised a method of using radioactive carbon as a means of
determining the approximate ages of fossils. The theory of carbon dating is based on the fact that the isotope carbon
14 is produced in the atmosphere by the action of cosmic radiation on nitrogen. The ratio of the amount of C-14 to
ordinary carbon in the atmosphere appears to be a constant, and as a consequence the proportionate amount of the
isotope present in all living organisms are the same as that in the atmosphere. When an organism dies, the absorption
of C-14, by either breathing or eating, ceases. Thus by comparing the proportionate amount of C-14 present, say, in a
fossil with the constant ratio found in the atmosphere, it is possible to obtain a reasonable estimation of the fossil’s
age. The method is based on the knowledge that the half-life of radioactive C-14 is approximately 5600 years. For his
work Libby won the Nobel Prize for chemistry in 1960. Libby’s method has been used to date wooden furniture in
Egyptian tombs, the woven üax wrappings of the Dead Sea scrolls, and the cloth of the enigmatic shroud of Turin.
Example. Age of Fossil. A fossilized bone is found to contain one-thousandth of the C-14 level found in living matter.
Estimate the age of the fossil.
NEWTON’S LAW OF COOLING/WARMING. The mathematical formulation of Newton’s empirical law of
cooling/warming of an object is given by the linear ûrst-order differential equation
čă
= ý(ă 2 ăă ),
čĆ
where ý is a constant of proportionality, ă(Ć) is the temperature of the object for Ć > 0, and ăă is the ambient
temperature 3 that is , the temperature of the medium around the object. Assume ăă is constant.
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January 2022 Denis CK Wong
Example. Cooling a Cake. When a cake is removed from an oven, its temperature is measured at 300° F. Three
minutes later its temperature is 200° F. How long will it take for the cake to cool off to a room temperature of 70° F?
SERIES CIRCUITS. For a series circuit containing only a resistor and an inductor, Kirchhoff’s second law states
Ăÿ
that the sum of the voltage drop across the inductor (Ā ( )) and the voltage drop access the resistor (ÿā) is the same
ĂĊ
as the impress voltage (ā(Ć)) on the circuit.
Thus, we obtain the linear differential equation for the current ÿ(Ć), Ā
Ăÿ
ĂĊ
+ āÿ = ā(Ć), where Ā and ā are constants
known as the inductance and the resistance, respectively. The current ÿ(Ć) is calle the response of the system. The
voltage drops across a capacitor with capacitance ÿ is given by
Kirchhoff’s second law gives āÿ +
1
�㔶
ă = ā(Ć).
But current ÿ and charge ă are related by ÿ =
Ăć
ĂĊ
ć(Ċ)
�㔶
, where ă is the charge on the capacitor. Hence,
1
, so we have āčă/čĆ + ă = ā(Ć).
�㔶
Example. Series circuit. A 12-volt battery is connected to a series circuit in which the inductance is 1/2 henry and
the resistance is 10 ohms. Determine the current ÿ if the initial current is zero.
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January 2022 Denis CK Wong
1.11 Nonlinear Models
Population Dynamics. If ÿ(Ć) denotes the size of a population at time Ć, the model for exponential growth begins
with the assumption that
Ăÿ
ĂĊ
= ýÿ for some ý > 0. In this model, the relative, or specific, growth rate defined by
Ăÿ/ĂĊ
ÿ
(*)
is a constant ý. True cases of exponential growth over long periods of time are hard to ûnd because the limited
resources of the environment will at some time exert restrictions on the growth of a population. Thus for other models,
(*) can be expected to decrease as the population ÿ increases in size. The assumption that the rate at which a population
grows (or decreases) is dependent only on the number ÿ present and not on any time-dependent mechanisms such as
seasonal phenomena can be stated as
Ăÿ/ĂĊ
ÿ
= ď(ÿ) or
Ăÿ
ĂĊ
= ÿď(ÿ) (**).
The DE in (**), which is widely assumed in models of animal populations, is called the density-dependent
hypothesis.
Logistic Equation. Suppose an environment is capable of sustaining no more than a ûxed number ÿ of individuals
in its population. The quantity ÿ is called the carrying capacity of the environment. Hence for the function f in (**)
we have ď(ÿ) = 0, and we simply let ď(0) = Ą.
The simplest assumption that we can make is that ď(ÿ) is linear 3 that is, ď(ÿ) = Č1 ÿ + Č2 . If we use the conditions
ď(0) = Ą and ď(ÿ) = 0, we find, in turn, Č2 = Ą and Č1 = 2Ą/ÿ, and so ď takes on the form
becomes
Ăÿ
ĂĊ
Ĉ
ď(ÿ) = Ą 3 (Ą/ÿ)ÿ (**)
= ÿ (Ą 2 ÿ). With constants relabeled, the nonlinear equation is the same as
�㔾
Ăÿ
ĂĊ
= ÿ(Ċ 2 ċÿ).
Around 1840 the Belgian mathematician-biologist P. F. Verhulst was concerned with mathematical models for
predicting the human populations of various countries. One of the equations he studied was (4), where a > 0 and b >
0. Equation (4) came to be known as the logistic equation, and its solution is called the logistic function. The graph
of a logistic function is called a logistic curve.
Example. Logistic Growth. Suppose a student carrying a üu virus returns to an isolated college campus of 1000
students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected
students but also to the number of students not infected, determine the number of infected students after 6 days if it is
further observed that after 4 days Ċ(4) = 50.
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January 2022 Denis CK Wong
1.12 Modeling with Systems of First-Order DEs
LINEAR/NONLINEAR SYSTEMS. We have seen that a single differential equation can serve as a mathematical
model for a single population in an environment. But if there are, say, two interacting and perhaps competing species
living in the same environment (for example, rabbits and foxes), then a model for their populations Ċ(Ć) and ċ(Ć)
might be a system of two ûrst-order differential equations such as
čĊ
= Đ1 (Ć, Ċ, ċ)
čĆ
čċ
= Đ2 (Ć, Ċ, ċ)
čĆ
When Đ1 and Đ2 are linear in the variables Ċ and ċ 3 that is, Đ1 and Đ2 have the forms
Đ1 (Ć, Ċ, ċ) = Č1 Ċ + Č2 ċ + ď1 (Ć) and Đ2 (Ć, Ċ, ċ) = Č3 Ċ + Č4 ċ + ď2 (Ć),
where the coefficients Čÿ could depend on Ć 3 then the system is said to be a linear system. A system of differential
equations that is not linear is said to be nonlinear.
RADIOACTIVE SERIES. The rate of radioactive decay is proportional to the number ý(Ć) of nuclei of the substance
present at time Ć. When a substance decays by radioactivity, it usually doesn’t just transmute in one step into a stable
substance; rather, the ûrst substance decays into another radioactive substance, which in turn decays into a third
substance, and so on. This process, called a radioactive decay series, continues until a stable element is reached.
Suppose a radioactive series is described schematically by
2ÿ1
2ÿ2
⏞ Ĉ→
⏞ĉ
ć→
where ý1 = ÿ1 < 0 and ý2 = ÿ2 < 0 are the decay constants for substance X and Y, respectively, and Z is a stable
element.
Suppose, too, that Ċ(Ć), ċ(Ć) and Č(Ć) denote amounts of substances X, Y, and Z, respectively, remaining at time Ć.
The decay of element X is described by
Ăþ
ĂĊ
Ăý
ĂĊ
= 2ÿ1 Ċ. The rate at which the second element Y decays is the net rate
= ÿ1 Ċ 2 ÿ2 ċ. Since Y is gaining atoms from the decay of X and at the same time losing atoms because of its own
decay. Since Z is stable element, it is simply gaining atoms from the decay of element Y:
Ăÿ
ĂĊ
= ÿ2 ċ. In other words, a
model of the radioactive decay series for three elements is the linear system of three ûrst-order differential equations
čĊ
= 2ÿ1 Ċ
čĆ
čċ
= ÿ1 Ċ 2 ÿ2 ċ
čĆ
čČ
= ÿ2 ċ
čĆ
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January 2022 Denis CK Wong
A PREDATOR-PREY MODEL. Suppose that two different species of animals interact within the same environment
or ecosystem, and suppose further that the ûrst species eats only vegetation and the second eats only the ûrst species.
In other words, one species is a predator and the other is a prey. For the sake of discussion, let us imagine that the
predators are foxes and the prey are rabbits.
Let Ċ(Ć) and ċ(Ć) denote the fix and rabbit populations, respectively, at time Ć. If there were no rabbits, then one might
expect that the foxes, lacking an adequate food supply, would decline in number according to
čĊ
= 2ĊĊ, Ċ > 0
čĆ
When rabbits are present in the environment, however, it seems reasonable that the number of encounters or
interactions between these two species per unit time is jointly proportional to their populations Ċ and ċ 2 that is,
proportional to the product Ċċ.
Thus when rabbits are present, there is a supply of food, so foxes are added to the system at a rate ċĊċ, ċ > 0. Adding
this last rate to previous DE gives a model for the fox population:
čĊ
= 2ĊĊ + ċĊċ
čĆ
On the other hand, if there were no foxes, then the rabbits would, with an added assumption of unlimited food supply,
grow at a rate that is proportional to the number of rabbits present at time Ć:
čċ
= čċ, č > 0
čĆ
But when foxes are present, a model for the rabbit population in above DE decreased by ČĊċ, Č > 0- that is, decreased
by the rate at which the rabbits are eaten during their encounters with the foxes:
čċ
= čċ 2 ČĊċ
čĆ
A system of nonlinear differential equations
čĊ
= 2ĊĊ + ċĊċ = Ċ(2Ċ + ċċ)
čĆ
čċ
= čċ 2 ČĊċ = ċ(č 2 ČĊ)
čĆ
where Ċ, ċ, Č and č are positive constants. This famous system of equations is known as the Lotka-Volterra predatorprey model.
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January 2022 Denis CK Wong
COMPETITION MODELS. Suppose two different species of animals occupy the same ecosystem, not as predator
and prey but rather as competitors for the same resources (such as food and living space) in the system. In the absence
of the other, let us assume that the rate at which each population grows is given by
Ăý
ĂĊ
= ĊĊ and
Ăþ
ĂĊ
= Čċ, respectively. (*)
Since the two species compete, another assumption might be that each of these rates is diminished simply by the
inüuence, or existence, of the other population. Thus, a model for the two populations is given by the linear system
Ăý
Ăþ
= ĊĊ 2 ċċ and = Čċ 2 čĊ, (**)
ĂĊ
where Ċ, ċ, Č, č are positive constant.
ĂĊ
Each growth rate in (*) should be reduced by a rate proportional to the number of interactions between the two species:
Ăý
Ăþ
= ĊĊ 2 ċĊċ and = Čċ 2 čĊy. (***)
ĂĊ
ĂĊ
Inspection shows that this nonlinear system is similar to the Lotka-Volterra predator prey model. Finally, it might be
more realistic to replace the rates in (*), which indicate that the population of each species in isolation grows
exponentially, with rates indicating that each population grows logistically (That is, over a long time the population
is bounded);
Ăý
Ăþ
= Ċ1 Ċ 2 ċ1 Ċ 2 and
= Ċ2 ċ 2 ċ2 ċ 2 . (****)
ĂĊ
ĂĊ
When these new rates are decreased by rates proportional to the number of interactions, we obtain another nonlinear
model:
Ăý
Ăþ
= Ċ1 Ċ 2 ċ1 Ċ 2 2 Č1 Ċċ = Ċ(Ċ1 2 ċ1 Ċ 2 Č1þ ) and = Ċ2 ċ 2 ċ2 ċ 2 2 Č2 Ċċ = ċ(Ċ2 2 ċ2 ċ 2 Č2 Ċ). (#)
ĂĊ
ĂĊ
where all coefûcients are positive. The linear system (**) and the nonlinear systems (***) and (#) are, of course,
called competition models.
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January 2022 Denis CK Wong
NETWORKS. An electrical network having more than one loop also gives rise to simultaneous differential equations.
As shown in Figure above, the current ÿ1 (Ć) splits in the directions shown at point þ1 , called a branch point of the
network. By Kirchhoff’s first law we can write ÿ1 (Ć) = ÿ2 (Ć) + ÿ3 (Ć). We can apply Kirchhoff’s second law to each
loop. For loop ý1 þ1 þ2 ý2 ý1 , summing the voltage drops across each part of the loop gives ā(Ć) = ÿ1 ā1 + Ā1
ÿ2 ā2 . Similarly for loop ý1 þ1 ÿ1 ÿ2 þ2 ý2 ý1 we find ā(Ć) = ÿ1 ā1 + Ā2
order equations
Ā1
equivalent to
Ăÿ3
ĂĊ
čÿ2
+ ā1 ÿ2 + ā1 ÿ3 = ā(Ć)
čĆ
Ā
čÿ1
+ āÿ2 = ā(Ć)
čĆ
āÿ
čÿ2
+ ÿ2 2 ÿ1 = 0
čĆ
The End
34
ĂĊ
+
. Upon elimination yields two linear first-
čÿ2
+ (ā1 + ā2 )ÿ2 + ā1 ÿ3 = ā(Ć)
čĆ
Ā1
Ăÿ2
January 2022 Denis CK Wong
Topic 2: Higher- Order Differential Equations
2.1 Initial-Value and Boundary-Value Problems
For a linear differential equation an nth-order initial-value problem is
Solve: ĊĄ (Ċ)
ĂĀþ
Ăý Ā
+ ĊĄ21 (Ċ)
Ă Ā21 þ
Ăý Ā21
+ ⋯ + Ċ1 (Ċ)
Ăþ
Ăý
+ Ċ0 (Ċ)ċ = Đ(Ċ)
(1)
Subject to: ċ(Ċ0 ) = ċ0 , ċ ′ (Ċ0 ) = ċ1 , & , ċ (Ą21) (Ċ0 ) = ċĄ21 .
Theorem 1. Existence of a unique solution. Let ĊĄ (Ċ), ĊĄ21 (Ċ), & , Ċ1 (Ċ), Ċ0 (Ċ) and Đ(Ċ) be continuous on an
interval �㔼 and let ĊĄ (Ċ) b 0 for every Ċ in this interval. If Ċ = Ċ0 is any point in this interval, then a solution ċ(Ċ)
of the initial-value problem (1) exists on the interval and is unique.
Example. The initial-value problem 3ċ999 + 5ċ99 2 ċ9 + 7ċ = 0, ċ(1) = 0, ċ9(1) = 0 and ċ99(1) = 0 has the trivial
solution ċ = 0.
Example. The function ċ = 2Ď 2ý + Ď 22ý 3 3Ċ is a solution of the IVP ċ99 2 4ċ = 12Ċ, ċ(0) = 4, ċ9(0) = 1.
Example. ċ = ČĊ 2 + Ċ + 3 is s solution of the initial-value problem Ċ 2 ċ99 2 2Ċċ9 + 2ċ = 6, ċ(0) = 3 , ċ9(0) =
1 on the interval (2∞, ∞) for any choice of the parameter Č.
Another type of problem consists of solving a linear differential equation of order two or greater in which the
dependent variable y or its derivatives are speciûed at different points. A problem such as
Solve: Ċ2 (Ċ)
Ă2 þ
Ăý 2
+ Ċ1 (Ċ)
Ăþ
Ăý
+ Ċ0(ý)þ = Đ(Ċ)
Subject to: ċ(Ċ) = ċ0 , ċ(ċ) = ċ1
is called a boundary-value problem (BVP). The values ċ(Ċ) = ċ0 , ċ(ċ) = ċ1 are called boundary conditions.
A solution of the foregoing problem is a function satisfying the differential equation on some interval �㔼, containing Ċ
and ċ, whose graph passes through the two points (Ċ, ċ0 ) and (ċ, ċ1 ).
For a second-order differential equation other pairs of boundary conditions could be
ċ′(Ċ) = ċ0 , ċ(ċ) = ċ1
ċ(Ċ) = ċ0 , ċ′(ċ) = ċ1
ċ′(Ċ) = ċ0 , ċ′(ċ) = ċ1
where ċ0 and ċ1 denote constants. These three pairs of conditions are just special cases of the general boundary
conditions
ÿ1 ċ(Ċ) + Ā1 ċ′(Ċ) = ā1
ÿ2 ċ(ċ) + Ā2 ċ′(ċ) = ā2
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January 2022 Denis CK Wong
Example: The two-parameter family of solutions of the differential equation Ċ99 + 16Ċ = 0 is Ċ = Č1 ČāąĆ 4Ć +
Č2 ąÿĀ 4Ć.
Homogeneous Equations. A linear Āth-order differential equation of the form
ĊĄ (Ċ)
ĂĀþ
Ăý Ā
+ ĊĄ21 (Ċ)
Ă Ā21 þ
Ăý Ā21
is said to be homogeneous, whereas an equation
ĊĄ (Ċ)
ĂĀ þ
Ăý Ā
+ ĊĄ21 (Ċ)
Ă Ā21 þ
Ăý Ā21
+ ⋯ + Ċ1 (Ċ)
+ ⋯ + Ċ1 (Ċ)
with Đ(Ċ) not identically zero, is said to be nonhomogeneous.
2ċ99 + 3ċ9 2 5ċ = 0
Ċ 3 ċ999 + 6ċ9 + 10ċ = Ď ý
Ăþ
+ Ċ0 (Ċ)ċ = 0
(#)
+ Ċ0 (Ċ)ċ = Đ(Ċ)
(##)
Ăý
Ăþ
Ăý
Homogeneous linear second-order differential equation
Nonhomogeneous linear third-order differential equation
To solve (##), we must first able to solve the associated homogeneous equation (#). We make the assumption that on
some interval �㔼, the coefficient function Ċÿ (Ċ), ÿ = 0,1,2, & , Ā and Đ(Ċ) are continuous; and ĊĄ (Ċ) b 0 for every
Ċ in �㔼.
Differential Operators. The symbol Ā, where Āċ =
Ăþ
differentiable function into another function. In general,
function.
, is called a differential operator because it transforms a
Ăý
ĂĀ þ
Ăý Ā
= ĀĄ ċ, where ċ represents a sufficiently differentiable
An Āth-order differential operator or polynomial operator is
Ā = ĊĄ (Ċ)Ā Ą + ĊĄ21 (Ċ)ĀĄ21 + ⋯ + Ċ1 (Ċ)Ā + Ċ0 (Ċ). (###)
Ā(Čď(Ċ)) = Č Ā(ď(Ċ)), Č is a constant
Ā(ď(Ċ) + Đ(Ċ)) = Āď(Ċ) + ĀĐ(Ċ)
The differential operator Ā possesses linearity property, that is, Ā(ÿ ď(Ċ) + Ā Đ(Ċ)) = ÿ Ā(ď(Ċ)) + Ā Ā (Đ(Ċ)),
where ÿ, Ā are constants and so the Āth-order differential operator Ā is a linear operator.
Example. ċ99 + 5ċ9 + 6ċ = 5Ċ 2 3 can be written as (Ā 2 + 5 Ā + 6) ċ = 5Ċ 3 3 and so we have Ā(ċ) = 5Ċ 2
3.
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January 2022 Denis CK Wong
Theorem. Superposition principle-Homogeneous Equations. Let ċ1 , ċ2 , & , ċā be solutions of the homogeneous
Āth-order differential equation (#) on the interval �㔼. Then the linear combination
Ĉ = Č1 ċ1 (Ċ) + Č2 ċ2 (Ċ) + ⋯ + Čā ċā (Ċ),
where the Čÿ , ÿ = 1,2, & , ý are arbitrary constants, is also a solution on the interval.
Corollary.
(a) A constant multiple ċ = Č1 ċ1 (Ċ) of a solution ċ1 (Ċ) of a homogeneous linear differential equation is also a
solution.
(b) A homogeneous linear differential equation always possesses the trivial solution ċ = 0.
Example. The functions ċ1 = Ċ 2 and ċ2 = Ċ 2 are both solution of the homogeneous linear equation Ċ 3 ċ999 2 2Ċċ9 +
4ċ = 0 on the interval (0, ∞).
Example. The function ċ = Ď 7ý is a solution of ċ99 + 9ċ9 + 14ċ = 0.
Definition. A set of functions ď1 (Ċ), ď2 (Ċ), & , ďĄ (Ċ) is said to be linearly dependent on an interval �㔼 if there exist
constants Č1 , Č2 , & , ČĄ , not all zero such that Č1 ď1 (Ċ) + Č2 ď2 (Ċ) + ⋯ + ČĄ ďĄ (Ċ) = 0 for every Ċ in the interval. If
the set of functions is not linearly dependent on the interval, it is said to be linearly independent.
Example. The set of functions ď1 (Ċ) = cos 2 Ċ , ď2 (Ċ) = ąÿĀ2 Ċ, ď3 (Ċ) = ąĎČ 2 Ċ, ď4 (Ċ) = ĆĊĀ2 Ċ is linearly
�㔋 �㔋
dependent on the interval (2 , ).
2 2
Example. The set of functions ď1 (Ċ) = √Ċ + 5 , ď2 (Ċ) = √Ċ + 5Ċ , ď3 (Ċ) = Ċ 3 1, ď4 (Ċ) = Ċ 2 is linearly
dependent on (0, ∞).
Definition. Suppose each of the functions ď1 (Ċ), ď2 (Ċ), & , ďĄ (Ċ) possesses at least Ā 2 1 derivatives. The determinant
ď1
ď1 ′
Ć(ď1 , ď2 , & , ďĄ ) = | &
(Ą21)
ď1
the functions.
ď2
ď2 ′
&
(Ą21)
ď2
&
&
&
&
ďĄ
ďĄ ′
& |, where the primes denote derivatives, is called the Wronskian of
(Ą21)
ďĄ
Theorem. Criterion for Linearly Independent Solutions. Let ċ1 , ċ2 , & , ċĄ be Ā solutions of the homogeneous linear
nth-order differential equation (#) on an interval �㔼. Then the set of solutions is linearly independent on �㔼 iff
Ć(ċ1 , ċ2 , & , ċĄ ) b 0 for every Ċ in the interval.
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Definition. Any set ċ1 , ċ2 , & , ċĄ of n linearly independent solutions of the homogeneous linear nth-order differential
equation (#) on an interval �㔼 is said to be a fundamental set of solutions on the interval.
Theorem. Existence of a Fundamental Set. There exists a fundamental set of solutions for the homogeneous linear
nth-order differential equation (6) on an interval �㔼.
Theorem. General Solution 4Homogeneous Equations. Let ċ1 , ċ2 , & , ċĄ be a fundamental set of solutions of the
homogenous linear Āth-order differential equation (#) on �㔼. Then the general solution the equation on the interval is
ċ = Č1 ċ1 (Ċ) + Č2 ċ2 (Ċ) + ⋯ + ČĄ ċĄ (Ċ), where Čÿ , ÿ = 1,2, & , Ā are arbitrary constants.
Example. The functions ċ1 = Ď 3ý and ċ2 = Ď 23ý are solutions of ċ99 3 9ċ = 0 on the interval (2∞, ∞).
Example. The function ċ = 4 ąÿĀ/ 3Ċ 3 5 Ď 3ý is a solution of ċ99 2 9ċ = 0.
Example. The functions ċ1 = Ď ý , ċ2 = Ď 2ý and ċ3 = Ď 3ý satisfy the third-order equation ċ999 2 6ċ99 + 11ċ9 2
6ċ = 0.
Nonhomogeneous Equations. Any function ċĆ , free of arbitrary parameters, that satisfies (##) is said to be a
particular solution or particular integral of the equation.
Theorem. General solution 3 Nonhomogeneous Equations. Let ċĆ be any particular solution of the
nonhomogeneous linear Ā th -order differential equation (##) on an interval �㔼, and let ċ1 , ċ2 , & , ċĄ be a fundamental
set of solutions of the associated homogeneous differential equation (#) on �㔼. Then the general solution of the equation
on the interval is ċ = Č1 ċ1 (Ċ) + Č2 ċ2 (Ċ) + ⋯ + ČĄ ċĄ (Ċ) + ċĆ , where the Čÿ , ÿ = 1,2 & , Ā are arbitrary constants.
The general solution of a nonhomogeneous linear equation consists of the sum of two functions:
ċ = Č1 ċ1 (Ċ) + Č2 ċ2 (Ċ) + ⋯ + ČĄ ċĄ (Ċ) + ċĆ (Ċ) = ċā (Ċ) + ċĆ (Ċ)
The linear combination ċā (Ċ) = Č1 ċ1 (Ċ) + Č2 ċ2 (Ċ) + ⋯ + ČĄ ċĄ (Ċ), which is the general solution of (#), is called the
complementary function for (##). In other words, to solve a nonhomogeneous linear differential equation, we ûrst
solve the associated homogeneous equation and then ûnd any particular solution of the nonhomogeneous equation.
Example. Solve the nonhomogeneous equation ċ999 2 6ċ99 + 11ċ9 2 6ċ = 3Ċ.
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Theorem. Superposition Principle- Nonhomogeneous Equations.
Let ċĆ1 , ċĆ2 , & , ċĆ�㕘 be k particular solutions of the non homogeneous linear nth-order differential equation (##) on an
interval �㔼, in turn, to ý distinct function Đ1 , Đ2 , & , Đā . That is, suppose ċĆ�㕖 denotes a particular solution of the
corresponding differential equation
ĊĄ (Ċ)ċ (Ą) + ĊĄ21 (Ċ)ċ (Ą21) + ⋯ + Ċ1 (Ċ)ċ ′ + Ċ0 (Ċ)ċ = Đÿ (Ċ)
where ÿ = 1,2, & , ý. Then
ċĆ = ċĆ1 (Ċ) + ċĆ2 (Ċ) + ⋯ + ċĆ�㕘 (Ċ)
is a particular solution of ĊĄ (Ċ)ċ (Ą) + ĊĄ21 (Ċ)ċ (Ą21) + ⋯ + Ċ1 (Ċ)ċ ′ + Ċ0 (Ċ)ċ = Đ1 (Ċ) + Đ2 (Ċ) + ⋯ + Đā (Ċ).
Example. Verify that ċ = 24Ċ 2 + Ď 2ý + ĊĎ ý is a solution of ċ99 2 3ċ9 + 4ċ = 216Ċ 2 + 24Ċ 2 8 + 2Ď 2ý +
2ĊĎ 2ý 2 Ď ý .
2.2 Reduction of Order
Reduction of Order. The general solution of a homogeneous linear second-order differential equation
Ċ2 (Ċ)ċ ′′ + Ċ1 (Ċ)ċ ′ + Ċ0 (Ċ)ċ = 0 (1)
is a linear combination ċ = Č1 ċ1 + Č2 ċ2 , where ċ1 and ċ2 are solutions that constitute a linearly independent set
on some interval �㔼.
Equation (1) can be reduced to a linear ûrst-order DE by means of a substitution involving the known solution ċ1 . A
second solution ċ2 of (1) is apparent after this ûrst-order differential equation is solved. Suppose that ċ1 denoted a
nontrivial solution of (1) and that ċ1 is defined on an interval �㔼. We seek a second solution ċ2 so that the set consisting
of ċ1 and ċ2 is linearly independent on �㔼.
þ (ý)
þ
If ċ1 and ċ2 are linearly independent, then 2 which is nonconstant on �㔼 3 that is, 2(ý) = ć(Ċ). The function u(x) can
þ1
þ1
be found by substituting ċ2 (Ċ) = ć(Ċ)ċ1 (Ċ) into the given differential equation. This method is called reduction of
order because we must solve a linear first -order DE to find ć.
Example. Given that ċ1 = Ď ý is a solution of ċ99 2 ċ = 0 on the interval (2∞, ∞), use reduction of order to find a
second solution ċ2 .
Suppose we divide by Ċ2 (Ċ) to put equation (1) in the standard form
ċ99 + ÿ(Ċ)ċ9 + Ā(Ċ)ċ = 0 , (2)
where ÿ(Ċ) and Ā(Ċ) are continuous on some �㔼.
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January 2022 Denis CK Wong
Suppose also that ċ1 (Ċ) is a known solution of (2) on �㔼 and that ċ1 (Ċ) b 0 for every Ċ in the interval. If we define
ċ = ć(Ċ)ċ1 (Ċ), it follows that
ċ ′ = ćċ1′ + ċ1 ć′
ċ ′′ = ćċ1′′ + 2ċ1′ ć′ + ċ1 ć′′
ċ ′′ + ÿċ ′ + Āċ = ć[ċ1′′ + ÿċ1′ + Āċ1 ] + ċ1 ć′′ + (2ċ1′ + ÿċ1 )ć′ = 0.
ċ1 ć′′ + (2ċ1′ + ÿċ1 )ć′ = 0
ċ1 ĉ′ + (2ċ1′ + ÿċ1 )ĉ = 0, where we have let ĉ = ć9.
The last equation is both linear and separable.
Separating variables and integrating, we obtain a second solution of (2) is
ċ2 = ċ1 (Ċ) ∫
ă 2 ∫ �㕝(ý)�㕑ý
þ12 (ý)
čĊ.
Example. The function ċ1 = Ċ 2 is a solution of Ċ 2 ċ99 2 3Ċċ9 + 4ċ = 0. Find the general solution of the differential
equation on the interval (0, ∞).
2.3 Homogeneous Linear Equations with Constant Coefficients.
Homogeneous Linear Equations with Constant Coefficients. Consider a homogeneous linear equation Ċċ ′ + ċċ =
0 where the coefficients Ċ b 0 and ċ are constants.
This type of equation can be solved either by separation of variables or with the aid of an integrating factor, but there
is another solution method, one that uses only algebra. Before illustrating this alternative method, we make one
observation:
Solving Ċċ ′ + ċċ = 0 for ċ9 yields ċ9 = ýċ, where ý is a constant.
This observation reveals the nature of the unknown solution y; the only nontrivial elementary function whose
derivative is a constant multiple of itself is an exponential function Ď ăý . Now the new solution method: If we substitute
ċ = Ď ăý and ċ ′ = ÿĎ ăý into Ċċ ′ + ċċ = 0, we get Ċÿ + ċ = 0. For this single value of ÿ, ċ = Ď ăý is a solution
of the DE.
Consider the second-order equation Ċċ ′′ + ċċ ′ + Čċ = 0, where Ċ, ċ and Č are constants. If we try to find a solution
of the form ċ = Ď ăý , then we obtain Ċÿ2 + ċÿ + Č = 0. This equation is called the auxiliary equation of the DE.
Two roots ÿ1 =
(2Ā+√Ā 2 24ÿā)
2ÿ
and ÿ2 =
(2Ā2√Ā 2 24ÿā)
2ÿ
.
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January 2022 Denis CK Wong
Case1: Distinct Real Roots. ÿ1 and ÿ2 real and distinct (ċ 2 2 4ĊČ > 0).
Two solutions, ċ1 = Ď ă1 ý and ċ2 = Ď ă2 ý both are linearly independent on (2∞, ∞) and hence form a fundamental
set. A general solution is ċ = Č1 Ď ă1 ý + Č2 Ď ă2 ý .
Case 2: Repeated Real Roots. ÿ1 and ÿ2 real and equal (ċ 2 2 4ĊČ = 0).
Ā
Only one solution ċ1 = Ď ă1 ý and obtained ÿ1 = 2 .
ă1 ý
ă 2ÿ1 ý
A second solution is ċ2 = Ď
∫ ă 2ÿ1ý čĊ = ĊĎ
A general solution is ċ = Č1 Ď ă1 ý + Č2 ĊĎ ă1ý .
ă1 ý
2ÿ
.
Case 3: Conjugate Complex Roots. ÿ1 and ÿ2 conjugate complex numbers (ċ 2 2 4ĊČ < 0).
ÿ1 = ÿ + ÿĀ and ÿ2 = ÿ 2 ÿĀ, where ÿ and Ā > 0 are real and ÿ 2 = 21.
A solution ċ = ÿ1 Ď (ÿ+ÿĀ)ý + ÿ2 Ď (ÿ2ÿĀ)ý .
Using the Euler’s formula: Ď ÿ�㔃 = cos �㔃 + ÿąÿĀ�㔃 , we have Ď ÿĀý = cos ĀĊ + ÿąÿĀĀĊ and Ď 2ÿĀý = cos ĀĊ 2 ÿąÿĀĀĊ .
Thus, Ď ÿĀý + Ď 2ÿĀý = 2 cos ĀĊ and Ď ÿĀý 2 Ď 2ÿĀý = 2ÿ sin ĀĊ.
Since ċ = ÿ1 Ď (ÿ+ÿĀ)ý + ÿ2 Ď (ÿ2ÿĀ)ý
is a solution, by choosing ÿ1 = ÿ2 = 1 (ċ1 = Ď ÿý (Ď ÿĀý + Ď 2Āý ) =
2Ď ÿý cos ĀĊ) and by choosing ÿ1 = 1 and ÿ2 = 21 (ċ2 = Ď ÿý (Ď ÿĀý 2 Ď 2Āý ) = 2Ď ÿý sin ĀĊ).
Ď ÿý cos ĀĊ and Ď ÿý sin ĀĊ are real solutions and form a fundamental set on (2∞, ∞).
The general solution is ċ = Č1 Ď ÿý cos ĀĊ + Č2 Ď ÿý sin ĀĊ.
Example. Solve the following differential equations.
(a) 2ċ99 2 5ċ9 2 3ċ = 0
(b) ċ99 2 10ċ9 + 25ċ = 0
(c) ċ99 + 4ċ9 + 7ċ = 0
Example. Solve 4ċ99 + 4ċ9 + 17ċ = 0, ċ(0) = 21, ċ9(0) = 2.
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In general, to solve an Āth-order differential equation, where the Ċÿ , ÿ = 0,1, & , Ā are real constant, we must solve an
Āth-degree polynomial equation
ĊĄ ÿĄ + ĊĄ21 ÿĄ21 + ⋯ + Ċ2 ÿ2 + Ċ1 ÿ + Ċ0 = 0
If all the roots are real and distinct, then the general solution is
ċ = Č1 Ď ă1 ý + Č2 Ď ă2 ý + ⋯ + ČĄ Ď ăĀ ý .
If ÿ1 is a root of multiplicity ý of an Āth-degree auxiliary equation (that is, ý roots are equal to ÿ1 ), then the general
solution is
ċ = Č1 Ď ă1 ý + Č2 ĊĎ ă1ý + ⋯ + Čā Ċ ā21 Ď ă1ý .
Example. Solve ċ999 + 3ċ99 2 4ċ = 0.
Example. Solve
Ă4þ
Ăý 4
+2
Ă2þ
Ăý 2
+ ċ = 0.
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2.4 Undetermined Coefficients 3 Superposition Approach.
The first of two ways we shall consider for obtaining a particular solution ċĆ for a nonhomogeneous linear DE is
called the method of undetermined coefficients. The underlying idea behind this method is a conjecture about the
form of ċĆ , and educated guess really, that is, motivated by the kinds of functions that make up the input function
Đ(Ċ).
The general method is limited to linear DEs of the form
•
•
ĊĄ ċ (Ą) + ĊĄ21 ċ (Ą21) + ⋯ + Ċ1 ċ ′ + Ċ0 ċ = Đ(Ċ), where
the coefficients Ċÿ , ÿ = 0,1, & , Ā are constants and
Đ(Ċ) is a constant k, a polynomial function, an exponential function Ď ÿý , a since or cosine function ąÿĀ ĀĊ
or Čāą ĀĊ, or finite sums and products of these functions.
That is, Đ(Ċ) is a linear combination of functions of the type
ÿ(Ċ) = ĊĄ Ċ Ą + ĊĄ21 Ċ Ą21 + ⋯ + Ċ1 Ċ + Ċ0 , ÿ(Ċ)Ď ÿý , ÿ(Ċ)Ď ÿý ąÿĀ ĀĊ and ÿ(Ċ)Ď ÿý Čāą ĀĊ, where Ā is a
nonnegative integer and ÿ and Ā are real numbers.
The method of undetermined coefficients is not applicable to when
1
Đ(Ċ) = þĀ Ċ, Đ(Ċ) = , Đ(Ċ) = ĆĊĀ Ċ, Đ(Ċ) = ąÿĀ21 Ċ and so on.
ý
Example. Solve ċ99 + 4ċ9 2 2ċ = 2Ċ 2 2 3Ċ + 6.
Example. Find a particular solution of ċ99 2 ċ9 + ċ = 2 ąÿĀ 3Ċ.
Example. Solve ċ99 2 2ċ9 2 3ċ = 4Ċ 2 5 + 6ĊĎ 2ý .
Example. Find a particular solution of ċ99 2 5ċ9 + 4ċ = 8Ď ý .
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Example. Determine the form of a particular solution of
(a) ċ99 2 8ċ9 + 25ċ = 5Ċ 3 Ď 2ý 3 7Ď 2ý
(b) ċ99 + 4ċ = Ċ Čāą Ċ.
Example. Determine the form of a particular solution of ċ99 2 9ċ9 + 14ċ = 3Ċ 2 3 5 ąÿĀ2Ċ + 7ĊĎ 6ý .
Example. Find a particular solution of ċ99 2 2ċ9 + ċ = Ď ý .
Example. Solve ċ99 + ċ = 4Ċ + 10 sin Ċ , ċ(�㔋) = 0, ċ9(�㔋) = 2.
Example. Solve ċ99 2 6ċ9 + 9ċ = 6Ċ 2 + 2 2 12Ď 3ý .
Example. Solve ċ999 + ċ99 = Ď ý Čāą Ċ
Example. Determine the form of a particular solution of ċ (4) + ċ999 = 1 2 Ċ 2 Ď 2ý .
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2.5 Undetermined coefficients 3 Annihilator Approach.
Āth-order differential equation can be written
where Ā ā ċ =
Ă �㕘 þ
Ăý �㕘
ĊĄ ĀĄ ċ + ĊĄ21 ĀĄ21 ċ + ⋯ + Ċ1 Āċ + Ċ0 ċ = Đ(Ċ), (1)
, ý = 0,1, & , Ā. We always write Ā(ċ) = Đ(Ċ), where Ā denotes the linear nth-order differential,
or polynomial, operator
ĊĄ ĀĄ + ĊĄ21 Ā Ą21 + ⋯ + Ċ1 Ā + Ċ0 . (2)
Factoring operators. When the coefficients Ċÿ , ÿ = 0,1, & , Ā are real constants, a linear differential operator (1) can
be factored whenever the characteristic polynomial ĊĄ ÿĄ + ĊĄ21 ÿĄ21 + ⋯ + Ċ1 ÿ + Ċ0 factors. In other words, if
Ą1 is a root of the auxiliary equation ĊĄ ÿĄ + ĊĄ21 ÿĄ21 + ⋯ + Ċ1 ÿ + Ċ0 = 0, then Ā = (Ā 2 Ą1 )ÿ(Ā), where the
polynomial expression ÿ(Ā) is a linear differential operator of order Ā 2 1.
<Factors of a linear differential operator with constant coefficients commute=.
Annihilator Operator. If Ā is a linear differential operator with constant coefûcients and ď is a sufûciently
differentiable function such that Ā(ď(Ċ)) = 0, then Ā is said to be an annihilator of the function. The differential
operator Ā Ą annihilates each of the functions 1, Ċ , Ċ 2 , & , Ċ Ą21 . Thus together with the fact that differentiation can be
done term by term, a polynomial Č0 + Č1 Ċ + Č2 Ċ 2 + ⋯ + ČĄ21 Ċ Ą21 can be annihilated by finding an operator
that annihilates the highest power of Ċ.
The functions that are annihilated by a linear Āth-order differential operator Ā are simply those functions that can be
obtained from the general solution of the homogeneous differential equation Ā(ċ) = 0.
The differential operator (Ā 2 ÿ)Ą annihilates each of the functions
Ď ÿý , ĊĎ ÿý , Ċ 2 Ď ÿý , & , Ċ Ą21 Ď ÿý .
To see this, note that the auxiliary equation of the homogeneous equation (Ā 2 ÿ)Ą ċ = 0 is (ÿ 2 ÿ)Ą = 0. Since ÿ
is a root of multiplicity Ā, the general solution is
ċ = Č1 Ď ÿý + Č2 ĊĎ ÿý + ⋯ + ČĄ Ċ Ą21 Ď ÿý .
Example. Find a differential operator that annihilates
(a) 1 2 5Ċ 2 + 8Ċ 3
(b) Ď 23ý
(c) 4Ď 2ý 2 10ĊĎ 2ý
When ÿ and Ā > 0 are real numbers, the quadratic formula reveals that [ÿ2 2 2ÿÿ + (ÿ 2 + Ā 2 )]Ą = 0 has complex
roots ÿ + ÿĀ, ÿ 2 ÿĀ, both of multiplicity Ā.
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The differential operator [Ā2 3 2ÿ Ā + (ÿ 2 + Ā 2 )]Ą annihilates each of the functions
Ď ÿý cos ĀĊ , ĊĎ ÿý cos ĀĊ , Ċ 2 Ď ÿý cos ĀĊ , & , Ċ Ą21 Ď ÿý cos ĀĊ
Ď ÿý sin ĀĊ , ĊĎ ÿý sin ĀĊ , Ċ 2 Ď ÿý sin ĀĊ , & , Ċ Ą21 Ď ÿý sin ĀĊ
Example. Find a differential operator that annihilates 5Ď 2ý cos 2Ċ 2 9Ď 2ý sin 2Ċ.
Undetermined Coefficients. Suppose that Ā(ċ) = Đ(Ċ) is a linear differential equation with constant coefficients
and that the input Đ(Ċ) consists of finite sums and products of functions listed previously, that is, Đ(Ċ) is a linear
combination of functions of the form
ý(ČāĀąĆĊĀĆ), Ċ ă , Ċ ă Ď ÿý , Ċ ă Ď ÿý cos ĀĊ ĊĀč Ċ ă Ď ÿý sin ĀĊ,
where ÿ is a nonnegative integer and ÿ and Ā are real numbers.
We now know that such a function Đ(Ċ) can be annihilated by a differential operator Ā1 of lowest order, consisting of
a product of the operator ĀĄ , (Ā 2 ÿ)Ą , and (Ā2 2 2ÿ Ā + ÿ 2 + Ā 2 )Ą . Applying Ā1 to both sides of the equation
Ā(ċ) = Đ(Ċ) yields Ā1 Ā(ċ) = Ā1 (Đ(Ċ)) = 0. By solving the homogeneous higher-order equation Ā1 Ā(ċ) = 0, we
can discover the form of a particular solution ċĆ for the original nonhomogeneous equation Ā(ċ) = Đ(Ċ). We then
substitute this assumed form into Ā(ċ) = Đ(Ċ) to find an explicit particular solution. This procedure for determining
ċĆ , called the method of undetermined coefficients.
The general solution of a nonhomogeneous linear differential equation Ā(ċ) = Đ(Ċ) is ċ = ċā + ċĆ , where ċā is
the complementary function-that is, the general solution of the associated homogeneous equation Ā(ċ) = 0. The
general solution of each equation Ā(ċ) = Đ(Ċ) is defined on the interval (2∞, ∞).
Example. Solve ċ99 + 3ċ9 + 2ċ = 4Ċ 2 .
Example. Solve ċ99 2 3ċ9 = 8Ď 3ý + 4 ąÿĀĊ.
Example. Solve ċ99 + ċ = Ċ Čāą Ċ 3 Čāą Ċ.
Example. Determine the form of a particular solution for ċ99 2 2ċ9 + ċ = 10Ď 22ý Čāą Ċ.
Example. Determine the form of a particular solution for ċ999 2 4ċ99 + 4ċ9 = 5Ċ 2 2 6Ċ + 4Ċ 2 Ď 2ý + 3Ď 5ý .
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2.6 Variation of Parameters
Variation of Parameters. To adapt the method of variation of parameters to a linear second-order differential
equation
Ċ2 (Ċ)ċ ′′ + Ċ1 (ċ)ċ ′ + Ċ0 (Ċ)ċ = Đ(Ċ), (1)
we begin by putting the equation into the standard form
ċ ′′ + ÿ(Ċ)ċ ′ + Ā(Ċ)ċ = ď(Ċ) (2)
by dividing through by the lead coefficient Ċ2 (Ċ). Equation (2) is the second-order analogue of the standard form of
a linear first-order equation:
some common interval �㔼.
Ăþ
Ăý
+ ÿ(Ċ) ċ = ď(Ċ). In (2), we suppose that ÿ(Ċ), Ā(Ċ) and ď(Ċ) are continuous on
Example. Solve ċ99 2 4ċ9 + 4ċ = (Ċ + 1)2 Ċ.
Example. Solve 4ċ99 + 36ċ = ČąČ3Ċ.
Constants of Integration.
1
Example. ċ99 2 ċ = .
ý
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2.7 Cauchy-Euler Equation
Cauchy-Euler Equation. A linear differential equation of the form
ĊĄ Ċ Ą
čĄ ċ
č Ą21 ċ
čċ
Ą22
+
Ċ
Ċ
+ ⋯ + Ċ1 Ċ
+ Ċ0 ċ = Đ(Ċ),
Ą21
Ą
Ą21
čĊ
čĊ
čĊ
where the coefficients ĊĄ , ĊĄ21 , & , Ċ0 are constants, is known as a Cauchy-Euler equation.
The observable characteristic of this type of equation is that the degree ý = Ā, Ā 2 1, & , 1, 0 of the monomial
coefficients Ċ ā matches the order ý of differentiation
Ă �㕘 þ
Ăý �㕘
.
Method of Solution. We try a solution of the form ċ = Ċ ă , where ÿ is to be determined. Analogous to what
happened when we substituted Ď ăý into a linear equation with constant coefficients, when we substitute Ċ ă , each
term of a Cauchy-Euler equation becomes a polynomial in m times Ċ ă , since
čā ċ
= Ċā Ċ ā ÿ(ÿ 2 1)(ÿ 2 2) & (ÿ 2 ý + 1)Ċ ă2ā = Ċā ÿ(ÿ 2 1)(ÿ 2 2) & (ÿ 2 ý + 1)Ċ ă
čĊ ā
For instance, when substitute ċ = Ċ ă , the second-order equation becomes
ĊĊ 2
Ă2ý
Ăþ 2
+ ċĊ
Ăþ
Ăý
+ Čċ = [Ċÿ(ÿ 2 1) + ċÿ + Č]Ċ ă .
Thus ċ = Ċ ă is a solution of the differential equation whenever m is a solution of the auxiliary equation
Ċÿ(ÿ 2 1) + ċÿ + Č = 0 or Ċÿ2 + (ċ 2 Ċ)ÿ + Č = 0. (1)
Case 1: Distinct Real Roots. Let ÿ1 and ÿ2 denote the real roots of (1) such that ÿ1 b ÿ2 . Then ċ1 = Ċ ă1 and
ċ2 = Ċ ă2 form a fundamental set of solutions. Here the general solution is ċ = Č1 Ċ ă1 + Č2 Ċ ă2 . (2)
Example. Solve Ċ 2
Ă2þ
Ăý^2
2 2Ċ
Ăþ
Ăý
2 4ċ = 0.
Case 2: Repeated Real Roots. If the roots of (1) are repeated (that is, ÿ1 = ÿ2 ), then we obtain only one solution
3 namely, ċ = Ċ ă1 . When the roots of the quadratic equation Ċÿ2 + (ċ 2 Ċ)ÿ + Č = 0 are equal, the discriminant
of the coefficients is necessarily zero. It follows that ÿ1 = 2
Ā2ÿ
2ÿ
.
We now construct a second solution ċ2 . We first write the Cauchy-Euler equation in the standard form
and make the identifications Ă(Ċ) =
Ā
ÿý
č2ċ
ċ čċ
Č
+
+ 2ċ = 0
2
čĊ
ĊĊ čĊ ĊĊ
Ā
Ā
and ∫ ( ) čĊ = ln Ċ. Thus ċ2 = Ċ
ÿý
The general solution is then ċ = Č1 Ċ ă1 + Č2 Ċ ă1 ln Ċ.
ÿ
49
ă1
∫
ă
Ā
2(ÿ) ln ý
ý 2ÿ1
čĊ = Ċ ă1 ln Ċ.
January 2022 Denis CK Wong
Example. Solve 4Ċ 2
Ă2 þ
Ăý 2
+ 8Ċ
Ăþ
Ăý
+ ċ = 0.
For higher-order equations, if ÿ1 is a root of multiplicity ý, then it can be shown that
Ċ ă1 , Ċ ă1 ln Ċ , Ċ ă1 (ln Ċ )2 , & ,
Ċ ă1 (ln Ċ)ā21
are ý linearly independent solutions. Thus, the general solution of the differential equation must then contain a linear
combination of these ý solutions.
Case 3: Conjugate Complex Roots. If the roots of (1) are the conjugate pair ÿ1 = ÿ + ÿ Ā , ÿ2 = ÿ 3 ÿ Ā, where
ÿ and Ā > 0 are real, then a solution is ċ = ÿ1 Ċ ÿ+ÿĀ + ÿ2 Ċ ÿ2ÿĀ . Note that Ċ ÿĀ = Ď ÿĀ ln ý , and by Euler’s formula,
we have
Ċ ÿĀ = cos(Ā ln Ċ) + ÿ sin(Ā ln Ċ) and Ċ ÿĀ = cos(Ā ln Ċ) + ÿ sin(Ā ln Ċ) and Ċ 2ÿĀ = cos(Ā ln Ċ) 2 ÿ sin(Ā ln Ċ)
Thus, Ċ ÿĀ + Ċ 2ÿĀ = 2 cos(Ā ln Ċ) and Ċ ÿĀ 2 Ċ 2ÿĀ = 2ÿ sin(Ā ln Ċ).
From ċ = ÿ1 Ċ ÿ+ÿĀ + ÿ2 Ċ ÿ2ÿĀ , we see, in turn, for ÿ1 = ÿ2 = 1 and ÿ1 = 1, ÿ2 = 21 that
ċ1 = Ċ ÿ (Ċ ÿĀ + Ċ 2ÿĀ ) and ċ2 = Ċ ÿ (Ċ ÿĀ 2 Ċ 2ÿĀ ) or
solutions.
ċ1 = 2Ċ ÿ cos(Ā ln Ċ) and ċ2 = 2ÿĊ ÿ sin(Ā ln Ċ) are also
Since Ć(Ċ ÿ cos(Ā ln Ċ) , Ċ ÿ sin(Ā ln Ċ)) = ĀĊ 2ÿ21 b 0, Ā > 0 on the interval (0, ∞), we conclude that
ċ1 = Ċ ÿ cos(Ā ln Ċ) and ċ2 = Ċ ÿ sin(Ā ln Ċ) constitute a fundamental set of real solutions of the differential equation.
Hence the general solution is ċ = Ċ ÿ [Č1 cos(Ā ln Ċ) + Č2 sin(Ā ln Ċ)].
1
Example. Solve 4Ċ 2 ċ99 + 17ċ = 0, ċ(1) = 21 , ċ9(1) = 2 .
Example. Solve Ċ 3
Ă3þ
Ăý 3
+ 5Ċ 2
Ă2þ
Ăý 2
+ 7Ċ
Ăþ
Ăý
2
+ 8ċ = 0.
Example. Solve Ċ 2 ċ99 2 3Ċċ9 + 3ċ = 2Ċ 4 Ď ý .
Reduction to Constant Coefficients.
Example. Solve Ċ 2 ċ99 3 Ċċ9 + ċ = þĀ Ċ.
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January 2022 Denis CK Wong
2.8 Solving Systems of Linear DEs by Elimination
System Elimination. The elimination of an unknown in a system of linear differential equations is expedited by
rewriting each equation in the system in differential operator notation. Recall a single linear equation
ĊĄ ċ (Ą) + ĊĄ21 ċ Ą21 + ⋯ + Ċ1 ċ ′ + Ċ0 ċ = Đ(Ć),
where the Ċÿ , ÿ = 0,1,2, & , Ā are constant, can be rewrite as
(ĊĄ Ā Ą + ĊĄ21 ĀĄ21 + ⋯ + Ċ1 Ā + Ċ0 )ċ = Đ(Ć).
If the Āth-order differential operator ĊĄ ĀĄ + ĊĄ21 Ā Ą21 + ⋯ + Ċ1 Ā + Ċ0 factors into differential operators of lower
order, then the factors commute.
Solution of a System. A solution of a system of DEs is a set of sufficiently differentiable Ċ = �㔙1 (Ć), ċ = �㔙2 (Ć), Č =
�㔙3 (Ć) , and so on that satisfies each equation in the system on some common interval �㔼.
Method of Solution. Consider the simple system of linear first-order equations
or
čċ
čĊ
= 3ċ,
= 2Ċ
čĆ
čĆ
ĀĊ 2 3ċ = 0 (1), 2Ċ 2 Āċ = 0
(2).
Ā × (1) add 23 × (2) to obtain Ā2 Ċ 3 6Ċ = 0 with roots ÿ1 = √6 and ÿ2 = 2√6 and so
Ċ(Ć) = Č1 Ď 2√6Ċ + Č2 Ď √6Ċ . (2)
2 × (1) 2 Ā × (2) gives Ā2 ċ 3 6ċ = 0 and so
ċ(Ć) = Č3 Ď 2√6Ċ + Č4 Ď √6Ċ . (3)
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January 2022 Denis CK Wong
Topic 3: Series Solutions of Second Order Linear Equations
A power series in Ċ 2 Ċ is an infinite series of the form
2
Ą
∑∞
Ą=0 ČĄ (Ċ 2 Ċ) = Č0 + Č1 (Ċ 2 Ċ) + Č2 (Ċ 2 Ċ) + ⋯.
Such a series is also said to be a power series centered at Ċ.
∞
Convergence. A power series ∑Ą=0
ČĄ (Ċ 2 Ċ)Ą is convergent at a specific value of Ċ if its sequence of partial sum
þ
{Ăþ (Ċ)} converges - that is, lim Ăþ (Ċ) = lim ∑Ą=0
ČĄ (Ċ 2 Ċ)Ą exists. If the limit does not exist at Ċ, then the series
is said to be divergent.
þ→∞
þ→∞
Interval of Convergence. Every power series has an interval of convergence. The interval of convergence is the set
of all real numbers x for which the series converges.
Radius of Convergence. Every power series has a radius of convergence ā. If ā > 0, then the power series
∞
∑Ą=0
ČĄ (Ċ 2 Ċ)Ą converges for |Ċ 2 Ċ| < ā and diverges for |Ċ 2 Ċ| > ā. If the series converges only at its center
a, then ā = 0. If the series converges for all Ċ, then we write ā = ∞. A power series might or might not converge at
the endpoints Ċ 2 ā and Ċ + ā of the interval.
Absolute Convergence. Within its interval of convergence, a power series converges absolutely. In other words, if x
is a number in the interval of convergence and is not an endpoint of the interval, then the series of absolute values
Ą
∑∞
Ą=0 |ČĄ (Ċ 2 Ċ) | converges.
ā
(ý2ÿ)Ā+1
ā
Ratio Test. Suppose ČĄ b 0 for all n and that lim | Ā+1(ý2ÿ)Ā | = |Ċ 2 Ċ| lim | Ā+1 | = Ā.
Ą→∞
If Ā < 1, the series converges absolutely.
If Ā > 1, the series diverges.
If Ā = 1, the test is inconclusive.
āĀ
Ą→∞
āĀ
∞
A Power Series Defined a Function. A power series defined a function ď(Ċ) = ∑Ą=0
ČĄ (Ċ 2 Ċ)Ą whose domain is
the interval of convergence of the series. If ā > 0, ď is continuous, differentiable, and integrable on (Ċ 2 ā, Ċ + ā).
ď ′(Ċ) and ∫ ď(Ċ)čĊ can be found by term-by-term differentiation and integration. Convergence at an endpoint may
be either lost by differentiation or gained through integration.
∞
∞
∞
If ċ = ∑Ą=0
ČĄ Ċ Ą , then ċ′ = ∑Ą=1
ČĄ ĀĊ Ą21 and ċ′′ = ∑Ą=2
ČĄ Ā(Ā 2 1)Ċ Ą22 .
∞
Identity Property. If ∑Ą=0
ČĄ (Ċ 2 Ċ)Ą = 0, ā > 0 for all number Ċ in the interval of convergence, then ČĄ = 0 for
all Ā.
Analytic at a Point. A function ď is analytic at a point Ċ if it can be represented by a power series in Ċ 2 Ċ with
positive or infinite radius of convergence.
Ď ý , Čāą Ċ, ąÿĀ Ċ, þĀ(1 2 Ċ), and so on can be represented by Taylor series
Ďý = 1 +
Ċ Ċ2
+ +⋯
1! 2!
sin Ċ = Ċ 2
cos Ċ = 1 2
Ċ3 Ċ5
+ 2⋯
3! 5!
Ċ2 Ċ4 Ċ6
+ 2 &
2! 4! 6!
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January 2022 Denis CK Wong
for |Ċ| < ∞ . These Taylor series centered at 0, called Maclaurin series, show that Ď ý , ąÿĀ Ċ and Čāą Ċ are analytic at
Ċ = 0.
Arithmetic of Power Series.
Ď ý sin Ċ = (1 +
Ċ Ċ2
Ċ3 Ċ5
Ċ3 Ċ5
+ + ⋯ ) (Ċ 2 + 2 ⋯ ) = Ċ + Ċ 2 + 2
2⋯
1! 2!
3! 5!
3 30
Since the power series for Ď ý and ąÿĀ Ċ converge for |Ċ| < ∞, the product series converges on the same interval.
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January 2022 Denis CK Wong
Power Series Solutions. Suppose the linear second-order differential equation
is put into standard form
Ċ2 (Ċ)ċ99 + Ċ1 (Ċ)ċ9 + Ċ0 (Ċ) ċ = 0 (*)
ċ99 + ÿ(Ċ)ċ9 + Ā(Ċ)ċ = 0 (**)
by dividing by the leading coefficient Ċ2 (Ċ).
Definition. A point Ċ0 is said to be an ordinary point of (*) if both ÿ(Ċ) and Ā(Ċ) in (**) are analytic at Ċ0 . A point
that is not an ordinary point is said to be a singular point of the equation.
Theorem. Existence of power series solutions. If Ċ = Ċ0 is an ordinary point of the (*), we can always find two
∞
linearly independent solutions in the form of power series centered at Ċ0 , that is, ċ = ∑Ą=0
ČĄ (Ċ 2 Ċ0 )Ą . A series
solution converges at least on some interval defined by |Ċ 2 Ċ0 | < ā, where ā is the distance from Ċ0 to the closest
singular point.
∞
A solution of the form ċ = ∑Ą=0
ČĄ (Ċ 2 Ċ0 )Ą is said to be a solution about the ordinary point Ċ0 . The distance ā is
the minimum value or the lower bound for the radius of convergence of series solutions of the DE about Ċ0 .
Example. The complex number 1 ± 2ÿ are singular points of the differential equation (Ċ 2 2 2Ċ + 5)ċ99 + Ċċ9 3 ċ =
0.
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January 2022 Denis CK Wong
Finding a power series solution (with polynomial coefficients).
Example. Solve ċ99 + Ċċ = 0
Example. Solve (Ċ^2 + 1)ċ99 + Ċċ9 2 ċ = 0
Example. Solve ċ99 2 (1 + Ċ)ċ = 0
With nonpolynomial coefficients.
Example. Solve ċ99 + (Čāą Ċ) ċ = 0
Solutions about singular points. A singular point Ċ0 of a linear differential equation
Ċ2 (Ċ)ċ99 + Ċ1 (Ċ)ċ9 + Ċ0 (Ċ)ċ = 0 (1)
is further classified as either regular or irregular. The classification again depends on the function ÿ and Ā in the
standard form
ċ99 + ÿ(Ċ)ċ9 + Ā(Ċ)ċ = 0. (2)
Definition. A singular point Ċ0 is said to be a regular singular point of (1) if Ă(Ċ) = (Ċ 3 Ċ0 )ÿ(Ċ) and g(Ċ) =
(Ċ 2 Ċ0 )2 Ā(Ċ) are both analytic at Ċ0 . A singular point that is not regular is said to be an irregular singular point of
the equation.
Example. Find and test the singular point(s) of (Ċ 2 2 4)2 ċ99 + 3(Ċ 2 2)ċ9 + 5ċ = 0.
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January 2022 Denis CK Wong
Theorem. Frobenius’ Theorem. If Ċ = Ċ0 is a regular singular point of (1), then there exists at least one solution of
the form
∞
∞
ċ = (Ċ 2 Ċ0 )Ĉ ∑Ą=0
ČĄ (Ċ 2 Ċ0 )Ą = ∑Ą=0
ČĄ (Ċ 2 Ċ0 )Ą+Ĉ (4)
where the number Ą is a constant to be determined. The series will converge at least on some interval 0 < Ċ 3 Ċ0 <
ā.
Example. Use the Frobenius’ Theorem to find a solution for 3Ċċ99 + ċ9 2 ċ = 0.
Example. Solve 2Ċċ99 + (1 + Ċ)ċ9 + ċ = 0.
Example. Solve Ċċ99 + ċ = 0.
Three Cases. For the sake of discussion let us again suppose that Ċ = 0 is a regular singular point of equation (1) and
that the indicial roots Ą1 and Ą2 of the singularity are real. When using the method of Frobenius, we distinguish three
cases corresponding to the nature of the indicial roots Ą1 and Ą2 . In the first two cases, Ą1 denotes the largest of two
distinct roots, that is, Ą1 > Ą2 . In the last case, Ą1 = Ą2 .
Case 1: If Ą1 and Ą2 are distinct and Ą1 3 Ą2 is not a positive integer, then there exist two linearly in dependent
solutions of (1) of the form
∞
ċ1 (Ċ) = ∑ ČĄ Ċ Ą+Ĉ1 , Č0 b 0
Ą=0
∞
ċ2 (Ċ) = ∑ ċĄ Ċ Ą+Ĉ2 , ċ0 b 0
Ą=0
Case 2: If Ą1 and Ą2 are distinct and Ą1 3 Ą2 is a positive integer, then there exist two linearly independent solutions
of (1) of the form
∞
ċ1 (Ċ) = ∑ ČĄ Ċ Ą+Ĉ1 , Č0 b 0
Ą=0
∞
ċ2 (Ċ) = ÿċ1 (Ċ) ln Ċ + ∑ ċĄ Ċ Ą+Ĉ2 , ċ0 b 0
Ą=0
where ÿ is a constant that could be zero.
Case 3: If Ą1 and Ą2 are equal, then there always exist two linearly independent solutions of (1) of the form
∞
ċ1 (Ċ) = ∑ ČĄ Ċ Ą+Ĉ1 , Č0 b 0
Ą=0
∞
ċ2 (Ċ) = ċ1 (Ċ) ln Ċ + ∑ ċĄ Ċ Ą+Ĉ1 , ċ0 b 0
Ą=1
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January 2022 Denis CK Wong
Finding a Second Solution. When the difference Ą1 3 Ą2 is a positive integer (Case 2), we may or may not be able to
∞
find two solutions having the form ċ(Ċ) = ∑Ą=0
ČĄ Ċ Ą+Ĉ . This is something that we do not know in advance but is
determined after we have found the indicial roots and have carefully examined the recurrence relation that defined the
coefficients ČĄ . We just might be lucky enough to find two solutions that involve only powers of Ċ, that is,
Ą+Ĉ1
Ą+Ĉ2
ċ1 (Ċ) = ∑∞
, and ċ2 (Ċ) = ∑∞
.
Ą=0 ČĄ Ċ
Ą=0 ċĄ Ċ
On the other hand, we see that the difference of the indicial roots is a positive integer and the method of Frobenius
fail to give a second series solution.
One way to obtain the second solution with the logarithm term is to use the fact that
ċ2 (Ċ) = ċ1 (Ċ) ∫
Ď ∫ ÿ(ý)Ăý
čĊ
ċ12 (Ċ)
is also a solution of ċ99 + ÿ(Ċ)ċ9 + Ā(Ċ) ċ = 0 whenever ċ1 (Ċ) is a known solution.
Example. Find the general solution of Ċċ99 + ċ = 0.
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January 2022 Denis CK Wong
Topic 4: Laplace Transform
Ā
Integral Transform. A definite integral such as ∫ÿ ÿ(ą, Ć) ď(Ć) čĆ transforms a function ď of the variable Ć into a
function Ă of the variable ą. We are particularly interested in an integral transform, where the interval of integration
∞
is the unbounded interval [0, ∞). If f(t) is defined for Ć g 0, then the improper integral ∫0 ÿ(ą, Ć) ď(Ć) čĆ is defined
as a limit:
∞
Ā
∫0 ÿ(ą, Ć) ď(Ć) čĆ = lim ∫0 ÿ(ą, Ć) ď(Ć) čĆ. (1)
Ā→∞
If the limit in (1) exists, then we say that the integral exists or is convergent; if the limit does not exist, the integral
does not exist and is divergent. The function ÿ(ą, Ć) in (1) is called the kernel of the transform. When ÿ(ą, Ć) = Ď 2ĉĊ
as the kernel gives an especially important integral transform.
Definition. Let ď be a function defined for Ć g 0. Then the integral
∞
ℒ{ď(Ć)} = ∫0 Ď 2ĉĊ ď(Ć) čĆ (2)
is the Laplace transform of ď, provided that the integral converges.
Example. Evaluate ℒ{1}.
Example. Evaluate ℒ{Ć}.
Example. Evaluate ℒ{Ď 23Ċ }.
Example. Evaluate ℒ{sin 2Ć}.
Remark. ℒ is a linear transform.
Example. Evaluate ℒ{1 + 5Ć}.
Example. Evaluate ℒ{4Ď 23Ċ 2 10 sin 2Ć}.
Theorem. Transforms of some basic functions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
1
ℒ{1} = .
ĉ
Ą!
ℒ{Ć Ą } = Ā+1 , Ā = 1,2,3, &.
ĉ
ℒ{Ď ÿĊ } =
1
ĉ2ÿ
.
ℒ{sin ýĆ} = 2
ā
ĉ +ā 2
ĉ
ℒ{cos ýĆ} = 2
.
ĉ +ā 2
ā
ℒ{sinh ýĆ} = 2
.
ĉ 2ā 2
ĉ
ℒ{cosh ýĆ} = 2
.
ĉ 2ā 2
.
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January 2022 Denis CK Wong
Sufficient conditions for existence of �㗛{�㖇(þ)}.
Definition. A function ď is said to be of exponential order Č if there exist constants Č, ā > 0, and ă > 0 such that
|ď(Ć)| f āĎ āĊ for all Ć > ă.
If ď is increasing function, then the condition |ď(Ć)| f āĎ āĊ , Ć > ă, state that the graph of f on the interval (ă, ∞)
does not grow faster than the graph of the exponential function āĎ āĊ , where c is a positive constant.
Theorem. Sufficient Conditions for Existence. If ď is piecewise continuous on [0, ∞) and of exponential order Č,
then ℒ{ď(Ć)} exists for ą > Č.
Example. Evaluate ℒ{ď(Ć)} where
0,
ď(Ċ) = {
2,
0fĆ<3
Ċg3
Theorem. Behavior of F(s) as ý → ∞. If ď is piecewise continuous on (0, ∞) and of exponential order and Ă(ą) =
ℒ{ď(Ć)}, then lim Ă(ą) = 0
ĉ→∞
Inverse Transforms.
The Inverse Problem. If Ă(ą) represents the Laplace transform of a function ď(Ć), that is, Ă(ą) = ℒ{ď(Ć)}, we then
say that ď(Ć) is the inverse Laplace transform of Ă(ą) and we write ď(Ć) = ℒ 21 {Ă(ą)}.
Theorem. Some Inverse transforms
(a)
(b)
(c)
(d)
(e)
(f)
(g)
1
1 = ℒ 21 { }.
ĉ
Ą!
Ć Ą = ℒ 21 { Ā+1} , Ā = 1,2,3, &.
ĉ
Ď ćĊ = ℒ 21 {
1
}.
ĉ2ÿ
ā
sin ýĆ = ℒ 21 { 2 2 }.
ĉ +ā
ĉ
21
cos ýĆ = ℒ
sinh ýĆ = ℒ
cosh ýĆ = ℒ
{2
ĉ +ā 2
ā
21
21
{2
}.
ĉ 2ā 2
ĉ
{2
ĉ 2ā 2
}.
}.
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January 2022 Denis CK Wong
Example. Evaluate
1
(a) ℒ 21 { 5}
ĉ
1
(b) ℒ 21 { 2 }
ĉ +7
Remark. ℒ 21 is a linear transform.
22ĉ+6
Example. Evaluate ℒ 21 { 2
Example. Evaluate ℒ 21 {
ĉ +4
}.
ĉ 2 +6ĉ+9
}.
(ĉ21)(ĉ22)(ĉ+4)
Transform a Derivative. Our immediate goal is to use the Laplace transform to solve differential equations. To that
Ăþ
Ă2þ
end, we need to evaluate quantities such as ℒ { } and ℒ {
Ăý
Ăý 2
}.
Theorem. If ď, ď9, & , ď (Ą21) are continuous on [0, ∞) and are of exponential order and if ď (Ą) (Ć) is piecewise
continuous on [0, ∞), then ℒ{ď (Ą) (Ć)} = ą Ą Ă(ą) 2 ą Ą21 ď(0) 2 ą Ą22 ď ′ (0) 2 ⋯ 2 ď (Ą21) (0), where Ă(ą) = ℒ{ď(Ć)}.
Solving Linear ODEs.
Example. Use the Laplace transform to solve the initial-value problem
Example. Solve ċ ′′ 2 3ċ ′ + 2ċ = Ď 24Ċ , ċ(0) = 1, ċ ′ (0) = 5.
60
Ăþ
ĂĊ
+ 3ċ = 13 sin 2Ć , ċ(0) = 6.
January 2022 Denis CK Wong
Translation on the ý 2axis.
Theorem. First Translation Theorem. If ℒ{ď(Ć)} = Ă(ą) and a is any real number, then ℒ{Ď ÿĊ ď(Ć)} = Ă(ą 2 Ċ).
Example. Evaluate (a) ℒ{Ď 5Ċ Ć 3 } (b) ℒ{Ď 22Ċ cos 4Ć}
Remark. ℒ 21 {Ă(ą 2 Ċ)} = Ď ÿĊ ď(Ć), where ď(Ć) = ℒ 21 {Ă(ą)}.
2ĉ+5
ĉ/2+5/3
Example. (a) ℒ 21 {(ĉ23)2 } (b) ℒ 21 { 2
ĉ +4ĉ+6
}
Example. Solve ċ ′′ 2 6ċ ′ + 9ċ = Ć 2 Ď 3Ċ , ċ(0) = 2, ċ ′ (0) = 17.
Example. Solve ċ ′′ + 4ċ ′ + 6ċ = 1 + Ď 2Ċ , ċ(0) = 0, ċ ′ (0) = 0.
Translation on the þ 2axis.
Theorem. Unit Step Function. The unit step function �㖰(Ć 2 Ċ) is defined to be
0,
�㖰(Ć 2 Ċ) = {
1,
0fĆ<Ċ
ĆgĊ
For a general piecewise-defined function of the type
Đ(Ć),
ď(Ć) = {
/(Ć),
0fĆ<Ċ
ĆgĊ
0,
Đ(Ć),
ď(Ć) = {
0,
0fĆ<Ċ
ĊfĆ<ċ
Ćgċ
is the same as ď(Ć) = Đ(Ć) 2 Đ(Ć)�㖰(Ć 2 Ċ) + /(Ć)�㖰(Ć 2 Ċ).
Similarly, a function of the type
can be written ď(Ć) = Đ(Ć)[�㖰(Ć 2 Ċ) 2 �㖰(Ć 2 ċ)].
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January 2022 Denis CK Wong
Example. Express
ď(Ć) = {
20Ć, 0 f Ć < 5
in terms of unit step functions. Graph.
0, Ć g 5
Theorem. Second Translation Theorem. If ℒ{ď(Ć)} = Ă(ą) and a>0, then ℒ{ď(Ć 2 Ċ)�㖰(Ć 2 Ċ)} = Ď 2ÿĉ Ă(ą).
Inverse form. If ď(Ć) = ℒ 21 {Ă(ą)} and a>0, then ℒ 21 {Ď 2ÿĉ Ă(ą)} = ď(Ć 2 Ċ)�㖰(Ć 2 Ċ).
Example. Evaluate (a) ℒ 21 {
1
ĉ24
ĉ
Ď 22ĉ } (b) ℒ 21 { 2
ĉ +9
�㔋ý
Ď2 2 }
Alternative Form. ℒ{Đ(Ć)�㖰(Ć 2 Ċ)} = Ď 2ÿĉ ℒ{Đ(Ć + Ċ)}.
Example. Evaluate ℒ{cos Ć �㖰(Ć 2 �㔋)}.
Derivatives of a Transform.
Multiplying a function by þ�㖏 .
Theorem. Derivatives of Transforms
If ℒ{ď(Ć)} = Ă(ą) and Ā = 1,2,3, &, then ℒ{Ć Ą ď(Ć)} = (21)Ą
Example. Evaluate ℒ{Ć sin ýĆ }.
Example. Solve Ċ ′′ + 16Ċ = cos 4Ć , Ċ(0) = 0, Ċ ′ (0) = 1.
62
ĂĀ
Ăĉ Ā
Ă(ą).
January 2022 Denis CK Wong
Transforms of Integrals.
Convolution. If ď and Đ are piecewise continuous on [0, ∞), then the convolution of ď and Đ is defined by the integral
Ċ
ď ∗ Đ = ∫0 ď(�㔏)Đ(Ć 2 �㔏)č�㔏.
Remark. ď ∗ Đ = Đ ∗ ď.
Theorem. Convolution Theorem. If ď and Đ are piecewise continuous on [0, ∞) and of exponential order, then
ℒ{ď ∗ Đ} = ℒ{ď(Ć)}ℒ{Đ(Ć)} = Ă(ą)ă(ą).
Ċ
Example. Evaluate ℒ{∫0 Ď �㔏 sin(Ć 2 �㔏) č�㔏 }.
Inverse form. ℒ 21 {Ă(ą)ă(ą)} = ď ∗ Đ.
Example. Evaluate ℒ 21 {(ĉ2
1
+ā 2 )2
}.
1
Transform of an Integral. When Đ(Ć) = 1 and ℒ{Đ(Ć)} = ă(ą) = , the convolutional theorem implies that the
Laplace transform of the integral of ď is
Ċ
�㔹(ĉ)
The inverse form is ∫0 ď(�㔏)č�㔏 = ℒ 21 {
ĉ
Ċ
�㔹(ĉ)
.
ℒ {∫0 ď(�㔏)č�㔏 } =
ĉ
ĉ
}.
Ċ
Example. Volterra Integral Equation. Solve ď(Ć) = 3Ć 2 2 Ď 2Ċ 2 ∫0 ď(�㔏)Ď Ċ2�㔏 č�㔏 for ď(Ć).
If a periodic function has period ă, ă > 0, then ď(Ć + ă) = ď(Ć).
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January 2022 Denis CK Wong
Theorem. Transform of a Periodic Function. If ď(Ć) is piecewise continuous on [0, ∞) and of exponential order,
and periodic with period ă, then ℒ{ď(Ć) } =
ă
1
∫ Ď 2ĉĊ ď(�㔏)čĆ.
12ă 2ý�㕇 0
Example. Find the Laplace transform of the periodic function shown in figure below.
Example. The differential equation for the current ÿ(Ć) in a single-loop LR series circuit is Ā
Ăÿ
ĂĊ
+ āÿ = ā(Ć).
Determine the current ÿ(Ć) when ÿ(0) = 0 and ā(Ć) is the square wave function shown in pervious example.
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January 2022 Denis CK Wong
System of Linear Differential Equations. Two masses ÿ1 and ÿ2 are connected to two springs A and B of negligible
mass having spring constants ý1 and ý2 , respectively. In turn the two springs are attached as shown below.
Let Ċ1 (Ć) and Ċ2 (Ć) denote the vertical displacements of the masses from their equilibrium positions. When the system
is in motion, spring B is subject to both an elongation and a compression; hence its net elongation is Ċ2 3 Ċ1 . Therefore,
it follows from Hooke’s law that springs A and B exert forces 2ý1 Ċ1 and ý2 (Ċ2 2 Ċ1 ), respectively, on ÿ1 . If no
external force is impressed on the system and if no damping force is present, then the net force on ÿ1 is 2ý1 Ċ1 +
ý2 (Ċ2 2 Ċ1 ). By Newton’s second law we can write ÿ1
Ă 2 ý1
ĂĊ 2
= 2ý1 Ċ1 + ý2 (Ċ2 2 Ċ1 ). Similarly, the net force exerted
on mass ÿ2 is due solely to the net elongation of B; that is, 2ý2 (Ċ2 2 Ċ1 ). Hence, we have ÿ2
Ă 2 ý2
ĂĊ 2
= 2ý2 (Ċ2 2 Ċ1 ).
In other words, the motion of the coupled system is represented by the system of simultaneous second-order
differential equations
ÿ1 Ċ1′′ = 2ý1 Ċ1 + ý2 (Ċ2 2 Ċ1 )
Example. Solve
ÿ2 Ċ2′′ = 2ý2 (Ċ2 2 Ċ1 )
Ċ1′′ + 10Ċ1 2 4Ċ2 = 0
24Ċ1 + Ċ2′′ + 4Ċ2 = 0
subject to Ċ1 (0) = 0, Ċ1′ (0) = 1, Ċ2 (0) = 0, Ċ2′ (0) = 21.
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January 2022 Denis CK Wong
Networks. The currents ÿ1 (Ć) and ÿ2 (Ć) in the network, containing an inductor, a resistor, and a capacitor, were
governed by the system of first-order differential equations
Ā
čÿ1
+ āÿ2 = ā(Ć)
čĆ
āÿ
čÿ2
+ ÿ2 2 ÿ1 = 0
čĆ
Example. Solve the system under the conditions ā(Ć) = 60 ą, Ā = 1 /, ā = 50 Ω, ÿ = 1024 ď, and the currents ÿ1 and
ÿ2 are initially zero.
The End
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January 2022 Denis CK Wong
Topic 5: Systems of First Order ODEs
Normal form of a first-order system of linear equations:
čĊ1
= Ċ11 (Ć)Ċ1 + Ċ12 (Ć)Ċ2 + ⋯ + Ċ1Ą (Ć)ĊĄ + ď1 (Ć)
čĆ
čĊ2
= Ċ21 (Ć)Ċ1 + Ċ22 (Ć)Ċ2 + ⋯ + Ċ2Ą (Ć)ĊĄ + ď2 (Ć)
čĆ
…..
čĊĄ
= ĊĄ1 (Ć)Ċ1 + ĊĄ2 (Ć)Ċ2 + ⋯ + ĊĄĄ (Ć)ĊĄ + ďĄ (Ć)
čĆ
We refer to this system as linear system. We assume that the coefficients ĊÿĀ as well as the functions ďÿ are continuous
on a common interval �㔼. When ďÿ (Ć) = 0, ÿ = 1,2, & , Ā , the linear system is said to be homogeneous; otherwise, it is
nonhomogeneous. If ć, ý(Ć) and Ă(Ć) denote the respective matrices
Ċ11 (Ć)
Ċ1 (Ć)
Ċ21 (Ć)
Ċ2 (Ć)
.
.
ć=
, ý(Ć) =
.
.
.
.
(ĊĄ (Ć))
(ĊĄ1 (Ć)
Ċ12 (Ć)
Ċ22 (Ć)
.
.
.
ĊĄ2 (Ć)
&
&
&
Ċ1Ą (Ć)
ď1 (Ć)
Ċ2Ą (Ć)
ď2 (Ć)
.
.
, Ă(Ć) =
,
.
.
.
.
(ďĄ (Ć))
ĊĄĄ (Ć))
then the system of linear first-order differential equations can be written as
If the system is homogeneous, its matrix form is
ć ′ = ýć + Ă. (*)
ć ′ = ýć. (**)
Definition. A solution vector on �㔼 is any column matrix
Ċ1 (Ć)
Ċ2 (Ć)
.
ć=
.
.
(ĊĄ (Ć))
whose entries are differentiable functions satisfying the system (*) on the interval.
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January 2022 Denis CK Wong
Example.
(a) Find the matrix form of the homogeneous system
čĊ
= 3Ċ + 4ċ
čĆ
čċ
= 5Ċ 2 7ċ
čĆ
22Ċ
6Ċ
1
3
(b) Verify that on the interval (2∞, ∞), ć1 = ( ) Ď 22Ċ = ( Ď 22Ċ ) and ć1 = ( ) Ď 6Ċ = (3Ď 6Ċ ) are solutions
21
5
2Ď
5Ď
1 3
′
of ć = (
) ć.
3 5
ā1
Ċ1 (Ć0 )
ā2
Ċ2 (Ć0 )
.
.
Initial-Value Problem. Let Ć0 denote a point on an interval �㔼 and ć(Ć0 ) =
and ć0 = . , where the
.
.
.
(
ā
(ĊĄ (Ć0 ))
Ą)
āÿ , ÿ = 1,2, & , Ā are given constants. Then the problem
Solve: ć ′ = ý(Ć)ć + Ă(Ć)
Subject to: ć(Ć0 ) = ć0
is an initial-value problem on �㔼.
Theorem. Existence of a Unique Solution. Let the entries of the matrices ý(Ć) and Ă(Ć) be functions continuous on
a common interval �㔼 that contains the point Ć0 . Then there exists a unique solution of the initial-value problem on the
interval.
Theorem. Superposition Principle. Let ć1 , ć2 , & , ćā be a set of solution vectors of the homogeneous system ć9 =
ýć on an interval �㔼. Then the linear combination ć = Č1 ć1 + Č2 ć2 + ⋯ + Čā ćā , where the Čÿ , ÿ = 1,2 & , ý are
arbitrary constants, is also a solution on the interval.
Definition. Let ć1 , ć2 , & , ćā be a set of solution vectors of the homogeneous system ć9 = ýć on an interval �㔼. We
say that the set is linearly dependent on the interval if there exist constants Č1 , Č2 , & , Čā , not all zero, such that
Č1 ć1 + Č2 ć2 + ⋯ + Čā ćā = 0 for every Ć in the interval. Otherwise, it is said to be linearly independent.
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January 2022 Denis CK Wong
Theorem. Criterion for linearly independent solutions.
Ċ11
Ċ12
Ċ1Ą
Ċ21
Ċ22
Ċ2Ą
.
.
.
Let ć1 =
, ć2 =
, & , ćĄ =
be Ā solution vectors of the homogeneous system ć9 = ýć on an
.
.
.
.
.
.
(ĊĄ1 )
(ĊĄ2 )
(ĊĄĄ )
interval �㔼. Then the set of solution vectors is linearly independent on �㔼 iff the Wronskian
Ċ11
Ċ21
| .
Ć(ć1 , ć2 , & , ćĄ ) = .
|
.
ĊĄ1
Ċ12
Ċ22
.
.
.
ĊĄ2
&
&
.
.
.
&
Ċ1Ą
Ċ2Ą
. |
. | b 0 for every Ć in the interval.
.
ĊĄĄ
Definition. Any set ć1 , ć2 , & , ćĄ of Ā linearly independent solution vectors of the homogeneous system ć9 = ýć on
an interval �㔼 is said to be a fundamental set of solutions on the interval.
Theorem. Existence of a Fundamental Set. There exists a fundamental set of solutions for the homogeneous system
ć9 = ýć on the interval �㔼.
Theorem. General solution 3 Homogeneous Systems. Let ć1 , ć2 , & , ćĄ be a fundamental set of solutions of the
homogeneous system ć9 = ýć on �㔼. Then the general solution of the system on �㔼 is ć = Č1 ć1 + Č2 ć2 + ⋯ +
ČĄ ćĄ , where the Čÿ , ÿ = 1,2 & , Ā are arbitrary constants.
Theorem. General Solution 3 Nonhomogeneous Systems. Let ćĆ be given a solution of the nonhomogeneous
system ć9 = ýć + Ă on an interval �㔼 and let ćā = Č1 ć1 + Č2 ć2 + ⋯ + ČĄ ćĄ denote the general solution on the
same interval of the associated homogeneous system ć9 = ýć. Then the general solution of the nonhomogeneous
system on the interval is ć = ćā + ćĆ . The general solution ćā of the associated homogeneous system ć9 = ýć is
called the complementary function of the nonhomogeneous system ć9 = ýć + Ă.
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January 2022 Denis CK Wong
Homogeneous Linear Systems. Can we find a solution of the form
ý1
ý2
.
ć=
Ď ÿĊ = ÿĎ ÿĊ (1)
.
.
( ýĄ )
for the general homogeneous linear first-order system
where ý is an Ā × Ā matrix of constants?
ć9 = ýć, (2)
If (1) is to be a solution vector of (2), then ć ′ = ÿÿĎ ÿĊ , so the system becomes ÿÿĎ ÿĊ = ýÿĎ ÿĊ . Upon simplify,
(ý 2 ÿ�㔼)ÿ = 0. (3)
To find a nontrivial solution ć of (2), we must find a nontrivial vector K that satisfies (3). For (3) to have solutions,
we must have čĎĆ(ý 2 ÿ �㔼) = 0. This polynomial equation in ÿ is called the characteristic equation of ý; its solution
are the eigenvalues of ý. A solution ÿ b 0 of (3) corresponding to an eigenvalue ÿ is called an eigenvector of ý. A
solution of the homogeneous system (2) is then ć = ÿĎ ÿĊ
Distinct Real Eigenvalues.
Theorem. General Solution 3 Homogeneous systems. Let ÿ1 , ÿ2 , & , ÿĄ be n distinct eigenvalues of the coefficient
matrix ý of the homogeneous system (2) and let ÿ1 , ÿ2 , & , ÿĄ be the corresponding eigenvectors. Then the general
solution of (2) on the interval (2∞, ∞) is given by ć = Č1 ÿ1 Ď ÿ1Ċ + Č2 ÿ2 Ď ÿ2Ċ + ⋯ + ČĄ ÿĄ Ď ÿĀĊ .
Example. Solve
Ăý
ĂĊ
Ăþ
ĂĊ
= 2Ċ + 3ċ
= 2Ċ + ċ
Example. Solve
Ăý
ĂĊ
Ăþ
ĂĊ
Ăÿ
ĂĊ
= 24Ċ + ċ + Č
= Ċ + 5ċ 2 Č
= ċ 2 5Č
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January 2022 Denis CK Wong
Repeated Eigenvalues.
If ÿ is positive integer and (ÿ 2 ÿ1 )ă is a factor of the characteristic equation while (ÿ 2 ÿ1 )ă+1 is not a factor, then
ÿ1 is an eigenvalue of multiplicity ÿ.
For some Ā × Ā matrices ý it may be possible to find m linearly independent eigenvectors ÿ1 , ÿ2 , & , ÿă
corresponding to an eigenvalue ÿ1 of multiplicity ÿ f Ā. The general solution of the system contains
the linear combination Č1 ÿ1 Ď ÿ1Ċ + Č2 ÿ2 Ď ÿ1Ċ + ⋯ + Čă ÿă Ď ÿ1Ċ .
If there is only one eigenvector corresponding to the eigenvalue ÿ1 of multiplicity ÿ, then ÿ linearly
independent solutions of the form
(i)
(ii)
ć1 = ÿ11 Ď ÿ1Ċ
ć2 = ÿ21 ĆĎ ÿ1Ċ + ÿ22 Ď ÿ1Ċ
…
ćă = ÿă1
Ć ă21
Ć ă22
Ď ÿ1Ċ + ÿă2
Ď ÿ1Ċ + ⋯ + ÿăă Ď ÿ1Ċ
(ÿ 2 1)!
(ÿ 2 2)!
where ÿÿĀ are column vectors, can always be found.
1
Example. Solve ć ′ = (22
2
22
1
22
2
22) ć.
1
Suppose ÿ1 is an eigenvalue of multiplicity two and that there is only one eigenvector associated with this value. A
second solution is
Ă1
ý1
Ă2
ý2
where ÿ = ( ) and ÿ = ( & ).
&
ĂĄ
ýĄ
ć2 = ÿĆĎ ÿ1Ċ + ÿĎ ÿ1Ċ , (#)
Example. Find the general solution of the system given by ć ′ = (
3
2
218
) ć.
29
When the coefficient matrix ý has only one eigenvector associated with an eigenvalue ÿ1 of multiplicity three, we can
find a second solution of the form of (#) and a third solution of the form
Ă1
ă1
ý1
Ă
ă2
ý
2
ć3 = ÿ Ď ÿ1Ċ + ÿĆĎ ÿ1Ċ + ĀĎ ÿ1Ċ , where ÿ = ( 2 ) , ÿ = ( & ) and Ā = ( & ).
2
&
ĂĄ
ăĄ
ýĄ
Ċ2
2
Example. Solve ć ′ = (0
0
1
2
0
6
5) ć.
2
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January 2022 Denis CK Wong
Complex Eigenvalues.
Theorem. Let ý be the coefficient matrix having real entries of the homogeneous system (2), and let ÿ1 be an
̅̅̅̅
eigenvector corresponding to the complex eigenvalue ÿ1 = ÿ + �㔼 Ā, ÿ, Ā real. Then ÿ1 Ď ÿ1Ċ and ̅̅̅
ÿ1 Ď ÿ1Ċ are solutions
of (2).
Theorem. Real solutions corresponding to a complex eigenvalue.
Let ÿ1 = ÿ + ÿ Ā be a complex eigenvalue of the coefficient matrix A in the homogeneous system (2) and let þ1 and
1
ÿ
þ2 denote the column vectors defined by þ1 = (ÿ1 + ̅̅̅
ÿ1 ) and þ2 = (2ÿ1 + ̅̅̅
ÿ1 ). Then
2
2
ć1 = [þ1 cos ĀĆ 2 þ2 sin ĀĆ]Ď ÿĊ
ć2 = [þ2 cos ĀĆ + þ1 sin ĀĆ]Ď ÿĊ
are linearly independent solutions of (2) on (2∞, ∞).
Example. Solve the initial-value problem ć ′ = (
2
21
8
2
) ć, ć(0) = ( ).
22
21
Nonhomogeneous Linear Systems.
Undetermined Coefficients
21
Example. Solve the system ć ′ = (
21
6
Example. Solve the system ć ′ = (
4
28
2
) ć + ( ) on (2∞, ∞).
3
1
6Ć
1
)ć + (
) on (2∞, ∞).
210Ć + 4
3
Example. Determine the form of a particular solution vector ćĆ of the system
čĊ
= 5Ċ + 3ċ 2 2Ď 2Ċ + 1
čĆ
čċ
= 2Ċ + ċ + Ď 2Ċ 2 5Ć + 7
čĆ
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January 2022 Denis CK Wong
Variation of Parameters. If ć1 , ć2 , & , ćĄ is a fundamental set of solutions of the homogeneous system ć9 = ýć on
an interval �㔼, then its general solution on the interval is the linear combination ć = Č1 ć1 + Č2 ć2 + ⋯ + ČĄ ćĄ or
Ċ11
Ċ12
Ċ12
Č1 Ċ11 + Č2 Ċ12 + ⋯ + ČĄ Ċ1Ą
Ċ21
Ċ22
Ċ22
Č1 Ċ21 + Č2 Ċ22 + ⋯ + ČĄ Ċ2Ą
ć = Č1 ( & ) + Č2 ( & ) + ⋯ + ČĄ ( & ) = (
). (1)
&
ĊĄ1
ĊĄ2
ĊĄ2
Č1 ĊĄ1 + Č2 ĊĄ2 + ⋯ + ČĄ ĊĄĄ
The last matrix in (1) is recognized as the product of an Ā × Ā matrix with an Ā × 1 matrix. The general solution (1)
can be written as the product
ć = Φ(Ć)ÿ, (2)
where ÿ is a n Ā × 1 column vector of arbitrary constants Č1 , Č2 , & , ČĄ and the Ā × Ā matrix, whose columns consist
Ċ11 Ċ12 & Ċ1Ą
Ċ21 Ċ22 & Ċ2Ą
.
.
.
of the entries of the solution vectors of the system ć9 = ýć, Φ(Ć) =
is called a fundamental
.
.
.
.
.
.
(ĊĄ1 ĊĄ2 & ĊĄĄ )
matrix of the system on the interval.
Two properties of a fundamental matrix:
1. Φ(Ć) is nonsingular.
2. If Φ(Ć) is a fundamental matrix of the system ć9 = ýć, then
Φ′(Ć) = ýΦ(Ć). (3)
We ask whether it is possible to replace the matrix of constant ÿ in (2) by a column matrix of functions
ć1 (Ć)
ć2 (Ć)
Ą(Ć) = (
) so ćĆ = Φ(Ć)Ą(Ć)
&
ćĄ (Ć)
(4)
is a particular solution of the nonhomogeneous system
ć9 = ýć + Ă(Ć).
(5)
It can be show that a particular solution of (5) is ćĆ = Φ(Ć) ∫ Φ21 (Ć)Ă(Ć)čĆ.
Thus, the general solution of the system (5) is ć = ćā + ćĆ = Φ(Ć)ÿ + Φ(Ć) ∫ Φ21 (Ć)Ă(Ć)čĆ.
Example. Solve the system ć ′ = (
23
2
1
3Ć
) ć + ( 2Ċ ) on (2∞, ∞).
Ď
24
The End
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January 2022 Denis CK Wong
Tutorial 1
1. State the order of the given ordinary differential equation. Determine whether the equation is linear
or nonlinear.
(a) (1 2 Ċ)ċ99 2 4Ċċ9 + 5ċ = cos Ċ
Ă3 þ
Ăþ 4
2
(
) +ċ =0
Ăý 3
Ăý
5 (4)
3 ′′
(b) Ċ
(c) Ć ċ
2 Ć ċ + 6ċ = 0
Ă2 ċ
Ăċ
(d) ĂĈ2 + ĂĈ + ć = cos(Ą + ć)
Ă2 þ
Ăþ 2
(e) Ăý 2 = √1 + (Ăý )
(f)
(g)
ā
Ă2 ā
=2 2
ā
ĂĊ 2
′′′
(sin
�㔃)ċ
2 (cos �㔃)ċ ′ = 2
2. Verify that the indicated family of function is a solution of the given differential equation. Assume an
appropriate interval �㔼 of definition for each solution.
Ăÿ
ā ăþ
(a) ĂĊ = ÿ(1 2 ÿ); ÿ = 1+ā1 ă þ
1
2 ý
2
2
Ăþ
(b) Ăý + 2Ċċ = 1; ċ = Ď 2ý ∫0 Ď Ċ čĆ + Č1 Ď 2ý
Ă2 þ
Ăþ
(c) Ăý 2 2 4 Ăý + 4ċ = 0; ċ = Č1 Ď 2ý + Č2 ĊĎ 2ý
Ă3 þ
Ă2 þ
Ăþ
(d) Ċ 3 3 + 2Ċ 2 2 2 Ċ + ċ = 12Ċ 2 ; ċ = Č1 Ċ 21 + Č2 Ċ + Č3 Ċ þĀ Ċ + 4Ċ 2
Ăý
Ăý
Ăý
3. Verify that the piecewise-defined function is a solution of the differential equation Ċċ9 2 2ċ = 0 on
(2∞, ∞).
ċ={
2Ċ 2 ,
Ċ2,
Ċ<0
Ċg0
4. Find values of ÿ so that the function ċ = Ċ ă is a solution of the differential equation Ċ 2 ċ ′′ 2 7Ċċ ′ +
15ċ = 0
5. Use the concept that ċ = Č , 2∞ < Ċ < ∞, is a constant function if and only if ċ9 = 0 to determine
whether the differential equation 3Ċċ9 + 5ċ = 10 possesses constant solutions.
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January 2022 Denis CK Wong
6. Find a linear second-order differential equation Ă(Ċ, ċ, ċ9, ċ99) = 0 for which ċ = Č1 Ċ + Č2 Ċ 2 is a
two-parameter family of solutions. Make sure that your equation is free of the arbitrary parameters Č1
and Č2 .
7. Ċ = Č1 cos Ć + Č2 sin Ć is a two-parameter family solutions of the second-order differential equation
Ċ99 + Ċ = 0. Find a solution of the second-order IVP consisting of this differential equation and the given
initial conditions.
(a) Ċ(0) = 21, Ċ9(0) = 8
�㔋
2
�㔋
2
(b) Ċ ( ) = 0, Ċ9 ( ) = 1
8. ċ = Č1 Ď ý + Č2 Ď 2ý is a two-parameter family solutions of the second-order differential equation
ċ99 2 ċ = 0. Find a solution of the second-order IVP consisting of this differential equation and the
given initial conditions.
(a) ċ(0) = 1, ċ9(0) = 2
(b) ċ(1) = 0, ċ9(1) = Ď
9.
(a) Verify that 3Ċ 2 2 ċ 2 = Č is a one-parameter family of solutions of the differential equation
Ăþ
ċ Ăý = 3Ċ.
(b) Sketch the graph of the implicit solution 3Ċ 2 3 ċ 2 = 3. Find all explicit solutions ċ = �㔙(Ċ) of
the DE in part (a) defined by this relation. Give the interval �㔼 of definition of the explicit solution.
(c) The point (22,3) is on the graph of 3Ċ 2 3 ċ 2 = 3 but which of the explicit solutions in part (b)
satisfies ċ(22) = 3?
10. Suppose that a model of the growing population of a small community is given by the initial-value
Ăÿ
problem ĂĊ = 0.15 ÿ(Ć) + 20, ÿ(0) = 100, where ÿ is the number of individuals in the community
and time Ć is measured in years. How fast is the population increasing at Ć = 0? How fast is the population
increasing when the population is 500.
11. Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students.
Determine a differential equation for the number of people Č(Ć) who have contracted the flu if the rate
at which the disease spreads is proportional to the number of interactions between the number of
students who have flu and the number of students who have not yet been exposed to it.
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January 2022 Denis CK Wong
12. At a time denoted as Ć = 0 a technological innovation is introduced into a community that has a fixed
population of Ā people. Determine a differential equation for the number of people Ċ(Ć) who have
adopted the innovation at time Ć if it is assumed that the rate at which the innovations spread through
the community is jointly proportional to the number of people who have adopted it and the number of
people who have not adopted it.
13. Solve the following differential equation by separation of variables.
Ăþ
(a) Ăý = sin 5Ċ
Ăþ
(b) Ăý = (Ċ + 1)2
(c) čĊ + Ď 3ý čċ = 0
(d) čċ 2 (ċ 2 1)2 čĊ = 0
Ăþ
= 4ċ
Ăý
Ăþ
(f) Ăý + 2Ċċ 2 = 0
Ăþ
(g) Ď ý ċ Ăý = Ď 2þ + Ď 22ý2þ
Ăý
þ+1 2
(h) ċ ln Ċ Ăþ = ( ý )
3
(e) Ċ
(i) sin 3Ċ čĊ + 2 ċ cos 3Ċ čċ = 0
1
1
(j) Ċ(1 + ċ 2 )2 čĊ = ċ(1 + Ċ 2 )2 čċ
ĂĀ
(k) ĂĊ = ý(Ā 2 70)
ýþ+3ý2þ23
Ăþ
= ýþ22ý+4þ28
Ăý
Ăþ
(m) Ăý = Ċ√1 2 ċ 2
Ăþ
(n) (Ď ý + Ď 2ý ) Ăý = ċ 2
(l)
14. Find an explicit solution of the following initial-value problem.
Ăý
(a) ĂĊ = 4(Ċ 2 + 1), Ċ(�㔋) = 1
Ăþ
(b) Ăý =
Ăþ
þ 2 21
, ċ(2) = 2
ý 2 21
(c) Ċ 2 Ăý = ċ 2 Ċċ, ċ(21) = 21
(d) (1 + Ċ 4 )čċ + Ċ(1 + 4ċ 2 )čĊ = 0, ċ(1) = 0
15. Find the general solution of the following differential equation. Give the largest interval �㔼 over which
the general solution is defined.
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January 2022 Denis CK Wong
Ăþ
= 5ċ
Ăý
Ăþ
(b)
+ ċ = Ď 3ý
Ăý
Ăþ
(c) 3 Ăý + 12 ċ = 4
′
2
2
(a)
(d) ċ + 3Ċ ċ = Ċ
(e) Ċ 2 ċ ′ + Ċċ = 1
Ăþ
(f) Ċ Ăý 2 ċ = Ċ 2 sin Ċ
Ăþ
(g) Ċ Ăý + 2ċ = 3
Ăþ
(h) Ċ Ăý + 4ċ = Ċ 3 2 Ċ
(i) Ċ 2 ċ ′ + Ċ(Ċ + 2)ċ = Ď ý
(j) Ċċ ′ + (1 + Ċ)ċ = Ď 2ý sin 2Ċ
Ăþ
(k) (Ċ + 2)2 Ăý = 5 2 8ċ 2 4Ċċ
Ăþ
(l) (Ċ 2 2 1) Ăý + 2ċ = (Ċ + 1)2
16. Solve the initial-value problem. Give the largest interval I over which the solution is defined.
(a) Ċċ ′ + ċ = Ď ý , ċ(1) = 2
Ăÿ
(b) Ā ĂĊ + āÿ = ā, ÿ(0) = ÿ0
Ăă
(c) ĂĊ = ý(ă 2 ăă ); ă(0) = ă0
(d) (Ċ + 1)
Ăþ
+ ċ = ln Ċ, ċ(1) = 10.
Ăý
17. Solve the given initial-value problem.
1, 0 f Ċ f 3
Ăþ
(a) Ăý + 2ċ = ď(Ċ), ċ(0) = 0, ĉ/ĎĄĎ ď(Ċ) = {
0, Ċ > 3
Ċ, 0 f Ċ < 1
Ăþ
(b) Ăý + 2Ċċ = ď(Ċ), ċ(0) = 2, ĉ/ĎĄĎ ď(Ċ) = {
0, Ċ g 1
18. Determine whether the given differential equation is exact. If it is exact, solve it.
(a) (2Ċ 2 1)čĊ + (3ċ + 7)čċ = 0
(b) (2Ċ + ċ)čĊ 2 (Ċ + 6ċ)čċ = 0
(c) (2Ċċ 2 2 3)čĊ + (2Ċ 2 ċ + 4)čċ = 0
þ
(d) (1 + ln Ċ + ý ) čĊ = (1 2 ln Ċ) čċ
(e) (Ċ 2 ċ 3 + ċ 2 sin Ċ) čĊ = (3Ċċ 2 + 2ċ cos Ċ )čċ
1
(f) (ċ ln ċ 2 Ď 2ýþ )čĊ + (þ + Ċ ln ċ) čċ = 0
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January 2022 Denis CK Wong
Ăþ
= 2ĊĎ ý 2 ċ + 6Ċ 2
Ăý
3
Ăþ
3
(h) (1 2 þ + Ċ) + ċ = 2 1
Ăý
ý
′
(5ċ
(g) Ċ
(i)
2 2Ċ)ċ 2 2ċ = 0
(j) (tan Ċ 2 sin Ċ sin ċ)čĊ + cos Ċ cos ċ čċ = 0
19. Solve the following initial-value problem.
(a) (Ċ + ċ)2 + (2Ċċ + Ċ 2 2 1)čċ = 0, ċ(1) = 1
Ċ
3þ 2 2Ċ 2 Ăþ
) ĂĊ + 2þ4 = 0, ċ(1) = 1
þ5
1
Ăþ
( 2 + cos Ċ 2 2Ċċ) = ċ(ċ + sin Ċ), ċ(0) = 1
1+þ
Ăý
(b) (
(c)
20. Verify that the given differential equation is not exact. Multiply the given differential equation by the
indicated integrating factor Ā(Ċ, ċ), and verify that the new equation is exact. Solve.
(a) (2Ċċ sin Ċ + 2ċ cos Ċ)čĊ + 2Ċ cos Ċ čċ = 0 ; Ā(Ċ, ċ) = Ċċ.
(b) (Ċ 2 + 2Ċċ 2 ċ 2 )čĊ + (ċ 2 + 2Ċċ 2 Ċ 2 )čċ = 0 ; Ā(Ċ, ċ) = (Ċ + ċ)22 .
21. Solve the given differential equation by finding an appropriate integrating factor.
(a) (2ċ 2 + 3Ċ)čĊ + 2Ċċčċ = 0
(b) ċ(Ċ + ċ + 1)čĊ + (Ċ + 2ċ)čċ = 0
(c) (10 2 6ċ + Ď 23ý )čĊ 2 2čċ = 0
22. Solve the given differential equation by using an appropriate substitution.
(a) (Ċ 2 ċ)čĊ + Ċčċ = 0
(b) (Ċ + ċ)čĊ + Ċ čċ = 0
(c) Ċ čĊ + (ċ 2 2Ċ)čċ = 0
(d) ċ čĊ = 2(Ċ + ċ)čċ
(e) (ċ 2 + ċĊ)čĊ + Ċ 2 čċ = 0
(f)
þ2ý
Ăþ
= þ+ý
Ăý
(g) 2ċ čĊ + (Ċ + √Ċċ)čċ = 0
Ăþ
(h) Ċ Ăý = ċ + √Ċ 2 2 ċ 2 , Ċ > 0.
(i)
(j)
Ăþ
12ý2þ
= ý+þ
Ăý
Ăþ
= sin(Ċ + ċ)
Ăý
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January 2022 Denis CK Wong
23. Solve the given initial-value problem.
Ăþ
(a) Ċċ 2 Ăý = ċ 3 2 Ċ 3 , ċ(1) = 2
Ăý
(b) (Ċ 2 + 2ċ 2 ) Ăþ = Ċċ , ċ(21) = 1
Ăþ
1
(c) Ċ 2 Ăý 2 2Ċċ = 3ċ 4 , ċ(1) = 2
1
(d) ċ 2
(e)
3
Ăþ
2 = 1, ċ(0) = 4
+
ċ
Ăý
�㔋
Ăþ
= cos(Ċ + ċ) , ċ(0) = 4
Ăý
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January 2022 Denis CK Wong
Tutorial 2
1. Use the Euler’s method to obtain a four decimal approximation of the indicated value. Carry out
the recursion by hand, first using / = 0.1 and then using / = 0.05.
(a)
ċ ′ = Ċ + ċ 2 , ċ(0) = 0; ċ(0.2)
(b)
ċ ′ = 2Ċ 2 3ċ + 1, ċ(1) = 5; ċ(1.2)
2. The population of a community is known to increase at a rate proportional to the number of
people present at time Ć. If an initial population ÿ0 has doubled in 5 years, how long will it take to
triple? To Quadruple?
3. The population of a town grows at a rate proportional to the population present at time Ć. The
initial population of 500 increases by 15% in 10 years. What will be the population in 30 years?
How fast is the population growing at Ć = 30?
4. When interest is compounded continuously, the amount of money increases at a rate
ĂĂ
proportional to the amount Ă present at time Ć, that is, ĂĊ = ĄĂ, where r is the annual rate of
interest. Find the amount of money accrued at the end of 5 years when RM5000 is deposited in a
savings account drawing 5.75% annual interest compounded continuously.
5. A model that describes the population of a fishery in which harvesting takes place at a constant
rate is given by
Ăÿ
ĂĊ
= ýÿ 2 /, where ý and / are positive constants. Solve the differential
equation subject to ÿ(0) = ÿ0 . Describe the behavior of the population ÿ(Ć) for increasing time.
6. The number Ă(Ć) of supermarkets throughout the country that are using a computerized
Ăþ
checkout system is described by the initial-value problem ĂĊ = Ă(1 2 0.0005 Ă), Ă(0) = 1.
Solve the initial-value problem.
Ăÿ
7. A model for population ÿ(Ć) in a suburb of a large city is given by the initial-value problem ĂĊ =
ÿ(1021 2 1027 ÿ), ÿ(0) = 5000, where Ć is measured in months. What is the limiting value of
the population? At what time will the population be equal to one-half of this limiting value?
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January 2022 Denis CK Wong
8. The given family of functions is the general solution of the differential equation on the indicated
interval. Find a member of the family that is a solution of the initial value problem.
(a)
ċ = Č1 Ď ý + Č2 Ď 22 , (2∞, ∞); ċ ′′ 2 ċ = 0, ċ(0) = 0, ċ ′ (0) = 1
(b)
ċ = Č1 Ċ + Č2 Ċ ln Ċ , (0, ∞); ċ ′′ 2 3ċ ′ 2 4ċ = 0, ċ(0) = 1, ċ ′ (0) = 2
9. Given that Ċ(Ć) = Č1 cos ÿĆ + Č2 sin ÿĆ is the general solution of Ċ ′′ + ÿ2 Ċ = 0 on the interval
(2∞, ∞), show that a solution satisfying the initial conditions Ċ(0) = Ċ0 , Ċ9(0) = Ċ1 is given by
ý
Ċ(Ć) = Ċ0 cos ÿĆ + ÿ1 sin ÿĆ.
10. Determine whether the given set of functions is linearly independent on the interval (2∞, ∞).
(a)
ď1 (Ċ) = Ċ, ď2 (Ċ) = Ċ 2 , ď3 (Ċ) = 4Ċ 2 3Ċ 2
(b)
ď1 (Ċ) = 0, ď2 (Ċ) = Ċ, ď3 (Ċ) = Ď ý
(c)
ď1 (Ċ) = 5, ď2 (Ċ) = cos2 Ċ , ď3 (Ċ) = sin2 Ċ
(d)
ď1 (Ċ) = 2 + Ċ, ď2 (Ċ) = 2 + |Ċ|
11. Verify that the given functions form a fundamental set of solutions of the differential equation on
the indicated interval. Form the general solution.
(a) ċ ′′ 2 ċ ′ 2 12ċ = 0; Ď 23ý , Ď 4ý , (2∞, ∞)
(b) ċ ′′ 2 4ċ = 0; cosh 2Ċ , sinh 2Ċ , (2∞, ∞)
(c) 4ċ ′′ 2 4ċ ′ + ċ = 0; Ď ý/2 , ĊĎ ý/2 , (2∞, ∞)
(d) ċ (4) + ċ ′′ = 0; 1, Ċ, cos Ċ , sin Ċ , (2∞, ∞)
12. Verify that the given two-parameter family of functions is the general solution of the
nonhomogeneous differential equation on the indicated interval.
(a)
ċ ′′ 2 7ċ ′ + 10ċ = 24Ď ý ; ċ = Č1 Ď 2ý + Č2 Ď 5ý + 6Ď ý , (2∞, ∞)
(b)
�㔋 �㔋
ċ ′′ + ċ = sec Ċ; ċ = Č1 cos Ċ + Č2 sin Ċ + Ċ sin Ċ + cos Ċ ln(cos Ċ) , (2 2 , 2 )
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January 2022 Denis CK Wong
13. The indicated function ċ1 (Ċ) is a solution of the given differential equation. Use reduction of
order to find a second solution ċ2 (Ċ)
(a)
ċ ′′ 2 4ċ ′ + 4ċ = 0; ċ1 = Ď 2ý
(b)
ċ ′′ + 2ċ ′ + ċ = 0, ċ1 = ĊĎ 2ý
(c)
ċ ′′ + 16ċ = 0; ċ1 = cos 4Ċ
(d)
ċ ′′ 2 ċ = 0; ċ1 = cosh Ċ
(e)
Ċċ ′′ + ċ ′ = 0; ċ1 = ln Ċ
(f)
Ċ 2 ċ ′′ 2 3Ċċ ′ + 5ċ = 0; ċ1 = Ċ 2 cos(ln Ċ)
14. The indicated function ċ1 (Ċ) is a solution of the associated homogeneous equation. Use the
method of reduction of order to find a second solution ċ2 (Ċ) and a particular solution of the given
nonhomogeneous equation.
(a)
ċ ′′ 2 4ċ = 2; ċ1 = Ď 22ý
(b)
ċ ′′ 2 3ċ ′ + 2ċ = 5Ď 3ý ; ċ1 = Ď ý
15. Find the general solution of the following second-order differential equation.
(a)
4ċ ′′ + ċ ′ = 0
(b)
ċ99 2 ċ9 2 6ċ = 0
(c)
ċ ′′ + 8ċ ′ + 16ċ = 0
(d)
12ċ99 2 5ċ9 2 2ċ = 0
(e)
ċ ′′ + 9ċ = 0
(f)
3ċ ′′ + 2ċ ′ + ċ = 0
16. Find the general solution of the given higher-order differential equation.
(a)
ċ ′′′ 2 4ċ ′′ 2 5ċ ′ = 0
(b)
ċ ′′′ + 3ċ ′′ 2 4ċ ′ 2 12ċ = 0
(c)
ċ ′′′ + 3ċ ′′ + 3ċ ′ + ċ = 0
(d)
ċ (4) 2 2ċ ′′ + ċ = 0
(e)
Ă5 ý
Ă4 ý
Ă3 ý
Ă2 ý
2 Ăĉ5 2 7 Ăĉ4 + 12 Ăĉ3 + 8 Ăĉ2 = 0
17. Solve the given initial-value problem
(a)
ċ ′′ + 16ċ = 0, ċ(0) = 2, ċ ′ (0) = 22
(b)
4ċ ′′ 2 4ċ ′ 2 3ċ = 0, ċ(0) = 1, ċ ′ (0) = 5
(c)
ċ ′′′ + 12ċ ′′ + 36ċ ′ = 0, ċ(0) = 0, ċ ′ (0) = 1, ċ ′′ (0) = 27
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January 2022 Denis CK Wong
18. Solve the given boundary-value problem.
(a)
ċ ′′ 2 10ċ ′ + 25ċ = 0, ċ(0) = 1, ċ(1) = 0
(b)
�㔋
ċ ′′ + ċ = 0, ċ ′ (0) = 0, ċ ′ ( 2 ) = 0
19. Solve the given differential equation by undetermined coefficients
(a)
ċ ′′ + 3ċ ′ + 2ċ = 6
(b)
4ċ ′′ + 9ċ = 15
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
1 ′′
ċ + ċ ′ + ċ = Ċ 2 2 2Ċ
4
′′
′
2
ċ 2 8ċ + 20ċ = 100Ċ 2 26ĊĎ ý
ċ ′′ 2 ċ ′ = 2Ď 4ý
ċ ′′ + 4ċ = 3 sin 2Ċ
ċ ′′ + ċ = 2Ċ sin Ċ
ċ ′′ 2 5ċ ′ = 2Ċ 3 2 4Ċ 2 2 Ċ + 6
ċ ′′ 2 2ċ ′ + 5ċ = Ď ý cos 2Ċ
ċ ′′′ 2 6ċ ′′ = 2 2 cos Ċ
ċ ′′′ 2 2ċ ′′ + 3ċ ′ 2 ċ = Ċ 2 4Ď ý
20. Solve the given initial-value problem
(a)
(b)
(c)
(d)
�㔋
8
1
�㔋
ċ ′′ + 4ċ = 22, ċ ( ) = 2 , ċ ′ ( 8 ) = 2
5ċ ′′ + ċ ′ = 26Ċ, ċ(0) = 0, ċ ′ (0) = 210
ċ ′′ + 4ċ ′ + 5ċ = 35Ď 24ý , ċ(0) = 23, ċ ′ (0) = 1
ċ ′′ 2 ċ = cosh Ċ , ċ(0) = 2, ċ ′ (0) = 12
21. Solve the boundary -value problem
(a)
ċ ′′ + ċ = Ċ 2 + 1, ċ(0) = 5, ċ(1) = 0
(b)
ċ ′′ 2 2ċ ′ + 2ċ = 2Ċ 2 2, ċ(0) = 0, ċ(�㔋) = �㔋
(c)
ċ ′′ + 3ċ = 6Ċ, ċ(0) = 0, ċ(1) + ċ ′ (1) = 0
22. Write the given differential equation in the form Ā(ċ) = Đ(Ċ), where Ā is a linear differential
operator with constant coefficients. If possible, factor Ā.
(a)
9ċ ′′ 2 4ċ = sin Ċ
(b)
ċ ′′ 2 4ċ ′ 2 12ċ = Ċ 2 6
(c)
ċ ′′′ + 10ċ ′′ + 25ċ ′ = Ď ý
(d)
ċ ′′′ + 4ċ ′′ + 3ċ ′ = Ċ 2 cos Ċ 2 3Ċ
(e)
ċ (4) + 8ċ ′ = 4
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January 2022 Denis CK Wong
23. Verify that the given differential operator annihilates the indicated functions.
Ā 2 + 64; ċ = 2 cos 8Ċ 2 5 sin 8Ċ
24. Find a linear differential operator that annihilates the given function.
(a)
1 + 6Ċ 2 2Ċ 3
(b)
Ċ 3 (1 2 5Ċ)
(c)
1 + 7Ď 2ý
(d)
1 + sin Ċ
(e)
13Ċ + 9Ċ 2 3 ąÿĀ 4Ċ
(f)
Ď 2ý + 2ĊĎ ý 2 Ċ 2 Ď ý
(g)
Ď 2ý sin Ċ 2 Ď 2ý cos Ċ
25. Find linearly independent functions that are annihilated by the give n differential operator.
(a)
Ā5
(b)
(Ā 2 6)(2Ā + 3)
(c)
Ā2 + 5
(d)
Ā 3 2 10Ā2 + 25Ā
26. Solve the following by undetermined coefficients
(a)
ċ ′′ 2 9ċ = 54
(b)
ċ ′′ + ċ ′ = 3
(c)
ċ ′′ + 4ċ ′ + 4ċ = 2Ċ + 6
(d)
ċ ′′ + 3ċ ′ = 4Ċ 2 5
(e)
ċ ′′′ + ċ ′′ = 8Ċ 2
(f)
ċ ′′ 2 ċ ′ 2 12ċ = Ď 4ý
(g)
ċ ′′ + 6ċ ′ + 8ċ = 3Ď 22ý + 2Ċ
(h)
ċ ′′ + 3ċ ′ 2 10ċ = Ċ(Ď ý + 1)
(i)
ċ ′′ 2 2ċ ′ + 5ċ = Ď ý sin Ċ
(j)
ċ ′′ + 4ċ = cos 2 Ċ
(k)
2ċ ′′′ 2 3ċ ′′ 2 3ċ ′ + 2ċ = (Ď ý + Ď 2ý )2
27. Solve the initial-value problem
(a)
ċ ′′ 2 64ċ = 16, ċ(0) = 1, ċ ′ (0) = 0
(b)
ċ ′′ + 5ċ ′ 2 6ċ = 10Ď 2ý , ċ(0) = 1, ċ ′ (0) = 1
(c)
ċ99 2 4ċ9 + 8ċ = Ċ 3 , ċ(0) = 2, ċ ′ (0) = 4
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January 2022 Denis CK Wong
28. Solve each differential equation by variation of parameters
(a)
ċ ′′ + ċ = sec Ċ
(b)
ċ ′′ + ċ = sin Ċ
(c)
ċ ′′ 2 ċ = cosh Ċ
ċ ′′ 2 4ċ =
(d)
ă 2ý
ý
1
ċ ′′ + 3ċ ′ + 2ċ = 1+ă ý
(e)
ăý
ċ ′′ 2 2ċ ′ + ċ = 1+ý2
(f)
ċ ′′ + 2ċ ′ + ċ = Ď 2Ċ ln Ć
2ċ ′′ + 2ċ ′ + ċ = 4√Ċ
ċ ′′ + 2ċ ′ 2 8ċ = 2Ď 22ý 2 Ď 2ý subject to ċ(0) = 1, ċ ′ (0) = 0
(g)
(h)
(i)
29. Solve the following
(a)
Ċ 2 ċ ′′ 2 2ċ = 0
(b)
Ċċ99 + ċ9 = 0
(c)
Ċ 2 ċ ′′ + Ċċ ′ + 4ċ = 0
(d)
25Ċ 2 ċ ′′ + 25Ċċ ′ + 4ċ = 0
(e)
3Ċ 2 ċ ′′ + 6Ċċ ′ + ċ = 0
(f)
Ċ 3 ċ ′′′ 2 6ċ = 0
(g)
Ċ 3 ċ ′′′ + Ċċ ′ 2 ċ = 0
(h)
Ċċ ′′ 2 4ċ ′ = Ċ 4
(i)
2Ċ 2 ċ ′′ + 5Ċċ ′ + ċ = Ċ 2 2 Ċ
(j)
Ċ 2 ċ ′′ + Ċċ ′ 2 ċ = ln Ċ
(k)
Ċ 2 ċ ′′ + 3Ċċ ′ = 0, ċ(1) = 0, ċ ′ (1) = 4
1
Ċċ ′′ + ċ ′ = Ċ, ċ(1) = 1, ċ ′ (1) = 2 2
(l)
30. Solve the given system of differential equations by systematic elimination
Ăý
(a) ĂĊ = 2Ċ 2 ċ ;
Ăý
(b) ĂĊ = 2ċ + Ć ;
Ăþ
=Ċ
ĂĊ
Ăþ
=Ċ2Ć
ĂĊ
(c) (Ā 2 + 5)Ċ 2 2ċ = 0 ; 22Ċ + (Ā 2 + 2)ċ = 0
(d) (Ā + 1)Ċ + (Ā 2 1)ċ = 2 ; 3Ċ + (Ā + 2)ċ = 21
Ă2 ý
(e) ĂĊ 2 = 4ċ + Ď Ċ ;
Ă2 þ
= 4Ċ 2 Ď Ċ
ĂĊ 2
(f) ĀĊ + Ā 2 ċ = Ď 3Ċ ; (Ā + 1)Ċ + (Ā 2 1)ċ = 4Ď 3Ċ
(g) Ā 2 Ċ 2 Āċ = Ć; (Ā + 3)Ċ + (Ā + 3)ċ = 2
(h) (2Ā 2 2 Ā 2 1)Ċ 2 (2Ā + 1)ċ = 1 ; (Ā 2 1)Ċ + Āċ = 21
Ăý
Ăþ
(i) 2 ĂĊ 2 5Ċ + ĂĊ = Ď Ċ ;
Ăý
Ăþ
2 Ċ + ĂĊ = 5Ď Ċ
ĂĊ
(Ā 2
(j) (Ā 2 1)Ċ + (Ā 2 + 1)ċ = 1 ;
(k) Ā 2 Ċ 2 2(Ā 2 + Ā)ċ = sin Ć ;
(l) ĀĊ = ċ ; Āċ = Č ; ĀČ = Ċ
2 1)Ċ + (Ā + 1)ċ = 2
Ċ + Āċ = 0
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January 2022 Denis CK Wong
Ăþ
Ăÿ
Ăý
= 6ċ ;
= Ċ+Č ;
=Ċ+ċ
ĂĊ
ĂĊ
ĂĊ
Ăý
Ăþ
Ăÿ
(n)
= 2Ċ + Č ; ĂĊ = 2ċ + Č ; ĂĊ = 2Ċ + ċ
ĂĊ
(m)
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January 2022 Denis CK Wong
Tutorial 3
1. Solve the given differential equation, find a lower bound for the radius of convergence of
power series solutions about the ordinary point Ċ = 0. About the ordinary point Ċ = 1.
(a) (Ċ 2 2 25)ċ ′′ + 2Ċċ ′ + ċ = 0
(b) (Ċ 2 2 2Ċ + 10)ċ ′′ + Ċċ ′ 2 4ċ = 0
2. Find the two power series solutions of the given differential equation about the ordinary
point Ċ = 0.
(a) ċ ′′ 2 Ċċ = 0
(b) ċ ′′ 2 2Ċċ ′ + ċ = 0
(c) ċ ′′ + Ċ 2 ċ ′ + Ċċ = 0
(d) (Ċ 2 1)ċ ′′ + ċ ′ = 0
(e) (Ċ 2 + 1)ċ ′′ 2 6ċ = 0
(f) (Ċ 2 2 1)ċ ′′ + Ċċ ′ 2 ċ = 0
3. Use the power series method to solve the given initial-value problem
(a) (Ċ 2 1)ċ ′′ 2 Ċċ ′ + ċ = 0, ċ(0) = 22, ċ ′ (0) = 6
(b) ċ ′′ 2 2Ċċ ′ + 8ċ = 0, ċ(0) = 3, ċ ′ (0) = 0
4. Determine the singular points of the given differential equation. Classify each singular point
as regular or irregular.
(a)
Ċ 3 ċ ′′ + 4Ċ 2 ċ ′ + 3ċ = 0
(b)
Ċ(Ċ + 3)2 ċ ′′ 2 ċ = 0
(Ċ 2 2 9)2 ċ ′′ + (Ċ + 3)ċ ′ + 2ċ = 0
(c)
(d)
Ċ 2 (Ċ 2 5)2 ċ ′′ + 4Ċċ ′ + (Ċ 2 2 25)ċ = 0
(Ċ 3 2 2Ċ 2 + 3Ċ)2 ċ ′′ + Ċ(Ċ 2 3)2 ċ ′ 2 (Ċ + 1)ċ = 0
(e)
5. In the following question Ċ = 0 is a regular singular point of the given differential equation.
Show that the indicial roots of the singularity do not differ by an integer. Use the method of
Frobenius to obtain two linearly independent series solution about Ċ = 0. Form the general
solution on (0, ∞).
(a)
2Ċċ ′′ 2 ċ ′ + 2ċ = 0
(b)
(c)
(d)
1
4Ċċ99 + 2 ċ9 + ċ = 0
3Ċċ99 + (2 2 Ċ)ċ9 2 ċ = 0
4
Ċ 2 ċ′′ + Ċċ ′ + (Ċ 2 2 9) ċ = 0
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January 2022 Denis CK Wong
Tutorial 4
1. Find ℒ{ď(Ć)}.
2. Find the given inverse Laplace transform
21, 0 f Ć < 1
(a) ď(Ć) = {
1, Ć g 1
4, 0 f Ć < 2
(b) ď(Ć) = {
0, Ć g 2
sin Ć , 0 f Ć < �㔋
(c) ď(Ć) = {
0, Ć g 0
�㔋
0, 0 f Ć < 2
(d) ď(Ć) = {
�㔋
cos Ć , Ć g 2
(e) ď(Ć) = Ď Ċ+7
(f) ď(Ć) = ĆĎ 4Ċ
(g) ď(Ć) = Ď 2Ċ sin Ć
(h) ď(Ć) = Ć cos Ć
(i) ď(Ć) = 2Ć 4
(j) ď(Ć) = 4Ć 2 10
(k) ď(Ć) = (Ć + 1)3
(l) ď(Ć) = 4Ć 2 2 5 sin 3Ć
(m) ď(Ć) = Ď Ċ sinh Ć
(n) ď(Ć) = Ć 5
(o) ď(Ć) = 7Ć + 3
(p) ď(Ć) = Ć 2 2 Ď 29Ċ + 5
(q) ď(Ć) = (Ď Ċ 2 Ď 2Ċ )2
(r) ď(Ć) = cos 5Ć + sin 2Ć
(s) ď(Ć) = Ď 2Ċ cosh Ć
(t) ď(Ć) = sin 2Ć cos 2Ć
(u) ď(Ć) = sin (4Ć + 5)
1
ĉ
1
48
(b) ℒ 21 {ĉ2 2 ĉ5 }
3
21 (ĉ+1)
(a) ℒ 21 { 3 }
(c) ℒ
{
3. Use Laplace transform to solve the given initial-value problem
Ăþ
2 ċ = 1, ċ(0) = 0
(a)
ĂĊ
Ăþ
(b) 2 + ċ = 0, ċ(0) = 23
ĂĊ
(c) ċ ′ + 6ċ = Ď 4Ċ , ċ(0) = 2
(d) ċ ′ 2 ċ = 2 cos 5Ć, ċ(0) = 0
(e) ċ ′′ 2 4ċ ′ = 6Ď 3Ċ 2 3Ď 2Ċ , ċ(0) = 1, ċ ′ (0) = 21
(f) ċ ′′ + ċ = √2 sin √2 Ć, ċ(0) = 10, ċ ′ (0) = 0
4. Find
(a) ℒ{ĆĎ 10Ċ }
(b) ℒ{Ć 3 Ď 22Ċ }
(c) ℒ{Ć(Ď Ċ + Ď 2Ċ )2 }
(d) ℒ{Ď Ċ sin 3Ć}
(e) ℒ{(1 2 Ď Ċ + 3Ď 24Ċ ) cos 5Ć}
1
(f) ℒ 21 {(ĉ+2)3 }
1
(g) ℒ 21 {ĉ2 26ĉ+10}
ĉ
}
ĉ +4ĉ+5
(h) ℒ 21 { 2
}
ĉ4
1
1
21 1
(d) ℒ { 2 2 ĉ + ĉ22}
ĉ
1
(e) ℒ 21 {4ĉ+1}
5
(f) ℒ 21 {ĉ2 +49}
4ĉ
(g) ℒ 21 {4ĉ2 +1}
2ĉ26
(h) ℒ 21 {ĉ2 +9}
1
(i) ℒ 21 {ĉ2 +3ĉ}
ĉ+1
(j) ℒ 21 {ĉ2 24ĉ}
1
(k) ℒ 21 {ĉ2 +ĉ220}
ĉ23
(l) ℒ 21 {(ĉ2 3)(ĉ+ 3)}
√
√
ĉ2 +1
21
(m) ℒ {(ĉ22)(ĉ23)(ĉ26)}
1
(n) ℒ 21 {ĉ3 +5ĉ}
6ĉ+3
(o) ℒ 21 {ĉ4 +5ĉ2 +4}
88
January 2022 Denis CK Wong
ĉ
(i) ℒ 21 {(ĉ+1)2 }
2ĉ21
(j) ℒ 21 {ĉ2 (ĉ+1)3 }
(ĉ+1)2
(k) ℒ 21 {(ĉ+2)4 }
5. Use the Laplace transform to solve the given Initial-value problem.
(a) ċ ′ + 4ċ = Ď 24Ċ , ċ(0) = 2
(b) ċ ′ 2 ċ = 1 + ĆĎ Ċ , ċ(0) = 0
(c) ċ99 + 2ċ9 + ċ = 0, ċ(0) = 1, ċ 8(0) = 1
(d) ċ ′′ 2 4ċ ′ + 4ċ = Ć 3 Ď 2Ċ , ċ(0) = 0, ċ ′ (0) = 0
(e) ċ ′′ 2 ċ ′ = Ď Ċ cos Ć , ċ(0) = 0, ċ ′ (0) = 0
(f) ċ ′′ 2 2ċ ′ + 5ċ = 1 + Ć, ċ(0) = 0, ċ ′ (0) = 4
6. Find
(a) ℒ{(Ć 2 1)Ą(Ć 2 1)}
(b) ℒ{ĆĄ(Ć 2 2)}
(c) ℒ{cos 2Ć Ą(Ć 2 �㔋)}
ă 22ý
}
ĉ3
2�㔋ý
ă
ℒ 21 { ĉ2 +1 }
ă 2ý
ℒ 21 {ĉ(ĉ+1)}
(d) ℒ 21 {
(e)
(f)
7. Use Laplace Transform to solve the initial-value problem.
0, 0 f Ć < 1
(a) ċ ′ + ċ = ď(Ć), ċ(0) = 0, ĉ/ĎĄĎ ď(Ć) = {
5, Ć g 1
Ć, 0 f Ć < 1
′
(b) ċ + 2ċ = ď(Ć), ċ(0) = 0, ĉ/ĎĄĎ ď(Ć) = {
0, Ć g 1
1, 0 f Ć < 1
′
(c) ċ ′ + 4ċ = ď(Ć), ċ(0) = 0, ċ ′ (0) = 21 ĉ/ĎĄĎ ď(Ć) = {
0, Ć g 1
′′
′ (0)
(d) ċ + 4ċ = sin Ć Ą(Ć 2 2�㔋), ċ(0) = 1, ċ
=0
(e) ċ ′′ 2 5ċ ′ + 6ċ = Ą(Ć 2 1), ċ(0) = 0, ċ ′ (0) = 1
(f) ċ ′ + ċ = Ć sin Ć , ċ(0) = 0
(g) ċ ′′ + 9ċ = cos 3Ć , ċ(0) = 2, ċ ′ (0) = 5
(h) ċ ′′ + ċ = sin Ć , ċ(0) = 1, ċ ′ (0) = 21
8. Evaluate
(a) ℒ{1 ∗ Ć 3 }
(b) ℒ{Ď 21 ∗ Ď Ċ cos Ć}
Ċ
(c) ℒ {∫0 Ď �㔏 č�㔏}
9. Use the Laplace Transform to solve
Ċ
(a) ď(Ć) + ∫0 (Ć 2 �㔏)ď(�㔏)č�㔏 = Ć
(b)
Ċ
ď(Ć) = 2Ć 2 4 ∫0 sin �㔏 ď(Ć 2 �㔏)č�㔏
Ċ
(c) ď(Ć) = ĆĎ Ċ + ∫0 �㔏ď(Ć 2 �㔏)č�㔏
Ċ
(d) ℒ {∫0 Ď �㔏 cos �㔏 č�㔏}
Ċ
(e) ℒ {∫0 �㔏Ď Ċ2�㔏 č�㔏}
Ċ
(f) ℒ {Ć ∫0 sin �㔏 č�㔏}
1
(g) ℒ 21 {ĉ(ĉ21)}
1
(h) ℒ 21 {ĉ3 (ĉ21)}
Ċ
(d) ď(Ć) + ∫0 ď(�㔏)č�㔏 = 1
89