CHAPTER 1
Introduction to
Differential Equations
Chapter Contents
1.1 Definitions and Terminology
1.2 Initial-Value Problems
1.3 Differential Equations as Mathematical Methods
Lecturer: Prof. Hsin-Lung Wu
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1.1 Definitions and Terminology
Introduction: differential equations means that
equations contain derivatives, eg:
dy/dx = 0.2xy
(1)
Definition 1.1.1 Differential equation
An equation contains the derivates of one or more
dependent variables with respect to one or more
independent variables (DE).
Ordinary DE: An equation contains only ordinary
derivates of one or more dependent variables of a
single independent variable.
eg: dy/dx + 5y = ex, (dx/dt) + (dy/dt) = 2x + y (2)
Lecturer: Prof. Hsin-Lung Wu
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Partial DE: An equation contains partial derivates of
one or more dependent variables of tow or more
independent variables.
2
2
2
2
 u  u
 u  u
u
(3)

 0,

2
x
2
y
2
x
2
t
2
t
Notations: Leibniz notation dy/dx, d2y/ dx2
prime notation y’, y”, …..
subscript notation ux, uy, uxx, uyy, uxy , ….
Order: highest order of derivatives
second order


2
d y
dx
Lecturer: Prof. Hsin-Lung Wu
first order
2
3
 dx 
x
 5
  4y  e
 dy 
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General form of n-th order ODE:
F ( x, y, y ',  , y
(n)
(4)
)0
Normal form of (4)
(5)
n
d y
dx
n
 f ( x, y, y ',  , y
( n 1 )
)
eg: normal form of 4xy’ + y = x, is
y’ = (x – y)/4x
Linearity: An n-th order ODE is linear if F is linear
in y, y’, y”, …, y(n). It means when (4) is linear, we
have
n
n 1
d y
d y
dy
(6)
a ( x)
a
   a ( x)
 a ( x) y  g ( x)
n
dx
n
n 1
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dx
n 1
1
dx
0
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The following cases are for n = 1 and n = 2
dy
and
a1 ( x)
 a0 ( x) y  g ( x)
dx
2
a2 ( x )
d y
dx
2
 a1 ( x)
dy
dx
 a0 ( x) y  g ( x)
(7)
Two properties of a linear ODE:
(1) y, y’, y”, … are of the first degree.
(2) Coefficients a0, a1, …, are at most on x
Nonlinear examples:
(1  y ) y '
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sin y
y
2
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Definition 1.1.2 Solution of an ODE
Any function , defined on an interval I, possessing at
least n derivatives that are continuous on I, when
replaced into an n-th order ODE, reduces the equation
into an identity, it said to be a solution of the equation
on the interval.
That is, a solution of (4) is a function  possesses at
least n derivatives and
F(x, (x), ’(x), …, (n)(x)) = 0 for all x in I,
where I is the interval  is defined on.
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Example 1 Verification of a Solution
Verify the indicated function is a solution of the given
ODE on (－, )
x
1/2
4



y

2
y

y

0
;
y

xe
(a) dy/dx = xy ; y = x /16 (b)
Solution:
3
3
dy
x
x
 4

(a) Left-hand side:
dx
Right-hand side: xy 1 / 2
then left = right
(b) Left-hand side:
16
4 1/ 2
2
3
 x4 
x
x

 x  
 x

4
4
 16 
y   2 y   y  ( xe  2 e )  2 ( xe  e )  xe
x
x
x
x
x
0
Right-hand side: 0
then left = right
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Note: y = 0 is also the solution of example 1, called
trivial solution
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Example 2 Function vs. Solution
y = 1/x, is the solution of xy’ + y = 0, however, this
function is not differentiable at x = 0. So, the interval of
definition I is (－, 0), (0, ).
Fig 1.1.1 Ex 2 illustrates the difference between
the function y = 1/x and the solution y = 1/x
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Explicit solution: dependent variable is expressed
solely in terms of independent variable and constants.
Eg: solution is y = (x).
Definition 1.3 Implicit solution of an ODE
G(x, y) = 0 is said to be an implicit solution of (4) on I,
provided there exists at least one function y = (x)
satisfying the relationship as well as the DE on I.
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Example 3 Verification of an Implicit
Solution
x2 + y2 = 25 is an implicit solution of
dy/dx = −x/y
on the interval -5 < x < 5.
Since
dx2/dx + dy2/dx = (d/dx)(25)
then
2x + 2y(dy/dx) = 0 and dy/dx = -x/y
solution curve is shown in Fig 1.1.2
Lecturer: Prof. Hsin-Lung Wu
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Fig 1.1.2 An Implicit solution and two
explicit solutions in Ex 3
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Families of solutions: A solution containing an
arbitrary constant represents a set G(x, y) = 0 of
solutions is called a one-parameter family of
solutions. A set G(x, y, c1, c2, …, cn) = 0 of solutions
is called a n-parameter family of solutions.
Particular solution: A solution free of arbitrary
parameters. eg: y = cx – x cos x is a solution of xy’ –
y = x2 sin x on (-, ), y = x cos x is a particular
solution according to c = 0. See Fig 1.1.3.
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Fig 1.1.3 Some solution of xy’-y=x2 sinx
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Example 4 Using Different Symbols
x = c1cos 4t and x = c2 sin 4t are solutions of
x + 16x = 0
we can easily verify that x = c1cos 4t + c2 sin 4t is also a
solution.
Lecturer: Prof. Hsin-Lung Wu
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Example 5 A piecewise-Defined Solution
We can verify y = cx4 is a solution of xy – 4y = 0 on
(-, ). See Fig 1.1.4(a).
The piecewise-defined function
 x , x  0
y
4
x
, x0

4
is a particular solution where we choose c = −1 for x <
0 and c = 1 for x  0. See Fig 1.1.4(b).
Lecturer: Prof. Hsin-Lung Wu
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Fig 1.1.4 Some solution of xy’-4y=0 in Ex 5
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Singular solution: A solution can not be obtained by
particularly setting any parameters.
y = (x2/4 + c)2 is the family solution of dy/dx = xy1/2 ,
however, y = 0 is a solution of the above DE.
We cannot set any value of c to obtain the solution y
= 0, so we call y = 0 is a singular solution.
Lecturer: Prof. Hsin-Lung Wu
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System of DEs: two or more equations involving of
two or more unknown functions of a single
independent variable.
dx/dt = f(t, x, y)
dy/dt = g(t, x, y)
(9)
Lecturer: Prof. Hsin-Lung Wu
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1.2 Initial-value Problems
Introduction
A solution y(x) of a DE satisfies an initial condition.
Example
On some interval
I containing xo,
n
d y
( n 1 )
solve:

f
(
x
,
y
,
y
'
,

,
y
)
n
dx
subject to:
y ( x 0 )  y 0 , y ' ( x 0 )  y1 ,  , y
( n 1 )
( x 0 )  y n 1
(1)
This is called an Initial-Value Problem (IVP).
y(xo) = yo , y(xo) = y1 , y ( n 1) ( x 0 )  y n 1
are called initial conditions.
Lecturer: Prof. Hsin-Lung Wu
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First and Second IVPs
solve :
dy
 f ( x, y )
dx
subject
(2)
to : y ( x 0 )  y 0
and
2
solve :
d y
dx
subject
2
 f ( x, y, y ')
(3)
to : y ( x 0 )  y 0 , y ' ( x 0 )  y 1
are first and second order initial-value problems,
respectively. See Fig 1.2.1 and 1.2.2.
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Lecturer: Prof. Hsin-Lung Wu
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Example 1 First-Order IVPs
We know y = cex is the solutions
of y’ = y on (-, ). If y(0) = 3,
then 3 = ce0 = c. Thus y = 3ex is
a solution of this initial-value
problem
y’ = y, y(0) = 3.
If we want a solution pass
through (1, -2), that is y(1) = -2,
-2 = ce, or c = -2e-1. The
function y = -2ex-1 is a solution
of the initial-value problem
y’ = y, y(1) = -2.
Lecturer: Prof. Hsin-Lung Wu
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Example 2 Interval / of Definition of a
Solution
In Problem 6 of Exercise 2.2, we have the solution of
y’ + 2xy2 = 0 is y = 1/(x2 + c). If we impose y(0) = -1, it
gives c = -1. Consider the following distinctions.
1) As a function, the domain of y = 1/(x2 - 1) is the
set of all real numbers except x = -1 and 1. See Fig
1.2.4(a).
2) As a solution, the intervals of definition are
(-, 1), (-1, 1), (1, )
3) As a initial-value problem, y(0) = -1, the interval
of definition is (-1, 1). See Fig 1.2.4(b).
Lecturer: Prof. Hsin-Lung Wu
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Fig 1.2.4 Graph of the function and solution
of IVP in Ex 2
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Example 3 Second-Order IVP
In Example 4 of Sec 1.1, we saw x = c1cos 4t + c2sin 4t
is a solution of
x + 16x = 0
Find a solution of the following IVP:
x + 16x = 0, x(/2) = −2, x(/2) = 1
(4)
Solution:
Substitute x(/2) = − 2 into x = c1cos 4t + c2sin 4t, we
find c1 = −2. In the same manner, from x(/2) = 1 we
have c2 = ¼.
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Existence and Uniqueness:
Does a solution of the IVP exist?
If a solution exists, is it unique?
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Example 4 An IVP Can Have Several
Solution
Since y = x4/16 and y = 0 satisfy the DE
dy/dx = xy1/2 , and also initial-value y(0) = 0, this DE
has at least two solutions, See Fig 1.2.5.
Lecturer: Prof. Hsin-Lung Wu
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Theorem 1.2.1 Existence of a Unique Solution
Let R be the region defined by a  x  b, c  y  d that
contains the point (x0, y0) in its interior. If f(x, y) and
f/y are continuous in R, then there exists some
interval I0: (x0- h, x0 + h), h > 0, contained in
[a, b] and a unique function y(x) defined on I0 that
is a solution of the IVP (2).
Lecturer: Prof. Hsin-Lung Wu
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Fig 1.2.6 Rectangular region R
The geometry of Theorem 1.2.1 shows in Fig 1.2.6.
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Example 5 Example 3 Revisited
For the DE: dy/dx = xy1/2 , inspection of the functions
f ( x , y )  xy
1/ 2
and
f
y

x
2y
1/ 2
we find they are continuous in y > 0. From Theorem
1.2.1, we conclude that through any point (x0, y0), y0 > 0,
there is some interval centered at x0 on which this DE
has a unique solution.
Lecturer: Prof. Hsin-Lung Wu
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Interval of Existence / Uniqueness
Suppose y(x) is a solution of IVP (2), the following
sets may not be the same:
the domain of y(x),
the interval of definition of y(x) as a solution,
the interval I0 of existence and uniqueness.
Lecturer: Prof. Hsin-Lung Wu
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1.3 DEs as Mathematical Models
Introduction
Mathematical models are mathematical descriptions
of something.
Level of resolution
Make some reasonable assumptions about the system.
The steps of modeling process are as following.
Lecturer: Prof. Hsin-Lung Wu
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Assumptions
Express assumptions in terms
of differential equations
If necessary,
alter assumptions
or increase resolution
of the model
Check model
Predictions
with know
facts
Lecturer: Prof. Hsin-Lung Wu
Mathematics
formulation
Solve the DEs
Display model predictions,
e.g., graphically
Obtain
solution
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Population Dynamics
If P(t) denotes the total population at time t, then
dP/dt  P or dP/dt = kP
(1)
where k is a constant of proportionality, and k > 0.
Radioactive Decay
If A(t) denotes the substance remaining at time t, then
dA/dt  A or dA/dt = kA
(2)
where k is a constant of proportionality, and k < 0.
A single DE can serve as a mathematical model for
many different phenomena.
Lecturer: Prof. Hsin-Lung Wu
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Newton’s Law of Cooling/Warming
If T(t) denotes the temperature of a body at time t, Tm
the temperature of surrounding medium, then
dT/dt  T - Tm or dT/dt = k(T - Tm)
(3)
where k is a constant of proportionality.
Lecturer: Prof. Hsin-Lung Wu
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Spread of a Disease
If x(t) denotes the number of people who have got the
disease and y(t) the number of people who have not
yet, then
dx/dt = kxy
(4)
where k is a constant of proportionality.
From the above description, suppose a small
community has a fixed population on n, If one
inflected person is introduced into this community,
we have x + y = n +1, and
dx/dt = kx(n+1-x)
(5)
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Chemical Reactions
Inspect the following equation
CH3Cl + NaOH  CH3OH + NaCl
Assume X is the amount of CH3OH,  and  are the
amount of the first two chemicals, then the rate of
formation is
dx/dt = k( - x)( - x)
(6)
Lecturer: Prof. Hsin-Lung Wu
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Mixtures
See Fig 1.3.1. If A(t) denotes the amount of salt in the
tank at time t, then
dA/dt = (input rate) – (output rate) = Rin - Rout (7)
We have Rin = 6 lb/min, Rout = A(t)/100 (lb/min), then
dA/dt = 6 – A/100 or dA/dt + A/100 = 6
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(8)
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Fig 1.3.1 Mixing tank
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Draining a Tank
Referring to Fig 1.3.2 and from Torricelli’s Law, if
V(t) denotes the volume of water in the tank at time t,
dV
dt
  Ah
2 gh
(9)
From (9), since we have V(t) = Awh, then
dh
dt
Lecturer: Prof. Hsin-Lung Wu

Ah
Aw
2 gh
(10)
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Fig 1.3.2 Water draining from a tank
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Series Circuits
Look at Fig 1.3.3.
From Kirchhoff’s second law, we have
2
L
d q
dt
2
R
dq
dt

1
q  E (t )
(11)
C
where q(t) is the charge and dq(t)/dt = i(t), which is
the current.
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Fig 1.3.3 Current i(t) and charge q(t) are
measured in amperes (A) and coulumbs (C)
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Falling Bodies
Look at Fig1.22.
From Newton’s law, we have
2
2
m
d s
dt
  mg
2
or
d s
dt
2
 g
(12)
Initial value problem
2
d s
dt
2
 g,
Lecturer: Prof. Hsin-Lung Wu
s (0)  s0 ,
s ' (0)  v0
(13)
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Fig 1.3.4 Position of rock measured from
ground level
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Falling Bodies and Air Resistance
From Fig 1.3.5.
We have the DE
m
dv
(14)
 mg  kv
dt
and can be written as
2
m
d s
dt
2
 mg  k
Lecturer: Prof. Hsin-Lung Wu
ds
dt
2
or
m
d s
dt
2
k
ds
 mg
(15)
dt
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Fig 1.3.5 Falling body of mass m
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A Slipping Chain
From Fig 1.3.6.
We have
L d x
2
32 dt
2
2
 2 x
Lecturer: Prof. Hsin-Lung Wu
or
d x
dt
2

64
x0
(16)
L
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Fig 1.3.6 Chain slipping from frictionless peg
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Suspended Cables
From Fig1.25.
We have
dy/dx = W/T1
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(17)
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Fig 1.3.7 Cables suspended between vertical
supports
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Fig 1.3.8 Element of cable
Fig 1.3.8 explains the Element of cable.
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Evaluation
One midterm 40%
Final 40%
Quiz & Homework 20%
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Some information
Textbook: Advanced Engineering Mathematics 4th-ed
by Zill and Wright
Teaching Website: Please link 數位學苑
Lecturer: Prof. Hsin-Lung Wu
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