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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
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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)
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Partial DE: An equation contains partial derivates of
one or more dependent variables of tow or more
independent variables.
 2u  2u
 2u  2u
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

first order

3
d y  dx 
x

5

4
y

e


dx 2
 dy 
2
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General form of n-th order ODE:
(4)
F ( x, y, y' , , y ( n ) )  0
Normal form of (4)
dny
( n1)

f
(
x
,
y
,
y
'
,

,
y
)
n
dx
(5)
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
dny
d n1 y
dy
(6)
a ( x)
a
   a ( x)  a ( x) y  g ( x)
n
dx n
n1
dx n1
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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
d2y
dy
a2 ( x) 2  a1 ( x)  a0 ( x) y  g ( x)
dx
dx
(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 '
sin y
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y2
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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:
dy
x3 x3
 4 
(a) Left-hand side:
dx
16 4 1/ 2
4
2
3


x
x
x
Right-hand side: xy1/ 2  x     x  
4
4
 16 
then left = right
(b) Left-hand side:
y  2 y  y  ( xe x  2e x )  2( xe x  e x )  xe 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
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(8)
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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.
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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 4 , x  0
y 4
 x , x0
is a particular solution where we choose c = −1 for x <
0 and c = 1 for x  0. See Fig 1.1.4(b).
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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.
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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)
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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 )  y0 , y ' ( x0 )  y1 , , y ( n1) ( x0 )  yn1
(1)
This is called an Initial-Value Problem (IVP).
y(xo) = yo , y(xo) = y1 , y ( n1) ( x0 )  yn1
are called initial conditions.
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First and Second IVPs
dy
solve :
 f ( x, y )
dx
subject to : y ( x0 )  y0
(2)
d2y
solve :
 f ( x, y , y ' )
2
dx
subject to : y ( x0 )  y0 , y ' ( x0 )  y1
(3)
and
are first and second order initial-value problems,
respectively. See Fig 1.2.1 and 1.2.2.
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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.
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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).
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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.
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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).
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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 )  xy1/ 2
and
f
x
 1/ 2
y 2 y
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.
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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.
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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.
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Assumptions
Express assumptions in terms
of different equations
If necessary,
alter assumptions
or increase resolution
of the model
Check model
Predictions
with know
facts
Mathematics
formulation
Solve the DEs
Display model predictions,
e.g., graphically
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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.
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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.
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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)
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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
  Ah 2 gh
dt
(9)
From (9), since we have V(t) = Awh, then
Ah
dh

2 gh
dt
Aw
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(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
d 2q
dq 1
L 2  R  q  E (t )
dt
dt C
(11)
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
m
d s
 mg or
2
dt
2
d s
 g
2
dt
(12)
Initial value problem
d 2s
 g,
2
dt
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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
dv
m  mg  kv
dt
(14)
and can be written as
d 2s
ds
m 2  mg  k
dt
dt
or
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d 2s
ds
m 2  k  mg
dt
dt
(15)
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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 2 x
 2 x
2
32 dt
or
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d 2 x 64
 x0
2
dt
L
(16)
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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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