Chapter 2. Introduction to Differential
Equations
Le Cong Nhan
Faculty of Applied Sciences
HCMC University of Technology and Education
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Contents
1
Definition and Terminology
2
Initial-Value Problems
3
Differential Equations as Mathematical Models
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2.1 Definition and Terminology
Definition 1 (Differential Equations, DEs)
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:
(1) Type (loại):
Ordinary differential equation (ODE):
dy
= 3x 2 + x,
dx
d 2x
dx
+3
+ x = sin t,
dt 2
dt
dx
dy
+
= 2x + 3y .
dt
dt
Partial differential equation (PDE)
∂2u
∂2u
+
= 0,
2
∂x
∂y 2
∂u
∂2u
∂u
=
−
,
∂t
∂x 2
∂x
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∂u
∂v
=
.
∂x
∂y
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(2) Order (cấp): The order of a differential equation (ODE or PDE) is the order
of the highest derivative in the equation.
dy
= 3x 2 + x,
dx
d 2x
dx
+3
+ x = sin t,
dt 2
dt
2
∂4u
∂2u
+
=0
∂x 4
∂t 2
An nth-order ODE can be expressed by the general form (dạng tổng quát)
F x, y , y 0 , ..., y (n) = 0,
(1)
or by the normal form (dạng chuẩn)
d ny
0
(n−1)
x,
y
,
y
,
...,
y
.
=
f
dx n
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(3) Linearity (Tính tuyến tính): The nth-order linear ODE
an (x)
d ny
d n−1 y
dy
+ a0 (x)y = g (x).
+ an−1 (x) n−1 + · · · + a1 (x)
n
dx
dx
dx
(3)
Examples for linear equations
(y − x)dx + 4xdy = 0,
y 00 − 2y 0 + y = 0,
x3
d 3y
dy
− 5y = e x .
+ 3x
dx 3
dx
and nonlinear equations
(1 − y )y 0 + 2y = e x ,
d 2y
+ sin y = 0,
dx 2
d 4y
+ y 2 = 0.
dx 4
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Definition 2 (Solution of an ODE)
A solution of an nth-order ordinary differential equation (1) is a function φ that
possesses at least n derivatives and
F x, φ(x), φ0 (x), ..., φ(n) (x) = 0 for all x ∈ I .
(4)
Terminology:
Interval of definition (Domain of solution)
Solution curve
Explicit and Implicit Solutions
Families of solutions
Singular solution
Systems of differential equations
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Example 3
The function y = − x1 is a solution of the differential equation
xy 0 + y = 0
on the interval (−∞, 0) or (0, ∞).
Domain of the function y = − x1 is D = R\{0}
Interval of definition I = (−∞, 0) or I = (0, ∞).
Distinguish function and solution.
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Example 4
The relation x 2 + y 2 = 25 is an implicit solution of the differential equation
dy
x
=−
dx
y
(5)
on the interval I = (−5, 5). This is because we can find at least one function
y = y (x) that satisfies the relation x 2 + y 2 = 25 and the equation on I = (−5, 5).
More precisely, there are two functions
p
p
y = 25 − x 2 and y = − 25 − x 2 (explicit solutions)
Remark:
The relation x 2 + y 2 − c = 0 satisfies (5) for any constant c. However, it is
understood that the relation should always make sense in the real number
system.
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Example 5
The one-parameter family y = 14 x 2 + c
solutions of the differential equation
2
is a one-parameter family of
1
dy
= xy 2
dx
(6)
on the interval (−∞, ∞).
General solution: (nghiệm tổng quát)
2
y = 41 x 2 + c
Particular solution: (nghiệm đặc biệt)
1 4
y = 16
x by choosing c = 0.
Singular solution: (nghiệm kỳ dị) y = 0.
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2.2 Initial-Value Problems
On some interval I containing x0 , the problem
d ny
Solve :
= f x, y , y 0 , ..., y (n−1)
n
dx
0
Subject to : y (x0 ) = y0 , y (x0 ) = y1 , ..., y
(n−1)
(7)
(x0 ) = yn−1 ,
where y0 , y1 , ..., yn−1 are arbitrarily specified real constants, is called an initial-value
problem (IVP).
First-order initial-value problems:
Solve :
dy
= f (x, y )
dx
(8)
Subject to : y (x0 ) = y0 ,
Hình: First-order IVP
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Second-order initial-value problems:
Solve :
d 2y
= f (x, y , y 0 )
dx 2
Subject to : y (x0 ) = y0 , y 0 (0) = y1
Hình: Second-order IVP
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Example 6
Let y = ce x be the one-parameter family of solutions of the simple first-order
equation y 0 = y on the interval (−∞, ∞).
Then the function y = 3e x is a solution
of the initial-value problem
y0 = y,
y (0) = 3.
(9)
and the function y = −2e x−1 is a solution
of the initial-value problem
y0 = y,
y (1) = −2.
(10)
Hình: Solutions of IVPs
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Let y = 1/(x 2 + c) be a one-parameter family of
solutions of the first-order equation
y 0 + 2xy 2 = 0.
Imposing the initial condition y (0) = −1 then
y=
1
x2 − 1
.
We distinguish the following:
a function with domain R\{±1}.
a solution of y 0 + 2xy 2 = 0 on the intervals
(−∞, −1), (−1, 1) and (1, ∞).
a solution of the initial-value problem
y 0 + 2xy 2 = 0,
y (0) = −1
on interval of definition I = (−1, 1).
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Existence and Uniqueness
Consider an initial-value problem
d ny
0
(n−1)
=
f
x,
y
,
y
,
...,
y
dx n
y (x0 ) = y0 , y 0 (x0 ) = y1 , ..., y (n−1) (x0 ) = yn−1 ,
There are two fundamental questions:
Does a solution of the problem exist? If a solution exists, is it unique?
1 4
The functions y = 0 and y = 16
x are
two solutions of the IVP
dy
= xy 1/2 ,
dx
y (0) = 0.
(11)
Hình: Two solutions of the same IVP
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Consider the first-order initial-value problems:
dy
= f (x, y )
dx
(12)
y (x0 ) = y0 ,
Hình: Rectangular region R
Theorem 7 (Existence of a Unique Solution)
Let R be a rectangular region containing (x0 , y0 ) in its interior given by
R = {(x, y ) : a ≤ x ≤ b, c ≤ y ≤ d} .
If f (x, y ) and ∂f /∂y (x, y ) are continuous on R, then there exists some interval
I0 : (x0 − h, x0 + h) ⊂ [a, b], (h > 0) and a unique function y (x) defined on I0 that
is a solution of the initial-value problem (12).
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Example 8
By using Theorem 7 prove that the IVP
dy
= xy 1/2 ,
dx
y (2) = 1
(13)
has a unique solution on I = (2 − h, 2 + h) for all h > 0.
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2.3 Differential Equations as Mathematical Models
A mathematical model is a mathematical description (often by means of a function
or an equation) of a real-world phenomenon. such as:
the size of a population
the demand for a product
the speed of a falling object, etc.
Hình: The modeling process
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Population Dynamics
How to model human population growth? The size of population at the time t?
Mathematical model
Variables: P(t) denotes the total population at time t.
Assumptions and hypotheses:
Thomas Malthus’s rule (1978): The rate at which a population of a
country grows at a certain time is proportional to the total population of the
country at that time.
Then P holds
dP
= kP,
dt
(14)
where k is a constant of proportionality.
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Radioactive Decay
The amount of the radioactive substance remaining at time t?
Mathematical model
Variables: A(t) is the amount of the substance remaining at time t.
Assumptions and hypotheses:
The rate dA/dt at which the nuclei of a substance decays is proportional to
the amount (more precisely, the number of nuclei) A(t) of the substance
remaining at time t.
Then A holds
dA
= kA,
dt
(15)
where k is a constant of proportionality.
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Newton’s Law of Cooling/Warming
The temperature of a body at time t?
Mathematical model
Variables: T (t) the temperature of a body at time t.
Assumptions and hypotheses:
Newton’s empirical law of cooling/warming: The rate at which the
temperature of a body changes is proportional to the difference between the
temperature of the body and the temperature of the surrounding medium.
Then T holds
dT
= k(T − Tm ),
dt
(16)
where Tm is the temperature of the surrounding medium.
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Spread of a Disease
A spreading disease, (for example, a flu virus) is spread throughout a community
by people coming into contact with other people?
Mathematical model
Variables: Let x(t) denote the number of people who have contracted the
disease and y (t) the number of people who have not yet been exposed.
Assumptions and hypotheses:
The rate dx/dt at which the disease spreads is proportional to the number of
encounters or interactions between these two groups of people. Then x holds
dx
= kxy ,
dt
(17)
where k is the usual constant of proportionality.
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Draining a Tank
Suppose a tank filled with water is allowed to drain through a hole under the
influence of gravity. We would like to find the depth h of water remaining in the
tank at time t.
Torricelli’s law: the speed v of efflux of water
through a sharp-edged hole at the bottom of a
tank filled to a depth h is the same as the speed
that a body (in this case a drop of water) would
acquire in falling freely from a height h; that is,
p
v = 2gh,
where g is the acceleration due to gravity. This
last expression comes from the equation
Hình: Water draining from a tank
kinetic energy = potential energy
that is, 12 mv 2 = mgh.
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Series Circuits
Consider the single-loop LRC-series circuit containing an inductor, resistor, and
capacitor.
The current in a circuit after a switch is closed is denoted by i(t) and the charge
on a capacitor at time t denoted by q(t).
Voltage drops across
di
an inductor L: henrys, (h) is L dt
.
a resistor R: ohms, (Ω) is iR.
a capacitor C : farads, (f) is C1 q.
Kirchhoff’s second law:
Hình: LRC-series circuit
E (t) = L
di
1
dq
+ iR + q, where i =
.
dt
C
dt
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Falling Bodies
Now suppose a rock is tossed upward from a roof of a building. What is the
position s(t) of the rock relative to the ground at time t?
Assume that the upward direction is positive,
then by Newton’s second law of motion gives
m
d 2s
= −mg
dt 2
or
d 2s
= −g
dt 2
This is because the net force is F = F1 = −W
and the magnitude of the weight is W = mg .
Hình: Position of rock measured
from ground level
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Falling Bodies and Air Resistance
Now suppose a feather is dropped from a roof of a building. What is the position
s(t) of the feather relative to the ground at time t?
Assume that the downward direction is positive,
then by Newton’s second law of motion the
net force acting on the mass is
F = F1 + F2 = mg − kv
where the weight F1 = mg and air resistance
F 2 = −kv is a force, called viscous damping, or
drag.
Hình: Falling body of mass m
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Suspended Cables
Suppose a flexible cable, wire, or heavy rope is suspended between two vertical
supports, for example: telephone wire, suspension bridge, etc. Our goal is to
construct a mathematical model that describes the shape that such a cable
assumes.
Hình: Telephone wire
Hình: Suspension bridge
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Let us examine only a portion or element of the cable between its lowest point
P1 and any arbitrary point P2 .
the tensions T1 and T2 in the cable with
magnitude T1 = |T1 | and T2 = |T2 |
T1 = −T1 i,
T2 = T2 cos(θ)i + T2 sin(θ)j
the portion W of the total vertical load
between the points P1 and P2 with
magnitude W = |W|.
Hình: Element of cable
W = −W j
Because of static equilibrium, we obtain
T1 = T2 cos(θ),
W = T2 sin(θ)
Since dy /dx = tan(θ), we arrive at
dy
W
=
.
dx
T1
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Parabolic cable
The vertical load is only a uniform roadbed distributed along the horizontal axis
(e.g. Suspension bridge). In other words, it is assumed that the weight of all
cables is negligible in comparison to the weight of the roadbed.
If the weight per unit length of the roadbed (say, pounds per horizontal foot) is a
constant ρ and x is the horizontal distance of P1 P2 , then W = ρx. This gives us
dy
ρ
=
x.
dx
T1
Hình: Shape of a cable
(19)
Hình: Suspension bridge
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Catenary cable
The shape of a uniform flexible cable hanging under its own weight, such as a wire
strung between two telephone posts, electric wire. The vertical load is now
distributed uniformly along the length of the cable.
If ρ is the linear density of the wire (measured, say, in lb/ft) and s is the length of
the segment P1 P2 , then
(20)
W = ρs,
where the arc length between points P1 and P2 is given by
s
2
Z x
dy
s=
1+
dx.
dx
0
This leads to the the nonlinear second-order equation
s
2
ρ ds
d 2y
ρ
dy
d 2y
=
or
=
1+
.
2
2
dx
T1 dx
dx
T1
dx
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