MATH168 - DIFFERENTIAL EQUATIONS I
1
This is the first of series of lecture
notes primarilly taken from the text
book (Elementary differential Equation
and boundary value problems by Boyce
and Diprima. After going through this
lecture notes, you would be able to:
1
basic concepts of differential equations
Lecturer: Dr. Peter Amoako-Yirenkyi
Recommended Textbook:Elementary differential Equation2
The laws of the universe are written largely in the language of mathematics. Algebra is sufficient to solve many static problems, but the
most interesting naturally phenomena involve change and are best described by equations that relate changing quantities. Many important
and significant problems in engineering, the physical sciences, and the
social sciences such as economics and business when formulated in
mathematical terms require the determination of a function satisfying
an equation containing the derivatives of unknown function. Such
equations are called differential equation3 .
• know what is a differential equation,
and state the difference between
the independent and dependent
variables.
• classify differential equations in
terms of types, order, degree and
linearity.
W.W. Boyce & R.C. DiPrima. Elementary Differential Equations and Boundary
Value Problems. John Wiley & Sons,
Inc., tenth edition, March 2012. ISBN
978-0-470-45832-7
2
Introduction
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One of the most familiar example of differential equation is Newton’s
law:
m
d2 x
= F.
dt2
(1)
for the position x (t) of a particle acted on by a force F. In general F
dx
will be a function oftime t, the position x , and the velocity
. To
dt
determine the motion of a particle acted on by a given force F it is
necessary to find a function x (t) satisfying Eq. (1). If the force is that
due to gravity, then F = −mg and
m
d2 x
= −mg.
dt2
(2)
On integrating Eq. (2) we have:
dx
= − gt + c1
dt
1
x (t) = − gt2 + c1 t + c2
2
(3)
where c1 and c1 are constants. To determine x (t) completely it is necessary to specify two additional conditions, such as the position and
velocity of the particle at some instant of time. These conditions can
be used to determine the constants c1 and c1 . This unit introduces
us to the differential equation in general and helps to differentiate
between various kinds and associated solutions.
Notation 1 The expressions y0 , y00 ,
y000 , y4 ,..., yn are often used to represent,
respectively, the first, second, third, fourth,
..., nth derivatives of y with respect to the
independent variable under consideration.
d2 y
if the independent
Thus y00 represents
dx2
d2 y
variable is x , but represents
if the
dp2
independent variable is p. Observe that
parentheses are used in y(4) and y(n)
distinguish it from the nth power, y4
and yn respectively. In mechanics, if the
independent variable is time, usually
denoted by t , primes are often replaced by
...
dy d2 y
,
,
dots. Thus, ẏ, ÿ , and y represent
dt dt2
d3 y
and 3 , respectively
dt
math168 - basic concepts of differential equations
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Differential Equations
Differential equations 4 are an important part of the calculus, the
fundamentals of which are presented here. 5 A differential equation, shortly DE, is a relationship between a finite set of functions and
its derivatives. In developing the theory of differential equations in
a systematic manner it is helpful to classify different types of equations.
There are two types of derivatives which usually are (and always can
be) interpreted as rates. For example, the ordinary derivative dy/dx is
the rate of change of y with respect to x (independent variable), and
the partial derivative ∂u/∂x is the rate of change of u with respect to
x when all independent variables except x are given fixed values.
Example 1 The following are examples of differential equations6 involving
their respective unknown functions:
Definition of Differential Equation
A differential equation is an equation
which involves one or more derivatives,
or differentials of an unknown function.
5
The solution of differential equations
plays an important role in the study
and modeling population growth
and hybrid selection, radioactive
element and chemical reaction, money,
modeling advertising awareness,
reservoir simulation, disease modeling
and drug discovery,etc.
4
When an equation involves one
or more derivatives with respect to
a particular variable, that variable
is called an independent variable.
A variable is called dependent if a
derivative of that variable occurs in the
equation.
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dy
dx
dR
+ kR
dt
d2 y
+ k2 y
dx2
( x2 + y2 )dx − 2xydy
∂v
∂t
d2 i
di
1
+R + i
2
dt C
dt
∂2 v
∂2 v
+
∂x2
∂y2
d2 w 3
dw
− xw
+w
2
dx
dx
d3 x
dx
+ x − 4xy
dy
dy3
L
d2 y
dx2
d3 y
dx3
d2 y
g
+ siny
l
dx2
dy 7
d2 y 3
dy 3
+ 3y
+ y3
dx
dx
dx2
= 3x2
(4)
= 0
(5)
= 0
(6)
= 0
= h2
(7)
∂2 v
∂x2
+
∂2 v ∂y2
= Eω cos ωt
(8)
(9)
= 0
(10)
= 0
(11)
For example looking at equation (9) i
would be the dependent variable, t the
independent varible, and L , R , C , E ,
and ω are called parameters.
Again, in equation (10) has one dependent variable v, and two independent
variables; x and y.
Finally since equation (7) may be
written either in the form:
x2 + y2 − 2xy
or
= 0
=
=
r
dy 2
m
1+
H
dx
h dy 3
i 32
+ ex
dx
(12)
(13)
dy
=0
dx
x 2 + y2
dx
dy
− 2xy = 0
we may consider(if no further information is given) either variable to be
dependent and the other being the
independent one
(14)
= 0
(15)
= 5x
(16)
Exercise 1 Identify the independent variables, the dependent variables, and the parameters in the equations given as
examples in this lecture notes.
math168 - basic concepts of differential equations
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Classification of Differential Equations
Recall that a differential equation is an equation (has an equal sign,
but not an identity) that involves derivatives. Just as biologists have
a classification system for organisms, mathematicians have a classification system for differential equations. 7 They can be further
described by attributes that best classifies them. In fact methods for
solving Differential equations and the nature of the solutions depend
heavily on the class of equation being solved.
Differential equations are classified
generally by:
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• type (ordinary or partial)
• order
• degree and
• linearity
Classification by type
One important classification is based on whether the unknown function depends on a single independent variable or on several independent variables. In the first case8 , only ordinary derivatives appear in
the differential equation, and it is said to be an ordinary differential
equation.
This allows us to place all differential equations into two types:
ordinary differential equations and partial differential equations.
Examples are equations: (4),(5),(6),(7),(9),(11), (12),(13),(14),(15),and
(16). So also is each of the two equations with more than one dependent variable:
dx
dy
+b
=c
dt
dt
dx
dy
d
+e
= f
dt
dt
a
(17)
(18)
A Partial Differential Equation9 is one involving partial derivatives of
one or more dependent variables with respect to one or more of the
independent variables. Examples are equation: (8) and (10)
Classification by Order
The order10 of a differential equation depends on the order of the
derivative that are present in the entire differential equation. For
instance,
d2 y
+y = 0
+ 2b
dx
dx2
dy An Ordinary Differential Equation,
(ODE) is one containing ordinary
derivatives of one or more unknown
functions (dependent variable) with respect to a single independent variable.
8
Note:
The systematic treatment of partial
differential equation lies beyond the
scope of this course.
9
10
(19)
Definition 1 The order of a differential
equation is the order of the highest-ordered
derivative appearing in the equation.
is an equation of “order two“. It is also referred at as a ”second-order
ordinary differential equation.“ More generally, the equation11
F x, y, y0 , y00 , ..., y(n) = 0
(20)
is called an “nth-order” ordinary differential equation. Equation
(22) represents a relation between the n + 2 variables x, y, y0 , ..., yn
Note: Equation (20) is called the
general form. Note also that Equation
(22) is an nth order differential equation
because:
11
yn =
dn y
dx n
(21)
math168 - basic concepts of differential equations
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which under suitable conditions can be solved for yn in terms of the
other variables:
yn = f x, y, y0 , y00 , ..., y(n−1)
(22)
For the purpose of this course we shall assume that this is always
possible. Otherwise, an equation of the form of equation12 (20) may
actually represent more than one equation of the form of equation
(22).
For example, the equation x2 (y0 )2 − 3y0 + 2x = 0 actually represents
the two different equations,
√
√
3 + 9 − 8x3
3 − 9 − 8x3
0
0
y =
or y =
2x2
2x2
12
Example 2 (classified by order)
L
di
+ Ri = E (order1)
dt
yy0 = x (order2)
∂2 z
∂2 z
+ 3 2 = 0 (order2)
2
∂x
∂y
(23)
(24)
(25)
Classification by Degree
The degree13 of the differential equation is the power (exponent) or
the index that its highest ordered derivative is raised, if the equation
is rationalized or cleared of fractions with regard to the dependent
variable and its derivatives involve in it.
From equation (11), squaring both sides results in:
d3 y 2
dx3
=
h dy 3
dx
3 2
+ ex
i3
5
13
Definition 2 The degree of a differential equation is the power of the highest
derivative term.
, degree 2
y
= e x is an ordix2 + 1
nary differential equation of order three and degree two.
The differential equation
d y
dx3
+
d2 y
dx2
+
Exercise 2 State the order and degree of equations:
(4),(5),(6),(7),(8),(9),(10),(11), (12),(13),(14),(15),and (16).
Classification by linearity
Generally, a differential equation may be classified as either linear14 or non-linear. A linear differential equations have homogeneous solutions which can be added together to form other homogeneous solutions. A linear differential equation can also be ordinary or
partial. The homogeneous solutions to linear equations form a vector
space.
Definition 3 A differential equation is linear if it can be put in the form:
an x y(n) + an−1 x y(n−1) + ... + a2 x y00 + a1 x y0 + a0 x y = f ( x ) (26)
where an is not identically zero and also the subscripted (or indexed) a’s are
functions of independent variable ( x ) only.
The conditions for a linear differential
equation are as follows:
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1. The dependent variable and all its
derivatives occur only in the first
degree15 (or to the first power).
2. No product of the dependent
variable, say y , and/or any of its
derivatives present.
3. No transcendental function (trigonometric, logarithmic or exponential)
of the dependent variable and/or its
derivatives occurs.
15
Remark 1 A linear differential equation is always in the first degree of the
dependent variable (variables) and the
derivatives.
math168 - basic concepts of differential equations
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dy
+ y = x2
dx
It is linear, it does not matter that the independent variable x is raised to
the power 2, the dependent and the derivative are notes
Example 3 ((linearity)) 1.
2. 3x2 y00 + 2In( x )y0 + e x y = 3x cos x
This is a second order linear ordinary differential equation
Example 4 ((Non-linear)) 1. 4yy00 − x3 y0 + cos y = e2x
This is not a linear differential equation because of the 4yy00 and the cos y
terms. Other examples are:
dy
+ y2 = 0
dx
dy 2
ii
+ 3y = 0
dx
d3 y d2 y 3 dy
iii
+
−
= ex
dx
dx3
dx2
i
Solution of a Differerntial Equation
Unlike algebra, in which we seek the unknown numbers that satisfy
an equation such as x3 + 7x2 − 11x + 41 = 0. In solving16
a differential equation we are challenged to find the unknown functions,
say y = f ( x ), for which an identity such as f 0 ( x ) − 2x f ( x ) = 0 (or in
dy
Leibniz notation of
− 2xy = 0) holds on some interval of numbers.
dx
Ordinarily, we will want to find all solutions of the differential equation if possible. The solution of differential equations
plays an important role in the study of the motions of heavenly
bodies such as planets, moons and artificial satellites. Two questions
that we will be asking repeatedly of a differential equation in this
course are:
1. Is there a solution to the differential equation?
2. Is the solution given unique?
Example 5 Show that for any values of the arbitrary constant c1 and c2
the function φ = c1 cos x + c2 sin x is a solution of the differential equation
d2 y
+ y = 0.
dx2
Solution:
We will need a second derivative of the solution function, φ. If φ is a
d2 y d2 φ solution to F y, 2 = 0, then F φ, 2 = 0. If φ proves otherwise
dx
dx
d2 y then it is not a solution to F y, 2 = 0. We differentiate φ twice to
dx
d2 φ
obtain
.
dx2
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Definition 4 A solution of a differential
equation17 is any function, say φ that
satisfies the given differential equation on a
specified interval, say φ.
A solution may be defined on
the whole real line (−∞, ∞) or on
only a part of the line often an interval ( a, b). The n derivatives of
the function must exist on the interval, say a < x < b such that
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φ(n) ( x ) = f x, φ( x ), φ0 ( x ), ..., φ(n−1) ( x )
for every x in a < x < b.
Thus if φ is a solution18 of some
first order differential equation, say
dy = 0 on an interval I(real),
F x, y,
dx
dφ then it implies F x, φ,
= 0.
dx
18
Note that to test whether a given
function solves a particular differential
equation, we substitute the function φ
and its derivatives into the differential
equation. If the equation reduces to
identity (0) , then the function φ solves
the equation otherwise it does not.
It is also important to note that since
solutions are often accompanied by intervals, these intervals can explain some
important information or behaviour
about the solution.
math168 - basic concepts of differential equations
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dφ
d2 φ
= −c1 sin x + c2 cos x ⇒ 2 = −c1 cos x − c2 sin x
dx
dx
d2 y
d2 φ
+y = 0 ⇒ 2 +φ = 0
dx2
dx
d2 y
+ y = (−c1 cos x − c2 sin x ) + (c1 cos x + c2 sin x ) = 0
dx2
This implies we have an identity. Hence φ is a solution to the given
ODE and that the formula φ = c1 cos x − c2 sin x gives all possible
d2 y
solution of the equation 2 + y = 0.
dx
Since sin x and cos x are continuous in the entire real line, the solution is defined in the entire real line (−∞, ∞) for any arbitrary
constant c1 and c2 .
1
is a solution of y0 + 2xy2 = 0 on
−1
I = (−1, 1) but not on any larger interval containing I
Example 6 Show that y =
x2
Solution:
We wil need a first derivative of the solution function.
1
−2x
y= 2
and y0 = 2
are well defined functions on (-1,1)
x −1
( x − 1)2
Imitating the LHS of the differential equation y0 + 2xy2 = 0, we
h 1 i2
2x
1
have: y0 + 2xy2 = − 2
+
2x
= 0. Thus, y = 2
is
( x − 1)2
x2 − 1
x −1
a solution of I = (−1, 1).
1
Note, however, that 2
is not defined at x = +1 and therefore
x −1
could not be a solution on any interval containing either of these two
points.
Example 7 Show that y = ln x is a solution of xy00 + y0 = 0 on I = (0, ∞)
but is not a solution on I = (−∞, ∞)
see 19 where the independent variable true for all positve values.
Solution:
We will need the first and second derivative of the soliution function.
1
1
y = ln x, y0 = and y00 = − 2 are well-defined functions on (0, ∞).
x
x
1 1
Imitating the LHS of xy00 + y0 = 0 ,we have: x − 2 + = 0.
x
x
Thus, y = ln x is a solution on (0, ∞). Note that y = ln x could not be
a solution on (−∞, ∞), since the logarithm is undefined for negative
numbers containing.
Example 9 Prove that y = e− x + sin x is a solution of
d2 y
+ y = 2e− x
dx2
Solution:
We will need a second derivative of solution function to do this.
y = e− x + sin x, y0 = −e− x + cos x and y00 = e− x − sin x Imitating the
d2 y
LHS of differential equation 2 + y = 2e− x
dx
e− x − sin x + e− x + sin x = 2e− x
19
3
Example 8 Show that y = x − 2 is a
solution of 4x2 y00 + 12xy0 + 3y = 0 for
x>0
Solution:
We will need the first and second
derivative of the solution function to
3
3 5
do this. y = x − 2 , y0 = − x − 2 and
2
15 − 7
y00 =
x 2 are well-defined functions
4
on x > 0 (0, ∞)
Imitating the LHS of 4x2 y00 + 12xy0 +
3y = 0, we have:
15 7 3 5
4x2
x − 2 + 12x − x − 2 + 3x = 0 (27)
4
2
3
Thus y = x − 2 is a solution of x > 0
3
1
Note, however, that x − 2 = √ could
x3
not be a solution on (−∞, 0], since zero
and any negative real number plug into
it would give an undefined number and
complex number respectively, which is
not what we are looking for.
math168 - basic concepts of differential equations
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Example 10 Show that the two functions y = c2 − x2
and y =
12
dy
− c2 − x2 are both solutions of the equation x + y
= 0, −c < x < c.
dx
Solution:
We will need a first derivative to do this
12
− 12
y = + − c2 − x 2 ,
y 0 = − + x c2 − x 2
h
ih
i
1
1
x + + −(c2 − x2 ) 2 − + x (c2 − x2 )− 2 = 0
(28)
(29)
References
W.W. Boyce & R.C. DiPrima. Elementary Differential Equations and
Boundary Value Problems. John Wiley & Sons, Inc., tenth edition,
March 2012. ISBN 978-0-470-45832-7.
7