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AMA126 Differential and
difference equations
University of Sheffield, Dept of Applied Mathematics
2007
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Revision of Integration (Indefinite Integration)
When a function f(x) is known we can differentiate it to obtain its derivative df/dx. The
reverse process is to obtain the function f(x) from knowledge of its derivative. This
process is called integration and it has numerous applications in all areas of sciences.
Suppose we differentiate the function y=x2. We obtain
dy
 2x
dx
Integration reverses this process and we say that the integral of 2x is x2. Schematically
we can regard the process of integration in the following way
differentiate
x2
2x
integrate
In reality the situation is a bit more complicated because differentiating any of
x 2  7, x 2  3, x 2  0.5
yields 2x. In fact, any function of the form x2+c, where c is a constant, will be the answer
to our question since when we differentiate the constant term we obtain zero.
Consequently, when we reverse the process of differentiation, we do not know what the
original constant term might have been. So we include in our answer an unknown
constant, c say, called the constant of integration and we state that the integral of 2x is
x2+c. This constant must always be included when finding an indefinite integral. The
solution containing the constant c defines a family of curves all being solutions of the
integral. If one of these curves is a solution of the integral, they all are. To select between
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them we need more information. This problem defines so called initial conditions, a topic
you will meet later when solving differential equations.
The indefinite integration is represented by the symbol ∫ known as an integral sign.
Accompanying the integral sign is always a term of the form dx, which indicates the
independent variable involved in the integration, in this case x. So we write
 2 x dx  x
2
c
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Example
(a) State the derivative of x3
(b) Hence find the indefinite integral of 3x2
Solution
(a) From our knowledge of differentiation, the derivative of x3 is 3x2.
(b) Indefinite integration reverses the process of differentiation, and so we write
 3x dx  x
2
3
c
We always include the additional constant of integration when finding indefinite
integrals. Note that our answer can be checked by differentiating x3+c to obtain 3x2.
More generally, we have the following relationship between derivatives and indefinite
integrals:
if
d
F x   f x  then
dx
 f x  dx  F x   c
In the expression ∫ f(x) dx, the function f(x) is referred to as the integrand. When we have
calculated ∫ f(x) dx, we say f(x) has been integrated with respect to x to yield F(x)+c.
In what follows we are going to discuss some techniques which can be used to evaluate
many types of integrals.
I. Integration by parts
If f and g are differentiable functions, then by the Product Rule
d
 f ( x) g ( x)  f ( x) dg ( x)  g ( x) df ( x)
dx
dx
dx
d
f ( x) g ' ( x)   f ( x) g ( x)  g ( x) f ' ( x)
dx
or equivalently
where the dash means the differentiation of the function with its argument. Integrating
both sides of the previous equation gives us
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 f ( x) g ' ( x)dx   dx  f ( x) g ( x)dx   g ( x) f ' ( x)dx
d
The simplest is to perform the first integral on the right side and the integral is f(x)g(x)+C
(why???). Since another constant of integration results from the second integral, it is
unnecessary to include C in the formula; that is
 f ( x) g ' ( x)dx 
f ( x) g ( x)   g ( x) f ' ( x)dx
If we let u=f(x) and v=g(x), so that du=f′(x)dx and dv=g′(x)dx, then the preceding
formula may be written (integration by part)
 udv  uv   vdu
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Example: Find
 xe
2x
dx
Solution
There are four possible choices for dv, namely dx, xdx, e2xdx or xe2xdx. If we let dv=
e2xdx, then the remaining part of the integrand is u; that is u=x. To find v we integrate dv
to obtain v=e2x/2. Note that a constant of integration is not added at this stage of the
solution. Since u=x we see that du=dx. For ease of reference it is convenient to display
these expressions as follows
ux
dv  e 2 x dx
du  dx v 
1 2x
e
2
Substituting these expressions in the definition equation, we obtain
 xe
2x
1
1

dx  x e 2 x    e 2 x dx
2
2

The integral on the right hand side may be found using the integrals of exponential
functions (see in the table below). This gives us
 xe
2x
1
 1
dx  x e 2 x   e 2 x  C
2
 4
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It takes considerable practice to become proficient in making a suitable choice for dv. To
illustrate, if we had chosen dv=xdx, then it would have been necessary to let u=e2x,
giving us
u  e2x
dv  xdx
du  2e 2 x dx v 
1 2
x
2
Integrating by parts we obtain
 xe
2x
dx 
1 2 2x
x e   x 2 e 2 x dx
2
Since the exponent associated with x has increased, the integral on the right is more
complicated than the given integral. This indicates an incorrect choice of dv.
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II. Trigonometric integrals
In many cases evaluating an integral we can meet trigonometric functions. For example,
integrals of the type ∫sinnxdx require new method of solving. If n is an odd positive
integer, we begin by writing
 sin
n
xdx   sin n 1 x sin xdx
Since the integer n-1 is even, we may then use the fact that sin2x=1-cos2x to obtain a form
which is easy to integrate.
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Example
Evaluate ∫sin5xdx.
Solution
As discussed earlier we have
 sin xdx   sin x sin xdx   (sin
  (1  2 cos x  cos x) sin xdx
5
4
2
2
x) 2 sin xdx   (1  cos 2 x) 2 sin xdx 
4
We next employ the method of substitution, letting u=cosx and du=-sinxdx. Thus
 sin
5
xdx    (1  2 cos 2 x  cos 4 x)(  sin x)dx    (1  2u 2  u 4 )du 
2
1
2
1
 u  u 3  u 5  C   cos x  cos 3 x  cos 5 x  C
3
5
3
5
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A similar technique can be employed for odd powers of cos x, specifically, we write
 cos
n
xdx   cos n 1 x cos xdx
and use the fact that cos2x=1-sin2x in order to obtain an integrable form. If the integrand
is sinnx or cosnx and n is even, then the half-angle formulas
1  cos 2 x
2
sin 2 x 
or cos 2 x 
1  cos 2 x
2
may be used to simplify the integrand.
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Example
Evaluate ∫sin4xdx
Solution
1
 1  cos 2 x 
2
 sin xdx   (sin x) dx    2  dx  4  (1  2 cos 2 x  cos 2 x)dx
2
4
2
2
We apply a half-angle formula again and write
cos 2 2 x 
1
1 1
(1  cos 4 x)   cos 4 x
2
2 2
Substituting in the last integral and simplifying gives us
 sin
4
xdx 
1 3
1
3
1
1

  2 cos 2 x  cos 4 x dx  x  sin 2 x  sin 4 x  C

4 2
2
8
4
32

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Integrals of the form ∫sinmxcosnxdx where m and n are positive integers may be found by
using variations of the previous techniques. If m and n are both even, then half-angle
formulas should be employed first. In n is odd, we can write
 sin
m
x cos n xdx   sin m x cos n 1 x cos xdx
and express cosn-1x in terms of sinx by using the identity cos2x=1-sin2x. The substitution
u=sinx then leads to an integrand which can be handled easily. A similar technique can
be used if m is odd.
III. Trigonometric substitutions
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If an integrand contains the expression
substitution x=asinθ leads to

a 2  x 2 where a>0, then the trigonometric

a 2  x 2  a 2  a 2 sin 2   a 2 1  sin 2   a 2 cos 2   a | cos  |
When making this substitution or the other trigonometric substitutions in the next
examples, we shall assume that θ is in the range of the corresponding inverse
trigonometric function. Thus, for the sine substitution above –π/2≤θ≤π/2. Consequently
cosθ≥0 and a 2  x 2  a cos . Of course, if
make the further restriction –π/2<θ<π/2.
a 2  x 2 occurs in a denominator we
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Example
Evaluate
1
 x 2 a 2  x 2 dx, a  0
Solution
Let x=asinθ , where –π/2<θ<π/2. It follows that


a 2  x 2  a 2  a 2 sin 2   a 2 1  sin 2   a cos 
Since x=asinθ, we have dx=acosθdθ. Substituting in the given integral
x
1
2
a x
2
2
dx  
1
(a sin  )a cos 
2
2
a cos  d 
1
1
1
d   2 cot   C
2 
2
a sin 
a
It is now necessary to return to the original variable of integration x. A simple method of
doing so is to use a geometrical approach. If 0<θ<π/2, then since sinθ=x/a, we may
interpret θ as an acute angle of a triangle having opposite side and hypotenuse of lengths
x and a, respectively. The length
the Pythagorean Theorem.
a 2  x 2 of the adjacent side is calculated by means of
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Referring to the triangle we see that
a2  x2
x
cot  
It can be shown that this formula is also valid if –π/2<θ<0. Thus the above figure can be
used whether θ is positive or negative. Substituting the new form of cotθ in the result
obtained for the integral, we have
x
1
2
a 2 x2
dx  
1
a2
a2  x2
a2  x2
C  
C
x
a2 x
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If an integrand contains a 2  x 2 , where a>0, then the substitution x=atanθ will
eliminate the radical sign. When using this substitution it will be assumed that θ is in the
range of the inverse tangent function; that is –π/2<θ<π/2. After making this substitution
and evaluating the resulting trigonometric integral, it is necessary to return to the original
variable, x. The preceding formulas show that
tan  
x
a
and sec  
a2  x2
a
For integrands containing x 2  a 2 we substitute x=asecθ, where θ is chosen in the
range of the inverse secant function; that is either 0≤θ≤π/2 or π≤θ<3π/2. After evaluating
the integral, we have to return to the original variable and we use
sec  
x
a
and
tan  
x2  a2
a
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Example
Evaluate

x2  9
dx
x
Solution
Let us substitute as follows
x  3 sec  , dx  3 sec  tan  d
Consequently
x 2  9  9 sec 2   9  3 sec 2   1  3 tan 
and therefore
x2  9
3 tan 
2
2
 x dx   3 sec 3 sec tan  d  3 tan  d  3 sec   1d 
 3tan      C
Since secθ=x/3 we may refer to the changing rule into the old variable and we write
 x2  9
x2  9
 x 
 x
dx  3
 sec 1    C  x 2  9  3 sec 1    C
x
3
3
 
3


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III. Partial Fractions
It is easy to verify that
2
1
1


x 1 x 1 x 1
2
The expression on the right side of the equation is called the partial fraction
decomposition of the expression on the left side. This decomposition may be used to find
the indefinite integral of the expression on the left side. We merely integrate each of the
fractions which make up the decomposition independently, obtaining
x
2
2
1
1
x 1
dx  
dx  
dx  ln | x  1 |  ln | x  1 | C  ln
C
x 1
x 1
x 1
1
It is theoretically possible to write any rational expression f(x)/g(x) as a sum of rational
expressions whose denominators involve powers of polynomials of degree not greater
than two. More specifically, if f(x) and g(x) are polynomials and the degree of f(x) is less
than the degree of g(x), then it follows from a theorem in algebra that
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f ( x)
 F1  F2    Fk
g ( x)
where each of Fi has one of the forms
A
 px  q 
m
or
ax
Cx  D
2
 bx  c

n
for some non-negative integers m and n, and when ax2+bx+c is irreductible, in the sense
that this quadratic expression has no real roots, that is b2-4ac<0.
To find the fraction decomposition of a rational expression f(x)/g(x) it is essential that f(x)
have lower degree than g(x). If this is not the case, then long division should be employed
to arrive at such expression. For example, given
x 3  6 x 2  5x  3
x2 1
we obtain, by long division,
x 3  6 x 2  5x  3
6x  9
 x6 2
2
x 1
x 1
The partial fraction decomposition is then found for the last term in the right side.
In order to obtain the decomposition of a rational expression of the form f(x)/g(x), we
begin by expressing the denominator g(x) as a product of factors px+q or irreductible
quadratic factors ax2+bx+c. Repeated factors are then collected so that g(x) is a product
of different factors of the form (px+q)m or (ax2+bx+c)n, where m and n are non-negative
integers and the quadratic forms are irreductible. We then apply the following rules
Rule 1. For each factor of the form (px+q)m where m≥1, the partial fraction
decomposition contains a sum of m partial fractions of the form
Am
A1
A2


2
px  q  px  q 
 px  q m
where each of Ai is a real number.
Rule 2. For each of the factor of the form (ax2+bx+c)n where n≥1 and b2-4ac<0, the
partial fraction decomposition contains n partial fractions of the form
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A1 x  B1
A2 x  B2

2
ax  bx  c ax 2  bx  c


2

An x  Bn
ax
2
 bx  c

n
where each of Ai and Bi are real numbers.
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Example
Evaluate

3x 3  18 x 2  29 x  4
x  1x  23
dx
Solution
By Rule 1, there is a partial fraction of the form A/(x+1) corresponding to the factor x+1
in the denominator of the integrand. For the factor (x-2)3 we apply Rule 1 (with m=3),
obtaining a sum of three partial fractions B/(x-2), C/(x-2)2, and D/(x-2)3. Consequently,
the partial fraction decomposition has the form
3x 3  18 x 2  29 x  4
x  1x  2
3

A
B
C
D



2
x  1 x  2 x  2
 x  2 3
Multiplying both sides by (x+1)(x-2)3 gives us
3x 3  18 x 2  29 x  4  A x  2   Bx  1x  2  C x  1x  2  Dx  1
3
2
Two of the unknown constants may be determined easily. If we let x=2 then
24-72+58-4=3D,
6=3D, and D=2
Similarly, letting x=-1 we obtain
-3-18-29-4=-27A, -54=-27A,
and A=2
The remaining constants may be found by comparing the coefficients. If the right side of
the decomposition is expanded and like powers of x collected, we see that the coefficient
of x3 is A+B. This must equal the coefficient of x3 on the left, that is
A+B=3
Since A=2, it follows that B=1. Finally, we compare the constant terms by letting x=0.
This gives us
-4=-8A+4B-2C+D
Substituting the values we have already found for A, B and D leads to
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-4=-16+4-2C+2
which has the solution C=-3. The partial fraction decomposition is, therefore
3x3  18 x 2  29 x  4
2
1
3
2




3
2
x  1 x  2 x  2 x  23
x  1x  2
To find the given integral we integrate each of the partial fractions on the right side of the
last equation. This gives us
3
1

E
x  2 ( x  2) 2
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Obviously, there are many characteristic types of indefinite integrals with their own rules.
To see them all, please consult any book on calculus.
2 ln | x  1 |  ln | x  2 | 
Differential equations
Definition: A differential equation is an equation which involves derivatives or
differentials. Since the equations we are going to study contain only one variable, these
equations are called ordinary differential equations. The primary objective of this lecture
is to develop techniques for solving certain basic types of ordinary differential equations.
Introduction
An equation of the form


F x,y,y' ,y' ' ,... y ( n )  0 ,
2.1
where F is a function of n+2 variables, y is a function of x, and y(k) denotes the k-th
derivative of y with respect to x, is called an ordinary differential equation of order n.
The following are examples of ordinary differential equations of orders 1, 2, 3, and 4,
respectively:
y'  2 x
3
d2y
 dy 
 x 2    15 y  0
2
dx
 dx 
 y' ' '4  x 2  y' '5  4 xy  xe x
2
d4y
dy
 4   1  x 3
dx
 dx 
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If, in (2.1), F is a polynomial function, then by definition the degree of the differential
equation is the greatest exponent associated with the highest order derivative y(n). The
degrees of the preceding differential equations are 1, 1, 4, and 2, respectively.
If a function f has the property that when f(x) is substituted for y in a differential equation,
the resulting expression is an identity for all x in some interval, then f(x) (or simply f) is
called a solution of the differential equation. For example, if C is any real number, then a
solution of y’=2x is
f x   x 2  C
because substitution of f(x) for y leads to the identity 2x=2x. We call x2+C the general
solution of y’=2x, since every solution has this form.
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Example
If C1 and C2 are any real numbers, prove that
f x   C1e 5 x  C 2 e 5 x
is a solution of y’’-25y=0.
Solution
Since
f ' x   5C1e 5 x  5C2 e 5 x
f ' ' x   25C1e 5 x  25C2 e 5 x  25 f x 
we have f″(x)-25f(x)=0. This shows that f(x) is a solution of the equation y″-25y=0.
The solution given in the above Example is called the general solution of y″-25y=0.
Observe that the differential equation is of the order 2 and the general solution contains
two arbitrary parameters C1 and C2 (for connections to integrals, see why these
parameters arise in the recap about integrals).
It can be shown that the general solution of the n-th order differential equation contains n
independent parameters C1, C2, …, Cn. A particular solution is obtained by assigning
specific values to the parameters.
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Example
Find the particular solution of y’=2x which satisfies the condition that y=5 if x=2.
Solution
Let us write the general solution of y’=2x in the form y=x2+C. If the given condition is to
be satisfied, then necessarily 5=4+C or C=1. Hence the desired particular solution is
y=x2+1.
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Conditions of the type stated in the previous Example are called boundary conditions
for the differential equation. If the general solution contains one parameter, one boundary
condition is sufficient to find the particular solution. If, as in the first Example, the
general solution contains two parameters, then two boundary conditions are needed for
particular solutions. Similar statements hold if more than two parameters are involved.
The solutions we have considered express y explicitly in terms of x. Solutions of certain
differential equations are stated implicitly. In this case, implicit differentiation is used to
check the solution, as illustrated in the following example
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Example
Show that x3+x2y-2y3=C is an implicit solution of
x
2

 6 y 2 y'3x 2  2 xy  0
Solution
If the first equation is satisfied by y=f(x), then differentiating implicitly,
3x 2  2 xy  x 2 y'6 y 2 y'  0 or
x
2

 6 y 2 y'3x 2  2 xy  0
Thus y=f(x) satisfies the differential equation.
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The simplest type of differential equation is one which can be written in the form
M x   N  y  y '  0
2.2
where M and N are continuous functions. If y=f(x) is a solution of the equation (2.2), then
M x  N  f x f ' x  0
If f′(x) is a continuous function, then indefinite integration leads to
 M x dx   N  f x  f ' x dx  C
or equivalently,
 M x dx   N  y dy  C
The last equation is an (implicit) solution of (2.2). A device which is useful for
remembering this method of solution is to write equation (2.2) as
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M x   N  y 
dy
0
dx
and then change to the following differential form
M xdx  N  y dy  0
2.3
The solution is then found by formally integrating each term. The differential equation
(2.2) is said to be separable, since the variables x and y may be separated as indicated in
(2.3).
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Example
Solve the differential equation


2 xy  6 x  x 2  4 y'  0.
Solution
The given equation may be written as

2 x  y  3  x 2  4
 dy
 0.
dx
Assuming that (y+3)(x2-4)≠0, we may divide both sides by this product and express the
equation in the differential form (2.3) as follows
2x
1
dx 
dy  0.
y3
x 4
2
Thus the given differential equation is separable, and integration gives us
ln x 2  4  ln y  3  C1

or

ln x 2  4  y  3  C1
This may also be written as
x
2

 4  y  3  e C1  C 2 .
It can be shown, furthermore, that the solution is
x
2

 4  y  3  C
for every nonzero C. In explicit form,
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y
- 16 -
C
 3.
x 4
2
Since it was assumed that (y+3)(x2-4)≠0, the cases x=±2 and y= -3 require special
attention. Direct substitution shows that y= -3 is a solution of the given differential
equation; however, y′ is undefined if x=±2. Consequently, the above solution given in
explicit form is the solution if C is any real number.
___________________________________________
Example
Prove that y=ex(C1cos x+C2sin x) is a solution of the differential equation
y ' '2 y '2 y  0
Example
Solve the differential equation
x 2 y' yx 2  y
Sometimes a substitution is necessary to turn an equation into separable variables, e.g.
use the substitution z=2x+y to solve the ordinary differential equation
dy
 tan( 2 x  y )  2
dx
Obviously, first we will need to see how the new variable (and its derivative) will look
like. For this we differentiate the new variable with respect to x
dy
dz
2
dx
dx
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So the original differential equation is transformed into
dz
 tan z
dx
which now can be solved by variable separation. Integrating we have
dz
 tan z   dx
or
cos z
 sin z dz   dx
and the solution can be found solving the following chain
ln(sin z )  x  C
ln(sin[ 2 x  y ])  x  C
sin( 2 x  y )  Ae x
2 x  y  sin 1 ( Ae x )
y  sin 1 ( Ae x )  2 x
A very important case (i.e. with many applications in various fields) is the differential
equation
dy
 ky
dt
This equation can be solved by variable separation. Supposing that y≠0, we have
dy
 kdt
y 
ln y  kt  C

y  e C  kt  Ae kt
The above equation is an example of an ordinary differential equation with constant
coefficients. This type of equation will play a very important role in the second chapter of
this course (population dynamics). Let us discuss some of the properties of this
differential equation
1. Set t=0, then y(0)=A, so y(t)=y(0)ekt
2. If k>0, we have an exponential growth, if k<0 we have an exponential decay
3. Half-life (when k<0)
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We set ek=2-1/T, in other words
k 
1
ln 2
ln 2 or T  
0
T
k
then
y (t )  y (0)2
t
T
y (T )  y (0)2 1 
y (2T )  y (0)2  2
y (0)
2
y (0) y (T )


4
2
T is called the “half-life” and it is widely used in sciences and engineering (e.g.
radioactive decay)
The snowfall problem: Before noon snow starts to fall at a steady rate. At noon a snow
plough begins ploughing forward along a straight road of constant width. It shifts a
constant volume of snow per unit time. It clears a two mile stretch by one o’clock and a
further one mile by two o’clock. At what time did the snow begin to fall?
Solution
Let the snow begin to fall at time T. Then the depth of snow at time t (≥T) is c(t-T). Let
the snow plough advance from x to x+Δx in the time interval t to t+Δt. The volume of
snow removed in this time interval is ωc(t-T)Δx, where ω is the width of the plough.
Thus the rate of snow removal is
 c(t  T )
x
dx
  c(t  T )
t
dt
which is a constant. Thus we have (t-T)dx/dt=C, or
dx
k

dt t  T
(t  T )
This differential equation can be solved by separating the variables as
x  k
dt
 k ln( t  T )  A
t T
which can be rewritten as
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xA
k
 x  A   1k  x  A k
t  T  exp 
  e  e

 k  
x
t  T  Ba
ln( t  T ) 
where B=e-A/k and a=e1/k. We have three unknown constants (T, B, a) which must be
determined. We have to use the initial conditions, i.e.
 x  0 at t  0

 x  2 at t  1
 x  3 at t  2

Using these conditions we have
0  T  B

2
1  T  Ba
2  T  Ba 3

From the first equation we have B=-T, which introduced back into the remaining two
equations yields
1  T (1  a 2 )

2  T (1  a 3 )
Calculating the ratio between these two expressions we have
1  a3 1  a  a2
2

1 a
1 a2
2  2a  1  a  a 2
a2  a 1 0
1 1 4 1 5

2
2
1 2 5  5 3  5
a2 

4
2
a
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Having this constant, we can determine T as
1
2
2
2
5 1
T




2
1 a
2  (3  5 )
1 5
5 1 5 1

2( 5  1)
( 5  1)

4
2
Calculating this ratio, we obtain T= –0.618 hours=-37.1 min. Accordingly, the snow
started to fall at 11.23 am.
HOMOGENEOUS DIFFERENTIAL EQUATIONS
A function f of two variables is said to be homogeneous of degree n if
f tx,ty  t n f x, y 
3.1
for every t>0 such that (tx, ty) is in the domain of f. For example, if
f x, y   2 x 4  x 2 y 2  5 xy3
then f is homogeneous of degree 4, since
f tx,ty  2tx  tx ty  5txty 
4

2
2
3

t 4 2 x 4  x 2 y 2  5 xy3  t 4 f x, y 
Similarly, if
f  x, y  
x
1
y
e
2
2
x y
then f is homogeneous of degree -2, since
tx
1
f tx,ty   2 2
e ty  t  2 f  x, y 
2 2
t x t y
A homogeneous differential equation is an equation which can be written in the form
Px, y dx  Qx, y dy  0
3.2
where P and Q are homogeneous functions of the same degree. Equations of this type can
be transformed into separable equations by means of substitutions
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y  xv, where v  vx
3.3
and
dy  vdx  xdv
Thus, substitution of xv for y in (3.2) yields
Px, xvdx  Qx, xvvdx  xdv  0
If P and Q are homogeneous functions of degree n, then
Px, xvdx  x n P1,v  and Qx, xv  x n Q1,v 
Substituting in the preceding differential equation and dividing both sides by x n we obtain
P1,v dx  Q1,v vdx  xdv   0
This equation can be written in the separable form
1
Q1,v 
dx 
dv  0
x
P1,v   vQ1,v 
3.4
provided non-zero denominators occur. We have proved that if y=xv is a solution of
(3.2), then v is a solution of (3.4). Conversely, if v is a solution of (3.4), then reversing
our argument shows that y=vx is a solution of (3.2). It is not advisable to memorize the
final form of (3.4). Instead, remember the substitution y=xv which is used to simplify the
homogeneous equation.
___________________________________________
Example
Solve the differential equation
y
2
 xy dx  x 2 dy  0.
Solution
If P(x,y)=y2-xy and Q(x,y)=x2, then the functions P and Q are both homogeneous of
degree 2. Hence the differential equation is homogeneous and we substitute
y  xv, dy  vdx  xdv.
This leads to the following chain of equations
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x v  x v dx  x vdx  xdv  0
x v  v dx  x vdx  xdv  0
v  v dx  vdx  xdv  0
2 2
2
2
2
2
2
2
v 2 dx  xdv  0
1
1
dx  2 dv  0
x
v
Integrating each term gives us
ln x 
1
x
x
 C1  ln | x |   C1  y 
v
y
ln | x | C1
Example
Solve the differential equation


2 xydx  y 2  x 2 dy  0
LINEAR DIFFERENTIAL EQUATIONS OF THE FIRST ORDER
A first-order linear differential equation is an equation of the form
y' P( x) y  Q( x)
3.5
where P and Q are continuous functions. If Q(x)=0 for all x, then equation (3.5) is
separable and we may write
y'
  P( x)
y
provided y≠0. Integrating, we obtain
ln y    P( x)dx  ln C .
We have expressed the constant of integration as ln|C| in order to change the form of the
last equation as follows:
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ln y  ln C    P( x)dx
ln
y
   P( x)dx
C
 P ( x ) dx
y
e 
C
ye 
P ( x ) dx
C
We next observe that
d   P ( x ) dx 
 P ( x ) dx  P( x) ye  P ( x ) dx
ye

y
'
e

dx 
 e
P ( x ) dx
y' yP( x).
Consequently, if we multiply both sides of (3.5) by e∫P(x)dx, then the resulting equation
may be written as
d   P ( x ) dx 
 P ( x ) dx .
ye

Q
(
x
)
e

dx 

This gives us the following (implicit) solution of (3.5)
ye
P ( x ) dx
  Q ( x )e 
P ( x ) dx
 D.
3.6
Solving this equation for y leads to an explicit solution. The expression e∫P(x)dx is called an
integrating factor of (3.5). We have shown that multiplications of both sides if (3.5) by
this expression leads to an equation which has the solution (3.6).
___________________________________________
Example
Solve the differential equation
x 2 y '5 xy  3 x 5  0
where x≠0.
Solution
In order to find the integrating factor we begin by expressing the given differential
equation in the “standardized” form (3.5), where the coefficient of y′ is 1. Thus, dividing
both sides by x2 we obtain
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y '
- 24 -
5
y  3x 3
x
which has the form (3.5) with P(x)=5/x and Q(x)=- 3x3. From the preceding discussion,
the required integrating factor is
e
P ( x ) dx
5
 e5 ln x  eln x  x .
5
If x>0 then |x|5=x5, whereas if x<0 then |x|5=-x5. In either case, multiplying both sides of
the standardized form by |x|5 gives us
x 5 y '5 x 4 y  3x 8
or
d 5
( x y )  3 x 8 .
dx
Thus a solution is
x9
x4 C .
 C or y    5
3
3 x
___________________________________________
x5 y  
A generalization of the (3.5) is the Bernoulli equation
y ' P( x) y  Q( x) y n ,
3.7
where n≠0. Evidently y=0 is a solution. If y≠0 we may divide both sides by yn, obtaining
y  n y ' P( x) y1 n  Q( x).
3.8
If we let w=y 1-n, then
w' 
dw
 (1  n) y  n y '
dx
and hence
y  n y' 
1
w'.
1 n
Replacing y-n y′ in (3.8) by the last expression gives us
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1
w' P( x) w  Q( x).
1 n
This first-order linear diff. eq. may be solved for w using the integrating factor technique.
After w was found, the solution (3.7) is given by y 1-n=w (and y=0).
__________________________________________
Example
Solve the differential equation
y '
2y
 x6 y 3.
x
Solution
The equation has the Bernoulli form (3.6) with n=3. If, as in the previous discussion, we
multiply both sides by y-3 and substitute w=y1-n=y-2 we obtain
2
 x6
2
xy
1
2w
 w'
 x6
2
x
4w
w'
 2 x 6
x
y 3 y '
Since the integrating factor for the last equation is
e
( 4 / x ) dx
 e 4 ln x  eln x
4
x
4
 x4
supposing that x>0 (in fact it can be shown that considering x<0, eventually we would
obtain the same result). Now we can write
x  4 w'4 x 5 w  2 x 2 .
Consequently
x4w  
2 x3
2 x7
 C or w  
 Cx 4 .
3
3
Finally, since w=y-2, the solution of the given equation is
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y 2  
- 26 -
 2x 7

2x 7
 Cx 4 or  
 Cx 4  y 2  1
3
 3

___________________________________________
Differential equations are indispensable in sciences (mathematics, physics, chemistry,
engineering, biology, etc). The application of the theory of differential equations for
physical problems will be solved in other modules and later in this module for population
dynamics.
Example
Solve the differential equations
xy   x  y  e x
y  y  y 2 e x
Linear differential equations of the second order
If f1, f2, … , fn and k are functions of one variable which have the same domain, then an
equation of the form
y n   f 1 x  y n 1    f n 1 x  y ' f n x  y  k x 
4.1
is called a linear differential equation of order n. If k(x)=0 for all x, the equation is said
to be homogeneous. Notice that this meaning of the word homogeneous is different from
that we met earlier. If k(x)≠0 for some x, then the equation (4.1) is said to be nonhomogeneous. We shall restrict our work to second-order equations in which f1 and f2 are
constant functions. First we are going to discuss the homogeneous case.
The general second-order homogeneous differential equation with constant coefficients
has the form
ay ' 'by 'cy  0
4.2
where a, b and c are constants. Before attempting to find particular solutions let us
establish the following result.
Theorem: (the superposition principle) If y=f(x) and y=g(x) are solutions of the
differential equation ay″+by′+cy=0, then
y  C1 f ( x)  C2 g ( x)
4.3
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is a solution for all real numbers C1 and C2.
Proof: By hypothesis
af ' ' ( x)  bf ' ( x)  cf ( x)  0

ag ' ' ( x)  bg ' ( x)  cg ( x)  0
If we multiply the first of these equations by C1, the second by C2, and add, the result is
aC1 f ' ' ( x)  C2 g ' ' ( x)  bC1 f ' ( x)  C2 g ' ( x)  cC1 f ( x)  C2 g ( x)  0
Thus C1f(x)+C2g(x) is a solution.
It can be shown that if the solutions f and g in above Theorem have the property that
f(x)≠Cg(x) for all real numbers C, and if g(x) is not identically 0, then y=C1f(x)+C2g(x) is
a general solution of ay″+by′+cy=0. Thus, to determine the general solution it is
sufficient to find two such functions f and g and employ Eq. (4.3).
In our search for solution of (4.2) we shall use y=Aemx as a trial solution, where A is an
arbitrary constant and we need to find m.
Then
dy
 Ame mx ;
dx
d2y
 Am 2 e mx
2
dx
Substituting in Eq. (4.2) gives
am
2

 bm  c Ae mx  0
or, since emx≠0 and we are interested in non-trivial solutions (A≠0)
am 2  bm  c  0
4.4
The equation (4.4) is called the auxiliary equation of eq. (4.2). It can be obtained from
this differential equation by replacing y″ by m2, y′ by m and y by 1. In simple cases, the
roots of the auxiliary equation can be found by factoring. If the factorisation is not
obvious, then applying the quadratic formula, we see that the roots of the auxiliary
equation are given by
m
 b  b 2  4ac
2a
4.5
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According to the sign of b2-4ac in eq. (4.5), we could have three different cases
(i)
(ii)
(iii)
Eq. (4.4) has distinct real roots (b2-4ac>0)
Eq. (4.4) has distinct complex roots (b2-4ac<0) (see later for this case)
Eq. (4.4) has repeated real roots (b2-4ac=0)
Theorem 1: If the roots m1 and m2 of the auxiliary equation are real and unequal, then the
general solution of ay″+by′+cy=0 is
y  C1e m1x  C 2 e m2 x
4.6
___________________________________________
Example
Solve the differential equation
y″+10y′+16y=0
Solution
The auxiliary equation is
m 2  10m  16  0 or (m  8)(m  2)  0
which means that m1=-8 and m2=-2. Since the roots of the auxiliary equations are real
and unequal, it follows from Theorem 1 that the general solution is
y  C1e 8 x  C 2 e 2 x
with C1 and C2 two arbitrary constants. To find the two constants we need two boundary
(or initial) conditions. For example, we may give two points the solution passes through
or one point and the gradient for some value of x, or two gradients. In this case, if y=0
and dy/dx=1 when x=0, using the general solution given above we have
C1  C 2  0

 8C1  2C 2  1
which eventually will lead to C1=-1/6 and C2=1/6. So, the final form of the solution is
y

1
 e 8 x  e  2 x
6

________________________________________________
Theorem 2: If the auxiliary equation has a double root m, then the solution of the
equation ay″+by′+cy=0 is
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y  C1e mx  C2 xemx  e mx C1  C2 x 
4.7
Proof: Using (4.5) with b2-4ac=0, we obtain m=-b/2a or 2am+b=0. Since m satisfies the
auxiliary equation, y=emx is a solution of the differential equation. According to the
remark following the proof of Theorem (4.3), it is sufficient to show that y=xemx is also a
solution. Substitution of xemx for y in ay″+by′+cy=0 gives us

 am
 

a 2me mx  m 2 xemx  b mxemx  e mx  cxemx 
2

 bm  c xemx  2am  b e mx 
 0 xemx  0e mx  0
which is what we wished to show.
________________________________________________
Example
Solve the differential equation
y″-6y′+9y=0
Solution
The auxiliary equation is m2-6m+9=0, or equivalently (m-3)2=0, has a double root 3.
Hence by Theorem (4.7), the general solution is
y  C1e 3 x  C 2 xe3 x  e 3 x C1  C 2 x 
Example:
Solve the differential equation:
y   5 y   6 y  0
Solution
The final case to consider is that in which the roots of the auxiliary equation are complex
numbers. For many of you the notion of complex numbers might sound a bit strange but
it is included in one of the module (SOM104) which runs parallel with this lecture.
The second case when deciding the nature of solution for a second order homogeneous
differential equation corresponds to the case when b2-4ac<0. In this case the auxiliary
equation is said to have complex roots.
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Complex numbers may be represented by expressions of the form a+ib, where a and b
are real numbers, and i is a symbol which may be manipulated in the same manner as a
real number, but has the additional property that i2=-1.
Two complex numbers a+ib and c+id are said to be equal, and we write a+ib=c+id, if
and only if a=c and b=d. Operations of addition, subtraction, multiplication and division
are defined just as in the case of real numbers; in the case of complex numbers all letters
denote real numbers, with the additional stipulation that whenever i2 occurs, it may be
replaced by –1. For example, the formulas for addition and multiplication of two complex
numbers a+ib and c+id are
a  ib   c  id   a  c   ib  d 
a  ib c  id   ac  bd   iad  bc 
We may regard the real numbers as a subset of the complex numbers by identifying the
real number a with the complex number a+i0. A complex number of the form 0+ib is
called an imaginary number.
Complex numbers are often required for solving equations of the form f(x)=0, where f(x)
is a polynomial. For example, if only real numbers are allowed, then the equation x2=-4
has no solutions. However, if complex numbers are allowed, then the equation has a
solution 2i, since
2i 2  2 2 i 2  4 1  4
Similarly, -2i is a solution of the equation x2=-4.
Since i2=-1, we sometimes use the symbol √(-1) in place of i and write
 13  i 13 , 2   25  2  i 25  2  5i
and so on.
A quadratic equation ax2+bx+c=0, where a, b and c are real numbers and a≠0, has roots
given by the quadratic formula
x
 b  b 2  4ac
.
2a
If b2-4ac<0, then the roots are complex numbers. To illustrate, if we apply the quadratic
formula to the equation x2-4x+13=0 we obtain
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x
- 31 -
4  16  52 4   36 4  6i


 2  3i
2
2
2
Thus the equation has two complex roots 2+3i and 2-3i.
The complex number a-bi is called the conjugate of the complex number a+bi. We see
from the quadratic formula that if a quadratic equation with real coefficients has complex
roots, then they must be conjugate complex numbers.
It follows from the previous discussion that if the auxiliary equation am2+bm+c=0 has
complex roots, then they are of the form
z1  s  it
z 2  s  it
and
where s and t are real numbers. We may anticipate, from Theorem (4.6), that the general
solution of the differential equation ay″+by′+cy=0 is
y  C1e z1x  C2 e z2 x  C1esit x  C2 esit x
4.8
In order to handle such complex exponents it is necessary to extend some of the concepts
of calculus to include functions whose domain includes complex numbers. Since a
complete development is beyond the scope of our work, we shall merely outline the main
ideas.
The Euler’s Theorem
For every complex number
e iz  cos z  i sin z
It can be shown that the Laws of Exponents are true for complex numbers. In addition,
formulas for derivatives developed earlier can be extended to functions of a complex
variable z. One such formula is dekz/dz=kekz, where k is a complex number.
Reminder: Laws of exponents
a m a n  a mn
am
 a mn
n
a
a 
m n
m
 a mn
am
a
   m
b
b
(ab) m  a m b m
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It can be proved that the general solution of ay″+by′+c=0, where the roots of the
auxiliary equation are complex numbers s±ti, is given by Eq. (4.8). The form of this
solution may be changed as follows:
y  C1e z1x  C2 e z2 x  C1e s it x  C2 e s it x
 C1e sxtxi  C2 e sxtxi
 C1e sx e txi  C2 e sx e txi
4.9
 e sx C1e itx  C2 e itx 
This can be further simplified by using Euler’s Formula. Specifically, we see from (4.8)
that
eitx  cos tx  i sin tx
e itx  cos tx  i sin tx
from which it follows that
cos tx 
e itx  e itx
e itx  e itx
; sin tx 
2
2i
4.10
If we let C1=C2=1/2 in (4.9) and use (4.10) we obtain the particular solution y=esx costx
of ay″+by′+cy=0. Letting C1= - i/2 and C2=i/2 gives us the particular solution y=esx
sintx. This is a particular proof of the next theorem.
________________________________________________
Example: Evaluate
e i  1
________________________________________________
Theorem
If the auxiliary equation am2+bm+c=0 has distinct complex roots s±ti, then the general
solution of ay″+by′+cy=0 is
y  e sx (C1 cos tx  C2 sin tx)
4.11
________________________________________________
Example
Solve the differential equation
y ' '10 y '41 y  0
Solution
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The roots of the auxiliary equation m2-10m+41=0 are
m
10  100  164 10   64 10  8i


 5  4i
2
2
2
Hence by Theorem (4.11), the general solution of the differential equation is
y  e5 x C1 cos 4 x  C2 sin 4 x 
________________________________________________
Example
Find the particular solution of the differential equation 2y″-3y′+2y=0 which satisfies the
boundary condition y=0 and y′=2 when x=0.
Solution
The roots of the auxiliary equation 2m2-3m+2=0 are
m
3  9  16 3   7 3  i 7


4
4
4
According to (4.11), the solution of the given differential equation can be written as
3x

7
7 
y  e 4  C1 cos
x  C 2 sin
x 
4
4


This solution contains two constants which can be determined using the boundary
conditions. It is obvious that y(0)=C1=0. In order to use the second boundary condition
we have differentiate the solution y(x) with respect to x
3x
3x

3 
7
7 
7
7
7
7 
y '  e 4  C1 cos
x  C 2 sin
x   e 4  
C1 sin
x
C 2 cos
x 
4 
4
4 
4
4
4
4


Taking into account that C1=0, we obtain that
y ' (0) 
7
C2  2
4
Solving this equation for C2, we obtain that
C2 
8 8 7

7
7
The final form of the solution will be
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- 34 3x
8 7 4
7
y
e sin
x
7
4
Example: Solve the differential equation:
y   6 y   13 y  0;
y (0)  2, y (0)  3
________________________________________________
A special second order homogeneous differential is the Euler-Cauchy differential
equation:
x 2 y   bxy   cy  0
where a and b are constants. Although this equation does not seem to be an equation with
constant coefficients, it can be reduced to that form using a new variable
x  et
or t  ln | x |
Since we changed the variable, we have to change y′ and y″, as well. To do so, we use the
chain rule :
dy dy dt dy 1 dy t



e
dx dt dx dt x dt
2
2
d 2 y dy d 2t d 2 y  dt 
dy  1  d 2 y 1
 2 t dy
2t d y


 e
e
  
  
dx 2
dt dx 2 dt 2  dx 
dt  x 2  dt 2 x 2
dt
dt 2
Introducing all derivatives back into the original equation, we have
d 2 y  2t
dy 2t
dy
e
e  e  2t
e  be t e t
 cy  0
2
dt
dt
dt
d2y
dy
 (b  1)  cy  0
2
dt
dt
2t
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which now is a second order homogeneous differential equation with constant
coefficients. Therefore, we use the previous results to solve this equation.
The auxiliary equation is
m 2  (b  1)m  c  0
and the solution of the Euler-Cauchy differential equation can be found according to the
nature of the roots of the auxiliary equation.
a. Real and distinct roots (m1 and m2)
y  A e m1t  Be m2t  Ae m1 ln|x|  Be m2 ln|x|  A | x |m1  B | x |m2
b. Real but equal roots
y  ( A  Bt )e mt  ( A  B ln | x |)e m ln| x|  ( A  B ln | x |) | x | m
c. Complex roots (m12=α±iβ)
y  e t (C1 cos  t  C2 sin  t )  e ln|x| (C1 cos  ln | x | C2 sin  ln | x |) 
| x | (C1 cos  ln | x | C2 sin  ln | x |)
Example
Solve the differential equation
x 2 y   xy   y  0 with
Solution
y(1)  1, y(e)  0
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Non-homogeneous linear differential equations
In this section we shall consider second-order non-homogeneous linear differential
equations with constant coefficients, that is, equations of the form
ay ' 'by 'cy  k ( x)
5.1
where a, b and c are constants and the function k(x) is continuous. The function k(x) is
often called the forcing term or the source term. It is convenient to use the differential
operator symbol D and D2 where if y=f(x), then
Dy  y'  f ' ( x), D 2 y  y' '  f ' ' ( x)
We shall also employ the linear differential operator
L  aD 2  bD  c
where by definition


L( y)  aD 2  bD  c y  aD 2 y  bDy  cy  ay' 'by'cy.
Using this notation, equation (5.1) can be written in the compact form L(y)=k(x). It is
easy to verify that for every real number C,
L(Cy)  CL( y )
5.2
Also, if y1=f1(x) and y2=f2(x), then it can be shown that
L( y1  y2 )  L( y1 )  L( y2 )
5.3
Given the differential equation (5.1), that is, L(y)=k(x), the corresponding homogeneous
equation L(y)=0 is called the complementary equation. Suppose that yp is a particular
solution of L(y)=k(x) and yc is any solution of the complementary equation. Since
L(yp)=k(x) and L(yc)=0
L( y p  y c )  L( y p )  L( yc )  k ( x)  0  k ( x)
which means that yp+yc is a solution of (5.1).
Theorem If yp is a particular solution of the differential equation L(y)=k(x) and if yc is a
general solution of the complementary equation L(y)=0, then the general solution of
L(y)=k(x) is y=yc+yp. (5.4)
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If we use the results of the previous section to find the solution yc of L(y)=0, then
according to the above theorem all that is needed to determine the general solution of
L(y)=k(x) is one particular solution yp.
Example
Solve the differential equation
y″-4y=6x-4x3.
Solution
We see by inspection that yp=x3 is a particular solution of the given equation. The
complementary equation y″-4y=0 which by (4.6), has solution
y  C1e 2 x  C2 e 2 x
Applying the theorem, the general solution of the given non-homogeneous differential
equation is
y  C1e 2 x  C2 e 2 x  x 3 .
In most cases a particular solution of (5.1) cannot be found by inspection as it was done
in the previous Example.
Given the differential equation
L( y)  ay' 'by'cy  e nx
where enx is not a solution of L(y)=0, it is reasonable to expect that there is a particular
solution of the form yp=Aenx, since enx is the result of finding ay″+by′+cy=0. This
suggests that we use Aenx as a trial solution in the given equation and attempt to find the
value of the coefficient A. This technique is called the method of undetermined
coefficients, and is illustrated in the next example
Example
Solve the differential equation
y″+2y′-8y=e3x.
Solution
Since the auxiliary equation m2+2m-8=0 of y″+2y′-8y=0 has roots 2 and -4 it follows
that the general solution of the complementary equation is
yc  C1e 2 x  C2 e 4 x .
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From the preceding remarks we seek a particular solution of the form yp=Ae3x. Since
yp′=3Ae3x and yp″=9Ae3x, substitution in the given equation leads to
9 Ae3 x  6 Ae3 x  8 Ae3 x  e 3 x .
Dividing both sides by e3x we obtain
9 A  6 A  8 A  1 or
1
A .
7
Thus yp=(1/7)e3x and by Theorem (5.4), the general solution is
1
y  C1e 2 x  C 2 e 4 x  e 3 x .
7
After all, the method of undetermined coefficients is based on ‘guessing’ the right form
for yp. The ‘guess’ is based on experience and comes down to a few rules:
1. If k(x) is a polynomial of degree q, try
y p  aq x q  aq1 x q1   a1 x  a0
i.e. a polynomial of the same degree.
2. If k(x)=acosmx+bsinmx (including the cases when a=0 or b=0), try
y p  A cos mx  B sin mx
4. If k(x)=aepx and p is not a root of the auxiliary equation, try
y p  Ae px
5. If k(x)=aepx and p is a root of the auxiliary equation, try
y p  Axe px
6. If k(x)=xepx and p is not a solution of the auxiliary equation
then try a particular solution of the form
y p  ( A  Bx )e px
7. If k(x)=esxsintx or k(x)=esxcostx and s+ti is not a solution of
the auxiliary equation, then try a particular solution of the form
y p  Aesx cos tx  Be sx sin tx.
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8. If k(x) is a sum of simpler functions in the list above, find
the particular integral for each of these simpler functions
and add them together.
Example
Find the general solution of the equation
d2y
dy
 13  12 y  2t  1
2
dt
dt
Solution
The characteristic equation is
m 2  13m  12  0
m  12m  1  0
or
m1  1, m2  12
which means that the complementary solution is
y c  Ae t  Be 12t
For the particular solution, try yp=at+b. Accordingly, yp′=a, yp″=0. We substitute back
into the equation
0  13a  12at  b   2t  1
12at  12b  13a   2t  1
a 1
6
b7
72
This means that general solution
y  Ae t  Be 12t 
t
7

6 72
Example: Solve the non-homogeneous differential equation
d2y
 y  1; y(0)  0,
dx 2
Solution:
y=1-sinx-cosx
 
y   0
2
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Example
Solve the differential equation
d2y
dy
 3  2 y  e 4t
2
dt
dt
Solution
The characteristic equation is
m 2  3m  2  0
m  2m  1  0
or
m1  1, m2  2
The complementary solution is
yc  Ae 2t  Be t
Since m≠4, we try a particular solution of the form yp=ae4t, i.e. yp′=4ae4t, yp″=16ae4t.
Introducing these expressions back into the original equation we have
16a  12a  2a e 4t
 e 4t , i.e. 6a  1, a  1
6
Finally, the general solution is
y  Ae 2t  Be t 
e 4t
6
Example
Solve the differential equation
d2y
dy

3
 2 y  e  2t given that y=0 and dy/dt=0 when t=0.
2
dt
dt
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Solution
The characteristic equation is
m 2  3m  2  0
m  2m  1  0 or
m1  1, m2  2 so
yc  Ae 2t  Be t
Here e-2t is part of the complementary solution so we must try a particular solution of the
form yp=ate-2t. Accordingly

 a 2e

y 'p  a e 2t  2te 2t  a1  2t e 2t
y 'p'
 2t
 
 21  2t e 2t  a  4e 2t  4te 2t

Introducing back into the original equation, the left hand side will look like


LHS  a  4e 2t  4te 2t  3e 2t  6te 2t  2te 2t  ae 2t
Note that the terms in te-2t cancel out, this will always happen. Comparing the two sides
of the equation we obtain that a=-1. So, the general solution of the equation is
y  Ae2t  Be t  te2t
Now we have to find the two constants, A and B, using the initial conditions given. When
t=0, y=0, i.e. 0=A+B. To use the other condition we calculate
y   2 Ae 2t  Be t  e 2t  2te 2t .
When t=0, we have 0=-2A-B-1. Combining the two relations obtained for the two
constants we obtain A=-1 and B=1. So, the final solution is
y  e2t  et  te2t
.
Example
Solve the differential equation
d2y
dy
 3  2 y  sin 2t and describe the behaviour of the solution as t→∞.
2
dt
dt
Solution
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Population dynamics
Population dynamics is concerned with numbers in populations and factors which cause
them to change. In addition to natural curiosity, population dynamics is useful in studying
problems involving economic and social situations, e.g. farming, fishing, spread of
epidemics, conservation (killing of whales, Amazon rain forests, etc. ).
In studying population dynamics we shall be concerned with either one species (e.g.
humans) or two interacting/competing species (rabbits and foxes).
Population dynamics tries to answer questions like:
 What will be the population of a certain country in 10 years?
 How are we protecting the resources from extinction?
To try to answer to these types of questions we construct simple mathematical models of
real life situation then use models to predict future changes.
Models simplify the real situation and aim to isolate the main factors and study their
effect on the main population. These models may be crude but they can give surprisingly
accurate predictions.
In order to be able to model these situations we have to make some assumptions:
1. the number of individuals in a given area can be represented by a single variable,
x, measured in terms of some chosen unit of population, so x is the population
density. By treating the population as a whole, we ignore age and sex differences.
2. x is a continuous and differential function of t, taken as a continuous variable.
This may seen a surprising assumption since we might expect to take x as an
integer, the population of any species changing by integer amounts. However if a
given population is very large and it is increased by a few in short time intervals,
those changes are very small compared to the total population and a graph of x
with respect to t looks like a continuous function and indeed a differentiable
function and as we assume that it is.
We have in mind populations such as the human population which has no fixed breeding
season. Some species such as sheep have a single breeding season and for these it may be
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more sensible to measure x at discrete time intervals, the ends of the breeding seasons, for
example. This leads to difference equation models which will be studied later.
One species models
(e.g. number of people in the world, number of bacteria in an experiment, number of trees
in the forest, etc.)
Let x(t) the number of members of a population. Large population if increase/decrease
not too drastically than it is reasonable to approximate the function x(t) by a continuous
and differentiable function of t.
Let the number, x(t), of members of a population at time t be large enough for x(t) to be
continuous. If x(t0) is known, can the number be predicted at subsequent times?
In general the time scale depends on the population type: for humans in years, for
bacteria in hours or days.
To model the growth we consider the rate at which the population changes. The
instantaneous growth rate at time t is dx/dt. The instantaneous relative growth (or rate of
change per individual) is
r (t ) 
1 dx
x dt
The population dynamics operate with a few models which are surprisingly accurate.
In order to model the problem, we are going to use two types of first order differential
equations already presented previously. In what follows we are revising the methods of
solving these types of equations
1. Separation of variables: if the differential equation can be written as
dx
 f (t ) g ( x) 
dt
dx
 g ( x)   f (t )dt
2. The integrating factor: if the differential equation can be written as
dx
 P(t ) x  Q(t )
dt
then the equation is solved by using the integrating factor

exp  P(t )dt

Example1
1. Solve the differential equation x′(t)=ax, where a is a constant for x≥0 such that
(i)
x=x0 when t=0
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(ii)
- 44 x=x0 when t=t0
Solution
This equation can be solved by variable separation

dx
 a  dt
x
Integrating both parts we have
ln x  at  A or
x  e at  A
or
x  Ce at
which is the general solution, A is an arbitrary constant. Now we are going to use the
initial conditions
(i)
of x=x0 when t=0, then we have
x0  Ce0  C i.e. x  x0eat
which is a particular solution
(ii)
if x=x0 when t=t0, then
x0  Ceat
0
i.e. C  x0e at
0
so x  x0ea t t
0

Example2
Find the solution of the differential equation x′(t)=ax(b-x), where a and b (b>0) are
constants for 0≤x≤b which satisfies x(0)=x0.
Solution
This equation can be solved again using variable separation. Accordingly we write
dx
 xb  x   a  dt
Now, the integrand of left-hand side can be decomposed as
1
A
B
 
xb  x  x b  x
or 1  Ab  x   Bx
This equation must be valid for all values of x. Putting first x=0 we have 1=Ab, so
A=1/b. Then we put x=b and we obtain 1=Bb, so B=1/b. Using these values, the
integrand can be decomposed as
1
1 1
1 
  

xb  x  b  x b  x 
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which means that the integral is
dx
1 dx 1 dx
1
1
 
 ln x  ln(b  x) 
x b bx b
b
 x(b  x)  b 
1  x 
 ln

b b  x
since 0≤x≤b. Returning to the differential equation obtained after separating the variables
we have
1  x 
 x 
ln
  at  A' or ln
  abt  A
b b  x
b  x
x

 Ce abt  Cect
bx
where A=A′b, A′ and c=ab are constants. If x=x0 when t=0
x0
x
x
C
 0 ect
b  x0
b  x b  x0
 xb  x0   x0 b  x ect
 x b  x0  x0ect   bx0ect
x
bx0
x0  b  x0 e  ct
Example
Find the general solution of the differential equation
dx
 ax  Aebt
dt
where a, b and A are constants.
Solution
This is a linear differential equation with P(t)=a and Q(t)=Aebt. Therefore the integrating
factor is
e
adt
 eat
Multiplying the original differential equation by the integrating factor gives
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eat
- 46 -
dx
 aeat x  Aebt eat  Ae( a  b )t
dt
Since we have multiplied the differential equation by the integrating factor, the left-hand
side of the above relation can be written as

d at
e x
dt

since
d at
e x   eat dx  x d eat   eat dx  aeat x
dt
dt
dt
dt
and so


d
xeat  Ae ( a b )t
dt
Integrating both sides with respect to t gives
xeat 
A ( a  b )t
e
B
ab
where B is an arbitrary constant. From above relation, x can be expressed as
x
A bt
e  Be at
ab
Exponential model (The Malthusian model)
Is the simplest model when we suppose that the instantaneous relative growth, r(t), is
constant (a)
1 dx
 a or
x dt
dx
 ax
dt
This model says that the rate of change of the population is proportional to the existing
population. If x=x0 when t=t0, we obtained
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x  x0ea t t
0

If a>0 then this result gives an exponential growth
a<0 then this result gives an exponential decay and a is called
growth (decay) rate
here
x(t)=P(t)
and
x0=2
and
a=0.2
A useful measure of the growth of a population is the doubling time. If t1-t0= time
required for population to double, then at t=t1, x=2x0 and so
2 x0  x0e a t
1 t0

 ln 2  at1  t0 
1
 t1  t0  ln 2
a
Example: (Application to world human population)
In 1961 the world population was 3.06x109. Over the next 10 years it increased at relative
growth rate of 2% per year. Calculate the doubling time.
Solution
Applying the formula, with t0=1961, a=0.02, x0=3.06x109, we have
x  3.06  109 e0.02t 1961
and the doubling time is ln2/0.02=34.7 years. Estimates from 1700 to 1961 suggest that
the doubling time was 35 years, which means that this model gives good agreement with
reality.
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What about future predictions: In 1991 a population of 5,575 million, in 2389 a
population of 15,969,164 million (the surface area of the Earth is approximately
16,700,500 million m2)
That means that the exponential model gives unreasonable predictions for the future. For
positive growth rate, when time tends to infinity, the population number tends to infinity
which is in contradiction with the available resources or environment limitations.
Therefore it is obvious that a new model has to be used which can give reasonable results
for large times.
The logistic model (Verhulst-Pearl model)
The exponential model (besides that it gives too large predictions for large times) does
not reflect the competition between individuals, the limits of the resources and food
supply. For many species with a finite food supply the population tends to an upper limit
which depends on the environment conditions. The idea behind the logistic model says
that while the population will continue to grow as time goes on, the rate at which it does
this growing gets smaller.
Let us denote by c the reproductive parameter, x(t) the population density and k the
carrying capacity of the environment (or a limiting size of population). Then the proposed
logistic equation (Verhulst 1844) is
dx
 x
 cx1  
dt
 k
In general, k is large. For small values of x, cx2/k is negligible compared to cx and
initially the population grows exponentially. When x is large, the cx2/k term is no longer
small and slows down the rate of increase of x.
The logistic equation can be interpreted if we expand out the logistic equation
dx
c
 cx  x 2
dt
k
which suggests a situation when the birth rate is still constant cx, but in which there is a
mortality term, cx2/k, that dominates when the population is high.
If x>k, then dx/dt<0 and the population decreases, when x=k then the population growth
is a constant.
Now
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dx c
 xk  x   axb  x 
dt k
putting a=c/k, b=k  ab=c, using Example2
x
kx0
x0  k  x0 e ct
Since
e ct  0 as t  
for c  0
it follows from the above form of the
solution that xk as t∞ for any x0.
(here c≡r)
Let us take the logistic equation
dx
 x
 cx1  
dt
 k
This equation has two population levels x=0 and x=k for which dx/dt=0. These are called
equilibrium levels. Suppose that the population is closed to one of these levels at some
time and we want to see whether it will remain close to that level for all t?
If it does, we say that the population is locally stable, if it does not remain close then it is
unstable. From the graph, the point x=k is locally stable but x=0 is unstable. We can gain
this information without sketching the graph as follows:
For x=k write x=k+X(t), where X(t) is a small quantity; then dx/dt=dX/dt, since
k=constant, and from the logistic equation
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dX
cX 2
 kX
 X
 ck  X 1 
  ck  X     cX 
dt
k 
k

 k
Since X(t) is a small quantity we can neglect the term containing the X2 term, so
dX
 cX  X  Ae ct
dt
Since c>0, X→ 0 as t→∞, i.e. x remains close to x=k and so is locally stable.
Harvesting
The population of fish in the North Sea can be described by the logistic equation
dx
 x
 cx1    cx  sx2
dt
 k
putting s=c/k a positive number.
Suppose that the fish are caught (harvested) at a constant rate h, i.e. in the absence of all
other factors
dx
 h
dt
To account for harvesting the logistic equation must be modified to
dx
 cx  sx2  h
dt
Let us first discuss the equilibrium levels of this model. The equilibrium levels are at
x=xe, such that dxe/dt=0, i.e.
sxe2  cxe  h  0 or
xe2  kxe 
h
0
s
Solving this second order equation we obtain
xe 
k  k2  4h
2
2
s k  k h
2
4 s
(*)
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Case (a)
If the determinant is positive, i.e. k2/4-h/s>0, then h  sk
2
(moderate
4
fishing).
In this case, there are two positive real roots (i.e. two equilibrium levels) x1 and x2 where
x1 
k
k2 h


2
4 s
k
k
k2 h
x2  

2
4 s
x2  x1  k
and
dx
h

  s x 2  kx     s x  x1  x  x 2 .
dt
s

Case (b) If the determinant is negative i.e. k2/4-h/s<0, then h  sk
2
(over intensive
4
fishing).
In this case there are no real roots, i.e. there are no equilibrium levels. Completing the
squares, dx/dt can be written as


(**)
2
dx
h
k  k 2 h
 2

  s x  kx     s  x       0
dt
s
2 


4
s 


0
 dx/dt< 0 for all values of x, i.e. since dx/dt never reaches 0, we have no
equilibrium levels.
Case (c) If the determinant is zero, i.e. . k2/4-h/s=0. Equation (*) has a repeated root
xe 
k
2
From equation (**)
2
dx
k
for all values of x except x=xe.

  s x    0
dt
2

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(ii) Local stability. First we consider case (a) where we have two equilibrium levels. Let
x=xe+X(t), where xe=x1 or x2. Then
dx dX

  sxe  X  x1 xe  X  x2 
dt
dt
When xe=x1
dX
  sX  x1  x2   X    s( x1  x2 ) X  OX 2 
dt
Neglecting terms proportional to X2, we arrive at a simple first order differential equation
which can be solved by separating the variables
X  Ae s  x  x
1
2
t
where A is a constant. Since x1>x2, X0 as t∞ and so x1 is locally stable.
When xe=x2
dX
  s x2  x1  X  X  Be s  x
dt
as t  
2
 x1 t
and so x2 is unstable.
(iii) Graph of x(t)
First of all we note that
dx
  s x  x1  x  x2 
dt
For
x>x1
x2<x<x1
0<x<x2
s  0 
dx/dt<0 => x decreasing
dx/dt>0 => x increasing
dx/dt<0 => x decreasing

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Here x0=x(0). If x0<x2, the graph intersects x=0 at a finite time T. From
dx
  s x  x1  x  x2 
dt
0
T
dx
  x  x  x  x    s  dt   sT
x
0
1
2
0
x0
1
dx
T  
s 0  x  x1  x  x2 
Case (b)
k2 h
sk 2

or h 
4 s
4
In this case we have no equilibrium levels, i.e. dx/dt <0 for all values of x.
In either case x becomes zero in a finite time.
Case (c) h=sk2/4. The graph can be obtained easily from case (a) when the two roots are
identical, i.e. x1=x2.
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Conclusions: Although limited, this modelling of harvesting indicates that there is a
maximum harvesting rate
h
sk 2 1 c 2 1

k  ck
4
4k
4
at which the population can sustain itself in equilibrium. Any greater rate leads to a
depletion of stock eventually to zero. At h=ck/4 the equilibrium level is k/2 (not the
equilibrium level k of the logistic model). Any harvesting of depleted species will result
in extinction.
Two species model. Predator-prey
Suppose an island supports a crop of foliage and populations of foxes and rabbits. The
foxes eat the rabbits. The rabbits eat the foliage (i.e. limited supply). What happens with
the populations???
Assuming no other interference with the system, we might expect the two populations to
oscillate.
Let x(t) be the population of rabbits at time t. With no foxes around, we assume that x(t)
obeys the logistic equation
dx
 ax  bx2 , a  0, b  0
dt
a/b being the carrying capacity or saturation level.
Let y(t) be the population of foxes at time t. Assuming that the number of encounters per
unit time between foxes and rabbits is proportional to x and y and that a certain
proportion of these results in rabbit being eaten, i.e. the rate of decrease in the rabbit
population is proportional to xy. In this case the differential equation becomes
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dx
 ax  bx 2  cxy, a, b, c  0
dt
(1)
If no rabbits are present, we assume foxes die out exponentially, i.e.
dy
  py,
dt
p0
With rabbits present, the rate of increase of foxes is proportional to the number of
successful encounters with rabbits and so the differential equation becomes
dy
  py  qxy,
dt
p, q  0
(2)
The two populations are therefore determined by the coupled pair of nonlinear ordinary
differential equations (1) and (2). Putting b=0 (=> an abundance of foliage) gives the
Lotka-Volterra (LV) equations
dx
 ax  cxy
dt
dy
  py  qxy
dt
(3)
(4)
Equilibrium populations. Let x=xe and y=ye the equilibrium populations and they are
given by the usual conditions dxe/dt=0, dye/dt=0, i.e.
xe a  cye   0
ye  p  qxe   0
(5)
(6)
Equation (5) is satisfied by xe=0 or ye=a/c.
If xe=0, then from Eq. (6) we obtain that ye=0.
If ye=a/c, then from Eq. (6) we obtain that -p+qxe=0 => xe=p/q.
The equilibrium populations are therefore
xe  0, ye  0 and
xe 
p
a
, ye 
q
c
Are these equilibrium populations stable?
Let us take the Lotka-Volterra equations (Eqs. (3) and (4)) and we consider
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x
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p
a
 X, y  Y
q
c
where the quantities X and Y are small increments. These expressions are introduced back
into the Lotka-Volterra equations and we obtain
dx dX  p
pc

a


   X   a  c  Y     Y
dt dt  q
q
c


dy dY  a
p
 qa


   Y    p  q  X   
X
dt dt  c

q
 c
=>
dX
pc
 Y
dt
q
dY qa

X
dt
c
(7)
(8)
This is a system of two coupled first order ordinary differential equations. In order to
solve this system, we use the usual procedure, i.e. first we differentiate Eq. (7) with
respect to t and we use the derivative of Y with respect to t from Eq. (8)
d2X
pc dY
pc  qa 

   X   apX , ap  0
2
q dt
q  c 
dt
(9)
To find the governing equation for Y, we follow the same way, i.e. we differentiate Eq.
(8) with respect to t and use Eq. (7)
d 2Y qa dX qa  pc 

   Y   apY
dt 2
c dt
c  q 
(10)
The general solution of (10) is
Y  Acost  B sin t
(11)
putting ω2=ap.
To find the solution for X, we use Eq. (8) which gives
qa
dY
X 
  A sin t  B cos t
c
dt
c
 A sin t  B cos t 
X 
aq
(12)
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[Using Eq. (9) to find X introduces extra constants which are connected to A and B via
Eq. (8)]
Suppose that at t=0, X=X0 and Y=Y0. Then A=Y0 and B=X0aq/cω, so that
X  X 0 cost 
c
Y0 sin t
aq
aq
Y  Y0 cost 
X 0 sin t
c
The above equations show that X and Y are bounded and so stay close to x=p/q, y=a/c for
all t, so this point is locally stable.
Also
Y  R sin  cos t  R cos  sin t  R sin t    (13)
putting
R sin   Y0 ,
X 
c
aq
R cos  
aq
X0
c
and
R sin  sin t  cos  cos t 

X 
c
aq
R cost   


Y  R cos t    
2

where we have used the trigonometrical identity
cos     cos cos  sin  sin   sin 
2
2
2


For the other equilibrium level (0,0), we put
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x=0+X=X, y=0+Y=Y
then from Eqs. (3) and (4) we have
dX
dY
 aX  cXY ,
  pY  qXY.
dt
dt
Having in mind that X and Y are small quantities, we can neglect the 2nd order terms XY
dX
 aX
dt

X  Aeat
 X grows exponentially in time since a>0, i.e. X∞ as tω
 the point (0,0) is unstable
N.B. Only need either X or Y to grow exponentially for an unstable point.
Summary
I. One species
I.a Exponential model:
dx
 ax  x  Aeat (comparison with data
dt
I.b Logistic model: limited resources, saturation level k
dx
 x
 cx1  
dt
 k
-solution
-graph
I.c Harvesting: constant rate h, the rate proportional to x and the
differential equation of the form similar to the logistic model
Equilibrium levels (x=xe): dxe/dt=0
Local stability (instability): take x=xe+X(t) and
neglect second order terms in X
II. Two species
II.a Predator-prey (notes)
II.b Competing species (example sheet 9, 11)
Equilibrium levels and stability
Discrete Mathematics and Difference Equations
Discrete mathematics is the branch of mathematics dealing with objects that can assume
only distinct, separated values. The term "discrete mathematics" is therefore used in
contrast with "continuous mathematics," which is the branch of mathematics dealing with
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objects that can vary smoothly (and which includes, for example, calculus). Whereas
discrete objects can often be characterized by integers, continuous objects require real
numbers.
The study of topics in discrete mathematics usually includes the study of algorithms, their
implementations, and efficiencies. Discrete mathematics is the mathematical language of
computer science, and as such, its importance has increased dramatically in recent
decades.
Mathematical computations frequently are based on equations that allow us to compute
the value of a function recursively from a given set of values. Such an equation is called a
“difference equation” or “recurrence equation”. These equations occur in numerous
settings and forms, both in mathematics itself and in its applications to statistics,
computing, electrical circuit analysis, dynamical systems, economics, biology, etc.
Example
In 1626, Peter Minuit purchased Manhattan Island for goods worth $24. If the $24 could
have been invested at an annual interest rate of 7% compounded quarterly, what would it
have been worth in 2004?
Solution
Let y(t) be the value of the investment after t quarters of a year. Then y(0)=24. Since the
interest rate is 1.75% per quarter, y(t) satisfies the difference equation
y (t  1)  y (t )  0.0175 y (t )  1.0175 y (t ) t  0,1,2, 
Computing y recursively, we have
y (1)  24(1.0175)
y (2)  24(1.0175) 2

y (t )  24(1.0175) t
In 378 years, or 1512 quarters, the value of the investment is
y(1512)=24(1.0175)1512≈5.9x1012 (about 5.9 trillion dollars!)
Example
It is observed that the decrease in the mass of a radioactive substance over a fixed time
period is proportional to the mass that was present at the beginning of the time period. If
the half life of radium is 1600 years, find a formula for its mass as a function of time.
Solution
Let m(t) represent the mass of the radium after t years. Then
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m(t+1)-m(t)=-km(t)
where k is a positive constant. Then
m(t+1)=(1-k)m(t), t=0,1,2…
Using iteration as in the previous example, we find
m(t )  m(0)(1  k ) t
Since the half life is 1600 years
1
1
 1 1600
m(1600)  m(0)(1  k )1600  m(0)  1  k   
2
2
so
t
 1 1600
m(t )  m(0) 
2
(In physics, this problem is usually solved using an integral of a differential equation. The
solution presented here is somewhat shorter and employs only elementary algebra)
The starting equations in the above examples were all difference equations. By definition,
an equation which expresses the value an of a sequence {an} as a function of the term an-1
is called first-order difference equation. If we can find a function f such that an=f(n),
n=1, 2, 3…, then we will have solved the difference equation.
Both examples had a general difference equation of the form
xn1  Axn ,
n  0,1, 2,
From the above examples, it is obvious that if x0 is the initial value, then
x n  A n x0
Given the constants A and B, a difference equation of the form
x n1  Axn  B,
n  0,1, 2, 
is called a first-order linear difference equation. Note that the linear difference equation
reduces to the difference equation if B=0. In order to solve this type of equation, we write
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x n  Ax n 1  B  A( Ax n  2  B)  B  A 2 x n  2  B( A  1)
 A 2 ( Ax n 3  B)  B( A  1)  A 3 x n 3  B( A 2  A  1)

 A n x 0  B( A n 1  A n  2   A 2  A  1).
Note that if A=1, this gives us
x n  x0  nB,
n  0,1, 2, 
as the solution of the difference equation xn+1=xn+B. If A≠1, we know that
A n 1  A n  2    A 2  A  1 
1  An
1 A
Hence
 1  An
x n  A x0  B
 1 A
n



is the solution of the first-order linear difference equation xn+1=Axn+B, when A≠1.
____________________________________________________
Example (Newton’s law of cooling)
If T0 represents the initial temperature of the object, S the constant temperature of the
surrounding environment and Tn the temperature of the object after n units of time, then
the change in the temperature over one unit of time is given by
Tn 1  Tn  k Tn  S 
where k is a constant which depends upon the object. This difference equation is known
as Newton’s law of cooling. Suppose a cup of tea, initially at a temperature of 82 C, is
placed in a room which is held at a constant temperature of 26 C. Moreover, suppose that
after one minute the tea has cooled to 80 C. What will the temperature be after 20
minutes?
Solution
If we let Tn be the temperature of the tea after n minutes and we let S be the temperature
of the room, then we have T0=82, T1=80, and S=26. Newton’s law of cooling states that
Tn 1  Tn  k Tn  26, n  0,1, 2, 
where k is the constant which we have to determine. To do so, we make use of the
information given about the change in the temperature of the tea during the first minute.
When n=0, we have
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T1  T0  k T0  26 or 80  82  k (82  26) so k  0.035
Thus, the original equation reduces to
Tn 1  Tn  0.035(Tn  26)  0.035Tn  0.91
so
Tn 1  Tn  0.035Tn  0.91  0.965Tn  0.91
Now the last equation is in the standard form of a first-order linear difference equation, so
from the solution (according to the method of solving presented earlier)
 1  0.965n 
  0.965n (82)  26(1  (0.965) n ) 
Tn  0.965n (82)  0.91
 1  0.965 
 26  56(0.965) n
In particular,
T20  26  56(0.965) 20  53.5
So after 20 minutes the tea has cooled to just below 54 C.
____________________________________________________
An equation of the type
au n  2  bu n 1  cu n  f (n)
where a, b and c are constants and f(n) a given function is called a second order constant
coefficient difference equation. The equation is said to be homogeneous if f(n)=0 and
inhomogeneous if f(n)≠0.
a. Homogeneous second order difference equations
We can have an idea on how these equations can be solved by looking first at first-order
equation with constant coefficient
u n 1  cu n  0
which has the solution
u n  Ac n
where A is any constant. With this in view, we attempt to find solutions of
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au n 2  bu n1  cu n  0
in the form
un  p n
where p is a constant. Thus
au n  2  bu n 1  cu n  ap n  2  bp n 1  cp n  (ap 2  bp  c) p n  0
for all n. The solution p=0 leads to the trivial solution un=0. We are interested in the
roots of the equation in the brackets which is called the characteristic equation.
According to the nature of the roots of the characteristic equation, we can have different
solutions.
Case 1: Distinct roots
The general solution of the second-order difference equation for distinct roots p1 and p2
of ap2+bp+c=0 is
u n  Ap1n  Bp 2n
for any constant A and B.
Example
Find the general solution of
u n  2  u n 1  6u n  0
Solution
The characteristic equation is p2-p-6=0 or (p-3)(p+2)=0. This means that the roots are
p1=3 and p2=-2. Hence the general solution is
u n  A  3n  B   2
n
Example
Find the solution of
un2  2un1  3un  0
that satisfies u0=1, u1=2.
Solution
The characteristic equation is
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p 2  2 p  3  0 or ( p  3)( p  1)  0
The roots are p1=-3 and p2=1. Hence the general solution is
u n  A 3  B  1n  A 3  B
n
n
Using the initial conditions
u 0  1  A  B, u1  2  3 A  B
Hence A=-1/4 and B=5/4. The required solution is u n   1  3n  5
4
4
Case 2: Equal roots(p1=p2=p)
The general solution of au n 2  bu n1  cu n  0 is
u n  ( A  Bn ) p n
(please note the difference in the form of the second-order difference equation)
Case 3: Complex roots
The general complex solution of au n 2  bu n1  cu n  0 , where b2<4ac, is
u n  A  i   B  i 
n
n
This solution can be written in other form by writing the complex numbers in polar
coordinates (r,θ). The connection between the two forms of writing a complex number is
r   2   2 , cos  

r
, sin  

r
, tan  


So the general solution of a second-order homogeneous difference equation with complex
roots can be also written as
u n  r n C cos  n  D sin  n 
Example
Obtain the general solution of
u n 2  u n  0
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Solution
The characteristic equation is p2+1=0, giving roots p1=i and p2=-i. Therefore
u n  Ai n  B i 
n
In polar form, r=1 and
sin   1    k

2
where k is an integer number. So, the solution is
u n  C cos n

2
 D sin n

2
b. Inhomogeneous second-order difference equations
The general inhomogeneous equation is
au n 2  bu n1  cu n  f (n)
We write un=vn+qn, where vn is the general solution of the corresponding homogeneous
equation. Substitute this form of un into the general equation
avn 2  q n 2   bvn1  q n1   cvn  q n   f (n)
or
avn2  bvn1  cvn   aqn2  bqn1  cqn  
f (n)
Since vn satisfies the homogeneous equation, it follows that
aqn 2  bqn1  cqn  f (n),
which means that qn must be a particular solution of the inhomogeneous equation. As in
the theory of differential equations, vn is known as the complementary function.
We construct particular solutions by appropriate choices of functions usually containing
adjustable parameters which are suggested by the form of the function f(n). In the
following table you will find the suggested forms of particular solutions
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k (a const )
C ; or Cn, if C fails ; Cn 2 , if C and Cn fail ; etc
kn
Ck n ; or Cnk n , if Ck n fails ; etc
n
C 0  C1 n
n p ( p int eger )
C 0  C1 n   C p n p
sin kn or cos kn
C1 cos kn  C 2 sin kn
Example
Obtain the general solution of
u n  2  u n 1  6u n  4
Solution
From the example solved earlier, the complementary solution is
v n  3 n A  (2) n B
For the particular solution, we try qn=C (see the table above). Then
q n  2  q n 1  6q n  4  C  C  6C  4  6C  4  0  C  
2
3
Hence qn=-2/3, and the general solution is
u n  3 n A  (2) n B 
2
3
Example
Obtain the general solution of
un2  2un1  3un  4
Solution
From a previous example, we know that the complementary solution is
vn  (3) n A  B
In this case we expect a particular solution of the form qn=C. However, if we introduce
this form into the general solution, we would obtain that the left-hand side of the equation
is identical zero. That means that this choice of the particular solution fails, so we are
choosing qn=Cn. Then
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qn2  2qn1  3qn  4  C (n  2)  2C (n  1)  3C  4  4C  4  0  C  1
Hence the general solution is
u n  (3) n A  B  n
Numerical methods for solving ordinary differential equations.
Most ordinary differential equations arising in real-world applications cannot be solved
exactly. These equations can be analyzed qualitatively. However, qualitative analysis
cannot give accurate answers. A numerical method can be used to get an accurate
approximate solution to a differential equation.
1. Euler’s method
We will focus on this method as the simplest numerical method used for solving ordinary
differential equations. In order to emphasize the importance of numerical methods, let us
take first an example: solve the equation y′=2x with y(0)=0.
It is obvious that this simple equation has as solution y=x2, i.e. a formulaic solution. On
the other hand, say we were to use a numerical technique. The resulting numerical
solution would be a table of values. To have a better idea of these two solutions, let us
compare them side by side, along with the graphs
Notice that the graph derived from the formulaic solution is smoothly continuous,
consisting of infinite number of points on the interval shown. On the other hand, the
graph based on the numerical solution consists of just a few points as the numerical
method apparently only found the value of the solution for x-increments of size 0.2.
The first question you could ask is “ then what good is the numerical solution if it leaves
out so much of the real answer?”. Well, we can answer to this question in several ways:
1. The numerical solution still looks like it is capturing the general trend of the ‘real’
solution => if we are looking for a qualitative view of the solution, we can still get it from
the numerical solution.
2. The numerical solution can be even improved by playing ‘join-the-dots’ with the set of
points it produces.
3. Using a numerical approach, we can find the value of the solution at a specific point.
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4. If a formulaic expression for solution is not possible, here the numerical solutions are
the only possibility to find solutions (errors)
In what follows we will concentrate mainly on first order equations only. In order to
develop a technique for solving first order differential equations when the initial values
are given (or initial value problems). We assume that the studied equation has the form
y′=f(x,y) with y(x0)=y0. Our goal is to find a numerical solution, i.e. we must find a set of
points which lie along the initial value problem’s solution. This means that we actually
know one point of the solution, i.e. (x0,y0)-initial value.
Now, remember we do not know the true solution of the problem, but let us act as if we
do know this elusive solution for a moment. Let us pretend that its ‘ghostly graph’ could
be superimposed onto our previous picture to get this
Since we are after a set of points which lie along the true solutions, we must now derive a
method of generating more solutions in addition to the initial condition. Let use the other
part of the problem, namely the form of the differential equation itself.
Remember that one interpretation of the quantity y′ appearing in this expression is as the
slope of the tangent line to the function y. But the function y is exactly what we are
looking for as solution of the problem, i.e.
slope of the function=f(x,y)
Now we have to think how can we use this slope to find subsequent points on the graph.
Actually, we can obtain the slope by substituting values for x and y into the function f.
These values, of course, must be the coordinates of a point lying on the solution’s
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graph—they cannot just be the coordinates of any point anywhere in the plane. Therefore
we use the initial value as starting point, i.e. we can find the slope of the graph at the
initial value
slope of the solution at (x0,y0)=f(x0,y0)
Remembering that this gives us the slope of the function’s tangent line at the initial point
we could put this together with the initial point itself to build the tangent line at the initial
point
Looking at the two graphs, we can observe that close to the initial value the two curves
are every close to each other. Let us say we move a short distance away, to a new xcoordinate of x1. Then we could locate the corresponding point lying on our tangent line.
It might look like this
Notice that our new point, called (x1,y1), is not too far away from the true value of the
solution at this x-coordinate, up on the curve. So we now have two points as part of our
numerical solution
a. (x0,y0): an exact value, known to lie on the solution curve
b. (x1,y1): an approximate value, known to lie on the solution curve’s tangent line through
(x0,y0).
We must now attempt to continue our quest for points on the solution curve. Next we will
repeat our last step, constructing a tangent line at our new point
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There is a problem here: since our new point did not actually lie on the true solution, we
cannot actually produce a tangent line to the solution at this point. But we can still
substitute our new point (x1,y1), into the formula
slope of the function=f(x,y)
to get the slope of a pseudo-tangent line to the curve at (x1,y1). We hope that our
approximate point, is close enough to the real solution that the pseudo-tangent line is
pretty close to the unknown real tangent line. We now attempt to use this new pseudotangent line to get another point in the approximate solution. As before, we move a short
distance away from our last point, to a new x-coordinate of x2. Then we locate the
corresponding point lying on our pseudo-tangent line. The result might look something
like this
We now have three points in our approximate solution:
a. (x0,y0): an exact value, known to lie on the solution curve
b. (x1,y1): an approximate value, known to lie on the solution curve’s tangent line through
(x0,y0)
c. (x2,y2): an approximate value, known to lie on the solution curve’s pseudo-tangent line
through (x1,y1).
This method can be repeated making new points for as long as we like.
Now let us see what the theoretical background of the Euler’s method is. We are solving
the initial value problem
y   f ( x, y),
y( x0 )  y 0
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As said before, the basic idea is to use a known point as a starter, and then use the tangent
line through this known point to jump to a new point. Let us use the names
(xn,yn) for the known point
(xn+1,yn+1) for the new point
Our picture, based on the previous experience, should look like something like this
Our task is to find formulas for the coordinates of the new point, xn+1. Clearly, it lies on
the tangent line, and this tangent line has a known slope, namely f(xn,yn). Let us mark the
sizes of the jumps in the x and y direction as we move from the known point to the new
point.
From the formula relating xn and xn+1 is obvious that x n 1  x n  h
Also we know from basic algebra that slope=rise/run, so applying this idea to the triangle
in our picture, the formula becomes
f x n , y n   y
h
which can be rearranged to solve for Δy giving
y  hf xn , yn 
But, we are really after a formula for yn+1. Looking at the picture, it is obvious that
yn+1=yn+Δy. Replacing Δy by our new formula, this becomes
y n 1  y n  hf x n , y n  .
Example
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Solve the initial value problem
y   x  2 y y (0)  0
numerically, finding a value for the solution at x=1, and using steps of size h=0.25.
Solution
We apply the theoretical formulas found before. The interval we want to find
intermediate points is [0,1]. In our case f(x,y)=x+2y and the starting point is x0=0, y0=0.
We generate values for x1, y1.
The x-iteration formula, with n=0 gives
x1  x0  h  0  0.25  0.25
The y-iteration formula, with n=0 gives
y1  y 0  hf x0 , y 0   y 0  h( x0  2 y 0 )  0  0.25(0  2 * 0)  0
So the second point in our numerical solution is x1=0.25, y1=0.
We move on to get the next point in the solution, (x2,y2), n=1.
x2  x1  h  0.25  0.25  0.5
y2  y1  hf x1 , y1   y1  h( x1  2 y1 )  0  0.25(0.25  2 * 0)  0.0625
So the third point is x2=0.5, y2=0.0625.
The x-iteration and y-iteration, with n=2 gives
x3  x2  h  0.5  0.25  0.75
y3  y2  hf x2 , y2   y2  h( x2  2 y2 ) 
0.0625  0.25(0.5  2 * 0.0625)  0.21875
So the fourth point is x3=0.75, y3=0.21875
Carrying on with the same method we obtain that the fifth point is x4=1, y4=0.515625.
We could summarise the results of all our calculations in a tabular form, as follows
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n xn
yn
0 0.00 0.000000
1 0.25 0.000000
2 0.50 0.062500
3 0.75 0.218750
4 1.00 0.515625
A question you should always ask yourself at this point using a numerical method to
solve a problem is “How accurate is my solution?” Sadly, the answer is “Not very!”. The
true solution of the given initial value problem turns out to be
y  0.25e 2 x  0.5x  0.25
If we use this formula to generate a table similar to the one above, we can see just how
poorly our numerical solution did
x
y
0.00 0.000000
0.25 0.037180
0.50 0.179570
0.75 0.495422
1.00 1.097264
We can get an even better feel for the inaccuracy we have incurred if we compare the
graphs of the numerical and true solutions, as shown here
The numerical solution gets worse and worse as we move further to the right. We might
even be prompted to ask the question “What good is a solution that is bad?” The answer
is “Very little good at all!”. So should we quit using this method? No! The reason our
numerical solution is so inaccurate is because our step-size is so large. To improve
the solution, we have to shrink the step-size. For h=0.02 we would get a much better
agreement (see the graph).
As you can see, the accuracy of this numerical solution is much higher than before, but so
is the amount of work needed.
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DIFFERENTIAL EQUATIONS AND MAPLE
Maple’s diversity makes it particularly well suited to performing many calculations
encountered when solving many ordinary and partial differential equations. In many
cases, Maple’s built-in functions can immediately solve a differential equation by
providing an explicit, implicit, or numerical solution. The advantages of using Maple in
the study of differential equations are numerous, but perhaps the most useful is that of
being able to produce the graphics associated with solutions of differential equations.
In what follows you will see how we can solve various types of differential equations
using Maple. By now you already have a vast knowledge of Maple (SOM104), so we can
start directly with applications without introducing basic concepts about Maple.
Generally, the command
dsolve -Solves ordinary differential equations (ODEs)
Calling sequences:
dsolve(ODE, y(x), extra_args)
dsolve({ODE, ICs}, y(x), extra_args)
dsolve({sysODE,ICs},{funcs}. extra_args)
Parameters
ODE
y(x)
ICs
{sysODE}
{funcs}
-an ordinary differential equation
- any indeterminate function of one variable
- initial conditions
- a set with a system of ODEs
- a set with indeterminate functions
attempts to solve the given differential equation for the specified function of one variable.
Example
Verify that the differential equation y′=-ycosx has infinitely many solutions
Solution
We use dsolve to solve the first-order linear equation and name the result sol. We
interpret the result to mean that if C is any number, a solution to the equation is y=Ce-sinx.
Thus, the equation has infinitely many solutions
>sol:=dsolve(diff(y(x),x)=-y(x)*cos(x),y(x));
sol:=y(x)=_C1e(-sin(x))
We can plot several solutions using the plot command
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>pict:={seq(subs(_C1=I,rhs(sol)), i=-5..5)}:
plot(pict,x=0..4*Pi);
In many applications we are given not only the differential equation to solve but also one
or more conditions that must be satisfied by the solution(s) as well. In the calling
sequence of dsolve, ICs means initial conditions.
The following commands illustrate how to plot some members of the family of solutions
by substituting various values of C into the general solution. We also plot the solution to
the problem
 dy
2
  3x  4 x
 dx
 y (1)  4
First we use seq to generate a table of functions x3-2x2+C for
C=-10, -8,…,8, 10,
naming the resulting set of function toplot. The set of functions toplot is not displayed
(for length reasons) because a colon (:) is included at the of the command
> cvals:=seq(2*I,i=-5..5):
toplot:={seq(x^3-2*x^2+c,c=cvals)}:
plot(toplot,x=-2..3,view=[-2..3,-15..15])
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Example (The Logistic Equation) The logistic equation is
y (t )  (r  ay (t )) y (t ),
where r and a are constants, subject to the condition y(0)=y0. This equation can be also
written as y′=ry-ay2, where the term –y2 represents an inhibitive factor. Under the
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assumptions made, the population is allowed neither to grow out of control not to grow or
decay constantly. The logistic equation is separable and, thus, can be solved by separation
of variables:
dy
 dt
(r  ay ) y
 a
 r 
 r  ay

 a


 r  ay
1 
r dy  dt
y 

1
dy  rdt
y 
 ln r  ay  ln y  rt  c  ln
y
 e rt c  Ke rt
r  ay

y
 rt  c
r  ay
1

y  r  e  rt  a 
K


1
Applying the initial condition y(0)=y0 and solving for K, we find
y0
K
r  ay 0
After substituting this value into the general solution and simplifying, the solution can be
written as
y
ry 0
ay 0  (r  ay 0 )e rt
We are also able to use dsolve to solve this initial-value problem. First, we use dsolve to
solve the initial-value problem, naming the resulting output Sol.
> y:=′y′:
Sol:=dsolve({diff(y(t),t)=(r-a*y(t))*y(t), y(0)=y0},y(t));
Sol : y (t ) 
a
e
(  rt )
r
(  r  y 0a )
y0
Then we use assign to name y(t) the result obtained in Sol and simplify to simplify y(t)
> assign(Sol): simplify(y(t));

y 0r
 y 0r  e r  e (  rt ) y 0a
(  rt )