University of zakho

Faculty of Education
Department of mathematics
Second stage
Semester 4

Systems of First Order Linear Differential Equations

Prepared by

KAMAL HUSEEN
AYOB ABDULRAZAQ
SHAKR RASHED

Supervised By
DAWOUD

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:Introduction
In the introduction to this section we briefly discussed how a system of
differential equations can arise from a population problem in which we
keep track of the population of both the prey and the predator. It makes
sense that the number of prey present will affect the number of the
predator present. Likewise, the number of predator present will affect
the number of prey present. Therefore the differential equation that
governs the population of either the prey or the predator should in some
way depend on the population of the other. This will lead to two
differential equations that must be solved simultaneously in order to
.determine the population of the prey and the predator
The whole point of this is to notice that systems of differential equations
can arise quite easily from naturally occurring situations. Developing an
effective predator-prey system of differential equations is not the
subject of this chapter. However, systems can arise from nthnth order
linear differential equations as well. Before we get into this however,
.let’s write down a system and get some terminology out of the way
We are going to be looking at first order, linear systems of differential
equations. These terms mean the same thing that they have meant up to
this point. The largest derivative anywhere in the system will be a first
derivative and all unknown functions and their derivatives will only occur
to the first power and will not be multiplied by other unknown functions.
Here is an example of a system of first order, linear differential
equations

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 Systems of First Order Linear Differential Equations

We will now turn our attention to solving systems of simultaneous

homogeneous first order linear differential equations. The solutions
of such systems require much linear algebra (Math 220). But since it
is not a prerequisite for this course, we have to limit ourselves to
the simplest instances: those systems of two equations and two
unknowns only. But first,we shall have a brief overview and learn
.some notations and terminology

A system of n linear first order differential equations in n

unknowns (an n × n system of linear equations) has the
general form:

Where the coefficients aij’s, and gi’s are arbitrary functions of t. If every
term gi is constant zero, then the system is said to be homogeneous.
Otherwise, it is a nonhomogeneous system if even one of the g’s is nonzero

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The system (*) is most often given in a shorthand format as a matrix
vector

:equation, in the form

'
X A Xg

Where the matrix of coefficients, A, is called the coefficient matrix of the

system. The vectors x′, x, and g are

For a homogeneous system, g is the zero vector. Hence it has the form

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.x′ = Ax

Fact: Every n-th order linear equation is equivalent to a system of n first

order linear equations. (This relation is not one-to-one. There are

multiple

).systems thus associated with each linear equation, for n > 1

Examples:
.The mechanical vibration equation m u″ + γ u′ + k u = F(t) is equivalent to )i(

'
x=
1 x2

' −k y F ( t)
x=
2 x−
1 x 2+
m m m

Note that the system would be homogeneous (respectively,

nonhomogeneous) if the original equation is homogeneous
.)respectively, nonhomogeneous(

' ''
y− ''
2y+ 3 y−' tnelaviuqe
4 y= 0si ¿. )ii(
'
x=
1 x2

'
x=
2 x2

x=
3 4 x−
1 3 x 2+ 2 x 3

This process can be easily generalized. Given an n-th order linear

equation

.any(n) + an−1 y(n−1) + an−2 y(n−2) + … + a2 y″ + a1 y′ + a0 y = g(t)

5
 Make the substitutions:
and xn′ = y(n).The
first n − 1 equations follow thusly. Lastly, substitute the
x’s into the original equation to rewrite it into the n-th
:equation and obtain the system of the form
'
x=
1 x2

'
x=
2 x3

'
x=
3 x4

:::

x n−' =
1 xn

' − a0 a1 a2 a g ( t)
x=
1 x−
1 x−
2 x−…
3 − n− 1 x n+
an an an an an

Note: The reverse is also true (mostly)*. Given an n × n system of linear

.equations, it can be rewritten into a single n-th order linear equation

Example : convert each linear equation into a system of first order

.equation
1 ¿ y −'' 4 y +' 5= 0
'' '
2¿ y − 5 y +'' 9 y=soc
t 2t
' ''
y (4+) 3 y − π y−' 6 y=11 )3
:Solution
'
1 ¿ ¿ x=
1 x2

'
x=−
2 5 x 1+ 4 x 2
'
2 ¿ ¿ x ¿=
1 x2

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 '
x=
2 x3

'
x=−
3 9 x 1+ 5 x 3+soc
t 2t
'
3 ¿ ¿ x=
1 x2

'
x=
2 x2

'
x=
3 4 x−
1 3 x 2+ 2 x 3

'
x=
4 6 x−
1 2 π x 2+ π x−
3 3 x 4+
11

:Review topics of matrcies

1. What is a matrix (pl. matrices)?
A matrix is a rectangular array of objects (called entries). Those
entries are usually numbers, but they can also include functions,
vectors, or even other matrices. Each entry’s position is addressed
by
the row and column (in that order) where it is located. For
example,

a52 represents the entry positioned at the 5th row and the 2nd
column of the matrix A
2. The size of a matrix.
The size of a matrix is specified by 2 numbers
[number of rows] × [number of columns].
Therefore, an m × n matrix is a matrix that contains m rows
and n
columns. A matrix that has equal number of rows and
columns is
called a square matrix. A square matrix of size n × n is usually
.referred to simply as a square matrix of size (or order) n
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 Notice that if the number of rows or columns is 1, the result (respectively, a
1 × n, or an m × 1 matrix) is just a vector. A 1 × n matrix is called a row
vector, and an m × 1 matrix is called a column vector. Therefore, vectors are
really just special types of matrices. Hence, you will probably notice the
similarities between many of the matrix operations defined below and vector
operations that you might be familiar with.

.Two special types of matrices .3

Identity matrices (square matrices only)

.The n × n identity matrix is often denoted by In

1 0 0
I=
2
1 0
[ ]
0 1
I=
3
[ ]
0 1 0
0 0 1

:Properties (assume A and I are of the same size)

AI = IA = A

In x = x , x = any n × 1 vector

.Zero matrices – matrices that contain all-zero entries

A+0=0+A=A
A0=0=0A
.Arithmetic operations of matrices .4
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.Addition / subtraction )i(

[ac bd ]± [ge hf=] [c±a± ge b± f

d±h ]
.Scalar Multiplication )ii(
k a b = ak bk rofyna
[ ][
c d ck dk
, ralacs ] k.

.Matrix multiplication )iii(

[ac bd ][ge hf=] eaec[ ++gbgd fa +hb
fc +hd ]
The matrix multiplication AB = C is defined only if there are as many

rows in B as there are columns in A. For example, when A is m × k

and B is k × n. The product matrix C is going to be of size m × n, and

whose ij-th entry, cij, is equal to the vector dot product between the ith

row of A and the j-th column of B. Since vectors are matrices, we

can also multiply together a matrix and a vector, assuming the above

-restriction on their sizes is met. The product of a 2 × 2 matrix and a 2

entry column vector is

[ac bd ][xy=] xaxc[ yb

yd ]
Note 1: Two square matrices of the same size can always be multiplied together.
Because, obviously, having the same number of rows and columns, they satisfy the size
.requirement outlined above

Note 2: In general, AB ≠ BA. Indeed, depending on the sizes of A and B, one product
. might not even be defined while the other product is

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 .Determinant (square matrices only) .5
For a 2 × 2 matrix, its determinant is given by the formula

ted [ac bd =ba

] −cb
Note: The determinant is a function whose domain is the set of all
square matrices of a certain size, and whose range is the set of all real

.numbers )or complex(

.Inverse matrix (of a square matrix) .6

Given an n × n square matrix A, if there exists a matrix B (necessarily

.of the same size) such that

𝑨B=BA= I n

Then the matrix B is called the inverse matrix of A , denoted

A−1 The inverse matrix, if it exists , is unique for each A. A
matrix is called invertible if it has an inverse matrix.

A= [ac bd ] Theorem: For any 2 × 2 matrix

.its inverse , if exists, is given by

1 d −b
1
A −= [
da −cb − c a ]
.Theorem: A square matrix is invertible if and only if its determinant is nonzero

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[15 −22na] dB= −[ 21 −43 ] = Examples: Let A
2− 2 − 4− −( 3 ) 0 − 1
(i 2) A− B= 2 1 − 2− 2 − 3 = ][
5 2[ ][
− 1 4 01 − −( 1 ) 4− 4
=
11 0 ][ ]

ii( BA
) = [15 −22 ][−21 −43 =] 01[ 2+−22 −51
− 3− 8
+8 ]
= [4 −11
8 −7 ]

:On the other hand

2 −3 1 −2 2−51 − 4−01 −31 −01
[
−1 4 5 2
=][
− 1+02 ][ 2+ 8
= ][
91 01 ] =AB

det (A) = -2 (-10) = 12 , det(B) = 8 – 3 = 5 )iii(

.Since neither is zero , as a result , they are both invertible matrices
1 1
vi( A−1
) =
1
2− −01
(
2 2
[=
1 2 2
] [
= 6
)− 5 1 21 − 5 1 − 5 1
6

21 21
] [ ]
.Systems of linear equations (also known as linear systems) .7
A system of linear (algebraic) equations, Ax = b, could have zero, exactly
one, or infinitely many solutions. (Recall that each linear equation has a
line as its graph. A solution of a linear system is common intersection point
of all the equations’ graphs − and there are only 3 ways a set of lines could
intersect. If the vector b on the right-hand side is the zero vector, then the
system is called homogeneous. A homogeneous linear system alway has a
solution, namely the all-zero solution (that is, the origin). This

solution is called the trivial solution of the system. Therefore, a system

homogeneous linear Ax = 0 could have either exactly one
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solution, or infinitely many solutions. There is no other possibility, since
it always has, at least, the trivial solution. If such a system has n equations

and exactly the same number of unknowns, then the number of

solution(s) the system has can be determined, without having to

:solve the system, by the determinant of its coefficient matrix

.Eigenvalues and Eigenvectors .8

Given a square matrix A, suppose there are a constant r and a nonzero vector
,x such that. Ax = r x

then r is called an Eigenvalue of A, and x is an Eigenvector of A corresponding

.to r

Rewrite the above equation, we get Ax − r x = 0. The next step would

be to factor out x. But doing so would give the expression :
( A – r )x = 0
.Notice that it requires us to subtract a number from an n × n matrix

That’s an undefined operation. Hence, we need to further refined it by

:rewriting the term r x = r I x, and then factoring out x, obtaining

x = 0 )A – r I (

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 This is an n × n system of homogeneous linear (algebraic)
equations, where the coefficient matrix is (A − r I). We are looking
for a nonzero solution x of this system. Hence, by the theorem we
have just seen, the necessary and sufficient condition for the
existence of such a nonzero solution, which will become an
eigenvector of A, is that the coefficient matrix (A − r I) must have
zero determinant. Set its determinant to zero and what we
get is a degree n polynomial equation in terms of r. The
:case of a 2 × 2 matrix is as follow

A−Ir= a b − r 1 0 = a− r b
c d [ ] [ ][
0 1 c d− r ]
Its determinant, set to 0, yields the equation

ted [a−c r b = a(− r )(d− r−cb

d− r] 2
) = r− a(= d )r+da( −cb =) 0

It is a degree 2 polynomial equation of r, as you can see

This polynomial on the left is called the characteristic polynomial of the

(original) matrix A, and the equation is the characteristic equation of A.
The root(s) of the characteristic polynomial are the eigenvalues of A. Since
any degree n polynomial always has n roots (real and/or complex; not
necessarily distinct), any n × n matrix always has at least one, and up to n
different eigenvalues. Once we have found the eigenvalue(s) of the given
matrix, we put each specific eigenvalue back into the linear system (A
.− r I) x = 0 find the corresponding eigenvectors

A= 2 eulavnegie
3
4 3[ ]
dniF srotcevnegie
(s∧) . :Examples
:Solution

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 A−Ir= [24 33 −] r [10 01=] [2−4 r 3
3− r ]
retcarahc
stI noitauqe si

ted [2−4 r 3
3− r ]
= 2(− r )(3− r−21
2
) = r− 5 r− 6

¿ (r+ 1 ) (r− 6=) 0

.The eigenvalues are, therefore, r = −1 and 6

Next, we will substitute each of the 2 eigenvalues into the matrix
.equation (A − r I) x = 0

For r = −1, the system of linear equations is

A
( −Ir )x= A( + I )x= [2+41 3
3+ 1] [ 34 ]x= [00 ]
x=
3
4

Notice that the matrix equation represents a degenerated system of 2

linear equations Both equations are constant multiples of the.
equation x1 + x2 = 0. There is now only 1 equation for the 2 unknowns,
therefore, there are infinitely many possible solutions. This is always the
case when solving for eigenvectors. Necessarily,there are infinitely many
.eigenvectors corresponding to each Eigenvalue

,Solving the equation x1 + x2 = 0, we get the relation x2 = − x1. Hence

the eigenvectors corresponding to r = −1 are all nonzero multiples of
1
k=
1
−1 [ ]
:Similarly , for r = 6 , the system of equation is

[2−4 6 3 x= − 4 3 x= 0
3− 6 ] [
4 −3 0 ] [ ] = x=(A – 6 I)x) A – r I(
Both equations in this second linear system are equivalent to

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.x1 − 3 x2 = 0. Its solutions are given by the relation 4 x1 = 3 x2 4

Hence, the eigenvectors corresponding to r = 6 are all nonzero

k=
2 [34 ] multiples of

.Note: Every nonzero multiple of an eigenvector is also an eigenvector

:Refernce

Differential Equations, Dynamical systems ₰ An introduction to chaos -1

.(second Edition) By Morris W.Hirsch ,Stephen Smale , RobertL.Devaney

ADVANCED ENGINEERING MATHEMATICS BY (ERWIN KREYSZIG, HERBERT -2

KREYSZIG, EDWARD J. NORMINTON)

ELEMENTARY DIFFERENTIAL EQUATIONSWITH BOUNDARY VALUE PROBLEM -3

)William F. Trench(
ORDINARY DIFFERENTIAL EQUATIONS -4
.)GABRIEL NAGY, Mathematics Department, Michigan State University, East Lansing, MI, 48824(

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