CHAPTER 1
System of First Order Differential Equations
In this chapter, we will discuss system of first order differential equations. There are many applications that involving find several unknown
functions simultaneously . Those unknown functions are related by a
set of equations that involving the unknown functions and their first
derivatives. For example, in Chapter Two, we studied the epidemic of
contagious diseases. Now if
• S(t) denotes number of people that is susceptible to the disease
but not infected yet.
• I(t) denotes number of people actually infected.
• R(t) denotes the number of people have recovered.
If we assume
• The fraction of the susceptible who becomes infected per unit
time is proportional to the number infected, b is the proportional number.
• A fixed fraction rS of the infected population recovers per unit
time, 0 ≤ r ≤ 1.
• A fixed fraction of the recovers g become susceptible and infected, 0 ≤ g ≤ 1. proportional function.
The system of differential equations model this phenomena are
S 0 = −bIS + gR
I 0 = bIS − rI
R0 = rI − gR
The numbers of unknown function in a system of differential equations can be arbitrarily large, but we will concentrate ourselves on 2 to
3 unknown functions.
1. Principle of superposition
Let aij (t), bj (t) i = 1, 2, · · · , n and j = 1, 2, · · · , n be known
function, and xi t, i = 1, 2, · · · , n be unknown functions, the linear first
1
2
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
order system of differential equation for xi (t) is the following,
x01 (t) = a11 (t)x1 (t) + a12 (t)x2 (t) + · · · + a1n (t)xn (t) + b1 (t)
x02 (t) = a21 (t)x1 (t) + a22 (t)x2 (t) + · · · + a2n (t)xn (t) + b2 (t)
x03 (t) = a31 (t)x1 (t) + a32 (t)x2 (t) + · · · + a3n (t)xn (t) + b3 (t)
..
.
x0n (t) = an1 (t)x1 (t) + an2 (t)x2 (t) + · · · + ann (t)xn (t) + f1 (t)
Let x(t) be the column vector of unknown functions xi t, i =
1, 2, · · · , n, A(t) = (aij (t), and b(t) be the column vector of known
functions bi t, i = 1, 2, · · · , n, we can write the first order system of
equations as
(1)
x0 (t) = A(t)x(t) + b(t)
• When n = 2, the linear first order system of equations for two
unknown functions in matrix form is,
· 0
¸ ·
¸·
¸ ·
¸
x1 (t)
a11 (t) a12 (t)
x1 (t)
b1 (t)
=
+
x02 (t)
a21 (t) a22 (t)
x2 (t)
b2 (t)
• When n = 3, the linear first order system of equations for
three unknown functions in matrix form is,
 0
 

 

x1 (t)
a11 (t) a12 (t) a13
x1 (t)
b1 (t)
 x02 (t)  =  a21 (t) a22 (t) a23   x2 (t)  +  b2 (t) 
x03 (t)
a31 (t) a32 (t) a33
x3 t
b3 (t)
A solution of equation (1) on the open interval I is a column vector function x(t) whose derivative (as a vector-values function) equals
A(t)x(t) + b(t). The following theorem gives existence and uniqueness
of solutions,
Theorem 1.1. If the vector-valued functions A(t) and b(t) are continuous over an open interval I contains t0 , then the initial value problem
½ 0
x (t) = A(t)x(t) + b(t)
x(t0 ) = x0
has an unique vector-values solution x(t) that is defined on entire interval I for any given initial value x0 .
When b(t) ≡ 0, the linear first order system of equations becomes
x0 (t) = A(t)x(t),
which is called a homogeneous equation.
As in the case of one equation, we want to find out the general
solutions for the linear first order system of equations. To this end, we
first have the following results for the homogeneous equation,
1. PRINCIPLE OF SUPERPOSITION
3
Theorem 1.2. Principle of Superposition Let x1 (t), bx2 (t), · · · , xn (t)
be n solutions of the homogeneous linear equation
x0 (t) = A(t)x(t)
on the open interval I. If c1 , c2 , · · · , cn are n constants, then the
linear combination
c1 x1 (t) + c2 x2 (t) + c3 x3 (t) + · · · + cn xn (t)
is also a solution on I.
Example 1.1. Let
·
¸
1 0
x (t) =
x(t)
0 −2
· t ¸
·
¸
e
0
, x1 (t) =
and x2 (t) =
are two solutions, as
0
e−2t
·
¸ · t ¸ ·
¸· t ¸
(et )0
e
1 0
e
0
bx1 (t) =
=
=
0
0
0 −2
0
0
and
·
bx02 (t)
=
0
(e−2t )0
¸
·
=
0
−2e−2t
¸
·
=
1 0
0 −2
¸·
0
¸
e−2t
By the Principle of Superposition, for any two constants c1 and c2
· t ¸
·
¸ ·
¸
e
0
c1 et
x(t) = c1 x1 (t) + c2 x2 (t) = c1
+ c2
=
0
e−2t
c2 e−2t
is also solution. We shall see that it is actually the general solution.
The next theorem gives the general solution of linear system of
equations,
Theorem 1.3.
- Let x1 (t), x2 (t), · · · , bxn (t) be n linearly independent (as vectors)
solution of the homogeneous system
x0 (t) = A(t)x(t),
then for any solution xc (t) there exists n constants c1 , c2 , · · · , cn such
that
xc (t) = c1 x1 (t) + c2 x2 (t) + · · · + cn xn (t).
We call xc (t) the general solution of the homogeneous system.
4
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
If xp (t) is a particular solution of the nonhomogeneous system,
x(t) = B(t)x(t) + b(t),
and xc (t) is the general solution to the associate homogeneous system,
x(t) = B(t)x(t)
then x(t) = xc (t) + xp (t) is the general solution.
Example 1.2. Let
·
¸
·
¸
4 −3
−4t2 + 5t
0
x (t) =
x(t) +
x(t)
6 −7
−6t2 + 7t + 1
·
¸
· −5t ¸
3e2t
e
, x1 (t) =
and x2 (t) =
are two linearly independent
2t
2e
3e−5t
· 2 ¸
t
solutions. and xp (t) =
is a particular solution. By Theorem
t
1.3,
·
¸
3c1 e2t + c2 e−5t + t2
(2) x(t) = c1 x1 (t) + c2 x2 (t) + xp (t) =
2c1 e2t + 3c2 e−5t + t
is the general solution. Now suppose we want to
· find ¸a particular so2
lution that satisfies the initial condition x(0) =
, then let t = 0
−1
in (2), we have
·
¸ ·
¸
3c1 + c2
2
x(0) =
=
,
2c1 + 3c2
−1
which can be written in matrix form,
·
¸·
¸ ·
¸
3 1
c1
2
=
,
2 3
c2
−1
·
¸
·
¸
c1
1
Solve this equation, we get
=
. So the particular
c2
−1
¸
·
3e2t − e−5t + t2
.
solution is x(t) =
2e2t − 3e−5t + t
From the above example, we can summarize the general steps in
find a solution to initial value problem,
½
x0 (t) = A(t)x(t) + b(t)
x(t0 ) = x0
2. HOMOGENEOUS SYSTEM
5
• Step One: Find the general solution xc = c1 x1 (t) + c2 x2 (t) +
· · · + cn xn (t), where x1 (t), x2 (t), · · · , xn (t) are a set of linearly independent solutions, to the associate homogeneous system, x0 (t) = A(t)x(t).
• Step Two: Find a particular solution xp (t)to the nonhomogeneous system, x0 (t) = A(t)x(t) + b(t).
• Step Three: Set x(t) = xc (t) + xp (t) and use the equation
x(t0 ) = x0 , to determine c1 , c2 , · · · , cn .
2. Homogeneous System
We will use a powerful method called eigenvalue method to solve
the homogeneous system
x0 (t) = Ax(t)
where A is a matrix with constant entry. We will present this method
for A is either a 2 × 2 or 3 × 3 cases. The method can be used for A
is an n × n matrix. The idea is to find solutions of form
(3)
x(t) = veλt ,
a straight line that passing origin in the direction v. Now taking derivative on x(t), we have
(4)
x0 (t) = λveλt
put (3) and (2.2) into the homogeneous equation, we get
x0 (t) = λveλt = Aveλt
So
Av = λv,
which indicates that λ must be an eigenvalue of A and v is an associate
eigenvector.
2.1. A is a 2 × 2 matrix. Suppose
·
¸
a11 a12
A=
a21 a22
Then the characteristic polynomial p(λ) of A is
p(λ) = |A−λI| = (a11 −λ)∗(a22 −λ)−a12 a21 = λ2 −(a11 +a22 )+(a11 a22 −a12 a22 .
So p(λ) is a quadratic polynomial of λ. From Algebra, we know that
p(λ) = 0 has either 2 distinct real solutions, or a double solution, or
2 conjugate complex solutions. The following theorem summarize the
solution to the homogeneous system,
6
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Theorem 2.1. Let p(λ) be the characteristic polynomial of A, for
x0 (t) = Ax(t),
Case 1: p(λ) = 0 has two
λ¸1 and λ2 .
· distinct
¸ real solutions
·
v11
v12
Suppose v 1 =
and v 2 =
are associate eigenv21
v22
vector (i.e, Av 1 = λ1 v 1 and Av 2 = λ2 v 2 ) Then the general
solution is
xc (t) = c1 v 1 eλ1 t + c2 v 2 eλ2 t
And
·
v11 eλ1 t v12 eλ2 t
v21 eλ1 t v22 eλ2 t
Φ(t) =
¸
is called the fundamental matrix(A fundamental matrix
is a square matrix whose columns are linearly independent solutions of the homogeneous system).
Case 2: p(λ) = 0 has a double solutions λ0 .
In this case p(λ) = (λ − λ0 )2 and λ0 is a zero of p(λ) with
multiplicity 2.
(1) λ0 has
eigenvectors:
· two¸ linearly independent
·
¸
v11
v12
Suppose v 1 =
and v 2 =
are associate linearly
v21
v22
independent eigenvectors. Then the general solution is
xc (t) = (c1 v 1 + c2 v 2 )eλ0 t
And
·
λ0 t
Φ(t) = e
v11 v12
v21 v22
¸
(2) λ0 has
· only¸ one associate eigenvector:
v11
Suppose v 1 =
is the only associated eigenvector and
v21
·
¸
v12
v2 =
is a solution of
v22
(λ0 I − A)v 2 = v 1 .
Then the general solution is,
xc (t) = (c1 v 1 + c2 (tv 1 + v 2 )eλ0 t
And
·
Φ(t) = e
λ0 t
v11 (v11 t + v12 )
v21 (v21 t + v22 )
is the fundamental solution matrix.
¸
2. HOMOGENEOUS SYSTEM
7
Case 3: p(λ) = 0 has two
solutions a + bi and a − bi.
· conjugate complex
¸
v11 + iv12
Suppose v =
is the associate complex eigenv21 + iv22
vector· with ¸respect to a· + bi, ¸then the general solution is,
v11
v12
v1 =
and v 2 =
v21
v22
xc (t) = [c1 (v 1 cos(bt) − v 2 sin(bt))c2 (v 2 cos(bt) + v 1 sin(bt))]eat .
And
at
Φ(t) = e
·
v11 cos(bt) − v12 sin(bt) v12 cos(bt) + v11 sin(bt)
v21 cos(bt) − v22 sin(bt) v22 cos(bt) + v21 sin(bt)
¸
is the fundamental matrix.
From Theorem 2.1, let Φ(t) be the fundamental
matrix,
the general
·
¸
c1
solution is given by xc (t) = Φ(t)c, with c =
and the solution
c2
that satisfies a given initial condition x(t0 ) = x0 is given by
x(t) = Φ(t)Φ(t )-t x
0
0
Example 2.1. ·
Two distinct
¸ eigenvalues case Find the general
2 −3
solution to x0 (t) =
x(t)
−1 −5
Solution
vecs()
Using Mathcad , functions eigenvals() and eigen-
In Mathcad , eigenvecs(M) Returns a matrix containing the eigenvectors. The
nth column of the matrix returned is an eigenvector corresponding to the nth
eigenvalue returned by eigenvals.
√
we find,λ1 = ·− 32 + 21 61
√
√
3
1
and
λ
=
−
−
61 with ¸associated eigen2
2· 2
¸
√
−7 − 61
−7 + 61
vectors v 1 =
and v 2 =
respectively. So
2
2
the fundamental matrix is
√
√
√
√
·
¸
3
− 12 61)t
(− 32 + 12 61)t
2
(−7 − 61)e√
(−7 + 61)e(−
√
Φ(t) =
3 1
3
1
2e(− 2 + 2 61)t
2e(− 2 − 2 61)t
·
¸
c1
and the general solution is, for c =
,
c2
xc (t) = Φ(t)c
a
8
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Example 2.2. One double eigenvalues with two linearly
in- ¸
·
2
0
dependent eigenvectors Find the general solution to x0 (t) =
x(t).
0 2
Solution
¸
·
¸
·The ¸eigenvalue is λ0 = 2 and associated
· eigenvectors
0
c1 e2t
1
and
, so the general solution is xc =
a
are
1
c2 e2t
0
Example 2.3. One double eigenvalues
·
¸with only one eigen·
¸
2
−3
2
0
vector Find the solution to x (t) =
x(t) and x(0) =
3 8
3
Solution
Using Mathcad , functions eigenvals() and
· eigen- ¸
−.707
vecs() we can find a double eigenvalue λ0 = 5 and eigenvector
.
.707
Notice, the symbolic operator →(bring
˙
up by either [Shift][Ctrl][.] or
[Ctrl][.]) will not work with eigenvecs() this time, but since multiply
an eigenvector
· by a¸ nonzero constant still get an eigenvector, we can
3
choose v 1 =
.
−3
To
(A − λ0 I)w = v 1 λ0 we will solve (A −
· find ¸w that
· satisfies
¸
w1
1
λ0 I)
=
. That is,
w2
−1
·
¸·
¸ ·
¸
−3 −3
w1
3
=
3
3
w2
−3
One solution is w1 = 1 and w2 = 0
So the fundamental matrix is
·
¸
−3 −3t + 1
5t
Φ(t) = e
3
3t
·
¸
c1
and the general solution is, c =
,
c2
·
Now, Φ(0) = e5(0)
·
¸
0
−1
1
.
3
−3 −3
xc (t) = Φ(t)c
¸
·
¸
−3 −3(0) + 1
−3 1
=
and Φ(0)-1 =
3
3(0)
3 0
Hence, the particular solution is x(t) = Φ(t)Φ(0)-1 x0 = e5t
·
−3t + 4
3t − 3
¸
a
2. HOMOGENEOUS SYSTEM
9
Example 2.4. Two conjugate
complex
eigenvalues case Find
¸
·
2
−3
the general solution to x0 (t) =
x(t)
1 2
Solution
Using Mathcad , functions eigenvals() and √eigenvecs() we find two conjugate complex eigenvalues,
1 = 2 + i 3 and
· √ λ¸
√
3
with respect to
λ2 = 2 − i 3 with associated eigenvector v 1 =
−i
√
λ1 . Compare this with the Theorem 2.1, we have a = 2, b = 3, v11 =
√
3, v21 = 0, v12 = 0, and v22 = −1.
So the fundamental matrix is
· √
¸
3
cos(bt)
−
sin(bt)
2t
√
Φ(t) = e
3 sin(bt) − cos(bt)
·
¸
c1
and the general solution is, c =
,
c2
√
√
¸·
¸
· √
3 cos(√ 3t) − sin( √3t)
c1
2t
√
xc (t) = Φ(t)c = e
c2
√3 sin( 3t) −
√cos(¸ 3t)
· √
3c cos(√ 3t) − c2 sin(√ 3t)
= e2t √ 1
3c1 sin( 3t) − c2 cos( 3t)
·
¸
1
Suppose we want to find a solution such that x(0) =
, then
2
x(t) = Φ(t)Φ(0)-1 x(0)
√
√
¸-1 ·
¸
· √
¸· √
1
3
cos(
3t)
−
sin(
3t)
3 0
2t
√
√
√
= e
2
· √3 sin( √3t) − cos(√ 3t) ¸ · 01 ¸ −1
√
3 cos(√ 3t) − sin( √3t)
3
= e2t √
−2
3
sin(
3t)
−
cos(
3t)
√
√
·
¸
cos( √
3t) + +2 sin(√ 3t)
2t
= e
− sin( 3t) + 2 cos( 3t)
a
2.2. A is a 3 × 3 matrix. Suppose


a11 a12 a13
A =  a21 a22 a23 
a31 a32 a33
Then the characteristic polynomial p(λ) of A given by
p(λ) = |A − λI|,
10
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
is a cubic polynomial of λ. From Algebra, we know that p(λ) = 0
has either 3 distinct real solutions, or 2 distinct solutions and one is a
double solution, or one real solution and 2 conjugate complex solutions,
or a triple solution. The following theorem summarize the solution to
the homogeneous system,
Theorem 2.2. Let p(λ) be the characteristic polynomial of A, for
x0 (t) = Ax(t),
Case 1: p(λ) = 0 has three
λ3 .


 λ1 , λ2 , and 
 distinct
 real solutions
v13
v12
v11
Suppose v 1 =  v21 , v 2 =  v22 , and v 3 =  v23 
v31
v32
v33
are associate eigenvector (i.e, Av 1 = λ1 v 1 , Av 2 = λ2 v 2 , and
Av 3 = λ3 v 3 ) Then the general solution is
xc (t) = c1 v 1 eλ1 t + c2 v 2 eλ2 t + c3 v 3 eλ3 t
And the fundamental matrix is


v11 eλ1 t v12 eλ2 t v13 eλ3 t
Φ(t) =  v21 eλ1 t v22 eλ2 t v23 eλ3 t  .
v31 eλ1 t v32 eλ2 t v33 eλ3 t
Case 2: p(λ) = 0 has a double solutions λ0 .
2
So p(λ)
 = (λ− λ0 ) (λ − λ1 ), and λ0 has multiplicity 2. Let
v12

v22  is the eigenvector associated with λ1 .
v3 =
v32
[1] λ0 has
eigenvectors:
· two¸linearly independent
·
¸
v11
v12
Suppose v 1 =
and v 2 =
are associate linearly
v21
v22
independent eigenvectors. Then the general solution is
xc (t) = (c1 v 1 + c2 v 2 )eλ0 t + c3 v 3 eλ1 t
And

v11 eλ0 t v12 eλ0 t v13 eλ1 t
Φ(t) =  v21 eλ0 t v22 eλ0 t v23 eλ1 t 
v31 eλ0 t v32 eλ0 t v33 eλ1 t

2. HOMOGENEOUS SYSTEM
11
[2] λ0 has
 one eigenvector:

v11
Suppose v 1 =  v21  is the associated eigenvector with rev31


v12
spect to λ0 and v 2 =  v22  is a solution of
v32
(λ0 I − A)v 2 = v 1 .
Then the general solution is,
xc (t) = (c1 v 1 + c2 (tv 1 + v 2 ))eλ0 t + c3 v 3 eλ1
And


v11 eλ0 t (v11 t + v12 )eλ0 t v13 eλ1
Φ(t) =  v21 eλ0 t (v21 t + v22 )eλ0 t v23 eλ1 
v31 eλ0 t (v31 t + v32 )eλ0 t v33 eλ1
is the fundamental solution matrix.
Case 3: p(λ) = 0 has two conjugate complex solutions a ± bi and a real
solution λ1 . 

v11 + iv12
Suppose v =  v21 + iv22  is the associate complex eigenv31 + iv32
vectorwith respect
to a + bi, then the general solution is, let

v13
v 3 =  v23 , are associated eigenvectors with respect to λ1 ,
V33
xc (t) = [c1 (v 1 cos(bt)−v 2 sin(bt))c2 (v 2 cos(bt)+v 1 sin(bt))]eat +c3 v 3 eλ1 .
And


v11 cos(bt) − v12 sin(bt) v12 cos(bt) + v11 sin(bt) v13 eλ1
Φ(t) = eat  v21 cos(bt) − v22 sin(bt) v22 cos(bt) + v21 sin(bt) v23 eλ1 
v31 cos(bt) − v32 sin(bt) v32 cos(bt) + v31 sin(bt) v33 eλ1
is the fundamental matrix.
Case 4: p(λ) = 0 has solution λ0 with multiplicity 3.
In this case, p(λ) = (λ − λ0 )3 .
[1] λ0has three
eigenvectors.



 linearly
 independent
v13
v12
v11
Let v 1 =  v21 , v 2 =  v22 , and v 3 =  v23  be
V33
V32
v31
the three linearly independent eigenvectors. Then the general
12
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
λ0 t
solution is xc (t) = (c1 v 1 + c2 v 2 + c3 v
and fundamental
3 )e
v11 v12 v13
 v21 v22 V23 



v
matrix is Φ(t) = eλ0 t 
31


 v32

V33
[2] λ0 has
two
linearly
eigenvectors.
independent



v12
v11

 are the linearly inde

v
v
, v2 =
Suppose v 1 =
22
21
V
v31
 32 
v13

v23  , then only one of the
pendent eigenvectors. Let v 3 =
V33
two equations, (A − λ0 I)v 3 = v 1 or (A − λ0 I)v 3 = v 2 can
has a solution that is linearly independent with v 1 , v 2 .
Suppose (A−λ0 I)v 3 = v 2 generates such a solution. Then
the general solution is xc (t) = [c1 v 1 +c2 v 2 + c3 (tv 2 + v 3 )]eλ0 t
v11 v12 tv12 + v13
and fundamental matrix is Φ(t) = eλ0 t  v21 v22 tv22 + V23 
v31 v32 tv32 + V33
[3] λ
has
only
one
eigenvector.
0

v11
Let v 1 =  v21  be the linearly independent eigenvectors.
v
 31 


v12
v13
Let v 2 =  v22  and v 3 =  v23  be two vectors that
V32
V33
satisfies
(A − λ0 I)v 2 = v 1
and (A − λ0 I)v 3 = v 2 .
Then the general solution is xc (t) = [c1 v 1 + c2 (tv 1 + v 2 ) +
c3 (t2v 1 + tv 2 + v 3 )]eλ0 t and fundamental matrix is Φ(t) =
v11 tv11 + v12 t2 v11 + tv12 + v13
eλ0 t  v21 tv21 + v22 t2 v21 + tv22 + V23 
v31 tv31 + v32 t2 v31 + tv32 + V33
Remark 2.1. Suppose A is an n × n matrix, for the homogeneous
system x0 (t) = Ax(t), three general case would happen
Case 1: A has n distinct eigenvalues λi , i = 1, 2, · · · , n with linearly
independent eigenvectors v i , i = 1, 2, · · · , n then the general
solution will be xc (t) = c1 v 1 eλ1 + c2 v 2 eλ2 + · · · + cn v n eλn
2. HOMOGENEOUS SYSTEM
13
Case2: A has m < n distinct eigenvalues, in this case some eigenvalues would have multiplicity greater than 1.
Suppose λr has multiplicity r. Depending on how many
linearly independent eigenvectors are associated with λr the
situation could be very complex. Let p be the number of linearly
eigenvectors associated with λr , then d = r − p is called the
deficit of λr . The simply cases are either d = 0 or d = r − 1.
When 0 < d < r − 1 the situation could be very complex.
Suppose d = r − 1 and v 1 is the only eigenvector associate
with λr , then one will have to solve r − 1 equations (A −
λr )i v i+1 = v i , i = 1, 2, · · · , r − 1. And the general solution
would contains terms like [c1 v 1 + c2 (v 1 t + v 2 ) + c3 (v 1 t2 + v 2 t +
v 3 ) + · · · + cr (v r1 + v 2 tr−1 + · · · + v r )]eλr .
Case 3: A complex root a+bi with associated eigenvector v a +iv b , then
the general solution contains term, [c1 (v a cos(bt)−v b sin(bt))+
c2 (v a sin(bt) + v b cos(bt))]eat .
Remark 2.2. Suppose x1 (t), x2 (t), x3 (t), · · · , xn (t) are n linearly
independent solution for n×n homogeneous system, x0 (t) = Ax(t), the
fundamental matrix Φ(t) is a matrix whose columns are xi (t), i =
1, 2, · · · , n.
Example 2.5. (Two distinct eigenvalues) Find the general solution to
x01 = 3x1 + 4x2 − 2x3
x02 = 2x1 + x2 − 4x3
x03 = x1 + 2x2




x1 (t)
3 4 −2
Solution Let x(t) =  x2 (t)  and  2 1 −4  The equax3 (t)
1 2 0
0
tions can be written in matrix form x (t) = Ax(t).
Using Mathcad , functions eigenvals() and eigenvecs()

 we find,λ1 =
−4
2 and λ2 = 1 with associated eigenvectors v 1 =  1  and v 2 =
−2


1
 0  respectively. Since λ1 has multiplicity 2 as 1 appeared twice
1
in the result of eigenvals() function, we need to solve the equation
(A − λ1 I)v 3 = v 2 .
14
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
To use Mathcad ,
(1) you first compute (A − λ1 I)v 3 using the following sequences of key stroke,
[*] type ([Ctrl][M], set the rows and
columns in the matrix definition popup menu,
input the data for A,
[*] type -[Ctrl][M] set the row and column number and input data for λ1 I,
[*] type )[Ctrl][M], now set 1 as column
number, enter a, b, c in the place holders,
[*] type [Ctrl][.] to compute symbolically and you get.
(2) Using the Given Find block to find a
solution. Type Given in a blank space, type
a+2b-c[Ctrl]= 1 and 2a-4c[Ctrl]=0 in two
rows, then type key word Find following by
typing (a,b)[Ctrl][.] you will get the solution
in terms of c.


4
Set c = 1, we get v 3 =  −1  .
1
So the fundamental matrix is


−4e2t et (t + 4)et
0
−et 
Φ(t) =  e2t
2t
t
−2e
e (t + 1)et
and the general solution is,
xc (t) = c1 v 1 e2t + c2 v 2 et + c3 (tv 2 + v 3 )et
a
Example
with deficit
1) Find the solution
 2.6. (One eigenvalue



3 0 1
2
to x0 (t) =  2 1 1  x(t) and x(0) =  3 
−4 0 −1
−4
Solution
Using Mathcad , functions eigenvals() (Notice the
eigenvecs() will not find a good result in this case due to the rounding
error.) we find, λ0 = 1 is the only eigenvalue. To find the associate
eigenvectors we compute (Using (A − λ0 I)v = 0)



2 0 1
v1
(A − λ0 I)v =  2 0 1   v2 
v3
 −4 0 −2

2v1 + v3
 2v1 + v3  = 0
−4v1 − 2v3
2. HOMOGENEOUS SYSTEM
15
We have only 2v1 + v3 = 0 for three variables v1 , v2 , v3 , this indicates


1
that v2 can be any value, and set v1 = 1 find v3 = −2, So v 1 =  0 
−2


1

1  are two eigenvectors.
and v 2 =
−2
To find the generalize eigenvector associated with λ0 we will have
to solve two equations

 

w1
1
(A − λ0 I)  w2  =  0  ,
w3
−2
and
From (2.2),

 

w1
1
(A − λ0 I)  w2  =  1  ,
w3
−2




·
¸
2 0 1
1
 2 0 1  w1 =  0  ,
w2
−4 0 −2
−2
we get two inconsistent equations 2w1 + w3 = 1 and 2w1 + w3 = 0. So
now solution can be found in this case.
From (2.2),




·
¸
2 0 1
1
 2 0 1  w1 =  1  ,
w2
−4 0 −2
−2
we get one equation 2w1 + w3 = 1 choose w3= 1 we
 get w1 = 0, since
0
w2 can be anything, we set w2 = 1. So v 3 =  1  and we can verify
1
that v 1 , v 2 , and v 3 are linearly independent.
So the fundamental matrix is


1 1
t
1
t+1 
Φ(t) = et  0
−2 −2 −2t + 1
and the general solution is,
xc (t) = [c1 v 1 + c2 v 2 + c3 (tv 2 + v 3 )]et
16
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS

1 1

0
1
Now, Φ(0) =
−2 −2
Hence, the particular


 

0
2
−1
1  and Φ(0)-1  3  =  3  .
1
−4
0
t
solution is x(t) = [v 1 + 3v 2 ]e
a
Example 2.7. (One
eigenvalue
 with deficit 2) Find the general

2 1 0
solution to x0 (t) =  1 4 1  x(t).
2 −2 3
Solution
Using Mathcad , functions eigenvals() (Notice the
eigenvecs() will not find a good result in this case due to the rounding
error.) we find, λ0 = 3 is the only eigenvalue. To find the associate
eigenvectors we compute (Using (A − λ0 I)v = 0)



−1 1 0
v1
1 1   v2 
(A − 3I)v =  1
2 −2 0
v3

−v1 + v2 + v3
 v1 + v2 + v3  = 0
2v1 − 2v2


1
We have only one eigenvector v 1 =  1  .
−2
To find the generalize eigenvector associated with λ0 we will have
to solve two equations
(A − 3I)v 2 = v 1 ,
and
(A − 3I)v 3 = v 2 ,
From (2.2),


 

−1 1 0
a
1
 1
1 1  b  =  1 ,
2 −2 0
c
−2
we have two equations
½
b−a
= 1
a+b+c=1


0
Choosing a = 0, we get b = 1, c = 0. Hence v 2 =  1 
0
2. HOMOGENEOUS SYSTEM
17
From (2.2),


 

−1 1 0
a
0
 1
1 1  b  =  1 ,
2 −2 0
c
0
we have two equations
½
b−a
= 0
a+b+c=1


0
Choosing a = 0, we get b = 0, c = 1. So v 3 =  0  and we can verify
1
that v 1 , v 2 , and v 3 are linearly independent.
So the fundamental matrix is


1
t
t2
t2 + t 
Φ(t) = et  1 1 + t
−2 −2t −2t2 + 1
and the general solution is,
xc (t) = [c1 v 1 + c2 (tv 1 + v 2 ) + c3 (t2 v 1 + tv 2 + v 3 )]e3t
a
Example 2.8. (Two conjugate
 complex eigenvalues case)
1 −1 3
Find the general solution to x0 (t) =  2 0 3  x(t)
1 0 1
Solution
Using Mathcad , functions eigenvals() and eigenvecs() we find two conjugate√complex eigenvalues and
√ one real eigenvalue, λ1 = 1, λ2 = 12 + i 12 19, and λ3 = 12 − i 12 19 with associ 




√
1
0
1 + i √19
ated eigenvector v 1 =  1  and v =  −2 + i 19  =  −2  +
2
1
2
 √ 
√ 19
i  19  with respect to λ3 . Compare this with the Theorem 1.3,
0
18
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS

we have a = 21 , b =
√
1
2

1
19, v 2 =  −2 (real part of v), and v 3 =
2
 √ 
√ 19
 19  (imaginary part of v).
0
The general solution is,
1
1
1√
1√
1√
1√
xc (t) = c1 v 1 et +c2 (v 2 cos(
19)−v 3 sin(
19))e 2 t +c3 (v 2 sin(
19)+v 3 cos(
19))e 2 t
2
2
2
2
a
3. Nonhomogeneous System of Equations
To find solutions to the initial value problem of nonhomogeneous
equations x0 (t) = Ax(t) + b(t), x(t0 ) = x0 we follow the steps below,
(1) Find the general solution xc (t) = Φ(t)c to homogeneous equation x0 (t) = Ax(t), where Φ(t) is the fundamental matrix.
(2) Find a particular solution xp to x0 (t) = Ax(t)
(3) The general solution to the nonhomogeneous equation x0 (t) =
Ax(t) is x(t) = xc (t) + xp (t). Using x(t0 ) = x0 to determine
the coefficient vector c.
The following theorem gives one way to find a particular solution based
on the fundamental matrix,
Theorem 3.1. Let Φ(t) be a fundamental matrix of x0 (t) = Ax(t),
a particular solution to x0 (t) = Ax(t) + b(t) is given by
Z
xp (t) = Φ(t) Φ(t)-1 b(t) dt.
Example 3.1. Find the general solution to
x01 = 3x1 + 4x2 − 2x3 + t2
x02 = 2x1 + x2 − 4x3
x03 = x1 + 2x2 − t2

Solution




x1 (t)
3 4 −2
Let x(t) =  x2 (t)  , A =  2 1 −4  , and
x3 (t)
1 2 0

t2
b(t) =  0  . The equations can be written in matrix form x0 (t) =
−t2
3. NONHOMOGENEOUS SYSTEM OF EQUATIONS
19
Ax(t)+b(t). From Example 2.5, we know that the fundamental matrix
to x(t) = Ax(t) is


−4e2t et (t + 4)et
0
−et 
Φ(t) =  e2t
2t
t
−2e
e (t + 1)et


− 52 t2 e−2t
t2 )e−t 
To find a particular solution, we first compute Φ-1 (t)b(t) =  −( 52 t3 + 11
5
2 2 −t
te
5
then we compute


R 2 2 −2t
−
t
e
dt
5
R
R
t2 )e−t dt 
Φ(t) Φ-1 (t)b(t) dt = Φ(t)  −( 25Rt3 + 11
5
2 2 −t
t e dt
5
 1 2
16
t
+
2t
+
5
5
7 
.
=  − 51 t2 − 35 t − 10
9 2
24
29
t
+
t
+
5
5
5
And so the general solution is,




−4
1
x(t) = c1  1  e2t + c2  0  et
−2  

1!
 1 2
à 
16
t
+
2t
+
1
4
5
5
7 
+ c3 t  0  +  −1  et +  − 15 t2 − 53 t − 10
9 2
24
29
1
1
t + 5t+ 5
5
The following is a screen shot that shows how to carry out the computation in Mathcad ,
To use Mathcad ,
(1) Define fundamental matrix A(t) and b(t) in the
same line (not as shown in
graph), and compute in the
next line A−1 b(t)
(2) type A(t)*[Ctrl][M] choose
column as 1, at each place
holder, type [Ctrl][I] to get
the indefinite integral,
(3) and put the corresponding
entry of A−1 b(t) in the integrant position.
(4) press [Shift][Ctrl][.] type
key work simplify
a
20
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Theorem 3.2.
for x0 (t) = Ax(t),
R -1If Φ(t) is the fundamental matrix
1
, and xp (t) = Φ (t)b(t) dt, then x(t) = Φ(t)Φ (t0 )(x0 − xp (t0 )) +
xp (t) is the solution to the nonhomogeneous initial value problem,
x0 (t) = Ax(t) + b(t),
x(t0 ) = x0
·
¸
2 −3
0
Example 3.2. Find the solution to x (t) =
x(t) +
3 8
·
¸
·
¸
e2t
2
and x(0) =
−2e2t
1
Solution
From Example 2.3 is the fundamental matrix is
·
¸
−3 −3t + 1
5t
Φ(t) = e
3
3t
¸
·
0 −1
1
1
and Φ(0) = 3
.
−3 −3
·
¸
R
e2t
Now b(t) =
, using the formula xp (t) = Φ(t) Φ-1 (t)b(t) dt
2t
−2e
·
¸
0
and Mathcad , we have xp (t) = 1 2t
e
3
Therefore,
¸
·
¸Ã·
¸ ·
¸! ·
1
0
0 −1
2
− 32
1
Φ(0) (x(0) − xp (0)) =
.
− 1
=
1
−8
3 −3 −3
3
and the solution is
x(t) = Φ(t)Φ(0)-1 (x(0) − xp (0)) + xp (t) = e5t
·
24t − 6
−24t − 2 + 31 e−3t
¸
a
4. Higher order differential equations
One can transform equations that involving higher order derivatives
of unknown functions to system of first order equations. For example,
suppose x(t) is an unknown scalar function that satisfies
mx00 (t) + cx0 (t) + kx(t) = f (t)
an equation can be used to model a spring system with external force
f (t) or an RCL electronic circuit with an energy source f (t).
4. HIGHER ORDER DIFFERENTIAL EQUATIONS
21
Now if we set x1 (t) = x(t) and x2 (t) = x0 (t) we then get an system
of first order equations
x01 (t) = x2 (t)
c
k
f (t)
(6)
x02 (t) = − x2 (t) − x1 (t) +
m
m
m
In general, if we have an differential equation that involving nth
order derivative x(n) (t) of unknown function x(t),
(5)
x(n) = a0 x(t) + a1 x0 (t) + · · · + an−1 x(n−1) + f (t),
we can transform it into an system of first order equations of n unknown
functions x1 (t) = x(t), x2 (t) = x0 (t), x3 (t) = x(2) (t), · · · , xn (t) =
x(n−1) (t), and using the eigenvalue method for system of differential
equation to solve the higher order equation.
Example 4.1. Transform the differential equation x(3) + 3x(2) −
7x (t) − 9x = sin(t) into system of first order equations.
0
Solution Here the highest order of derivative is third derivative
x(3) of x(t). So we transfer it into system of 3 equations.
Let x1 (t) = x(t), x2 (t) = x0 (t), x3 (t) = x00 (t), we have
x01 (t) = x2 (t)
x02 (t) = x3 (t)
x03 (t) = −3x3 (t) + 7x2 (t) + 9x1 (t) − sin(t)






x1 (t)
1 0 0
0
Let x(t) =  x2 (t)  , A =  0 1 0  , and b(t) =  0 
x3 (t)
9 7 −3
f (t)
we can write the system of equation in matrix form x0 (t) = Ax(t) +
b(t).
a
(7)
(8)
(9)
Example 4.2. Find the general solution for the 3rd order differential equation x(3) + 3x(2) − 7x0 (t) − 9x = sin(t).
From
previousexample, Example
4.1,
 Solution



 Let x(t) =
x1 (t)
1 0 0
0
 x2 (t)  , A =  0 1 0  , and b(t) =  0  we can write
x3 (t)
9 7 −3
f (t)
the system of equation in matrix form x0 (t) = Ax(t) + b(t).
√ Using
Mathcad we find the eigenvalues are λ1 =−1, λ
+ 10, λ3 = 
2 = −1 
1√
1
√
−1− 10 with associate eigenvectors, v 1 =  −1  , v 2 =  −1 + √10  ,
1
11 − 2 10
22
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS


1√
and v 3 =  −1 − √10  respectively (after multiply the results of
11 + 2 10
Mathcad by some constants). So the fundamental matrix is
√
√
 −t

(−1+ 10)t
(1+ 10)t
e
e
e√
√
√
√
Φ(t) =  −e−t (−1 + 10)e(−1+ √10)t −(1 + 10)e(1+ √10)t 
√
√
e−t (11 − 2 10)e(−1+ 10)t (11 + 2 10)e(1+ 10)t
From Φ(t) we find a particular solution
 3

5
Z
cos(t) − 39
sin(t)
52
1
35
cos(t) − 156
sin(t) 
xp (t) = Φ(t) Φ-1 (t)b(t) dt =  26
3
8
− 52 cos(t) − 39 sin(t)
Hence the general solution to the system is


x1 (t)
 x2 (t)  =
x (t)
√
√
 3
−t
(−1+ 10)t
(1+ 10)t

3
5
c1 e + c2 e
+ c3 e
+ 52
cos(t)
− 39
sin(t)
√
√
√
√
 −c1 e−t + c2 (−1 + 10e(−1+ 10)t − c3 (1 + 10)e(1+ 10)t + 1 cos(t) − 35 sin(t) 
26
156
√
√
√
√
3
8
c1 e−t + c2 (11 − 2 10)e(−1+ 10)t + c3 (11 + 2 10)e(1+ 10)t − 52
cos(t) − 39
sin(t)
√
√
3
5
and x1 (t) = c1 e−1t + c2 e(−1+ 10)t + c3 e(1+ 10)t + 52
cos(t) − 39
sin(t) is
the general solution to the third order ordinary differential equation
x(3) + 3x(2) − 7x0 (t) − 9x = sin(t)
.
a
Example 4.3. Find the solution to the initial value problem
x00 − 10x0 + 9x = tet , x(0) = 1, x0 (0) = −1.
Solution Since the given equation is of second order, we will have
two unknowns x1 (t) = x(t), x2 (t) = x0 (t) to transform the equation into
a system of first order equations,
x01 (t) = x2
x02 (t) = 10x2 − 9x1 + tet ,
and the initial conditions
are¸ x1 (0) =·x(0) = 1 ¸x2 (0) = x0 (0) =
·
· −1. ¸
x1 (t)
0 1
0
Now let x(t) =
,A=
, and b(t) =
.
x2 (t)
−9 10
tet
We have the matrix version of this equation, x0 (t) = Ax(t) + b(t)
4. HIGHER ORDER DIFFERENTIAL EQUATIONS
23
Using Mathcad , we find· the ¸eigenvalues
· λ1¸ = 1, λ1 = 9, and
1
1
associate eigenvectors v 1 =
, v2 =
. And fundamental
1
9
· t
¸
e e9t
matrix Φ(t) =
.
et 9e9t
From Φ(t) we find a particular solution
¸
·
Z
1
(32t2 + 8t + 1)et
− 512
1
xp (t) = Φ(t) Φ (t)b(t) dt =
1
− 512
(32t2 + 72t + 9)et
·
¸
1
The solution with initial values x0 =
is given by
−1
-1
x(t) = Φ(t)Φ
 µ (0)(x0 − b(0)) +¶b(t)

1
639
127 9t
1 2
t
− 16 t − 64 t + 512 e − 512 e 


.
µ
¶
= 

1 2
9
631
1143
− 16 t − 64 t + 512 et − 512 e9t
Hence the solution to the initial value
µ problem of the¶second order
1 2
1
639
differential equation is x(t) = x1 (t) = − 16
t − 64
t + 512
et − 127
e9t . a
512
Project
At beginning you should enter: Project title, your name, ss#, and
due date in the following format
Project One: Define and Graph Functions
John Doe
SS# 000-00-0000
Due: Mon. Nov. 23rd, 2003
You should format the text region so that the color of text is different
than math expression. You can choose color for text from Format–
>Style select normal and click modify, then change the settings for
font. You can do this for headings etc.
(1) Solutions To System of Equations
Finding solution to linear system using Mathcad and study
the long time dynamic behavior of the solutions.
• Find general solution to
½ 0
x = −y
y0 = x
24
1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Plot several solutions with different initial values in
[-] xt-plane, yt-plane
xy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graph
in xy-plane is called the trajectory. If this models the
movement of a satellite, what is its trajectory.
• Find general solution to
½ 0
x = −8y
y 0 = 18x
Plot several solutions with different initial values in
[-] xt-plane, yt-plane
xy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graph
in xy-plane is called the trajectory. If this models the
movement of a satellite, what is its trajectory.
• Find general solution to
½ 0
x = 2x − y
y 0 = y − 3x
Plot several solutions with different initial values in
[-] xt-plane, yt-plane
xy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graph
in xy-plane is called the trajectory. If this models the
movement of a system of two species, what is you conclusion about interdependency of these species? Can you
find initial value such that x(t) = 0 (distinct) for some t?
what about y(t).
(2) Solution of Higher order equation In general mx00 + cx0 +
kx = f (t) models a object with mass m attached to a spring
with constant k and damping force that is proportional to
the velocity x0 , c ≥ 0, k > 0. Suppose m = 1 and f (t) =
Ae−at sin(bt), that is the external force is oscillatory (b > 0)
and diminishing (a > 0) Find solutions and graph the solutions.
- c = b = 0 Find general solution and graph some particular
solutions.
- c = 20, k = 10, a = 0, b = 41 , A = 1
√
- c = 2, k = 3, A = 100, a = 2, b = 2