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CHAPTER 1
System of First Order Differential Equations
In this chapter, we will discuss system of first order differential equa-tions. There are many applications that involving find several unknownfunctions simultaneously . Those unknown functions are related by aset of equations that involving the unknown functions and their firstderivatives. For example, in Chapter Two, we studied the epidemic ofcontagious diseases. Now if
• S(t) denotes number of people that is susceptible to the diseasebut 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 unittime is proportional to the number infected, b is the propor-tional number.
• A fixed fraction rS of the infected population recovers per unittime, 0 ≤ r ≤ 1.
• A fixed fraction of the recovers g become susceptible and in-fected, 0 ≤ g ≤ 1. proportional function.
The system of differential equations model this phenomena are
S ′ = −bIS + gRI ′ = bIS − rIR′ = rI − gR
The numbers of unknown function in a system of differential equa-tions can be arbitrarily large, but we will concentrate ourselves on 2 to3 unknown functions.
1. Principle of superposition
Let aij(t), bj(t) i = 1, 2, · · · , n and j = 1, 2, · · · , n be knownfunction, and xit, 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,
x′1(t) = a11(t)x1(t) + a12(t)x2(t) + · · ·+ a1n(t)xn(t) + b1(t)x′2(t) = a21(t)x1(t) + a22(t)x2(t) + · · ·+ a2n(t)xn(t) + b2(t)x′3(t) = a31(t)x1(t) + a32(t)x2(t) + · · ·+ a3n(t)xn(t) + b3(t)
...x′n(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 xit, i =1, 2, · · · , n, A(t) = (aij(t), and b(t) be the column vector of knownfunctions bit, i = 1, 2, · · · , n, we can write the first order system ofequations as
x′(t) = A(t)x(t) + b(t)(1)
• When n = 2, the linear first order system of equations for twounknown functions in matrix form is,[
x′1(t)x′2(t)
]=
[a11(t) a12(t)a21(t) a22(t)
] [x1(t)x2(t)
]+
[b1(t)b2(t)
]
• When n = 3, the linear first order system of equations forthree unknown functions in matrix form is,
x′1(t)x′2(t)x′3(t)
=
a11(t) a12(t) a13
a21(t) a22(t) a23
a31(t) a32(t) a33
x1(t)x2(t)x3t
+
b1(t)b2(t)b3(t)
A solution of equation (1) on the open interval I is a column vec-tor function x(t) whose derivative (as a vector-values function) equalsA(t)x(t) + b(t). The following theorem gives existence and uniquenessof solutions,
Theorem 1.1. If the vector-valued functions A(t) and b(t) are con-tinuous over an open interval I contains t0, then the initial value prob-lem {
x′(t) = A(t)x(t) + b(t)x(t0) = x0
has an unique vector-values solution x(t) that is defined on entire in-terval I for any given initial value x0.
When b(t) ≡ 0, the linear first order system of equations becomes
x′(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, wefirst 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
x′(t) = A(t)x(t)
on the open interval I. If c1, c2, · · · , cn are n constants, then thelinear combination
c1x1(t) + c2x2(t) + c3x3(t) + · · ·+ cnxn(t)
is also a solution on I.
Example 1.1. Let
x′(t) =
[1 00 −2
]x(t)
, x1(t) =
[et
0
]and x2(t) =
[0
e−2t
]are two solutions, as
bx′1(t) =
[(et)′
0
]=
[et
0
]=
[1 00 −2
] [et
0
]
and
bx′2(t) =
[0
(e−2t)′
]=
[0
−2e−2t
]=
[1 00 −2
] [0
e−2t
]
By the Principle of Superposition, for any two constants c1 and c2
x(t) = c1x1(t) + c2x2(t) = c1
[et
0
]+ c2
[0
e−2t
]=
[c1e
t
c2e−2t
]
is also solution. We shall see that it is actually the general solution.
The next theorem gives the general solution of linear system ofequations,
Theorem 1.3.- Let x1(t), x2(t), · · · , bxn(t) be n linearly independent (as vectors)solution of the homogeneous system
x′(t) = A(t)x(t),
then for any solution xc(t) there exists n constants c1, c2, · · · , cn suchthat
xc(t) = c1x1(t) + c2x2(t) + · · ·+ cnxn(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
x′(t) =
[4 −36 −7
]x(t) +
[ −4t2 + 5t−6t2 + 7t + 1
]x(t)
, x1(t) =
[3e2t
2e2t
]and x2(t) =
[e−5t
3e−5t
]are two linearly independent
solutions. and xp(t) =
[t2
t
]is a particular solution. By Theorem
1.3,
x(t) = c1x1(t) + c2x2(t) + xp(t) =
[3c1e
2t + c2e−5t + t2
2c1e2t + 3c2e
−5t + t
](2)
is the general solution. Now suppose we want to find a particular so-
lution that satisfies the initial condition x(0) =
[2−1
], then let t = 0
in (2), we have
x(0) =
[3c1 + c2
2c1 + 3c2
]=
[2−1
],
which can be written in matrix form,[
3 12 3
] [c1
c2
]=
[2−1
],
Solve this equation, we get
[c1
c2
]=
[1−1
]. So the particular
solution is x(t) =
[3e2t − e−5t + t2
2e2t − 3e−5t + t
].
From the above example, we can summarize the general steps infind a solution to initial value problem,
{x′(t) = A(t)x(t) + b(t)x(t0) = x0
2. HOMOGENEOUS SYSTEM 5
• Step One: Find the general solution xc = c1x1(t)+ c2x2(t)+· · · + cnxn(t), where x1(t), x2(t), · · · , xn(t) are a set of lin-early independent solutions, to the associate homogeneous sys-tem, x′(t) = A(t)x(t).
• Step Two: Find a particular solution xp(t)to the nonhomo-geneous system, x′(t) = A(t)x(t) + b(t).
• Step Three: Set x(t) = xc(t) + xp(t) and use the equationx(t0) = x0, to determine c1, c2, · · · , cn.
2. Homogeneous System
We will use a powerful method called eigenvalue method to solvethe homogeneous system
x′(t) = Ax(t)
where A is a matrix with constant entry. We will present this methodfor A is either a 2 × 2 or 3 × 3 cases. The method can be used for Ais an n× n matrix. The idea is to find solutions of form
x(t) = veλt,(3)
a straight line that passing origin in the direction v. Now taking deriv-ative on x(t), we have
x′(t) = λveλt(4)
put (3) and (2.2) into the homogeneous equation, we get
x′(t) = λveλt = Aveλt
So
Av = λv,
which indicates that λ must be an eigenvalue of A and v is an associateeigenvector.
2.1. A is a 2× 2 matrix. Suppose
A =
[a11 a12
a21 a22
]
Then the characteristic polynomial p(λ) of A is
p(λ) = |A−λI| = (a11−λ)∗(a22−λ)−a12a21 = λ2−(a11+a22)+(a11a22−a12a22.
So p(λ) is a quadratic polynomial of λ. From Algebra, we know thatp(λ) = 0 has either 2 distinct real solutions, or a double solution, or2 conjugate complex solutions. The following theorem summarize thesolution to the homogeneous system,
6 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Theorem 2.1. Let p(λ) be the characteristic polynomial of A, forx′(t) = Ax(t),
Case 1: p(λ) = 0 has two distinct real solutions λ1 and λ2.
Suppose v1 =
[v11
v21
]and v2 =
[v12
v22
]are associate eigen-
vector (i.e, Av1 = λ1v1 and Av2 = λ2v2) Then the generalsolution is
xc(t) = c1v1eλ1t + c2v2e
λ2t
And
Φ(t) =
[v11e
λ1t v12eλ2t
v21eλ1t v22e
λ2t
]
is called the fundamental matrix(A fundamental matrixis a square matrix whose columns are linearly independent so-lutions 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) withmultiplicity 2.
(1) λ0 has two linearly independent eigenvectors:
Suppose v1 =
[v11
v21
]and v2 =
[v12
v22
]are associate linearly
independent eigenvectors. Then the general solution is
xc(t) = (c1v1 + c2v2)eλ0t
And
Φ(t) = eλ0t
[v11 v12
v21 v22
]
(2) λ0 has only one associate eigenvector:
Suppose v1 =
[v11
v21
]is the only associated eigenvector and
v2 =
[v12
v22
]is a solution of
(λ0I −A)v2 = v1.
Then the general solution is,
xc(t) = (c1v1 + c2(tv1 + v2)eλ0t
And
Φ(t) = eλ0t
[v11 (v11t + v12)v21 (v21t + v22)
]
is the fundamental solution matrix.
2. HOMOGENEOUS SYSTEM 7
Case 3: p(λ) = 0 has two conjugate complex solutions a+bi and a−bi.
Suppose v =
[v11 + iv12
v21 + iv22
]is the associate complex eigen-
vector with respect to a + bi, then the general solution is,
v1 =
[v11
v21
]and v2 =
[v12
v22
]
xc(t) = [c1(v1 cos(bt)− v2 sin(bt))c2(v2 cos(bt) + v1 sin(bt))]eat.
And
Φ(t) = eat
[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
solution is given by xc(t) = Φ(t)c, with c =
[c1
c2
]and the solution
that satisfies a given initial condition x(t0) = x0 is given by
x(t) = Φ(t)Φ(t0)-tx0
Example 2.1. Two distinct eigenvalues case Find the general
solution to x′(t) =
[2 −3−1 −5
]x(t)
Solution Using Mathcad , functions eigenvals() and eigen-vecs()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
+ 12
√61 and λ2 = −3
2− 1
2
√61 with associated eigen-
vectors v1 =
[−7−√61
2
]and v2 =
[−7 +
√61
2
]respectively. So
the fundamental matrix is
Φ(t) =
[(−7−√61)e(− 3
2+ 1
2
√61)t (−7 +
√61)e(− 3
2− 1
2
√61)t
2e(− 32+ 1
2
√61)t 2e(− 3
2− 1
2
√61)t
]
and the general solution is, for c =
[c1
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-
dependent eigenvectors Find the general solution to x′(t) =
[2 00 2
]x(t).
Solution The eigenvalue is λ0 = 2 and associated eigenvectors
are
[10
]and
[01
], so the general solution is xc =
[c1e
2t
c2e2t
]a
Example 2.3. One double eigenvalues with only one eigen-
vector Find the solution to x′(t) =
[2 −33 8
]x(t) and x(0) =
[23
]
Solution Using Mathcad , functions eigenvals() and eigen-
vecs() we can find a double eigenvalue λ0 = 5 and eigenvector
[ −.707.707
].
Notice, the symbolic operator →̇(bring up by either [Shift][Ctrl][.] or[Ctrl][.]) will not work with eigenvecs() this time, but since multiplyan eigenvector by a nonzero constant still get an eigenvector, we can
choose v1 =
[3−3
].
To find w that satisfies (A − λ0I)w = v1 λ0 we will solve (A −λ0I)
[w1
w2
]=
[1−1
]. That is,
[ −3 −33 3
] [w1
w2
]=
[3−3
]
One solution is w1 = 1 and w2 = 0So the fundamental matrix is
Φ(t) = e5t
[ −3 −3t + 13 3t
]
and the general solution is, c =
[c1
c2
],
xc(t) = Φ(t)c
Now, Φ(0) = e5(0)
[ −3 −3(0) + 13 3(0)
]=
[ −3 13 0
]and Φ(0)-1 =
13
[0 −1−3 −3
].
Hence, the particular solution is x(t) = Φ(t)Φ(0)-1x0 = e5t
[ −3t + 43t− 3
]a
2. HOMOGENEOUS SYSTEM 9
Example 2.4. Two conjugate complex eigenvalues case Find
the general solution to x′(t) =
[2 −31 2
]x(t)
Solution Using Mathcad , functions eigenvals() and eigen-vecs() we find two conjugate complex eigenvalues, λ1 = 2 + i
√3 and
λ2 = 2− i√
3 with associated eigenvector v1 =
[ √3
−i
]with respect to
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
Φ(t) = e2t
[ √3 cos(bt) − sin(bt)√3 sin(bt) − cos(bt)
]
and the general solution is, c =
[c1
c2
],
xc(t) = Φ(t)c = e2t
[ √3 cos(
√3t) − sin(
√3t)√
3 sin(√
3t) − cos(√
3t)
] [c1
c2
]
= e2t
[ √3c1 cos(
√3t)− c2 sin(
√3t)√
3c1 sin(√
3t)− c2 cos(√
3t)
]
Suppose we want to find a solution such that x(0) =
[12
], then
x(t) = Φ(t)Φ(0)-1x(0)
= e2t
[ √3 cos(
√3t) − sin(
√3t)√
3 sin(√
3t) − cos(√
3t)
] [ √3 0
0 −1
]-1 [12
]
= e2t
[ √3 cos(
√3t) − sin(
√3t)√
3 sin(√
3t) − cos(√
3t)
] [1√3
−2
]
= e2t
[cos(
√3t) + +2 sin(
√3t)
− sin(√
3t) + 2 cos(√
3t)
]
a
2.2. A is a 3× 3 matrix. Suppose
A =
a11 a12 a13
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) = 0has either 3 distinct real solutions, or 2 distinct solutions and one is adouble solution, or one real solution and 2 conjugate complex solutions,or a triple solution. The following theorem summarize the solution tothe homogeneous system,
Theorem 2.2. Let p(λ) be the characteristic polynomial of A, forx′(t) = Ax(t),
Case 1: p(λ) = 0 has three distinct real solutions λ1, λ2, and λ3.
Suppose v1 =
v11
v21
v31
, v2 =
v12
v22
v32
, and v3 =
v13
v23
v33
are associate eigenvector (i.e, Av1 = λ1v1, Av2 = λ2v2, andAv3 = λ3v3) Then the general solution is
xc(t) = c1v1eλ1t + c2v2e
λ2t + c3v3eλ3t
And the fundamental matrix is
Φ(t) =
v11eλ1t v12e
λ2t v13eλ3t
v21eλ1t v22e
λ2t v23eλ3t
v31eλ1t v32e
λ2t v33eλ3t
.
Case 2: p(λ) = 0 has a double solutions λ0.So p(λ) = (λ − λ0)
2(λ − λ1), and λ0 has multiplicity 2. Let
v3 =
v12
v22
v32
is the eigenvector associated with λ1.
[1] λ0 has two linearly independent eigenvectors:
Suppose v1 =
[v11
v21
]and v2 =
[v12
v22
]are associate linearly
independent eigenvectors. Then the general solution is
xc(t) = (c1v1 + c2v2)eλ0t + c3v3e
λ1t
And
Φ(t) =
v11eλ0t v12e
λ0t v13eλ1t
v21eλ0t v22e
λ0t v23eλ1t
v31eλ0t v32e
λ0t v33eλ1t
2. HOMOGENEOUS SYSTEM 11
[2] λ0 has one eigenvector:
Suppose v1 =
v11
v21
v31
is the associated eigenvector with re-
spect to λ0 and v2 =
v12
v22
v32
is a solution of
(λ0I −A)v2 = v1.
Then the general solution is,
xc(t) = (c1v1 + c2(tv1 + v2))eλ0t + c3v3e
λ1
And
Φ(t) =
v11eλ0t (v11t + v12)e
λ0t v13eλ1
v21eλ0t (v21t + v22)e
λ0t v23eλ1
v31eλ0t (v31t + v32)e
λ0t v33eλ1
is the fundamental solution matrix.Case 3: p(λ) = 0 has two conjugate complex solutions a± bi and a real
solution λ1.
Suppose v =
v11 + iv12
v21 + iv22
v31 + iv32
is the associate complex eigen-
vector with respect to a + bi, then the general solution is, let
v3 =
v13
v23
V33
, are associated eigenvectors with respect to λ1,
xc(t) = [c1(v1 cos(bt)−v2 sin(bt))c2(v2 cos(bt)+v1 sin(bt))]eat+c3v3eλ1 .
And
Φ(t) = eat
v11 cos(bt)− v12 sin(bt) v12 cos(bt) + v11 sin(bt) v13eλ1
v21 cos(bt)− v22 sin(bt) v22 cos(bt) + v21 sin(bt) v23eλ1
v31 cos(bt)− v32 sin(bt) v32 cos(bt) + v31 sin(bt) v33eλ1
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 linearly independent eigenvectors.
Let v1 =
v11
v21
v31
, v2 =
v12
v22
V32
, and v3 =
v13
v23
V33
be
the three linearly independent eigenvectors. Then the general
12 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
solution is xc(t) = (c1v1 + c2v2 + c3v3)eλ0t and fundamental
matrix is Φ(t) = eλ0t
v11 v12 v13
v21 v22 V23
v31
v32
V33
[2] λ0 has two linearly independent eigenvectors.
Suppose v1 =
v11
v21
v31
, v2 =
v12
v22
V32
are the linearly inde-
pendent eigenvectors. Let v3 =
v13
v23
V33
, then only one of the
two equations, (A − λ0I)v3 = v1 or (A − λ0I)v3 = v2 canhas a solution that is linearly independent with v1,v2.
Suppose (A−λ0I)v3 = v2 generates such a solution. Thenthe general solution is xc(t) = [c1v1 + c2v2 + c3(tv2 + v3)]e
λ0t
and fundamental matrix is Φ(t) = eλ0t
v11 v12 tv12 + v13
v21 v22 tv22 + V23
v31 v32 tv32 + V33
[3] λ0 has only one eigenvector.
Let v1 =
v11
v21
v31
be the linearly independent eigenvectors.
Let v2 =
v12
v22
V32
and v3 =
v13
v23
V33
be two vectors that
satisfies
(A− λ0I)v2 = v1 and (A− λ0I)v3 = v2.
Then the general solution is xc(t) = [c1v1 + c2(tv1 + v2) +c3(t
2v1 + tv2 + v3)]eλ0t and fundamental matrix is Φ(t) =
eλ0t
v11 tv11 + v12 t2v11 + tv12 + v13
v21 tv21 + v22 t2v21 + tv22 + V23
v31 tv31 + v32 t2v31 + tv32 + V33
Remark 2.1. Suppose A is an n× n matrix, for the homogeneoussystem x′(t) = Ax(t), three general case would happen
Case 1: A has n distinct eigenvalues λi, i = 1, 2, · · · , n with linearlyindependent eigenvectors vi, i = 1, 2, · · · , n then the generalsolution will be xc(t) = c1v1e
λ1 + c2v2eλ2 + · · ·+ cnvne
λn
2. HOMOGENEOUS SYSTEM 13
Case2: A has m < n distinct eigenvalues, in this case some eigenval-ues would have multiplicity greater than 1.
Suppose λr has multiplicity r. Depending on how manylinearly independent eigenvectors are associated with λr thesituation could be very complex. Let p be the number of linearlyeigenvectors associated with λr, then d = r − p is called thedeficit 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 v1 is the only eigenvector associatewith λr, then one will have to solve r − 1 equations (A −λr)
ivi+1 = vi, i = 1, 2, · · · , r − 1. And the general solutionwould contains terms like [c1v1 + c2(v1t+v2)+ c3(v1t
2 +v2t+v3) + · · ·+ cr(v
r1 + v2t
r−1 + · · ·+ vr)]eλr .
Case 3: A complex root a+bi with associated eigenvector va+ivb, thenthe general solution contains term, [c1(va cos(bt)−vb sin(bt))+c2(va sin(bt) + vb cos(bt))]eat.
Remark 2.2. Suppose x1(t),x2(t),x3(t), · · · ,xn(t) are n linearlyindependent solution for n×n homogeneous system, x′(t) = Ax(t), thefundamental matrix Φ(t) is a matrix whose columns are xi(t), i =1, 2, · · · , n.
Example 2.5. (Two distinct eigenvalues) Find the general so-lution to
x′1 = 3x1 + 4x2 − 2x3
x′2 = 2x1 + x2 − 4x3
x′3 = x1 + 2x2
Solution Let x(t) =
x1(t)x2(t)x3(t)
and
3 4 −22 1 −41 2 0
The equa-
tions can be written in matrix form x′(t) = Ax(t).Using Mathcad , functions eigenvals() and eigenvecs() we find,λ1 =
2 and λ2 = 1 with associated eigenvectors v1 =
−41−2
and v2 =
101
respectively. Since λ1 has multiplicity 2 as 1 appeared twice
in the result of eigenvals() function, we need to solve the equation(A− λ1I)v3 = v2.
14 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
To use Mathcad ,
(1) you first compute (A − λ1I)v3 using the fol-lowing sequences of key stroke,
[*] type ([Ctrl][M], set the rows andcolumns in the matrix definition popup menu,input the data for A,
[*] type -[Ctrl][M] set the row and col-umn number and input data for λ1I,
[*] type )[Ctrl][M], now set 1 as columnnumber, enter a, b, c in the place holders,
[*] type [Ctrl][.] to compute symboli-cally and you get.
(2) Using the Given Find block to find asolution. Type Given in a blank space, typea+2b-c[Ctrl]= 1 and 2a-4c[Ctrl]=0 in tworows, then type key word Find following bytyping (a,b)[Ctrl][.] you will get the solutionin terms of c.
Set c = 1, we get v3 =
4−11
.
So the fundamental matrix is
Φ(t) =
−4e2t et (t + 4)et
e2t 0 −et
−2e2t et (t + 1)et
and the general solution is,
xc(t) = c1v1e2t + c2v2e
t + c3(tv2 + v3)et
aExample 2.6. (One eigenvalue with deficit 1) Find the solution
to x′(t) =
3 0 12 1 1−4 0 −1
x(t) and x(0) =
23−4
Solution Using Mathcad , functions eigenvals() (Notice theeigenvecs() will not find a good result in this case due to the roundingerror.) we find, λ0 = 1 is the only eigenvalue. To find the associateeigenvectors we compute (Using (A− λ0I)v = 0)
(A− λ0I)v =
2 0 12 0 1−4 0 −2
v1
v2
v3
2v1 + v3
2v1 + v3
−4v1 − 2v3
= 0
2. HOMOGENEOUS SYSTEM 15
We have only 2v1 + v3 = 0 for three variables v1, v2, v3, this indicates
that v2 can be any value, and set v1 = 1 find v3 = −2, So v1 =
10−2
and v2 =
11−2
are two eigenvectors.
To find the generalize eigenvector associated with λ0 we will haveto solve two equations
(A− λ0I)
w1
w2
w3
=
10−2
,
and
(A− λ0I)
w1
w2
w3
=
11−2
,
From (2.2),
2 0 12 0 1−4 0 −2
[w1
w2
]=
10−2
,
we get two inconsistent equations 2w1 + w3 = 1 and 2w1 + w3 = 0. Sonow solution can be found in this case.
From (2.2),
2 0 12 0 1−4 0 −2
[w1
w2
]=
11−2
,
we get one equation 2w1 + w3 = 1 choose w3 = 1 we get w1 = 0, since
w2 can be anything, we set w2 = 1. So v3 =
011
and we can verify
that v1,v2, and v3 are linearly independent.So the fundamental matrix is
Φ(t) = et
1 1 t0 1 t + 1−2 −2 −2t + 1
and the general solution is,
xc(t) = [c1v1 + c2v2 + c3(tv2 + v3)]et
16 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Now, Φ(0) =
1 1 00 1 1−2 −2 1
and Φ(0)-1
23−4
=
−130
.
Hence, the particular solution is x(t) = [v1 + 3v2]et a
Example 2.7. (One eigenvalue with deficit 2) Find the general
solution to x′(t) =
2 1 01 4 12 −2 3
x(t).
Solution Using Mathcad , functions eigenvals() (Notice theeigenvecs() will not find a good result in this case due to the roundingerror.) we find, λ0 = 3 is the only eigenvalue. To find the associateeigenvectors we compute (Using (A− λ0I)v = 0)
(A− 3I)v =
−1 1 01 1 12 −2 0
v1
v2
v3
−v1 + v2 + v3
v1 + v2 + v3
2v1 − 2v2
= 0
We have only one eigenvector v1 =
11−2
.
To find the generalize eigenvector associated with λ0 we will haveto solve two equations
(A− 3I)v2 = v1,
and(A− 3I)v3 = v2,
From (2.2),
−1 1 01 1 12 −2 0
abc
=
11−2
,
we have two equations {b− a = 1a + b + c = 1
Choosing a = 0, we get b = 1, c = 0. Hence v2 =
010
2. HOMOGENEOUS SYSTEM 17
From (2.2),
−1 1 01 1 12 −2 0
abc
=
010
,
we have two equations
{b− a = 0a + b + c = 1
Choosing a = 0, we get b = 0, c = 1. So v3 =
001
and we can verify
that v1,v2, and v3 are linearly independent.So the fundamental matrix is
Φ(t) = et
1 t t2
1 1 + t t2 + t−2 −2t −2t2 + 1
and the general solution is,
xc(t) = [c1v1 + c2(tv1 + v2) + c3(t2v1 + tv2 + v3)]e
3t
a
Example 2.8. (Two conjugate complex eigenvalues case)
Find the general solution to x′(t) =
1 −1 32 0 31 0 1
x(t)
Solution Using Mathcad , functions eigenvals() and eigen-vecs() we find two conjugate complex eigenvalues and one real eigen-value, λ1 = 1, λ2 = 1
2+ i1
2
√19, and λ3 = 1
2− i1
2
√19 with associ-
ated eigenvector v1 =
011
and v =
1 + i√
19
−2 + i√
192
=
1−22
+
i
√19√190
with respect to λ3. Compare this with the Theorem 1.3,
18 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
we have a = 12, b = 1
2
√19, v2 =
1−22
(real part of v), and v3 =
√19√190
(imaginary part of v).
The general solution is,
xc(t) = c1v1et+c2(v2 cos(
1
2
√19)−v3 sin(
1
2
√19))e
12t+c3(v2 sin(
1
2
√19)+v3 cos(
1
2
√19))e
12t
a
3. Nonhomogeneous System of Equations
To find solutions to the initial value problem of nonhomogeneousequations x′(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 equa-tion x′(t) = Ax(t), where Φ(t) is the fundamental matrix.
(2) Find a particular solution xp to x′(t) = Ax(t)(3) The general solution to the nonhomogeneous equation x′(t) =
Ax(t) is x(t) = xc(t) + xp(t). Using x(t0) = x0 to determinethe coefficient vector c.
The following theorem gives one way to find a particular solution basedon the fundamental matrix,
Theorem 3.1. Let Φ(t) be a fundamental matrix of x′(t) = Ax(t),a particular solution to x′(t) = Ax(t) + b(t) is given by
xp(t) = Φ(t)
∫Φ(t)-1b(t) dt.
Example 3.1. Find the general solution to
x′1 = 3x1 + 4x2 − 2x3 + t2
x′2 = 2x1 + x2 − 4x3
x′3 = x1 + 2x2 − t2
Solution Let x(t) =
x1(t)x2(t)x3(t)
, A =
3 4 −22 1 −41 2 0
, and
b(t) =
t2
0−t2
. The equations can be written in matrix form x′(t) =
3. NONHOMOGENEOUS SYSTEM OF EQUATIONS 19
Ax(t)+b(t). From Example 2.5, we know that the fundamental matrixto x(t) = Ax(t) is
Φ(t) =
−4e2t et (t + 4)et
e2t 0 −et
−2e2t et (t + 1)et
To find a particular solution, we first compute Φ-1(t)b(t) =
−25t2e−2t
−(25t3 + 11
5t2)e−t
25t2e−t
then we compute
Φ(t)∫
Φ-1(t)b(t) dt = Φ(t)
∫ −25t2e−2t dt∫ −(2
5t3 + 11
5t2)e−t dt∫
25t2e−t dt
=
15t2 + 2t + 16
5−15t2 − 3
5t− 7
1095t2 + 24
5t + 29
5
.
And so the general solution is,
x(t) = c1
−41−2
e2t + c2
101
et
+ c3
(t
101
+
4−11
)et +
15t2 + 2t + 16
5−15t2 − 3
5t− 7
1095t2 + 24
5t + 29
5
The following is a screen shot that shows how to carry out the compu-tation in Mathcad ,
To use Mathcad ,
(1) Define fundamental ma-trix A(t) and b(t) in thesame line (not as shown ingraph), and compute in thenext line A−1b(t)
(2) type A(t)*[Ctrl][M] choosecolumn as 1, at each placeholder, type [Ctrl][I] to getthe indefinite integral,
(3) and put the correspondingentry of A−1b(t) in the in-tegrant position.
(4) press [Shift][Ctrl][.] typekey work simplify
a
20 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Theorem 3.2. If Φ(t) is the fundamental matrix for x′(t) = Ax(t),, and xp(t) =
∫Φ-1(t)b(t) dt, then x(t) = Φ(t)Φ-1(t0)(x0 − xp(t0)) +
xp(t) is the solution to the nonhomogeneous initial value problem,
x′(t) = Ax(t) + b(t), x(t0) = x0
Example 3.2. Find the solution to x′(t) =
[2 −33 8
]x(t) +
[e2t
−2e2t
]and x(0) =
[21
]
Solution From Example 2.3 is the fundamental matrix is
Φ(t) = e5t
[ −3 −3t + 13 3t
]
and Φ(0)-1 = 13
[0 −1−3 −3
].
Now b(t) =
[e2t
−2e2t
], using the formula xp(t) = Φ(t)
∫Φ-1(t)b(t) dt
and Mathcad , we have xp(t) =
[0
13e2t
]
Therefore,
Φ(0)-1(x(0)−xp(0)) =1
3
[0 −1−3 −3
] ( [21
]−
[013
] )=
[ −23−8
].
and the solution is
x(t) = Φ(t)Φ(0)-1(x(0)− xp(0)) + xp(t) = e5t
[24t− 6
−24t− 2 + 13e−3t
]
a
4. Higher order differential equations
One can transform equations that involving higher order derivativesof unknown functions to system of first order equations. For example,suppose x(t) is an unknown scalar function that satisfies
mx′′(t) + cx′(t) + kx(t) = f(t)
an equation can be used to model a spring system with external forcef(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) = x′(t) we then get an systemof first order equations
x′1(t) = x2(t)(5)
x′2(t) = − c
mx2(t)− k
mx1(t) +
f(t)
m(6)
In general, if we have an differential equation that involving nthorder derivative x(n)(t) of unknown function x(t),
x(n) = a0x(t) + a1x′(t) + · · ·+ an−1x
(n−1) + f(t),
we can transform it into an system of first order equations of n unknownfunctions x1(t) = x(t), x2(t) = x′(t), x3(t) = x(2)(t), · · · , xn(t) =x(n−1)(t), and using the eigenvalue method for system of differentialequation 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.
Solution Here the highest order of derivative is third derivativex(3) of x(t). So we transfer it into system of 3 equations.
Let x1(t) = x(t), x2(t) = x′(t), x3(t) = x′′(t), we have
x′1(t) = x2(t)(7)
x′2(t) = x3(t)(8)
x′3(t) = −3x3(t) + 7x2(t) + 9x1(t)− sin(t)(9)
Let x(t) =
x1(t)x2(t)x3(t)
, A =
1 0 00 1 09 7 −3
, and b(t) =
00
f(t)
we can write the system of equation in matrix form x′(t) = Ax(t) +b(t). a
Example 4.2. Find the general solution for the 3rd order differen-tial equation x(3) + 3x(2) − 7x′(t)− 9x = sin(t).
Solution From previous example, Example 4.1, Let x(t) =
x1(t)x2(t)x3(t)
, A =
1 0 00 1 09 7 −3
, and b(t) =
00
f(t)
we can write
the system of equation in matrix form x′(t) = Ax(t) + b(t). UsingMathcad we find the eigenvalues are λ1 = −1, λ2 = −1 +
√10, λ3 =
−1−√10 with associate eigenvectors, v1 =
1−11
, v2 =
1
−1 +√
10
11− 2√
10
,
22 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
and v3 =
1
−1−√10
11 + 2√
10
respectively (after multiply the results of
Mathcad by some constants). So the fundamental matrix is
Φ(t) =
e−t e(−1+√
10)t e(1+√
10)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
xp(t) = Φ(t)
∫Φ-1(t)b(t) dt =
352
cos(t)− 539
sin(t)126
cos(t)− 35156
sin(t)− 3
52cos(t)− 8
39sin(t)
Hence the general solution to the system is
x1(t)x2(t)x3(t)
=
c1e−t + c2e(−1+√
10)t + c3e(1+√
10)t + 352 cos(t)− 5
39 sin(t)−c1e
−t + c2(−1 +√
10e(−1+√
10)t − c3(1 +√
10)e(1+√
10)t + 126 cos(t)− 35
156 sin(t)c1e
−t + c2(11− 2√
10)e(−1+√
10)t + c3(11 + 2√
10)e(1+√
10)t − 352 cos(t)− 8
39 sin(t)
and x1(t) = c1e−1t + c2e
(−1+√
10)t + c3e(1+
√10)t + 3
52cos(t) − 5
39sin(t) is
the general solution to the third order ordinary differential equation
x(3) + 3x(2) − 7x′(t)− 9x = sin(t)
. a
Example 4.3. Find the solution to the initial value problem
x′′ − 10x′ + 9x = tet, x(0) = 1, x′(0) = −1.
Solution Since the given equation is of second order, we will havetwo unknowns x1(t) = x(t), x2(t) = x′(t) to transform the equation intoa system of first order equations,
x′1(t) = x2
x′2(t) = 10x2 − 9x1 + tet,
and the initial conditions are x1(0) = x(0) = 1 x2(0) = x′(0) = −1.
Now let x(t) =
[x1(t)x2(t)
], A =
[0 1−9 10
], and b(t) =
[0
tet
].
We have the matrix version of this equation, x′(t) = Ax(t) + b(t)
4. HIGHER ORDER DIFFERENTIAL EQUATIONS 23
Using Mathcad , we find the eigenvalues λ1 = 1, λ1 = 9, and
associate eigenvectors v1 =
[11
], v2 =
[19
]. And fundamental
matrix Φ(t) =
[et e9t
et 9e9t
].
From Φ(t) we find a particular solution
xp(t) = Φ(t)
∫Φ-1(t)b(t) dt =
[ − 1512
(32t2 + 8t + 1)et
− 1512
(32t2 + 72t + 9)et
]
The solution with initial values x0 =
[1−1
]is given by
x(t) = Φ(t)Φ-1(0)(x0 − b(0)) + b(t)
=
(− 1
16t2 − 1
64t + 639
512
)et − 127
512e9t
(− 1
16t2 − 9
64t + 631
512
)et − 1143
512e9t
.
Hence the solution to the initial value problem of the second order
differential equation is x(t) = x1(t) =
(− 1
16t2− 1
64t+ 639
512
)et− 127
512e9t. a
Project
At beginning you should enter: Project title, your name, ss#, anddue date in the following format
Project One: Define and Graph Functions
John DoeSS# 000-00-0000
Due: Mon. Nov. 23rd, 2003
You should format the text region so that the color of text is differentthan math expression. You can choose color for text from Format–>Style select normal and click modify, then change the settings forfont. You can do this for headings etc.
(1) Solutions To System of EquationsFinding solution to linear system using Mathcad and studythe long time dynamic behavior of the solutions.• Find general solution to{
x′ = −yy′ = x
24 1. SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS
Plot several solutions with different initial values in[-] xt-plane, yt-planexy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graphin xy-plane is called the trajectory. If this models themovement of a satellite, what is its trajectory.
• Find general solution to{x′ = −8yy′ = 18x
Plot several solutions with different initial values in[-] xt-plane, yt-planexy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graphin xy-plane is called the trajectory. If this models themovement of a satellite, what is its trajectory.
• Find general solution to{x′ = 2x− yy′ = y − 3x
Plot several solutions with different initial values in[-] xt-plane, yt-planexy-plane. Here you will need to define range variable
t = 0, 0.1 · · · 7 and set X := x(t), Y := y(t). The graphin xy-plane is called the trajectory. If this models themovement of a system of two species, what is you con-clusion about interdependency of these species? Can youfind initial value such that x(t) = 0 (distinct) for some t?what about y(t).
(2) Solution of Higher order equation In general mx′′+ cx′+kx = f(t) models a object with mass m attached to a springwith constant k and damping force that is proportional tothe velocity x′, 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 solu-tions.
- c = b = 0 Find general solution and graph some particularsolutions.
- c = 20, k = 10, a = 0, b = 14, A = 1
- c = 2, k = 3, A = 100, a = 2, b =√
2
