chapter 6: systems of linear dif- ferential equationsacaglar/math3321f17/ch6-slides-sp17in.pdf ·...

Chapter 6: Systems of Linear Dif-

ferential Equations

Let a11(t), a12(t), . . . , ann(t),

b1(t), b2(t), . . . , bn(t)

be continuous functions on the in-

terval I.

The system of n first-order differ-

ential equations

1

x′1 = a11(t)x1 + a12(t)x2 + · · ·+ a1n(t)xn + b1(t)

x′2 = a21(t)x1 + a22(t)x2 + · · ·+ a2n(t)xn + b2(t)

... ...

x′n = an1(t)x1 + an2(t)x2 + · · ·+ ann(t)xn + bn(t)

is called a first-order linear differ-

ential system. (For us, a linear

ddifferential system.)

The system is homogeneous if

b1(t) ≡ b2(t) ≡ · · · ≡ bn(t) ≡ 0 on I.

2

It is nonhomogeneous if the func-

tions bi(t) are not all identically

zero on I.

3

Set

A(t) =

a11(t) a12(t) · · · a1n(t)a21(t) a22(t) · · · a2n(t)

... ... ...an1(t) an2(t) · · · ann(t)

and

x =

x1x2...

xn

, b(t) =

b1(t)b2(t)

...bn(t)

.

The system can be written in the

vector-matrix form

x′ = A(t)x + b(t). (S)

c.f., Section 2.14

The matrix A(t) is called the ma-

trix of coefficients or the coeffi-

cient matrix.

The vector b(t) is called the non-

homogeneous term, or forcing func-

tion.

5

A solution of the linear differential

system (S) is a differentiable vector

function

x(t) =

x1(t)x2(t)

...xn(t)

that satisfies (S) on the interval I.

6

THEOREM: The initial-value prob-

lem

x′ = A(t)x + b(t), x(t0) = c

has a unique solution x = x(t).

7

Example 1:

x′1 = x1 + 2x2 + 2e4t

x′2 = 2x1 + x2 + e4t

Vector/matrix form

x1x2

′

=

1 22 1

x1x2

+

2e4t

e4t

or

x′ =

1 22 1

x +

2e4t

e4t

8

x(t) =

85 e4t

75 e4t

is a solution

9

x(t) = C1

e3t

e3t

+C2

e−t

−e−t

+

85 e4t

75 e4t

is a solution for any numbers C1, C2.

10

Example 2:

x′1 = 3x1 − x2 − x3

x′2 = −2x1 + 3x2 + 2x3

x′3 = 4x1 − x2 − 2x3

Vector/matrix form

x1x2x3

′

=

3 −1 −1−2 3 24 −1 −2

x1x2x3

or

x′ =

3 −1 −1−2 3 24 −1 −2

x

(a homogeneous system)

11

x(t) =

e3t

−e3t

e3t

= e3t

1−11

is a solution.

12

x(t) =

C1 e3t

1−11

+C2 e2t

101

+C3 e−t

1−37

is a solution for any numbers C1, C2, C3

13

Converting a linear equation to a

linear system

Consider the second order equation

y′′ + p(t)y′ + q(t)y = 0

Solve for y′′

y′′ = −q(t)y − p(t)y′

14

Introduce new dependent variables

x1, x2, as follows:

x1 = y

x2 = x′1 (= y′)

15

Vector-matrix form:

x1x2

′

=

0 1

−q −p

x1x2

+

0f

Note that this system is just a very

special case of the “general” system

of two, first-order differential equa-

tions:

x′1 = a11(t)x1 + a12(t)x2 + b1(t)

x′2 = a21(t)x1 + a22(t)x2 + b2(t)

16

Vector-matrix form:

x1x2

′

=

a11(t) a12(t)a21(t) a22(t)

x1x2

+

b1(t)b2(t)

or

x′ = A(t)x + b

Example 1:

y′′ − 5y′ + 6y = 4e4t

which can be written

y′′ = −6y + 5y′ + 4e4t

Set

17

y′′ − 5y′ + 6y = 4e4t

Characteristic equation:

Fundamental set:

Particular solution

General solution:

18

System:

x1x2

′

=

0 1

−6 5

x1x2

+

0

4e4t

y = e2t+2e4t is a soln. of equation

Corresponding solution of system????

19

Consider the third-order equation

y′′′ + p(t)y′′ + q(t)y′ + r(t)y = f(t)

or

y′′′ = −r(t)y − q(t)y′ − p(t)y′′ + f(t).

20

Introduce new dependent variables

x1, x2, x3, as follows:

x1 = y

x2 = x′1 (= y′)

x3 = x′2 (= y′′)

Then

y′′′ = x′3 = −r(t)x1−q(t)x2−p(t)x3+f(t)

The third-order equation can be writ-

ten equivalently as the system of

three first-order equations:

21

x′1 = x2

x′2 = x3

x′3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t).

That is

x′1 = 0x1 + 1x2 + 0x3 + 0

x′2 = 0x1 + 0x2 + 1x3 + 0

x′3 = −r(t)x1 − q(t)x2 − p(t)x3 + f(t).

22

Vector-matrix form:

x1x2x3

′

=

0 1 00 0 1

−r −q −p

x1x2x3

+

00f

23

Note that this system is just a very special

case of the “general” system of three, first-

order differential equations:

x′1 = a11(t)x1 + a12(t)x2 + a13(t)x3(t) + b1(t)

x′2 = a21(t)x1 + a22(t)x2 + a23(t)x3(t) + b2(t)

x′3 = a31(t)x1 + a32(t)x2 + a33(t)x3(t) + b3(t).

or in vector-matrix form:

x′ = A(t)x + b

24

Example 2:

y′′′ − 3y′′ − 4y′ + 12y = 6e−t.

which can be written

y′′′ = −12y + 4y′ + 3y′′ + 6e−t.

Set

x1 = y

x2 = x′1 (= y′)

x3 = x′2 (= y′′)

Then x′3 = −12x1+4x2+3x3+6e−t

25

Equivalent system:

x′1 = x2

x′2 = x3

x′3 = −12x1 + 4x2 + 3x3 + 6e−t.

Which is

x′1 = 0x1 + 1x2 + 0x3 + 0

x′2 = 0x1 + 0x2 + 1 x3 + 0

x′3 = −12x1 + 4x2 + 3x3 + 6e−t.

26

Vector-matrix form:

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3

+

00

6e−t

or

x′ = Ax + b

y′′′ − 3y′′ − 4y′ + 12y = 6e−t


Fundamental set:

Particular solution

General solution

27

y = e3t + 12 e−t is a solution of the

equation.

System:

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3

Recall:

x1x2x3

=

yy′

y′′

x =

e3t

3e3t

9e3t

+

12 e−t

−12 e−t

12 e−t

is a corresponding solution of the

system.

28

II. Homogeneous Systems: Gen-

eral Theory:

x′1 = a11(t)x1 + a12(t)x2 + · · · + a1n(t)xn(t)

x′2 = a21(t)x1 + a22(t)x2 + · · · + a2n(t)xn(t)

... ...

x′n = an1(t)x1 + an2(t)x2 + · · ·+ ann(t)xn(t)

x′ = A(t)x. (H)

Note: The zero vector z(t) ≡ 0 =

00...0

is a solution of (H). This so-

lution is called the trivial solution.

29

THEOREM: If v1 and v2 are

solutions of (H), then u = v1 + v2

is also a solutions of (H); the sum of

any two solutions of (H) is a solution

of (H).

THEOREM: If v is a solution of

(H) and α is any real number, then

u = αv is also a solution of (H);

any constant multiple of a solution

of (H) is a solution of (H).

30

In general,

THEOREM: If v1, v2, . . . , vk are

solutions of (H), and if c1, c2, . . . , ck

are real numbers, then

c1v1 + c2v2 + · · · + ckvk

is a solution of (H); any linear com-

bination of solutions of (H) is also

a solution of (H).

31

Linear Dependence/Independence

– in general

Let

v1(t) =

v11v21

...vn1

, v2(t) =

v12v22

...vn2

,

. . . , vk(t) =

v1kv2k

...vnk

be vector functions defined on some

interval I.

32

The vectors are linearly dependent

on I if there exist k real numbers

c1, c2, . . . , ck, not all zero, such

that

c1v1(t)+c2v2(t)+· · ·+ckvk(t) ≡ 0 on I.

Otherwise the vectors are linearly

independent on I.

33

THEOREM Let

v1(t), v2(t), . . . , vk(t)

be k, k-component vector func-

tions defined on an interval I. If

the vectors are linearly dependent,

then∣∣∣∣∣∣∣∣∣∣∣

v11 v12 · · · v1kv21 v22 · · · v2k... ... ... ...

vk1 vk2 · · · vkk

∣∣∣∣∣∣∣∣∣∣∣

≡ 0 on I.

34

The determinant

∣∣∣∣∣∣∣∣∣∣∣

v11 v12 · · · v1kv21 v22 · · · v2k... ... ... ...

vk1 vk2 · · · vkk

∣∣∣∣∣∣∣∣∣∣∣

is called the Wronskian of the vec-

tor functions v1, v2, . . . , vk.

35

Special case: n solutions of (H)

THEOREM Let v1, v2, . . . , vn

be n solutions of (H). Exactly one

of the following holds:

1. W (v1, v2, . . . , vn)(t) ≡ 0 on

I and the solutions are linearly de-

pendent.

2. W (v1, v2, . . . , vn)(t) 6= 0 for all

t ∈ I and the solutions are linearly

independent.36

THEOREM Let v1, v2, . . . , vn

be n linearly independent solutions

of (H). Let u be any solution of

(H). Then there exists a unique set

of constants c1, c2, . . . , cn such that

u = c1v1 + c2v2 + · · · + cnvn.

That is, every solution of (H) can

be written as a unique linear combi-

nation of v1, v2, . . . , vn.

37

A set of n linearly independent solu-

tions of (H) is called a fundamental

set of solutions. A fundamental

set is also called a solution basis

for (H).

38

Let v1, v2, . . . , vn be a fundamen-

tal set of solutions of (H). Then

x = c1v1 + c2v2 + · · · + cnvn,

c1, c2, . . . , cn arbitrary constants, is

the general solution of (H).

39

Example:

y′′′ − 3y′′ − 4y′ + 12y = 0

Fundamental set:

{

y1 = e3t, y2 = e2t, y3 = e−2t}

Vector-matrix form

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3

Solutions:

40

x1 =

e3t

3e3t

9e3t

, x2 =

e2t

2e2t

4e2t

x3 =

e−2t

−2e−2t

4e−2t

is a fundamental set of solutions of

the corresponding system

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3

41

III. Homogeneous Systems with

Constant Coefficients

x′1 = a11x1 + a12x2 + · · · + a1nxn

x′2 = a21x1 + a22x2 + · · · + a2nxn

... ...

x′n = an1x1 + an2x2 + · · · + annxn

where a11, a12, . . . , ann are constants.

42

The system in vector-matrix form is

x1x2−

xn

′

=

a11 a12 · · · a1na21 a22 · · · a2n− − − −

an1 an2 · · · ann

x1x2−

xn

or

x′ = Ax.

43

Solutions of x′ = Ax:

Example:

x1 =

e3t

3e3t

9e3t

= e3t

139

is a solution of

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3

What is

139

relative to

0 1 00 0 1

−12 4 3

?

44

0 1 00 0 1

−12 4 3

139

=

45

THAT IS:

3 is an eigenvalue of A and

v =

139

is a corresponding eigenvector.

46

NOTE:

y′′′ − 3y′′ − 4y′ + 12y = 0

Characteristic equation

r3−3r2−4r+12 = (r−3)(r−2)(r+2) = 0

Fundamental set:

{

y1 = e3t, y2 = e2t, y3 = e−2t}

47

Vector-matrix form

x1x2x3

′

=

0 1 00 0 1

−12 4 3

x1x2x3


det (A − λI) =

∣∣∣∣∣∣∣∣

−λ 1 00 −λ 1

−12 4 3 − λ

∣∣∣∣∣∣∣∣

= −λ3 + 3λ2 + 4λ − 12 = 0

or

λ3−3λ2−4λ+12 = (λ−3)(λ−2)(λ+2) = 0

48

Eigenvalues:

λ1 = 3, λ2 = 2, λ3 = −2

Eigenvectors:

49

Given the homogeneous system with

constant coefficients

x′ = Ax.

CONJECTURE: If λ is an eigen-

value of A and v is a correspond-

ing eigenvector, then

x = eλtv

is a solution.

50

Proof:

Let λ be an eigenvalue of A with

corresponding eigenvecvtor v.

Set x = eλtv

51

If λ1, λ2, · · · , λk are distinct eigen-

values of A with corresponding

eigenvectors v1, v2, · · · , vk, then

x1 = eλ1tv1, x2 = eλ2t

v2, · · · ,xk = eλktvk

are linearly independent solutions of

x′ = Ax.

52

In particular: If λ1, λ2, · · · , λn are

distinct eigenvalues of A with cor-

responding eigenvectors v1, v2, · · · , vn,

then

x1 = eλ1tv1, x2 = eλ2t

v2, · · · ,xn = eλktvn

form a fundamental set of solutions

of

x′ = Ax.

and x = C1x1+C2x2+ · · · +Cnxn

is the general solution.

53

1. Find the general solution of

x′ =

4 −12 1

x.

Step 1. Find the eigenvalues of A:

det(A − λI) =

∣∣∣∣∣∣

4 − λ −12 1 − λ

∣∣∣∣∣∣

= λ2 − 5λ + 6.


λ2 − 5λ + 6 = (λ − 2)(λ − 3) = 0.

Eigenvalues: λ1 = 2, λ2 = 3.

54

Step 2. Find the eigenvectors:

A − λI =

4 − λ −1

2 1 − λ

55

λ1 = 2, v1 =

12

; λ2 = 3, v2 =

11

.

Solutions:

Fundamental set of solution vectors:

x1 = e2t

12

, x2 = e3t

11

General solution of the system:

x = C1e2t

12

+ C2e3t

11

.

56

2. Solve x′ =

3 −1 −1−12 0 5

4 −2 −1

x.

Step 1. Find the eigenvalues of A:

det(A − λI) =

∣∣∣∣∣∣∣∣

3 − λ −1 −1−12 −λ 54 −2 −1 − λ

∣∣∣∣∣∣∣∣

= −λ3 + 2λ2 + λ − 2.


λ3−2λ2−λ+2 = (λ−2)(λ−1)(λ+1) = 0.

Eigenvalues:

λ1 = 2, λ2 = 1, λ3 = −1.

57

Step 2. Find the eigenvectors:

A − λI =

3 − λ −1 −1−12 −λ 54 −2 −1 − λ

λ1 = 2:

58

λ1 = 2 : v1 =

1−12

,

λ2 = 1 : v2 =

3−17

,

λ3 = −1 : v3 =

122

.

Fundamental set of solutions:

x1 = e2t

1−12

, x2 = et

3−17

,

x3 = e−t

122

.

59

The general solution of the system:

x = C1e2t

1−12

+C2et

3−17

+C3e−t

122

.

60

3. Solve the initial-value problem

x′ =

3 −1 −1−12 0 5

4 −2 −1

x, x(0) =

101

.

To find the solution vector satisfy-

ing the initial condition, solve

C1v1(0)+C2v2(0)+C3v3(0) =

101

which is:

C1

1−12

+C2

3−17

+C3

122

=

101

61

or

1 3 1−1 −1 22 7 2

C1C2C3

=

101

.

Augmented matrix:

1 3 1 1−1 −1 2 02 7 2 1

62

Solution:

C1 = 3, C2 = −1, C3 = 1.

The solution of the initial-value prob-

lem is:

x = 3e2t

1−12

−et

3−17

+e−t

122

.

63

TWO DIFFICULTIES:

I. A has complex eigenvalues and

complex eigenvectors.

II. A has an eigenvalue of multiplic-

ity greater than 1.

64

I. Complex eigenvalues/eigenvectors

Examples:

1. Find the general solution of

x′ =

−3 −24 1

x.

det(A−λI) =

∣∣∣∣∣∣

−3 − λ −24 1 − λ

∣∣∣∣∣∣

= λ2+2λ+5.

The eigenvalues are:

λ1 = −1 + 2i, λ2 = −1 − 2i.

65

A − λI =

−3 − λ −2

4 1 − λ

For λ1 = −1 + 2i: Solve

−2 − 2i −2 0

4 2 − 2i 0

→

66

The solution set is:

x2 = −(1 + i)x1, x1 arbitrary

Set x1 = 1. Then, for λ1 = −1+2i:

v1 =

1

−1 − i

=

1

−1

+i

0

−1

.

and, for λ2 = −1 − 2i:

v2 =

1

−1

− i

0

−1

.

67

Solutions

u1 = eλ1tv1

= e(−1+2i)t

[(

1−1

)

+ i

(

0−1

)]

= e−t(cos 2t + i sin 2t)

[(

1−1

)

+ i

(

0−1

)]

= e−t

[

cos 2t

(

1−1

)

− sin 2t

(

0−1

)]

+

i e−t

[

cos 2t

(

0−1

)

+ sin 2t

(

1−1

)]

.

68

u2 = eλ2tv2

= e(−1−2i)t

[(

1−1

)

− i

(

0−1

)]

= e−t(cos 2t + i sin 2t)

[(

1−1

)

+ i

(

0−1

)]

= e−t

[

cos 2t

(

1−1

)

− sin 2t

(

0−1

)]

−

i e−t

[

cos 2t

(

0−1

)

+ sin 2t

(

1−1

)]

.

69

Fundamental set:

x1 =u1 + u2

2

= e−t

cos 2t

1

−1

− sin 2t

0

−1

x2 =u1 + u2

2i

= e−t

cos 2t

0

−1

+ sin 2t

1

−1

General solution:

x = C1 e−t

cos 2t

1

−1

− sin 2t

0

−1

+

C2 e−t

cos 2t

0

−1

+ sin 2t

1

−1

70

Summary: x′ = Ax, A n×n const.

a+ib, a−ib complex eigenvalues.

→α +i

→β ,

→α −i

→β corresponding

eigenvectors.

Independent (complex-valued) solu-

tions:

u1 = e(a+ib)t(

→α +i

→β

)

u2 = e(a−ib)t(

→α −i

→β

)

71

Corresponding real-valued solu-

tions:

x1 = eat[

cos bt→α − sin bt

→β

]

x2 = eat[

cos bt→β +sin bt

→α

]

General solution:

x = C1 eat[

cos bt→α − sin bt

→β

]

+

C2eat[

cos bt→β +sin bt

→α

]

72

2. Determine a fundamental set

of solution vectors of

x′ =

1 −4 −13 2 31 1 3

x.

det(A−λI) =

∣∣∣∣∣∣∣∣

1 − λ −4 −13 2 − λ 31 1 3 − λ

∣∣∣∣∣∣∣∣

=

−λ3+6λ2−21λ+26 = −(λ−2)(λ2−4λ+13).

The eigenvalues are:

λ1 = 2, λ2 = 2 + 3i, λ3 = 2 − 3i.

73

A − λI =

1 − λ −4 −13 2 − λ 31 1 3 − λ

λ1 = 2: Solve

−1 −4 −1 03 0 3 01 1 1 0

→

v1 =

10

−1

.

74

A − λI =

1 − λ −4 −13 2 − λ 31 1 3 − λ

For λ2 = 2 + 3i: Solve

−1 − 3i −4 −1 03 −3i 3 01 1 1 − 3i 0

→

75

The solution set is:

x1 =

(

−5

2+

3

2i

)

x3, x2 =

(

3

2+

3

2i

)

x3, ,

x3 arbitrary.

v2 =

−5 + 3i3 + 3i

2

=

−532

+i

330

.

and

v3 =

−5 − 3i3 − 3i

2

=

−532

− i

330

.

76

Now

u1 = e(2+3i)t

−532

+ i

330

and

u2 = e(2−3i)t

−532

− i

330

convert to:

x1 = e2t

cos 3t

−532

− sin 3t

330

and

x2 = e2t

cos 3t

330

+ sin 3t

−532

77

Fundamental set of solution vectors:

x1 = e2t

10

−1

,

x2 = e2t

cos 3t

−532

− sin 3t

330

,

x3 = e2t

cos 3t

330

+ sin 3t

−532

.

General solution:

x = C1x1 + C2x2 + C3x3

78

II. Repeated eigenvalues

Examples:

1. Find a fundamental set of

solutions of

x′ =

1 −3 33 −5 36 −6 4

x.

det(A−λI) =

∣∣∣∣∣∣∣∣

1 − λ −3 33 −5 − λ 36 −6 4 − λ

∣∣∣∣∣∣∣∣

= 16+12λ−λ3 = −(λ−4)(λ+2)2.

Eigenvalues: λ1 = 4, λ2 = λ3 = −2

79

λ1 = 4 : (A − 4I) =

−3 −3 33 −9 33 6 0

80

λ2 = λ3 = −2:

A − (−2)I =

3 −3 33 −3 36 −6 6

81

which row reduces to

1 −1 1 00 0 0 00 0 0 0

.

Solution set:

x1 = a − b, x2 = a, x3 = b

a, b any real numbers.

Set a = 1, b = 0 : v2 =

110

;

Set a = 0, b = −1 : v3 =

10

−1

.

82

Fundamental set:

e4t

112

, e−2t

110

, e−2t

10

−1

.

83


solutions of

x′ =

5 6 20 −1 −81 0 −2

x.

det(A−λI) =

∣∣∣∣∣∣∣∣

5 − λ 6 20 −1 − λ −81 0 −2 − λ

∣∣∣∣∣∣∣∣

= −36+15λ+2λ2−λ3 = −(λ+4)(λ−3)2.

Eigenvalues: λ1 = −4, λ2 = λ3 = 3.

84

λ1 = −4 : v1 =

6−8−3

.

A − (−4)I =

9 6 20 3 −81 0 2

85

λ2 = λ3 = 3: Solve

A − 3I =

2 6 20 −4 −81 0 −5

86

2 6 2 00 −4 −8 01 0 −5 0

which row reduces to

1 0 −5 00 1 2 00 0 0 0

.

x1 = 5x3, x2 = −2x3, x3 arbitrary

Set x3 = 1: v2 =

5−21

Problem: Only one eigenvector!

87

Solutions:

x1 = e−4t

6−8−3

, x2 = e3t

5−21

.

We need a third solution x3 which

is independent of x1, x2.

88

3. y′′′ + y′′ − 8y′ − 12y = 0

Char.eqn.

r3+r2−8r−12 = (r−3)(r+2)2 = 0.

Fundamental set:{

e3t, e−2t, te−2t}

89

Equivalent system: x′ =

0 1 00 0 1

12 8 −1

x

det(A − λI) =

∣∣∣∣∣∣∣∣

−λ 1 00 λ 112 8 −1 − λ

∣∣∣∣∣∣∣∣

= −λ3 − λ2 + 8λ + 12λ

char. eqn.:

λ3 + λ2 − 8λ − 12 = (λ − 3)(λ + 2)2

Eigenvalues: λ1 = 3, λ2 = λ3 = −2

90

Fundamental set:

x1 = e3t

139

, x2 = e−2t

1−24

,

x3 = e−2t

01

−4

+ te−2t

1−24

Question: What is the vector

01

−4

?

91

[A − (−2)I]

01

−4

=

2 1 00 2 1

12 8 1

01

−4

=

92

A − (−2I) “maps”

01

−4

onto

the eigenvector

1−24

.

01

−4

is called a generalized eigenvec-

tor.

93

2. continued. Solve

(A − 3I)w =

5−21

2 6 2 50 −4 −8 −21 0 −5 1

94

→

1 0 −5 10 1 2 1/20 0 0 0

.

Solution set:

x1 = 1+5x3, x2 =1

2−2x3, x3 arbitrary

Set x3 = 0: w =

11/2

0

95

Fundamental set:

x1 = e−4t

6−8−3

, x2 = e3t

5−21

,

x3 = e3t

11/2

0

+ te3t

5−21

96

Eigenvalues of multiplicity 2:

Given x′ = Ax.

Suppose that A has an eigenvalue

λ of multiplicity 2. Then exactly

one of the following holds:

97

1. λ has two linearly independent

eigenvectors, v1 and v2. Corre-

sponding linearly independent solu-

tion vectors of the differential sys-

tem are

x1(t) = eλtv1 and x2(t) = eλt

v2.

(See Example 1.)

98

2. λ has only one (independent)

eigenvector v. (See Examples 2

and 3.) Then a linearly independent

pair of solution vectors is:

x1(t) = eλtv and x2(t) = eλt

w+teλtv

where w is a vector that satisfies

(A − λI)w = v.

The vector w is called a general-

ized eigenvector corresponding to

the eigenvalue λ.99

Examples:


solutions and the general solution of

x′ =

2 5

−1 4

x.

det (A − λ I) =

∣∣∣∣∣∣

2 − λ 5−1 4 − λ

∣∣∣∣∣∣

= λ2 − 6λ + 13

Eigenvalues: 3 + 2i, 3 − 2i

100

(A − λ I) =

2 − λ 5−1 4 − λ

λ1 = 3 + 2i, v1 =

11

+ i

−20

101

Fundamental set:

x1 = e3t

cos 2t

11

− sin 2t

−20

x2 = e3t

cos 2t

−20

− sin 2t

11

General solution:

x(t) = C1x1 + C2x2

102



x′ =

−4 1 −22 −3 22 −1 0

x.

HINT: −3 is an eigenvalue and −2

is an eigenvalue of multiplicity 2

Characteristic eqn: (λ+3)(λ+2)2=0

103

(A − λI) =

−4 − λ 1 −22 −3 − λ 22 −1 −λ

λ1 = −3 :

−111

104

(A − λI) =

−4 − λ 1 −22 −3 − λ 22 −1 −λ

λ2 = λ3 = −2 :

120

,

021

105

Fundamental set:

e−3t

−111

, e−2t

120

, e−2t

021

General solution:

x =

C1e−3t

−111

+C2e−2t

120

+C3e−2t

021

106



x′ =

−3 1 −1−7 5 −1−6 6 −2

x.

HINT: 4 is an eigenvalue and −2

is an eigenvalue of multiplicity 2

Characteristic eqn: (λ−4)(λ+2)2=0

107

(A−λI) =

−3 − λ 1 −1−7 5 − λ −1−6 6 −2 − λ

λ1 = 4 : v1 =

011

108

(A−λI) =

−3 − λ 1 −1−7 5 − λ −1−6 6 −2 − λ

λ2 = λ3 = −2 : v2 =

110

109

(A−λI) =

−3 − λ 1 −1−7 5 − λ −1−6 6 −2 − λ

[A − (−2)I]w =

110

110

Fund. Set: e4t

011

, e−2t

110

e−2t

11

−1

+ te−2t

110

General solution:

x = C1e4t

011

+ C2e−2t

110

+

C3

e

−2t

11

−1

+ te−2t

110

111

Eigenvalues of Multiplicity 3.

Given the differential system

x′ = Ax.

Suppose that λ is an eigenvalue of

A of multiplicity 3. Then exactly

one of the following holds:

112

1. λ has three linearly indepen-

dent eigenvectors v1, v2, v3. Then

three linearly independent solution

vectors of the system corresponding

to λ are:

x1(t) = eλtv1, x2(t) = eλt

v2,

x3(t) = eλtv3.

113

2. λ has two linearly independent

eigenvectors v1, v2. Then three

linearly independent solutions of the

system corresponding to λ are:

x1(t) = eλtv1, x2(t) = eλt

v2

and

x3(t) = eλtw + teλt

v

where v is an eigenvector corre-

sponding to λ and (A−λI)w = v.

That is: (A − λI)2w = 0.

114

3. λ has only one (independent)

eigenvector v. Then three linearly

independent solutions of the system

have the form:

x1 = eλtv, x2 = eλt

w + teλtv,

v3(t) = eλtz + teλt

w + t2eλtv

where

(A−λI)z = w & (A−λI)w = v, i.e.

(A − λI)3z = 0 & (A − λI)2w = 0

115

Example:

y′′′ − 6y′′ + 12y′ − 8y = 0

Char. eqn.: (r − 2)3 = 0

Char. roots: r1 = r2 = r3 = 2

Fundamental set:

{

e2t, te2t, t2e2t}

116

Corresponding system:

x′ =

0 1 00 0 18 −12 6

x

Fundamental set:

e2t

124

, e2t

014

+ te2t

124

,

e2t

002

+ te2t

028

+ t2e2t

124

117

Nonhomogeneous Linear Differential Sys-

tems

Let A(t) be the n × n matrix

A(t) =

a11(t) a12(t) · · · a1n(t)a21(t) a22(t) · · · a2n(t)

... ... ...an1(t) an2(t) · · · ann(t)

and let x and b(t) be the vectors

x =

x1x2...

xn

, b(t) =

b1(t)b2(t)

...bn(t)

.

Nonhomogeneous system:

x′ = A(t)x + b(t) (N)

Reduced system:

x′ = A(t)x (H)

118

Theorem 1. If z1(t) and z2(t) are

solutions of (N), then

x(t) = z1(t) − z2(t)

is a solution of (H). (C.f. Theorem 1, Sec-

tion 3.4, and Theorem 7, Section 6.1.)

Theorem 2. Let x1(t), x2(t), . . . ,xn(t)

be a fundamental set of solutions the re-

duced system (H) and let z = z(t) be a

particular solution of (N). If u = u(t) is

any solution of (N), then there exist con-

stants c1, c2, . . . , cn such that

u(t) = c1x1(t)+c2x2(t)+· · ·+cnxn(t)+z(t)

(C.f. Theorem 2, Section 3.4, and Theo-

rem 8, Section 6.1.)

119

General Solution of (N):

x = C1x1(t) + C2x2(t) + · · · + Cnxn(t)︸︷︷︸

general solution of (H)

+

z(t).︸︷︷︸

particular solution of (N)

120

Variation of Parameters

x1(t), x2(t), . . . ,xn(t)

a fundamental set of solutions of (H).

V (t) the corresponding fundamental ma-

trix.

x(t) = V (t)C where C =

C1C2

...Cn

.

is the general solution of (H).

V satisfies the matrix differential system

X′ = A(t)X.

That is, V ′(t) = A(t)V (t). (Exercises 6.3,

Problem 19.)

121

Replace the constant vector C by a vector

function u(t) which is to be determined

so that

z(t) = V (t)u(t)

is a solution of (N).

u′(t) = V −1(t)b(t)

u(t) =

∫

V −1(t)b(t) dt.

and

z(t) = V (t)∫

V −1(t)b(t) dt

is a solution of (N).

General solution of (N):

x(t) = V (t)C + V (t)∫

V −1(t)b(t) dt.

122

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