Solutions to Math 54 Problem Set #12

November 17, 2010

9.6.1 The eigenvalues of A =[

2 −42 −2

]are α ± iβ, where α = 0, β = 2. The corresponding

eigenvectors are a± ib, with a =[

11

]and b =

[10

]. The general solution of the system

x′ = Ax is then given by c1x1(t) + c2x2(t), where

x1(t) = cos(2t)[

11

]− sin(2t)

[10

]=[

cos(2t)− sin(2t)cos(2t)

]and

x2(t) = sin(2t)[

11

]+ cos(2t)

[10

]=[

sin(2t) + cos(2t)sin(2t)

].

9.6.4 The characteristic equation of A =

5 −5 −5−1 4 23 −5 −3

is −λ3 + 6λ2 − 13λ + 10 = 0, so

A has eigenvalues λ1 = 2 and λ2,3 = α ± iβ, where α = 2, β = 1. The corresponding

eigenvectors are u1 =

0−11

and u2,3 = a± ib, with a =

5−25

and b =

0−10

. The

general solution of the system x′ = Ax is then given by c1x1(t) + c2x2(t) + c3x3(t), where

x1(t) = e2t

0−11

=

0−e2te2t

,

x2(t) = e2t cos(t)

5−25

− e2t sin(t)

0−10

=

5e2t cos(t)−2e2t cos(t) + e2t sin(t)

5e2t cos(t)

,and

x3(t) = e2t sin(t)

5−25

+ e2t cos(t)

0−10

=

5e2t sin(t)−2e2t sin(t)− e2t cos(t)

5e2t sin(t)

.

9.6.9 The characteristic equation of A =

0 1 1−1 0 1−1 −1 0

is −λ3− 3λ = 0, so A has eigenvalues

λ1 = 0 and λ2,3 = α ± iβ, where α = 0, β =√

3. The corresponding eigenvectors are

1

u1 =

1−11

and u2,3 = a ± ib, with a =

−112

and b =

−√3−√

30

. The general

solution of the system x′ = Ax is then given by c1x1(t) + c2x2(t) + c3x3(t), where

x1(t) =

1−11

,

x2(t) = cos(√

3t)

−112

− sin(√

3t)

−√3−√

30

=

− cos(√

3t) +√

3 sin(√

3t)cos(√

3t) +√

3 sin(√

3t)2 cos(

√3t)

,and

x3(t) = sin(√

3t)

−112

+ cos(√

3t)

−√3−√

30

=

− sin(√

3t)−√

3 cos(√

3t)sin(√

3t)−√

3 cos(√

3t)2 sin(

√3t)

.A fundamental matrix for x′ = Ax is then 1 − cos(

√3t) +

√3 sin(

√3t) − sin(

√3t)−

√3 cos(

√3t)

−1 cos(√

3t) +√

3 sin(√

3t) sin(√

3t)−√

3 cos(√

3t)1 2 cos(

√3t) 2 sin(

√3t)

.

9.6.11 The characteristic equation of A =

0 1 0 00 0 1 00 0 0 1−2 2 −3 2

is λ4 − 2λ3 + 3λ2 − 2λ+ 2 = 0, so

A has eigenvalues λ1,2 = α ± iβ, where α = 1, β = 1 and λ3,4 = α′ ± iβ′, where α′ = 0,

β′ = 1. The corresponding eigenvectors are u1,2 = a± ib, with a =

−1024

, b =

−1−2−20

,

and u3,4 = a′ ± ib′, with a′ =

0−101

and b′ =

10−10

. The general solution of the

system x′ = Ax is then given by c1x1(t) + c2x2(t) + c3x3(t) + c4x4(t), where

x1(t) = et cos(t)

−1024

− et sin(t)

−1−2−20

=

−et cos(t) + et sin(t)

2et sin(t)2et cos(t) + 2et sin(t)

4et cos(t)

,

x2(t) = et sin(t)

−1024

+ et cos(t)

−1−2−20

=

−et sin(t)− et cos(t)

−2et cos(t)2et sin(t)− 2et cos(t)

4et sin(t)

,

2

x3(t) = cos(t)

0−101

− et sin(t)

10−10

=

− sin(t)− cos(t)sin(t)cos(t)

,

x4(t) = sin(t)

0−101

+ et cos(t)

10−10

=

cos(t)− sin(t)− cos(t)sin(t)

,A fundamental matrix for x′ = Ax is then

−et cos(t) + et sin(t) −et sin(t)− et cos(t) − sin(t) cos(t)2et sin(t) −2et cos(t) − cos(t) − sin(t)

2et cos(t) + 2et sin(t) 2et sin(t)− 2et cos(t) sin(t) − cos(t)4et cos(t) 4et sin(t) cos(t) sin(t)

.

9.6.18 The eigenvalues of A =[−1 −19 −1

]are λ1,2 = α ± iβ, where α = −1, β = 3. The

corresponding eigenvectors are u1,2 = a ± ib, with a =[

01

]and b =

[1/30

]. If we

worked over the complex numbers, the general solution would be c1tλ1u1 + c2tλ2u2. To get

a basis consisting of real valued functions, we shall use the following consequence of theEuler formula:

tα+iβ = tαei ln(t)β = tα(cos(β ln(t)) + i sin(β ln(t))).

We obtain

tλ1u1 = tα[cos(β ln(t)) + i sin(β ln(t))] · (a + ib)= tα cos(β ln(t))a− tα sin(β ln(t))b + i [tα sin(β ln(t))a + tα cos(β ln(t))b] .

The real and imaginary parts of tλ1u1, which we call x1 and x2, give a basis for the solutionset of our equation. More explicitly

x1(t) = t−1 cos(3t)[

01

]− t−1 sin(3t)

[1/30

]=[−t−1 sin(3t)/3t−1 cos(3t)

],

and

x2(t) = t−1 sin(3t)[

01

]+ t−1 cos(3t)

[1/30

]=[t−1 cos(3t)/3t−1 sin(3t)

].

A general solution of our equation is then given by c1x1(t) + c2x2(t).

9.7.3 The eigenvalues of A =

1 −2 2−2 1 22 2 1

are λ1 = −3, λ2 = 3, λ3 = 3, with corresponding

eigenvectors u1 =

−1−11

, u2 =

101

, u3 =

−110

. It follows that a fundamental

matrix for the homogeneous system x′ = Ax is given by

X(t) =

−e−3t e3t −e3t−e−3t 0 e3t

e−3t e3t 0

.3

We look for a particular solution xp to x′ = Ax+f of the form et ·a, and find a =

10−1

.

It follows that the general solution of x′ = Ax + f is given by

c1e−3t

−1−11

+ c2e3t

101

+ c3e3t

−110

+ et

10−1

.9.7.5 We look for a particular solution xp = e−2t(a + tb). This will work, since −2 is not an

eigenvalue of A, so none of the terms in xp can be a solution of the homogeneous equation.

9.7.10 We look for a particular solution xp = e−t(a + tb). This will work, since −1 is not aneigenvalue of A, so none of the terms in xp can be a solution of the homogeneous equation.

9.7.13 The eigenvalues of A =[

2 1−3 −2

]are λ1 = −1, λ2 = 1, with corresponding eigenvectors

u1 =[−13

], u2 =

[−11

]. A fundamental matrix is therefore

X(t) =[−e−t −et3e−t et

], with X−1(t) =

12

[et et

−3e−t −e−t].

We get

X−1(t) · f(t) =[

3e2t

−5

],

thus ∫X−1(t) · f(t)dt =

[3e2t/2−5t

],

and

X(t)∫X−1(t) · f(t)dt =

[−3et/2 + 5tet

9et/2− 5tet

].

It follows that the general solution is given by

c1

[−e−t3e−t

]+ c2

[−etet

]+[−3et/2 + 5tet

9et/2− 5tet

].

9.7.15 The eigenvalues of A =[−4 22 −1

]are λ1 = −5, λ2 = 0, with corresponding eigenvectors

u1 =[−21

], u2 =

[12

]. A fundamental matrix is therefore

X(t) =[−2e−5t 1e−5t 2

], with X−1(t) =

−15

[2e5t −e5t−1 −2

].

We get

X−1(t) · f(t) =[

4e5t/58/5 + t−1

],

4

thus ∫X−1(t) · f(t)dt =

[4e5t/25

8t/5 + ln(t)

],

and

X(t)∫X−1(t) · f(t)dt =

[−8/25 + 8t/5 + ln(t)4/25 + 16t/5 + 2 ln(t)

].

It follows that the general solution is given by

c1

[−2e−5t

e−5t

]+ c2

[12

]+[−8/25 + 8t/5 + ln(t)4/25 + 16t/5 + 2 ln(t)

].

9.7.27 The eigenvalues of A =

0 1 11 0 11 1 0

are λ1 = 2, λ2 = −1, λ3 = −1, with corresponding

eigenvectors u1 =

111

, u2 =

−101

, u3 =

−110

, so the general solution of the

homogeneous equation x′ = Ax is given by

c1e2t

111

+ c2e−t

−101

+ c3e−t

−110

.To find a particular solution xp of the nonhomogeneous equation, we use superpositionand find xp1 ,xp2 solutions of

x′ = Ax +

−1−10

, and x′ = Ax + e−t

0−1−2

.We take xp1 = a, xp2 = e−t(b + tc) and find a =

001

, b =

100

, c =

10−1

. Note

that we didn’t choose xp2 of the form e−t ·b, since −1 was an eigenvalue of A. The generalsolution of our equation can now be written as

c1e2t

111

+ c2e−t

−101

+ c3e−t

−110

+

001

+ e−t

100

+ te−t

10−1

.9.8.2 The matrix A =

[1 −11 3

]has only one eigenvalue, λ = 2, with multiplicity 2. It follows

by the Cayley-Hamilton theorem that (A− 2I)2 = 0. We get

eAt = e2t · e(A−2I)t = e2t · (I + (A− 2I)t)

= e2t ·([

1 00 1

]+[−t −tt t

])=[

(1− t)e2t −te2tte2t (1 + t)e2t

].

5

9.8.4 The matrix A =

2 1 30 2 −10 0 2

has only one eigenvalue, λ = 2, with multiplicity 3. It

follows by the Cayley-Hamilton theorem that (A− 2I)3 = 0. We get

eAt = e2t · e(A−2I)t = e2t · (I + (A− 2I)t+12!

(A− 2I)2t2)

= e2t ·

1 0 00 1 00 0 1

+

0 t 3t0 0 −t0 0 0

+

0 0 −t2/20 0 00 0 0

=

e2t te2t (3t− t2/2)e2t

0 e2t −te2t0 0 e2t

.

9.8.5 The matrix A =

−2 0 04 −2 01 0 −2

has only one eigenvalue, λ = −2, with multiplicity 3.

It follows by the Cayley-Hamilton theorem that (A + 2I)3 = 0 (in fact, in this case weactually have (A+ 2I)2 = 0). We get

eAt = e−2t · e(A+2I)t = e−2t · (I + (A+ 2I)t+12!

(A+ 2I)2t2)

= e−2t ·

1 0 00 1 00 0 1

+

0 0 04t 0 0t 0 0

+

0 0 00 0 00 0 0

=

e−2t 0 04te−2t e−2t 0te−2t 0 e−2t

.9.8.7 The matrix A =

[0 1−1 0

]has eigenvalues α±iβ, where α = 0, β = 1. The corresponding

eigenvectors are a ± ib, with a =[

01

]and b =

[−10

]. The general solution of the

system x′ = Ax is then given by c1x1(t) + c2x2(t), where

x1(t) = cos(t)[

01

]− sin(t)

[−10

]=[

sin(t)cos(t)

]and

x2(t) = sin(t)[

01

]+ cos(t)

[−10

]=[− cos(t)sin(t)

].

It follows that a fundamental matrix for x′ = Ax is given by

X(t) =[

sin(t) − cos(t)cos(t) sin(t)

].

We obtain

eAt = X(t) ·X−1(0) =[

cos(t) sin(t)− sin(t) cos(t)

].

6

Remark. Identifying a complex number a + bi with the matrix[

a b−b a

], the above

relation is nothing else than the special case of Euler’s relation ea+bi = ea(cos(b)+i sin(b)),where a = 0 and b = t.

7