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Part III Systems of Differential Equations
1010 Systems of LinearDifferential Equations
EXERCISES 10.1
Preliminary Theory
1. Let X =
(
x
y
)
. Then X′ =
(
3 −5
4 8
)
X.
2. Let X =
(
x
y
)
. Then X′ =
(
4 −7
5 0
)
X.
3. Let X =
xy
z
. Then X′ =
−3 4 −96 −1 0
10 4 3
X.
4. Let X =
xy
z
. Then X′ =
 1 −1 01 0 2
−1 0 1
X.
5. Let X =
xy
z
. Then X′ =
 1 −1 12 1 −1
1 1 1
X +
 0−3t2
t2
 +
 t0
−t
 +
−10
2
.
6. Let X =
xy
z
. Then X′ =
−3 4 05 9 0
0 1 6
X +
 e
−t sin 2t
4e−t cos 2t
−e−t
.
7.
dx
dt
= 4x + 2y + et;
dy
dt
= −x + 3y − et
8.
dx
dt
= 7x + 5y − 9z − 8e−2t; dy
dt
= 4x + y + z + 2e5t;
dz
dt
= −2y + 3z + e5t − 3e−2t
9.
dx
dt
= x− y + 2z + e−t − 3t; dy
dt
= 3x− 4y + z + 2e−t + t; dz
dt
= −2x + 5y + 6z + 2e−t − t
10.
dx
dt
= 3x− 7y + 4 sin t + (t− 4)e4t; dy
dt
= x + y + 8 sin t + (2t + 1)e4t
11. Since
X′ =
( −5
−10
)
e−5t and
(
3 −4
4 −7
)
X =
( −5
−10
)
e−5t
551
10.1 Preliminary Theory
we see that
X′ =
(
3 −4
4 −7
)
X.
12. Since
X′ =
(
5 cos t− 5 sin t
2 cos t− 4 sin t
)
et and
(−2 5
−2 4
)
X =
(
5 cos t− 5 sin t
2 cos t− 4 sin t
)
et
we see that
X′ =
(−2 5
−2 4
)
X.
13. Since
X′ =
( 3
2
−3
)
e−3t/2 and
(−1 14
1 −1
)
X =
( 3
2
−3
)
e−3t/2
we see that
X′ =
(−1 1/4
1 −1
)
X.
14. Since
X′ =
(
5
−1
)
et +
(
4
−4
)
tet and
(
2 1
−1 0
)
X =
(
5
−1
)
et +
(
4
−4
)
tet
we see that
X′ =
(
2 1
−1 0
)
X.
15. Since
X′ =
 00
0
 and
 1 2 16 −1 0
−1 −2 −1
X =
 00
0

we see that
X′ =
 1 2 16 −1 0
−1 −2 −1
X.
16. Since
X′ =
 cos t12 sin t− 12 cos t
− cos t− sin t
 and
 1 0 11 1 0
−2 0 −1
X =
 cos t12 sin t− 12 cos t
− cos t− sin t

we see that
X′ =
 1 0 11 1 0
−2 0 −1
X.
17. Yes, since W (X1,X2) = −2e−8t �= 0 the set X1, X2 is linearly independent on −∞ < t < ∞.
18. Yes, since W (X1,X2) = 8e2t �= 0 the set X1, X2 is linearly independent on −∞ < t < ∞.
19. No, since W (X1,X2,X3) = 0 the set X1, X2, X3 is linearly dependent on −∞ < t < ∞.
20. Yes, since W (X1,X2,X3) = −84e−t �= 0 the set X1, X2, X3 is linearly independent on −∞ < t < ∞.
21. Since
X′p =
(
2
−1
)
and
(
1 4
3 2
)
Xp +
(
2
−4
)
t +
( −7
−18
)
=
(
2
−1
)
552
10.1 Preliminary Theory
we see that
X′p =
(
1 4
3 2
)
Xp +
(
2
−4
)
t +
( −7
−18
)
.
22. Since
X′p =
(
0
0
)
and
(
2 1
1 −1
)
Xp +
(−5
2
)
=
(
0
0
)
we see that
X′p =
(
2 1
1 −1
)
Xp +
(−5
2
)
.
23. Since
X′p =
(
2
0
)
et +
(
1
−1
)
tet and
(
2 1
3 4
)
Xp −
(
1
7
)
et =
(
2
0
)
et +
(
1
−1
)
tet
we see that
X′p =
(
2 1
3 4
)
Xp −
(
1
7
)
et.
24. Since
X′p =
 3 cos 3t0
−3 sin 3t
 and
 1 2 3−4 2 0
−6 1 0
Xp +
−14
3
 sin 3t =
 3 cos 3t0
−3 sin 3t

we see that
X′p =
 1 2 3−4 2 0
−6 1 0
Xp +
−14
3
 sin 3t.
25. Let
X1 =
 6−1
−5
 e−t, X2 =
−31
1
 e−2t, X3 =
 21
1
 e3t, and A =
 0 6 01 0 1
1 1 0
 .
Then
X′1 =
−61
5
 e−t = AX1,
X′2 =
 6−2
−2
 e−2t = AX2,
X′3 =
 63
3
 e3t = AX3,
and W (X1,X2,X3) = 20 �= 0 so that X1, X2, and X3 form a fundamental set for X′ = AX on −∞ < t < ∞.
26. Let
X1 =
(
1
−1−√2
)
e
√
2 t,
X2 =
(
1
−1 +√2
)
e−
√
2 t,
553
10.1 Preliminary Theory
Xp =
(
1
0
)
t2 +
(−2
4
)
t +
(
1
0
)
,
and
A =
(−1 −1
−1 1
)
.
Then
X′1 =
( √
2
−2−√2
)
e
√
2 t = AX1,
X′2 =
( −√2
−2 +√2
)
e−
√
2 t = AX2,
X′p =
(
2
0
)
t +
(−2
4
)
= AXp +
(
1
1
)
t2 +
(
4
−6
)
t +
(−1
5
)
,
and W (X1,X2) = 2
√
2 �= 0 so that Xp is a particular solution and X1 and X2 form a fundamental set on
−∞ < t < ∞.
EXERCISES 10.2
Homogeneous Linear Systems
1. The system is
X′ =
(
1 2
4 3
)
X
and det(A− λI) = (λ− 5)(λ + 1) = 0. For λ1 = 5 we obtain(−4 2 0
4 −2 0
)
=⇒
(
1 − 12 0
0 0 0
)
so that K1 =
(
1
2
)
.
For λ2 = −1 we obtain (
2 2 0
4 4 0
)
=⇒
(
1 1 0
0 0 0
)
so that K2 =
(−1
1
)
.
Then
X = c1
(
1
2
)
e5t + c2
(−1
1
)
e−t.
2. The system is
X′ =
(
2 2
1 3
)
X
and det(A− λI) = (λ− 1)(λ− 4) = 0. For λ1 = 1 we obtain(
1 2 0
1 2 0
)
=⇒
(
1 2 0
0 0 0
)
so that K1 =
(−2
1
)
.
For λ2 = 4 we obtain (−2 2 0
1 −1 0
)
=⇒
(−1 1 0
0 0 0
)
so that K2 =
(
1
1
)
.
554
10.2 Homogeneous Linear Systems
Then
X = c1
(−2
1
)
et + c2
(
1
1
)
e4t.
3. The system is
X′ =
( −4 2
− 52 2
)
X
and det(A− λI) = (λ− 1)(λ + 3) = 0. For λ1 = 1 we obtain( −5 2 0
− 52 1 0
)
=⇒
(−5 2 0
0 0 0
)
so that K1 =
(
2
5
)
.
For λ2 = −3 we obtain ( −1 2 0
− 52 5 0
)
=⇒
(−1 2 0
0 0 0
)
so that K2 =
(
2
1
)
.
Then
X = c1
(
2
5
)
et + c2
(
2
1
)
e−3t.
4. The system is
X′ =
(− 52 2
3
4 −2
)
X
and det(A− λI) = 12 (λ + 1)(2λ + 7) = 0. For λ1 = −7/2 we obtain(
1 2 0
3
4
3
2 0
)
=⇒
(
1 2 0
0 0 0
)
so that K1 =
(−2
1
)
.
For λ2 = −1 we obtain (− 32 2 0
3
4 −1 0
)
=⇒
(−3 4 0
0 0 0
)
so that K2 =
(
4
3
)
.
Then
X = c1
(−2
1
)
e−7t/2 + c2
(
4
3
)
e−t.
5. The system is
X′ =
(
10 −5
8 −12
)
X
and det(A− λI) = (λ− 8)(λ + 10) = 0. For λ1 = 8 we obtain(
2 −5 0
8 −20 0
)
=⇒
(
1 − 52 0
0 0 0
)
so that K1 =
(
5
2
)
.
For λ2 = −10 we obtain (
20 −5 0
8 −2 0
)
=⇒
(
1 − 14 0
0 0 0
)
so that K2 =
(
1
4
)
.
Then
X = c1
(
5
2
)
e8t + c2
(
1
4
)
e−10t.
6. The system is
X′ =
(−6 2
−3 1
)
X
555
10.2 Homogeneous Linear Systems
and det(A− λI) = λ(λ + 5) = 0. For λ1 = 0 we obtain(−6 2 0
−3 1 0
)
=⇒
(
1 − 13 0
0 0 0
)
so that K1 =
(
1
3
)
.
For λ2 = −5 we obtain (−1 2 0
−3 6 0
)
=⇒
(
1 −2 0
0 0 0
)
so that K2 =
(
2
1
)
.
Then
X = c1
(
1
3
)
+ c2
(
2
1
)
e−5t.
7. The system is
X′ =
 1 1 −10 2 0
0 1 −1
X
and det(A− λI) = (λ− 1)(2− λ)(λ + 1) = 0. For λ1 = 1, λ2 = 2, and λ3 = −1 we obtain
K1 =
 10
0
 , K2 =
 23
1
 , and K3 =
 10
2
 ,
so that
X = c1
 10
0
 et + c2
 23
1
 e2t + c3
 10
2
 e−t.
8. The system is
X′ =
 2 −7 05 10 4
0 5 2
X
and det(A− λI) = (2− λ)(λ− 5)(λ− 7) = 0. For λ1 = 2, λ2 = 5, and λ3 = 7 we obtain
K1 =
 40
−5
 , K2 =
−73
5
 , and K3 =
−75
5
 ,
so that
X = c1
 40
−5
 e2t + c2
−73
5
 e5t + c3
−75
5
 e7t.
9. We have det(A− λI) = −(λ + 1)(λ− 3)(λ + 2) = 0. For λ1 = −1, λ2 = 3, and λ3 = −2 we obtain
K1 =
−10
1
 , K2 =
 14
3
 , and K3 =
 1−1
3
 ,
so that
X = c1
−10
1
 e−t + c2
 14
3
 e3t + c3
 1−1
3
 e−2t.
556
10.2 Homogeneous Linear Systems
10. We have det(A− λI) = −λ(λ− 1)(λ− 2) = 0. For λ1 = 0, λ2 = 1, and λ3 = 2 we obtain
K1 =
 10
−1
 , K2 =
 01
0
 , and K3 =
 10
1
 ,
so that
X = c1
 10
−1
 + c2 01
0
 et + c3
 10
1
 e2t.
11. We have det(A− λI) = −(λ + 1)(λ + 1/2)(λ + 3/2) = 0. For λ1 = −1, λ2 = −1/2, and λ3 = −3/2 we obtain
K1 =
 40
−1
 , K2 =
−126
5
 , and K3 =
 42
−1
 ,
so that
X = c1
 40
−1
 e−t + c2
−126
5
 e−t/2 + c3
 42
−1
 e−3t/2.
12. We have det(A− λI) = (λ− 3)(λ + 5)(6− λ) = 0. For λ1 = 3, λ2 = −5, and λ3 = 6 we obtain
K1 =
 11
0
 , K2 =
 1−1
0
 , and K3 =
 2−2
11
 ,
so that
X = c1
 11
0
 e3t + c2
 1−1
0
 e−5t + c3
 2−2
11
 e6t.
13. We have det(A− λI) = (λ + 1/2)(λ− 1/2) = 0. For λ1 = −1/2 and λ2 = 1/2 we obtain
K1 =
(
0
1
)
and K2 =
(
1
1
)
,
so that
X = c1
(
0
1
)
e−t/2 + c2
(
1
1
)
et/2.
If
X(0) =
(
3
5
)
then c1 = 2 and c2 = 3.
14. We have det(A− λI) = (2− λ)(λ− 3)(λ + 1) = 0. For λ1 = 2, λ2 = 3, and λ3 = −1 we obtain
K1 =
 5−3
2
 , K2 =
 20
1
 , and K3 =
−20
1
 ,
so that
X = c1
 5−3
2
 e2t + c2
 20
1
 e3t + c3
−20
1
 e−t.
557
10.2 Homogeneous Linear Systems
If
X(0) =
 13
0

then c1 = −1, c2 = 5/2, and c3 = −1/2.
15. X = c1
 0.3821750.851161
0.359815
 e8.58979t + c2
 0.405188−0.676043
0.615458
 e2.25684t + c3
−0.923562−0.132174
0.35995
 e−0.0466321t
16. X = c1

0.0312209
0.949058
0.239535
0.195825
0.0508861
 e5.05452t + c2

−0.280232
−0.836611
−0.275304
0.176045
0.338775
 e4.09561t + c3

0.262219
−0.162664
−0.826218
−0.346439
0.31957
 e−2.92362t
+c4

0.313235
0.64181
0.31754
0.173787
−0.599108
 e2.02882t + c5

−0.301294
0.466599
0.222136
0.0534311
−0.799567
 e−0.155338t
17. (a)
(b) Letting c1 = 1 and c2 = 0 we get x = 5e8t, y = 2e8t. Eliminating the parameter we find y = 25x, x > 0.
When c1 = −1 and c2 = 0 we find y = 25x, x < 0. Letting c1 = 0 and c2 = 1 we get x = e−10t, y = 4e−10t.
Eliminating the parameter we find y = 4x, x > 0. Letting c1 = 0 and c2 = −1 we find y = 4x, x < 0.
(c) The eigenvectors K1 = (5, 2) and K2 = (1, 4) are shown in the figure in part (a).
18. In Problem 2, letting c1 = 1 and c2 = 0 we get x = −2et, y = et.
Eliminating the parameter we find y = − 12x, x < 0. When c1 = −1
and c2 = 0 we find y = − 12x, x > 0. Letting c1 = 0 and c2 = 1 we
get x = e4t, y = e4t. Eliminating the parameter we find y = x, x > 0.
When c1 = 0 and c2 = −1 we find y = x, x < 0.
558
10.2 Homogeneous Linear Systems
In Problem 4, letting c1 = 1 and c2 = 0 we get x = −2e−7t/2, y =
e−7t/2. Eliminating the parameter we find y = − 12x, x < 0. When
c1 = −1 and c2 = 0 we find y = − 12x, x > 0. Letting c1 = 0 and
c2 = 1 we get x = 4e−t, y = 3e−t. Eliminating the parameter we find
y = 34x, x > 0. When c1 = 0 and c2 = −1 we find y = 34x, x < 0.
19. We have det(A− λI) = λ2 = 0. For λ1 = 0 we obtain
K =
(
1
3
)
.
A solution of (A− λ1I)P = K is
P =
(
1
2
)
so that
X = c1
(
1
3
)
+ c2
[(
1
3
)
t +
(
1
2
)]
.
20. We have det(A− λI) = (λ + 1)2 = 0. For λ1 = −1 we obtain
K =
(
1
1
)
.
A solution of (A− λ1I)P = K is
P =
(
0
1
5
)
so that
X = c1
(
1
1
)
e−t + c2
[(
1
1
)
te−t +
(
0
1
5
)
e−t
]
.
21. We have det(A− λI) = (λ− 2)2 = 0. For λ1 = 2 we obtain
K =
(
1
1
)
.
A solution of (A− λ1I)P = K is
P =
(− 13
0
)
so that
X = c1
(
1
1
)
e2t + c2
[(
1
1
)
te2t +
(− 13
0
)
e2t
]
.
22. We have det(A− λI) = (λ− 6)2 = 0. For λ1 = 6 we obtain
K =
(
3
2
)
.
A solution of (A− λ1I)P = K is
P =
( 1
2
0
)
559
10.2 Homogeneous Linear Systems
so that
X = c1
(
3
2
)
e6t + c2
[(
3
2
)
te6t +
( 1
2
0
)
e6t
]
.
23. We have det(A− λI) = (1− λ)(λ− 2)2 = 0. For λ1 = 1 we obtain
K1 =
 11
1
 .
For λ2 = 2 we obtain
K2 =
 10
1
 and K3 =
 11
0
 .
Then
X = c1
 11
1
 et + c2
 10
1
 e2t + c3
 11
0
 e2t.
24. We have det(A− λI) = (λ− 8)(λ + 1)2 = 0. For λ1 = 8 we obtain
K1 =
 21
2
 .
For λ2 = −1 we obtain
K2 =
 0−2
1
 and K3 =
 1−2
0
 .
Then
X = c1
 21
2
 e8t + c2
 0−2
1
 e−t + c3
 1−2
0
 e−t.
25. We have det(A− λI) = −λ(5− λ)2 = 0. For λ1 = 0 we obtain
K1 =
−4−5
2
 .
For λ2 = 5 we obtain
K =
−20
1
 .
A solution of (A− λ1I)P = K is
P =

5
2
1
2
0

560
10.2 Homogeneous Linear Systems
so that
X = c1
−4−5
2
 + c2
−20
1
 e5t + c3

−20
1
 te5t +

5
2
1
2
0
 e5t
 .
26. We have det(A− λI) = (1− λ)(λ− 2)2 = 0. For λ1 = 1 we obtain
K1 =
 10
0
 .
For λ2 = 2 we obtain
K =
 0−1
1
 .
A solution of (A− λ2I)P = K is
P =
 0−1
0

so that
X = c1
 10
0
 et + c2
 0−1
1
 e2t + c3

 0−1
1
 te2t +
 0−1
0
 e2t
 .
27. We have det(A− λI) = −(λ− 1)3 = 0. For λ1 = 1 we obtain
K =
 01
1
 .
Solutions of (A− λ1I)P = K and (A− λ1I)Q = P are
P =
 01
0
 and Q =

1
2
0
0

so that
X = c1
 01
1
 et + c2

 01
1
 tet +
 01
0
 et
 + c3

 01
1
 t22 et +
 01
0
 tet +

1
2
0
0
 et
 .
28. We have det(A− λI) = (λ− 4)3 = 0. For λ1 = 4 we obtain
K =
 10
0
 .
Solutions of (A− λ1I)P = K and (A− λ1I)Q = P are
P =
 01
0
 and Q =
 00
1

561
10.2 Homogeneous Linear Systems
so that
X = c1
 10
0
 e4t + c2

 10
0
 te4t +
 01
0
 e4t
 + c3

 10
0
 t22 e4t +
 01
0
 te4t +
 00
1
 e4t
 .
29. We have det(A− λI) = (λ− 4)2 = 0. For λ1 = 4 we obtain
K =
(
2
1
)
.
A solution of (A− λ1I)P = K is
P =
(
1
1
)
so that
X = c1
(
2
1
)
e4t + c2
[(
2
1
)
te4t +
(
1
1
)
e4t
]
.
If
X(0) =
(−1
6
)
then c1 = −7 and c2 = 13.
30. We have det(A− λI) = −(λ + 1)(λ− 1)2 = 0. For λ1 = −1 we obtain
K1 =
−10
1
 .
For λ2 = 1 we obtain
K2 =
 10
1
 and K3 =
 01
0

so that
X = c1
−10
1
 e−t + c2
 10
1
 et + c3
 01
0
 et.
If
X(0) =
 12
5

then c1 = 2, c2 = 3, and c3 = 2.
31. In this case det(A − λI) = (2 − λ)5, and λ1 = 2 is an eigenvalue of multiplicity 5. Linearly independent
eigenvectors are
K1 =

1
0
0
0
0
 , K2 =

0
0
1
0
0
 , and K3 =

0
0
0
1
0
 .
32. In Problem 20 letting c1 = 1 and c2 = 0 we get x = et, y = et. Eliminating the parameter we find y = x, x > 0.
When c1 = −1 and c2 = 0 we find y = x, x < 0.
562
10.2 Homogeneous Linear Systems
In Problem 21 letting c1 = 1 and c2 = 0 we get x = e2t, y = e2t. Eliminating the parameter we find y = x,
x > 0. When c1 = −1 and c2 = 0 we find y = x, x < 0.
In Problems 33-46 the form of the answer will vary according to the choice of eigenvector. For example, in
Problem 33, if K1 is chosen to be
(
1
2− i
)
the solution has the form
X = c1
(
cos t
2 cos t + sin t
)
e4t + c2
(
sin t
2 sin t− cos t
)
e4t.
33. We have det(A− λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
K1 =
(
2 + i
5
)
so that
X1 =
(
2 + i
5
)
e(4+i)t =
(
2 cos t− sin t
5 cos t
)
e4t + i
(
cost + 2 sin t
5 sin t
)
e4t.
Then
X = c1
(
2 cos t− sin t
5 cos t
)
e4t + c2
(
cos t + 2 sin t
5 sin t
)
e4t.
34. We have det(A− λI) = λ2 + 1 = 0. For λ1 = i we obtain
K1 =
(−1− i
2
)
so that
X1 =
(−1− i
2
)
eit =
(
sin t− cos t
2 cos t
)
+ i
(− cos t− sin t
2 sin t
)
.
Then
X = c1
(
sin t− cos t
2 cos t
)
+ c2
(− cos t− sin t
2 sin t
)
.
35. We have det(A− λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
K1 =
(−1− i
2
)
so that
X1 =
(−1− i
2
)
e(4+i)t =
(
sin t− cos t
2 cos t
)
e4t + i
(− sin t− cos t
2 sin t
)
e4t.
563
10.2 Homogeneous Linear Systems
Then
X = c1
(
sin t− cos t
2 cos t
)
e4t + c2
(− sin t− cos t
2 sin t
)
e4t.
36. We have det(A− λI) = λ2 − 10λ + 34 = 0. For λ1 = 5 + 3i we obtain
K1 =
(
1− 3i
2
)
so that
X1 =
(
1− 3i
2
)
e(5+3i)t =
(
cos 3t + 3 sin 3t
2 cos 3t
)
e5t + i
(
sin 3t− 3 cos 3t
2 sin 3t
)
e5t.
Then
X = c1
(
cos 3t + 3 sin 3t
2 cos 3t
)
e5t + c2
(
sin 3t− 3 cos 3t
2 sin 3t
)
e5t.
37. We have det(A− λI) = λ2 + 9 = 0. For λ1 = 3i we obtain
K1 =
(
4 + 3i
5
)
so that
X1 =
(
4 + 3i
5
)
e3it =
(
4 cos 3t− 3 sin 3t
5 cos 3t
)
+ i
(
4 sin 3t + 3 cos 3t
5 sin 3t
)
.
Then
X = c1
(
4 cos 3t− 3 sin 3t
5 cos 3t
)
+ c2
(
4 sin 3t + 3 cos 3t
5 sin 3t
)
.
38. We have det(A− λI) = λ2 + 2λ + 5 = 0. For λ1 = −1 + 2i we obtain
K1 =
(
2 + 2i
1
)
so that
X1 =
(
2 + 2i
1
)
e(−1+2i)t =
(
2 cos 2t− 2 sin 2t
cos 2t
)
e−t + i
(
2 cos 2t + 2 sin 2t
sin 2t
)
e−t.
Then
X = c1
(
2 cos 2t− 2 sin 2t
cos 2t
)
e−t + c2
(
2 cos 2t + 2 sin 2t
sin 2t
)
e−t.
39. We have det(A− λI) = −λ (λ2 + 1) = 0. For λ1 = 0 we obtain
K1 =
 10
0
 .
For λ2 = i we obtain
K2 =
−ii
1

so that
X2 =
−ii
1
 eit =
 sin t− sin t
cos t
 + i
− cos tcos t
sin t
 .
564
10.2 Homogeneous Linear Systems
Then
X = c1
 10
0
 + c2
 sin t− sin t
cos t
 + c3
− cos tcos t
sin t
 .
40. We have det(A− λI) = −(λ + 3)(λ2 − 2λ + 5) = 0. For λ1 = −3 we obtain
K1 =
 0−2
1
 .
For λ2 = 1 + 2i we obtain
K2 =
−2− i−3i
2

so that
X2 =
−2 cos 2t + sin 2t3 sin 2t
2 cos 2t
 et + i
− cos 2t− 2 sin 2t−3 cos 2t
2 sin 2t
 et.
Then
X = c1
 0−2
1
 e−3t + c2
−2 cos 2t + sin 2t3 sin 2t
2 cos 2t
 et + c3
− cos 2t− 2 sin 2t−3 cos 2t
2 sin 2t
 et.
41. We have det(A− λI) = (1− λ)(λ2 − 2λ + 2) = 0. For λ1 = 1 we obtain
K1 =
 02
1
 .
For λ2 = 1 + i we obtain
K2 =
 1i
i

so that
X2 =
 1i
i
 e(1+i)t =
 cos t− sin t
− sin t
 et + i
 sin tcos t
cos t
 et.
Then
X = c1
 02
1
 et + c2
 cos t− sin t
− sin t
 et + c3
 sin tcos t
cos t
 et.
42. We have det(A− λI) = −(λ− 6)(λ2 − 8λ + 20) = 0. For λ1 = 6 we obtain
K1 =
 01
0
 .
565
10.2 Homogeneous Linear Systems
For λ2 = 4 + 2i we obtain
K2 =
−i0
2

so that
X2 =
−i0
2
 e(4+2i)t =
 sin 2t0
2 cos 2t
 e4t + i
− cos 2t0
2 sin 2t
 e4t.
Then
X = c1
 01
0
 e6t + c2
 sin 2t0
2 cos 2t
 e4t + c3
− cos 2t0
2 sin 2t
 e4t.
43. We have det(A− λI) = (2− λ)(λ2 + 4λ + 13) = 0. For λ1 = 2 we obtain
K1 =
 28−5
25
 .
For λ2 = −2 + 3i we obtain
K2 =
 4 + 3i−5
0

so that
X2 =
 4 + 3i−5
0
 e(−2+3i)t =
 4 cos 3t− 3 sin 3t−5 cos 3t
0
 e−2t + i
 4 sin 3t + 3 cos 3t−5 sin 3t
0
 e−2t.
Then
X = c1
 28−5
25
 e2t + c2
 4 cos 3t− 3 sin 3t−5 cos 3t
0
 e−2t + c3
 4 sin 3t + 3 cos 3t−5 sin 3t
0
 e−2t.
44. We have det(A− λI) = −(λ + 2)(λ2 + 4) = 0. For λ1 = −2 we obtain
K1 =
 0−1
1
 .
For λ2 = 2i we obtain
K2 =
−2− 2i1
1

so that
X2 =
−2− 2i1
1
 e2it =
−2 cos 2t + 2 sin 2tcos 2t
cos 2t
 + i
−2 cos 2t− 2 sin 2tsin 2t
sin 2t
 .
566
10.2 Homogeneous Linear Systems
Then
X = c1
 0−1
1
 e−2t + c2
−2 cos 2t + 2 sin 2tcos 2t
cos 2t
 + c3
−2 cos 2t− 2 sin 2tsin 2t
sin 2t
 .
45. We have det(A− λI) = (1− λ)(λ2 + 25) = 0. For λ1 = 1 we obtain
K1 =
 25−7
6
 .
For λ2 = 5i we obtain
K2 =
 1 + 5i1
1

so that
X2 =
 1 + 5i1
1
 e5it =
 cos 5t− 5 sin 5tcos 5t
cos 5t
 + i
 sin 5t + 5 cos 5tsin 5t
sin 5t
 .
Then
X = c1
 25−7
6
 et + c2
 cos 5t− 5 sin 5tcos 5t
cos 5t
 + c3
 sin 5t + 5 cos 5tsin 5t
sin 5t
 .
If
X(0) =
 46
−7

then c1 = c2 = −1 and c3 = 6.
46. We have det(A− λI) = λ2 − 10λ + 29 = 0. For λ1 = 5 + 2i we obtain
K1 =
(
1
1− 2i
)
so that
X1 =
(
1
1− 2i
)
e(5+2i)t =
(
cos 2t
cos 2t + 2 sin 2t
)
e5t + i
(
sin 2t
sin 2t− 2 cos 2t
)
e5t.
and
X = c1
(
cos 2t
cos 2t + 2 sin 2t
)
e5t + c3
(
sin 2t
sin 2t− 2 cos 2t
)
e5t.
If X(0) =
(−2
8
)
, then c1 = −2 and c2 = 5.
567
10.2 Homogeneous Linear Systems
47.
48. (a) From det(A− λI) = λ(λ− 2) = 0 we get λ1 = 0 and λ2 = 2. For λ1 = 0 we obtain(
1 1 0
1 1 0
)
=⇒
(
1 1 0
0 0 0
)
so that K1 =
(−1
1
)
.
For λ2 = 2 we obtain(−1 1 0
1 −1 0
)
=⇒
(−1 1 0
0 0 0
)
so that K2 =
(
1
1
)
.
Then
X = c1
(−1
1
)
+ c2
(
1
1
)
e2t.
The line y = −x is not a trajectory of the
system. Trajectories are x = −c1 + c2e2t,
y = c1 +c2e2t or y = x+2c1. This is a family
of lines perpendicular to the line y = −x. All
of the constant solutions of the system do,
however, lie on the line y = −x.
(b) From det(A− λI) = λ2 = 0 we get λ1 = 0 and
K =
(−1
1
)
.
A solution of (A− λ1I)P = K is
P =
(−1
0
)
so that
X = c1
(−1
1
)
+ c2
[(−1
1
)
t +
(−1
0
)]
.
All trajectories are parallel to y = −x, but
y = −x is not a trajectory. There are con-
stant solutions of the system, however, that
do lie on the line y = −x.
568
10.2 Homogeneous Linear Systems
49. The system of differential equations is
x′1 = 2x1 + x2
x′2 = 2x2
x′3 = 2x3
x′4 = 2x4 + x5
x′5 = 2x5.
We see immediately that x2 = c2e2t, x3 = c3e2t, and x5 = c5e2t. Then
x′1 = 2x1 + c2e
2t so x1 = c2te2t + c1e2t,
and
x′4 = 2x4 + c5e
2t so x4 = c5te2t + c4e2t.
The general solution of the system is
X =

c2te
2t + c1e2t
c2e
2t
c3e
2t
c5te
2t + c4e2t
c5e
2t

= c1

1
0
0
0
0
 e2t + c2


1
0
0
0
0
 te2t +

0
1
0
0
0
 e2t

+ c3

0
0
1
0
0
 e2t + c4

0
0
0
1
0
 e2t + c5


0
0
0
1
0
 te2t +

0
0
0
0
1
 e2t

= c1K1e2t + c2
K1te2t +

0
1
0
0
0
 e2t

+ c3K2e2t + c4K3e2t + c5
K3te2t +

0
0
0
0
1
 e2t
 .
There are three solutions of the form X = Ke2t, where K is an eigenvector, and two solutions of the form
X = Kte2t + Pe2t. See (12) in the text. From (13) and (14) in the text
(A− 2I)K1 = 0
and
569
10.2 Homogeneous Linear Systems
(A− 2I)K2 = K1.
This implies 
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0


p1
p2
p3
p4
p5
 =

1
0
0
0
0
 ,
so p2 = 1 and p5 = 0, whilep1, p3, and p4 are arbitrary. Choosing p1 = p3 = p4 = 0 we have
P =

1
0
0
0
0
 .
Therefore a solution is
X =

1
0
0
0
0
 te2t +

1
0
0
0
0
 e2t.
Repeating for K3 we find
P =

0
0
0
0
1
 ,
so another solution is
X =

0
0
0
1
0
 te2t +

0
0
0
0
1
 e2t.
50. From x = 2 cos 2t− 2 sin 2t, y = − cos 2t we find x + 2y = −2 sin 2t. Then
(x + 2y)2 = 4 sin2 2t = 4(1− cos2 2t) = 4− 4 cos2 2t = 4− 4y2
and
x2 + 4xy + 4y2 = 4− 4y2 or x2 + 4xy + 8y2 = 4.
This is a rotated conic section and, from the discriminant b2 − 4ac = 16 − 32 < 0, we see that the curve is an
ellipse.
51. Suppose the eigenvalues are α ± iβ, β > 0. In Problem 36 the eigenvalues are 5 ± 3i, in Problem 37 they are
±3i, and in Problem 38 they are −1± 2i. From Problem 47 we deduce that the phase portrait will consist of a
family of closed curves when α = 0 and spirals when α �= 0. The origin will be a repellor when α > 0, and an
attractor when α < 0.
570
10.3 Solution by Diagonalization
52. (a) The given system can be written as
x′′1 = −
k1 + k2
m1
x1 +
k2
m1
x2, x
′′
2 =
k2
m2
x1 − k2
m2
x2.
In terms of matrices this is X′′ = AX where
X =
(
x1
x2
)
and A =
−
k1 + k2
m1
k2
m1
k2
m2
− k2
m2
 .
(b) If X = Keωt then X′′ = ω2Keωt and AX = AKeωt so that X′′ = AX becomes ω2Keωt = AKeωt or
(A− ω2I)K = 0. Now let ω2 = λ.
(c) When m1 = 1, m2 = 1, k1 = 3, and k2 = 2 we obtain A =
(−5 2
2 −2
)
. The eigenvalues and corresponding
eigenvectors of A are λ1 = −1, λ2 = −6, K1 =
(
1
2
)
, K2 =
(−2
1
)
. Since ω1 = i, ω2 = −i, ω3 =
√
6 i, and
ω4 = −
√
6 i a solution is
X = c1
(
1
2
)
eit + c2
(
1
2
)
e−it + c3
(−2
1
)
e
√
6 it + c4
(−2
1
)
e−
√
6 it.
(d) Using eit = cos t + i sin t and e
√
6 it = cos
√
6 t + i sin
√
6 t the preceding solution can be rewritten as
X = c1
(
1
2
)
(cos t + i sin t) + c2
(
1
2
)
(cos t− i sin t)
+ c3
(−2
1
)
(cos
√
6 t + i sin
√
6 t) + c4
(−2
1
)
(cos
√
6 t + i sin
√
6 t)
= (c1 + c2)
(
1
2
)
cos t + i(c1 − c2)
(
1
2
)
sin t
+ (c3 + c4)
(−2
1
)
cos
√
6 t + i(c3 − c4)
(−2
1
)
sin
√
6 t
= b1
(
1
2
)
cos t + b2
(
1
2
)
sin t + b3
(−2
1
)
cos
√
6 t + b4
(−2
1
)
sin
√
6 t
where b1 = c1 + c2, b2 = i(c1 − c2), b3 = c3 + c4, and b4 = i(c3 − c4).
EXERCISES 10.3
Solution by Diagonalization
1. λ1 = 7, λ2 = −4, K1 =
(
3
1
)
, K2 =
(−2
3
)
, P =
(
3 −2
1 3
)
;
X = PY =
(
3 −2
1 3
) (
c1e
7t
c2e
−4t
)
=
(
3c1e7t − 2c2e−4t
c1e
7t + 3c2e−4t
)
= c1
(
3
1
)
e7t + c2
(−2
3
)
e−4t
571
10.3 Solution by Diagonalization
2. λ1 = 0, λ2 = 1, K1 =
(
1
−1
)
, K2 =
(
1
1
)
, P =
(
1 1
−1 1
)
;
X = PY =
(
1 1
−1 1
) (
c1
c2e
t
)
=
(
c1 + c2et
−c1 + c2et
)
= c1
(
1
−1
)
+ c2
(
1
1
)
et
3. λ1 = 12 , λ2 =
3
2 , K1 =
(
1
−2
)
, K2 =
(
1
2
)
, P =
(
1 1
−2 2
)
;
X = PY =
(
1 1
−2 2
) (
c1e
t/2
c2e
3t/2
)
=
(
c1e
t/2 + c2e3t/2
−2c1et/2 + 2c2e3t/2
)
= c1
(
1
−2
)
et/2 + c2
(
1
2
)
e3t/2
4. λ1 = −
√
2 , λ2 =
√
2 , K1 =
( −1
1 +
√
2
)
, K2 =
( −1
1−√2
)
, P =
( −1 −1
1 +
√
2 1−√2
)
;
X = PY =
( −1 −1
1 +
√
2 1−√2
) (
c1e
−√2 t
c2e
√
2 t
)
=
( −c1e−√2 t − c2e√2 t
(1 +
√
2 )c1e−
√
2 t + (1−√2 )c2e
√
2 t
)
= c1
( −1
1 +
√
2
)
e−
√
2 t + c2
( −1
1−√2
)
e
√
2 t
5. λ1 = −4, λ2 = 2, λ3 = 6, K1 =
−11
0
, K2 =
 11
1
, K3 =
 00
1
, P =
−1 1 01 1 0
0 1 1
;
X = PY =
−1 1 01 1 0
0 1 1

 c1e
−4t
c2e
2t
c3e
6t
 =
−c1e
−4t + c2e2t
c1e
−4t + c2e2t
c2e
2t + c3e6t
 = c1
−11
0
 e−4t + c2
 11
0
 e2t + c3
 00
1
 e6t
6. λ1 = −1, λ2 = 1, λ3 = 4, K1 =
−10
1
, K2 =
 1−2
1
, K3 =
 11
1
, P =
−1 1 10 −2 1
1 1 1
;
X = PY =
−1 1 10 −2 1
1 1 1

 c1e
−t
c2e
t
c3e
4t
 =
 −c1 + c2e
t + c3e4t
−2c2et + c3e4t
c1e
−t + c2et + c3e4t
 = c1
−10
1
 e−t+c2
 1−2
1
 et+c3
 11
1
 e4t
7. λ1 = −1, λ2 = 2, λ3 = 2, K1 =
 11
1
, K2 =
−11
0
, K3 =
−10
1
, P =
 1 −1 −11 1 0
1 0 1
;
X = PY =
 1 −1 −11 1 0
1 0 1

 c1e
−t
c2e
2t
c3e
2t
 =
 c1e
−t − c2e2t − c3e2t
c1e
−t + c2e2t
c1e
−t + c3e2t

= c1
 11
1
 e−t + c2
−11
0
 e2t + c3
−10
1
 e2t
572
10.3 Solution by Diagonalization
8. λ1 = 0, λ2 = 0, λ3 = 0, λ4 = 4, K1 =

−1
1
0
0
, K2 =

−1
0
1
0
, K3 =

−1
0
0
1
, K4 =

1
1
1
1
,
P =

−1 −1 −1 1
1 0 0 1
0 1 0 1
0 0 1 1
;
X = PY =

−1 −1 −1 1
1 0 0 1
0 1 0 1
0 0 1 1


c1
c2
c3
c4e
4t
 =

−c1 − c2 − c3 + c4e4t
c1 + c4e4t
c2 + c4e4t
c3 + c4e4t

= c1

−1
1
0
0
 + c2

−1
0
1
0
 + c3

−1
0
0
1
 + c4

1
1
1
1
 e4t
9. λ1 = 1, λ2 = 2, λ3 = 3, K1 =
 11
1
, K2 =
 22
3
, K3 =
 34
5
, P =
 1 2 31 2 4
1 3 5
;
X = PY =
 1 2 31 2 4
1 3 5

 c1e
t
c2e
2t
c3e
3t
 =
 c1e
t + 2c2e2t + 3c3e3t
c1e
t + 2c2e2t + 4c3e3t
c1e
t + 3c2e2t + 5c3e3t
 = c1
 11
1
 et + c2
 22
3
 e2t + c3
 34
5
 e3t
10. λ1 = 0, λ2 = −2
√
2 , λ3 = 2
√
2 , K1 =
 10
−1
, K2 =
 1−√2
1
, K3 =
 1√2
1
, P =
 1 1 10 −√2 √2
−1 1 1
;
X = PY =
 1 1 10 −√2 √2
−1 1 1

 c1c2e−2√2 t
c3e
2
√
2 t
 =
 c1 + c2e
−2√2 t + c3e2
√
2 t
−√2 c2e−2
√
2 t +
√
2 c3e2
√
2 t
−c1 + c2e−2
√
2 t + c3e2
√
2 t

= c1
 10
−1
 + c2
 1−√2
1
 e−2√2 t + c3
 1√2
1
 e2√2 t
11. (a) Since M =
(
m1 0
0 m2
)
is a diagonal matrix with nonzero diagonal entries, it has an inverse. Writing the
system in the form
m1x
′′
1 + (k1 + k2)x1 − k2x2 = 0
m2x
′′
2 − k2x1 + k2x2 = 0
we see that K =
(
k1 + k2 −k2
−k2 k2
)
.
(b) Since M has an inverse, MX′′ + KX = 0 can be written as X′′ + M−1KX = 0 or X′′ + BX = 0 where
573
10.3 Solution by Diagonalization
B = M−1K =
 1m1 0
0
1
m2
 ( k1 + k2 −k2
−k2 k2
)
=

k1 + k2
m1
− k2
m1
− k2
m2
k2
m2
 .
(c) With m1 = 1,m2 = 1, k1 = 3, and k2 = 2 we have B =
(
5 −2
−2 2
)
. The eigenvalues of B are λ1 = 1 and
λ2 = 6 with corresponding eigenvectors
(
1
2
)
and
(−2
1
)
. Letting X = PY the system can be written
PY′′ + BPY = 0 or Y′′ + P−1BPY = 0 where
(
1 −2
2 1
)
and P−1BP =
(
1 0
0 6
)
. The system is then
Y′′ +
(
1 0
0 6
)
Y = 0, which is uncoupled and equivalent to y′′1 + y1 = 0 and y
′′
2 + 6y2 = 0. The solutions
are y1 = c1 cos t + c2 sin t and y2 = c3 cos
√
6 t + c4 sin
√
6 t.
(d) From
X = PY =
(
1 −2
2 1
) (
y1
y2
)
=
(
y1 − 2y2
2y1 + y2
)
we have
x1 = c1 cos t + c2 sin t− 2c3 cos
√
6 t− 2c4 sin
√
6 t
x1 = 2c1 cos t + 2c2 sin t + c3cos
√
6 t + c4 sin
√
6 t
which is the same as
X = c1
(
1
2
)
cos t + c2
(
1
2
)
sin t + c3
(−2
1
)
cos
√
6 t + c4
(−2
1
)
sin
√
6 t.
EXERCISES 10.4
Nonhomogeneous Linear Systems
1. Solving
det(A− λI) =
∣∣∣∣ 2− λ 3−1 −2− λ
∣∣∣∣ = λ2 − 1 = (λ− 1)(λ + 1) = 0
we obtain eigenvalues λ1 = −1 and λ2 = 1. Corresponding eigenvectors are
K1 =
(−1
1
)
and K2 =
(−3
1
)
.
Thus
Xc = c1
(−1
1
)
e−t + c2
(−3
1
)
et.
Substituting
Xp =
(
a1
b1
)
574
10.4 Nonhomogeneous Linear Systems
into the system yields
2a1 + 3b1 = 7
−a1 − 2b1 = −5,
from which we obtain a1 = −1 and b1 = 3. Then
X(t) = c1
(−1
1
)
e−t + c2
(−3
1
)
et +
(−1
3
)
.
2. Solving
det(A− λI) =
∣∣∣∣ 5− λ 9−1 11− λ
∣∣∣∣ = λ2 − 16λ + 64 = (λ− 8)2 = 0
we obtain the eigenvalue λ = 8. A corresponding eigenvector is
K =
(
3
1
)
.
Solving (A− 8I)P = K we obtain
P =
(
2
1
)
.
Thus
Xc = c1
(
3
1
)
e8t + c2
[(
3
1
)
te8t +
(
2
1
)
e8t
]
.
Substituting
Xp =
(
a1
b1
)
into the system yields
5a1 + 9b1 = −2
−a1 + 11b1 = −6,
from which we obtain a1 = 1/2 and b1 = −1/2. Then
X(t) = c1
(
3
1
)
e8t + c2
[(
3
1
)
te8t +
(
2
1
)
e8t
]
+
( 1
2
− 12
)
.
3. Solving
det(A− λI) =
∣∣∣∣ 1− λ 33 1− λ
∣∣∣∣ = λ2 − 2λ− 8 = (λ− 4)(λ + 2) = 0
we obtain eigenvalues λ1 = −2 and λ2 = 4. Corresponding eigenvectors are
K1 =
(
1
−1
)
and K2 =
(
1
1
)
.
Thus
Xc = c1
(
1
−1
)
e−2t + c2
(
1
1
)
e4t.
Substituting
Xp =
(
a3
b3
)
t2 +
(
a2
b2
)
t +
(
a1
b1
)
into the system yields
a3 + 3b3 = 2
3a3 + b3 = 0
a2 + 3b2 = 2a3
3a2 + b2 + 1 = 2b3
a1 + 3b1 = a2
3a1 + b1 + 5 = b2
575
10.4 Nonhomogeneous Linear Systems
from which we obtain a3 = −1/4, b3 = 3/4, a2 = 1/4, b2 = −1/4, a1 = −2, and b1 = 3/4. Then
X(t) = c1
(
1
−1
)
e−2t + c2
(
1
1
)
e4t +
(− 14
3
4
)
t2 +
( 1
4
− 14
)
t +
(−2
3
4
)
.
4. Solving
det(A− λI) =
∣∣∣∣ 1− λ −44 1− λ
∣∣∣∣ = λ2 − 2λ + 17 = 0
we obtain eigenvalues λ1 = 1 + 4i and λ2 = 1− 4i. Corresponding eigenvectors are
K1 =
(
i
1
)
and K2 =
(−i
1
)
.
Thus
Xc = c1
[(
0
1
)
cos 4t +
(−1
0
)
sin 4t
]
et + c2
[(−1
0
)
cos 4t−
(
0
1
)
sin 4t
]
et
= c1
(− sin 4t
cos 4t
)
et + c2
(− cos 4t
− sin 4t
)
et.
Substituting
Xp =
(
a3
b3
)
t +
(
a2
b2
)
+
(
a1
b1
)
e6t
into the system yields
a3 − 4b3 = −4
4a3 + b3 = 1
a2 − 4b2 = a3
4a2 + b2 = b3
−5a1 − 4b1 = −9
4a1 − 5b1 = −1
from which we obtain a3 = 0, b3 = 1, a2 = 4/17, b2 = 1/17, a1 = 1, and b1 = 1. Then
X(t) = c1
(− sin 4t
cos 4t
)
et + c2
(− cos 4t
− sin 4t
)
et +
(
0
1
)
t +
( 4
17
1
17
)
+
(
1
1
)
e6t.
5. Solving
det(A− λI) =
∣∣∣∣ 4− λ 139 6− λ
∣∣∣∣ = λ2 − 10λ + 21 = (λ− 3)(λ− 7) = 0
we obtain the eigenvalues λ1 = 3 and λ2 = 7. Corresponding eigenvectors are
K1 =
(
1
−3
)
and K2 =
(
1
9
)
.
Thus
Xc = c1
(
1
−3
)
e3t + c2
(
1
9
)
e7t.
Substituting
Xp =
(
a1
b1
)
et
into the system yields
3a1 +
1
3
b1 = 3
9a1 + 5b1 = −10
from which we obtain a1 = 55/36 and b1 = −19/4. Then
X(t) = c1
(
1
−3
)
e3t + c2
(
1
9
)
e7t +
( 55
36
− 194
)
et.
576
10.4 Nonhomogeneous Linear Systems
6. Solving
det(A− λI) =
∣∣∣∣−1− λ 5−1 1− λ
∣∣∣∣ = λ2 + 4 = 0
we obtain the eigenvalues λ1 = 2i and λ2 = −2i. Corresponding eigenvectors are
K1 =
(
5
1 + 2i
)
and K2 =
(
5
1− 2i
)
.
Thus
Xc = c1
(
5 cos 2t
cos 2t− 2 sin 2t
)
+ c2
(
5 sin 2t
2 cos 2t + sin 2t
)
.
Substituting
Xp =
(
a2
b2
)
cos t +
(
a1
b1
)
sin t
into the system yields
−a2 + 5b2 − a1 = 0
−a2 + b2 − b1 − 2 = 0
−a1 + 5b1 + a2 + 1 = 0
−a1 + b1 + b2 = 0
from which we obtain a2 = −3, b2 = −2/3, a1 = −1/3, and b1 = 1/3. Then
X(t) = c1
(
5 cos 2t
cos 2t− 2 sin 2t
)
+ c2
(
5 sin 2t
2 cos 2t + sin 2t
)
+
( −3
− 23
)
cos t +
(− 13
1
3
)
sin t.
7. Solving
det(A− λI) =
∣∣∣∣∣∣∣
1− λ 1 1
0 2− λ 3
0 0 5− λ
∣∣∣∣∣∣∣ = (1− λ)(2− λ)(5− λ) = 0
we obtain the eigenvalues λ1 = 1, λ2 = 2, and λ3 = 5. Corresponding eigenvectors are
K1 =
 10
0
 , K2 =
 11
0
 and K3 =
 12
2
 .
Thus
Xc = C1
 10
0
 et + C2
 11
0
 e2t + C3
 12
2
 e5t.
Substituting
Xp =
 a1b1
c1
 e4t
into the system yields
−3a1 + b1 + c1 = −1
−2b1 + 3c1 = 1
c1 = −2
577
10.4 Nonhomogeneous Linear Systems
from which we obtain c1 = −2, b1 = −7/2, and a1 = −3/2. Then
X(t) = C1
 10
0
 et + C2
 11
0
 e2t + C3
 12
2
 e5t +
−
3
2
− 72
−2
 e4t.
8. Solving
det(A− λI) =
∣∣∣∣∣∣∣
−λ 0 5
0 5− λ 0
5 0 −λ
∣∣∣∣∣∣∣ = −(λ− 5)2(λ + 5) = 0
we obtain the eigenvalues λ1 = 5, λ2 = 5, and λ3 = −5. Corresponding eigenvectors are
K1 =
 10
0
 , K2 =
 11
1
 and K3 =
 10
−1
 .
Thus
Xc = C1
 10
1
 e5t + C2
 11
1
 e5t + C3
 10
−1
 e−5t.
Substituting
Xp =
 a1b1
c1

into the system yields
5c1 = −5
5b1 = 10
5a1 = −40
from which we obtain c1 = −1, b1 = 2, and a1 = −8. Then
X(t) = C1
 10
1
 e5t + C2
 11
1
 e5t + C3
 10
−1
 e−5t +
−82
−1
 .
9. Solving
det(A− λI) =
∣∣∣∣−1− λ −23 4− λ
∣∣∣∣ = λ2 − 3λ + 2 = (λ− 1)(λ− 2) = 0
we obtain the eigenvalues λ1 = 1 and λ2 = 2. Corresponding eigenvectors are
K1 =
(
1
−1
)
and K2 =
(−4
6
)
.
Thus
Xc = c1
(
1
−1
)
et + c2
(−4
6
)
e2t.
Substituting
Xp =
(
a1
b1
)
578
10.4 Nonhomogeneous Linear Systems
into the system yields
−a1 − 2b1 = −3
3a1 + 4b1 = −3
from which we obtain a1 = −9 and b1 = 6. Then
X(t) = c1
(
1
−1
)
et + c2
(−4
6
)
e2t +
(−9
6
)
.
Setting
X(0) =
(−4
5
)
we obtain
c1 − 4c2 − 9 = −4
−c1 + 6c2 + 6 = 5.
Then c1 = 13 and c2 = 2 so
X(t) = 13
(
1
−1
)
et + 2
(−4
6
)
e2t +
(−9
6
)
.
10. (a) Let I =
(
i2
i3
)
so that
I′ =
(−2 −2
−2 −5
)
I +
(
60
60
)
and
Ic = c1
(
2
−1
)
e−t + c2
(
1
2
)
e−6t.
If Ip =
(
a1
b1
)
then Ip =
(
30
0
)
so that
I = c1
(
2
−1
)
e−t + c2
(
1
2
)
e−6t +
(
30
0
)
.
For I(0) =
(
0
0
)
we find c1 = −12 and c2 = −6.
(b) i1(t) = i2(t) + i3(t) = −12e−t − 18e−6t + 30.
11. From
X′ =
(
3 −3
2 −2
)
X +
(
4
−1
)
we obtain
Xc = c1
(
1
1
)
+ c2
(
3
2
)
et.
Then
Φ =
(
1 3et
1 2et
)
and Φ−1 =
( −2 3
e−t −e−t
)
so that
U =
∫
Φ−1F dt =
∫ ( −11
5e−t
)
dt =
( −11t
−5e−t
)
and
Xp = ΦU =
(−11
−11
)
t +
(−15
−10
)
.
579
10.4 Nonhomogeneous Linear Systems
12. From
X′ =
(
2 −1
3 −2
)
X +
(
0
4
)
t
we obtain
Xc = c1
(
1
1
)
et + c2
(
1
3
)
e−t.
Then
Φ =
(
et e−t
et 3e−t
)
and Φ−1 =
( 3
2e
−t − 12e−t
− 12et 12et
)
so that
U =
∫
Φ−1F dt =
∫ (−2te−t
2tet
)
dt =
(
2te−t + 2e−t
2tet − 2et
)
and
Xp = ΦU =
(
4
8
)
t +
(
0
−4
)
.
13. From
X′ =
(
3 −5
3
4 −1
)
X +
(
1
−1
)
et/2
we obtain
Xc = c1
(
10
3
)
e3t/2 + c2
(
2
1
)
et/2.
Then
Φ =
(
10e3t/2 2et/2
3e3t/2 et/2
)and Φ−1 =
( 1
4e
−3t/2 − 12e−3t/2
− 34e−t/2 52e−t/2
)
so that
U =
∫
Φ−1F dt =
∫ ( 3
4e
−t
− 134
)
dt =
(− 34e−t
− 134 t
)
and
Xp = ΦU =
(− 132
− 134
)
tet/2 +
(− 152
− 94
)
et/2.
14. From
X′ =
(
2 −1
4 2
)
X +
(
sin 2t
2 cos 2t
)
we obtain
Xc = c1
(− sin 2t
2 cos 2t
)
e2t + c2
(
cos 2t
2 sin 2t
)
e2t.
Then
Φ =
(−e2t sin 2t e2t cos 2t
2e2t cos 2t 2e2t sin 2t
)
and Φ−1 =
(− 12e−2t sin 2t 14e−2t cos 2t
1
2e
−2t cos 2t 14e
−2t sin 2t
)
so that
U =
∫
Φ−1F dt =
∫ ( 1
2 cos 4t
1
2 sin 4t
)
dt =
( 1
8 sin 4t
− 18 cos 4t
)
and
Xp = ΦU =
(− 18 sin 2t cos 4t− 18 cos 2t cos 4t
1
4 cos 2t sin 4t− 14 sin 2t cos 4t
)
e2t.
580
10.4 Nonhomogeneous Linear Systems
15. From
X′ =
(
0 2
−1 3
)
X +
(
1
−1
)
et
we obtain
Xc = c1
(
2
1
)
et + c2
(
1
1
)
e2t.
Then
Φ =
(
2et e2t
et e2t
)
and Φ−1 =
(
e−t −e−t
−e−2t 2e−2t
)
so that
U =
∫
Φ−1F dt =
∫ (
2
−3e−t
)
dt =
(
2t
3e−t
)
and
Xp = ΦU =
(
4
2
)
tet +
(
3
3
)
et.
16. From
X′ =
(
0 2
−1 3
)
X +
(
2
e−3t
)
we obtain
Xc = c1
(
2
1
)
et + c2
(
1
1
)
e2t.
Then
Φ =
(
2et e2t
et e2t
)
and Φ−1 =
(
e−t −e−t
−e−2t 2e−2t
)
so that
U =
∫
Φ−1F dt =
∫ (
2e−t − e−4t
−2e−2t + 2e−5t
)
dt =
(−2e−t + 14e−4t
e−2t − 25e−5t
)
and
Xp = ΦU =
(
1
10e
−3t − 3
− 320e−3t − 1
)
.
17. From
X′ =
(
1 8
1 −1
)
X +
(
12
12
)
t
we obtain
Xc = c1
(
4
1
)
e3t + c2
(−2
1
)
e−3t.
Then
Φ =
(
4e3t −2e−3t
e3t e−3t
)
and Φ−1 =
(
1
6e
−3t 1
3e
−3t
− 16e3t 23e3t
)
so that
U =
∫
Φ−1F dt =
∫ (
6te−3t
6te3t
)
dt =
(−2te−3t − 23e−3t
2te3t − 23e3t
)
and
Xp = ΦU =
(−12
0
)
t +
(− 43
− 43
)
.
581
10.4 Nonhomogeneous Linear Systems
18. From
X′ =
(
1 8
1 −1
)
X +
(
e−t
tet
)
we obtain
Xc = c1
(
4
1
)
e3t + c2
(−2
1
)
e−3t.
Then
Φ =
(
4e3t −2e3t
e3t e−3t
)
and Φ−1 =
(
1
6e
−3t 1
3e
−3t
− 16e3t 23e3t
)
so that
U =
∫
Φ−1F dt =
∫ ( 1
6e
−4t + 13 te
−2t
− 16e2t + 23 te4t
)
dt =
(− 124e−4t − 16 te−2t − 112e−2t
− 112e2t + 16 te4t − 124e4t
)
and
Xp = ΦU =
( −tet − 14et
− 18e−t − 18et
)
.
19. From
X′ =
(
3 2
−2 −1
)
X +
(
2
1
)
e−t
we obtain
Xc = c1
(
1
−1
)
et + c2
[(
1
−1
)
tet +
(
0
1
2
)
et
]
.
Then
Φ =
(
et tet
−et 12et − tet
)
and Φ−1 =
(
e−t − 2te−t −2te−t
2e−t 2e−t
)
so that
U =
∫
Φ−1F dt =
∫ (
2e−2t − 6te−2t
6e−2t
)
dt =
( 1
2e
−2t + 3te−2t
−3e−2t
)
and
Xp = ΦU =
( 1
2
−2
)
e−t.
20. From
X′ =
(
3 2
−2 −1
)
X +
(
1
1
)
we obtain
Xc = c1
(
1
−1
)
et + c2
[(
1
−1
)
tet +
(
0
1
2
)
et
]
.
Then
Φ =
(
et tet
−et 12et − tet
)
and Φ−1 =
(
e−t − 2te−t −2te−t
2e−t 2e−t
)
so that
U =
∫
Φ−1F dt =
∫ (
e−t − 4te−t
2e−t
)
dt =
(
3e−t + 4te−t
−2e−t
)
and
Xp = ΦU =
(
3
−5
)
.
582
10.4 Nonhomogeneous Linear Systems
21. From
X′ =
(
0 −1
1 0
)
X +
(
sec t
0
)
we obtain
Xc = c1
(
cos t
sin t
)
+ c2
(
sin t
− cos t
)
.
Then
Φ =
(
cos t sin t
sin t − cos t
)
and Φ−1 =
(
cos t sin t
sin t − cos t
)
so that
U =
∫
Φ−1F dt =
∫ (
1
tan t
)
dt =
(
t
− ln | cos t|
)
and
Xp = ΦU =
(
t cos t− sin t ln | cos t|
t sin t + cos t ln | cos t|
)
.
22. From
X′ =
(
1 −1
1 1
)
X +
(
3
3
)
et
we obtain
Xc = c1
(− sin t
cos t
)
et + c2
(
cos t
sin t
)
et.
Then
Φ =
(− sin t cos t
cos t sin t
)
et and Φ−1 =
(− sin t cos t
cos t sin t
)
e−t
so that
U =
∫
Φ−1F dt =
∫ (−3 sin t + 3 cos t
3 cos t + 3 sin t
)
dt =
(
3 cos t + 3 sin t
3 sin t− 3 cos t
)
and
Xp = ΦU =
(−3
3
)
et.
23. From
X′ =
(
1 −1
1 1
)
X +
(
cos t
sin t
)
et
we obtain
Xc = c1
(− sin t
cos t
)
et + c2
(
cos t
sin t
)
et.
Then
Φ =
(− sin t cos t
cos t sin t
)
et and Φ−1 =
(− sin t cos t
cos t sin t
)
e−t
so that
U =
∫
Φ−1F dt =
∫ (
0
1
)
dt =
(
0
t
)
and
Xp = ΦU =
(
cos t
sin t
)
tet.
583
10.4 Nonhomogeneous Linear Systems
24. From
X′ =
(
2 −2
8 −6
)
X +
(
1
3
)
1
t
e−2t
we obtain
Xc = c1
(
1
2
)
e−2t + c2
[(
1
2
)
te−2t +
( 1
2
1
2
)
e−2t
]
.
Then
Φ =
(
1 t + 12
2 2t + 12
)
e−2t and Φ−1 =
(−4t− 1 2t + 1
4 −2
)
e2t
so that
U =
∫
Φ−1F dt =
∫ (
2 + 2/t
−2/t
)
dt =
(
2t + 2 ln t
−2 ln t
)
and
Xp = ΦU =
(
2t + ln t− 2t ln t
4t + 3 ln t− 4t ln t
)
e−2t.
25. From
X′ =
(
0 1
−1 0
)
X +
(
0
sec t tan t
)
we obtain
Xc = c1
(
cos t
− sin t
)
+ c2
(
sin t
cos t
)
.
Then
Φ =
(
cos t sin t
− sin t cos t
)
t and Φ−1 =
(
cos t − sin t
sin t cos t
)
so that
U =
∫
Φ−1F dt =
∫ (− tan2 t
tan t
)
dt =
(
t− tan t
− ln | cos t|
)
and
Xp = ΦU =
(
cos t
− sin t
)
t +
( − sin t
sin t tan t
)
−
(
sin t
cos t
)
ln | cos t|.
26. From
X′ =
(
0 1
−1 0
)
X +
(
1
cot t
)
we obtain
Xc = c1
(
cos t
− sin t
)
+ c2
(
sin t
cos t
)
.
Then
Φ =
(
cos t sin t
− sin t cos t
)
and Φ−1 =
(
cos t − sin t
sin t cos t
)
so that
U =
∫
Φ−1F dt =
∫ (
0
csc t
)
dt =
(
0
ln | csc t− cot t|
)
and
Xp = ΦU =
(
sin t ln | csc t− cot t|
cos t ln | csc t− cot t|
)
.
584
10.4 Nonhomogeneous Linear Systems
27. From
X′ =
(
1 2
− 12 1
)
X +
(
csc t
sec t
)
et
we obtain
Xc = c1
(
2 sin t
cos t
)
et + c2
(
2 cos t
− sin t
)
et.
Then
Φ =
(
2 sin t 2 cos t
cos t − sin t
)
et and Φ−1 =
(
1
2 sin t cos t
1
2 cos t − sin t
)
e−t
so that
U =
∫
Φ−1F dt =
∫ ( 3
2
1
2 cot t− tan t
)
dt =
(
3
2 t
1
2 ln | sin t|+ ln | cos t|
)
and
Xp = ΦU =
(
3 sin t
3
2 cos t
)
tet +
(
cos t
− 12 sin t
)
et ln | sin t|+
(
2 cos t
− sin t
)
et ln | cos t|.
28. From
X′ =
(
1 −2
1 −1
)
X +
(
tan t
1
)
we obtain
Xc = c1
(
cos t− sin t
cos t
)
+ c2
(
cos t + sin t
sin t
)
.
Then
Φ =
(
cos t− sin t cos t + sin t
cos t sin t
)
and Φ−1 =
(− sin t cos t + sin t
cos t sin t− cos t
)
so that
U =
∫
Φ−1F dt =
∫ (
2 cos t + sin t− sec t
2 sin t− cos t
)
dt =
(
2 sin t− cos t− ln | sec t + tan t|
−2 cos t− sin t
)
and
Xp = ΦU =
(
3 sin t cos t− cos2 t− 2 sin2 t + (sin t− cos t) ln | sec t + tan t|
sin2 t− cos2 t− cos t(ln | sec t + tan t|)
)
.
29. From
X′ =
 1 1 01 1 0
0 0 3
X +
 e
t
e2t
te3t

we obtain
Xc = c1
 1−1
0
 + c2
 11
0
 e2t + c3
 00
1
 e3t.
Then
Φ =
 1 e
2t 0
−1 e2t 0
0 0 e3t
 and Φ−1 =

1
2 − 12 0
1
2e
−2t 1
2e
−2t 0
0 0 e−3t

so that
U =
∫
Φ−1F dt =
∫ 
1
2e
t − 12e2t
1
2e
−t + 12
t
 dt =

1
2e
t − 14e2t
− 12e−t + 12 t
1
2 t
2

585
10.4 Nonhomogeneous Linear Systems
and
Xp = ΦU =

− 14e2t + 12 te2t
−et + 14e2t + 12 te2t
1
2 t
2e3t
 .
30. From
X′ =
 3 −1 −11 1 −1
1 −1 1
X +
 0t
2et
we obtain
Xc = c1
 11
1
 et + c2
 11
0
 e2t + c3
 10
1
 e2t.
Then
Φ =
 e
t e2t e2t
et e2t 0
et 0 e2t
 and Φ−1 =
−e
−t e−t e−t
e−2t 0 −e−2t
e−2t −e−2t 0

so that
U =
∫
Φ−1F dt =
∫  te
−t + 2
−2e−t
−te−2t
 dt =
−te
−t − e−t + 2t
2e−t
1
2 te
−2t + 14e
−2t

and
Xp = ΦU =
−
1
2
−1
− 12
 t +
−
3
4
−1
− 34
 +
 22
0
 et +
 22
2
 tet.
31. From
X′ =
(
3 −1
−1 3
)
X +
(
4e2t
4e4t
)
we obtain
Φ =
(−e4t e2t
e4t e2t
)
, Φ−1 =
(− 12e−4t 12e−4t
1
2e
−2t 1
2e
−2t
)
,
and
X = ΦΦ−1(0)X(0) + Φ
∫ t
0
Φ−1F ds = Φ ·
(
0
1
)
+ Φ ·
(
e−2t + 2t− 1
e2t + 2t− 1
)
=
(
2
2
)
te2t +
(−1
1
)
e2t +
(−2
2
)
te4t +
(
2
0
)
e4t.
32. From
X′ =
(
1 −1
1 −1
)
X +
(
1/t
1/t
)
we obtain
Φ =
(
1 1 + t
1 t
)
, Φ−1 =
(−t 1 + t
1 −1
)
,
and
X = ΦΦ−1(1)X(1) + Φ
∫ t
1
Φ−1F ds = Φ ·
(−4
3
)
+ Φ ·
(
ln t
0
)
=
(
3
3
)
t−
(
1
4
)
+
(
1
1
)
ln t.
586
10.4 Nonhomogeneous Linear Systems
33. Let I =
(
i1
i2
)
so that
I′ =
(−11 3
3 −3
)
I +
(
100 sin t
0
)
and
Ic = c1
(
1
3
)
e−2t + c2
(
3
−1
)
e−12t.
Then
Φ =
(
e−2t 3e−12t
3e−2t −e−12t
)
, Φ−1 =
(
1
10e
2t 3
10e
2t
3
10e
12t − 110e12t
)
,
U =
∫
Φ−1F dt =
∫ (
10e2t sin t
30e12t sin t
)
dt =
(
2e2t(2 sin t− cos t)
6
29e
12t(12 sin t− cos t)
)
,
and
Ip = ΦU =
(
332
29 sin t− 7629 cos t
276
29 sin t− 16829 cos t
)
so that
I = c1
(
1
3
)
e−2t + c2
(
3
−1
)
e−12t + Ip.
If I(0) =
(
0
0
)
then c1 = 2 and c2 = 629 .
34. (a) The eigenvalues are 0, 1, 3, and 4, with corresponding eigenvectors
−6
−4
1
2
 ,

2
1
0
0
 ,

3
1
2
1
 , and

−1
1
0
0
 .
(b) Φ =

−6 2et 3e3t −e4t
−4 et e3t e4t
1 0 2e3t 0
2 0 e3t 0
, Φ−1 =

0 0 − 13 23
1
3 e
−t 1
3 e
−t −2e−t 83 e−t
0 0 23 e
−3t − 13 e−3t
− 13 e−4t 23 e−4t 0 13 e−4t

(c) Φ−1(t)F(t) =

2
3 − 13 e2t
1
3 e
−2t + 83 e
−t − 2et + 13 t
− 13 e−3t + 23e−t
2
3 e
−5t + 13 e
−4t − 13 te−3t
,
∫
Φ−1(t)F(t)dt =

− 16 e2t + 23 t
− 16 e−2t − 83 e−t − 2et + 16 t2
1
9 e
−3t − 23 e−t
− 215 e−5t − 112 e−4t + 127 e−3t + 19 te−3t
,
Xp(t) = Φ(t)
∫
Φ−1(t)F(t)dt =

−5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912
−2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536
− 32 e2t + 23 t + 29
−e2t + 43 t− 19
,
587
10.4 Nonhomogeneous Linear Systems
Xc(t) = Φ(t)C =

−6c1 + 2c2et + 3c3e3t − c4e4t
−4c1 + c2et + c3e3t + c4e4t
c1 + 2c3e3t
2c1 + c3e3t
,
X(t) = Φ(t)C + Φ(t)
∫
Φ−1(t)F(t)dt
=

−6c1 + 2c2et + 3c3e3t − c4e4t
−4c1 + c2et + c3e3t + c4e4t
c1 + 2c3e3t
2c1 + c3e3t
 +

−5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912
−2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536
− 32 e2t + 23 t + 29
−e2t + 43 t− 19

(d) X(t) = c1

−6
−4
1
2
 + c2

2
1
0
0
 et + c3

3
1
2
1
 e3t + c4

−1
1
0
0
 e4t
+

−5e2t − 15 e−t − 127 et − 19 tet + 13 t2et − 4t− 5912
−2e2t − 310 e−t + 127et + 19 tet + 16 t2et − 83 t− 9536
− 32 e2t + 23 t + 29
−e2t + 43 t− 19

35. λ1 = −1, λ2 = −2, K1 =
(
1
3
)
, K2 =
(
2
7
)
, P =
(
1 2
3 7
)
, P−1 =
(
7 −2
−3 1
)
, P−1F =
(
34
−14
)
;
Y′ =
(−1 0
0 −2
)
Y +
(
34
−14
)
y1 = 34 + c1e−t, y2 = −7 + c2e−2t
X = PY =
(
1 2
3 7
) (
34 + c1e−t
−7 + c2e−2t
)
=
(
20 + c1e−t + 2c2e−2t
53 + 3c1e−t + 7c2e−2t
)
= c1
(
1
3
)
e−t + c2
(
2
7
)
e−2t +
(
20
53
)
36. λ1 = −1, λ2 = 4, K1 =
(
3
−2
)
, K2 =
(
1
1
)
, P =
(
3 1
−2 1
)
, P−1 =
1
5
(
1 −1
2 3
)
, P−1F =
(
0
et
)
;
Y′ =
(−1 0
0 4
)
Y +
(
0
et
)
y1 = c1e−t, y2 = −13e
t + c2e4t
X = PY =
(
3 1
−2 1
) (
c1e
−t
− 13et + c2e4t
)
=
(− 13et + 3c1e−t + c2e4t
− 13et − 2c1e−t + c2e4t
)
= c1
(
3
−2
)
e−t + c2
(
1
1
)
e4t − 1
3
(
1
1
)
et
37. λ1 = 0, λ2 = 10, K1 =
(
1
−1
)
, K2 =
(
1
1
)
, P =
(
1 1
−1 1
)
, P−1 =
1
2
(
1 −1
1 1
)
, P−1F =
(
t− 4
t + 4
)
;
Y′ =
(
0 0
0 10
)
Y +
(
t− 4
t + 4
)
y1 =
1
2
t2 − 4t + c1, y2 = − 110 t−
41
100
+ c2e10t
588
10.5 Matrix Exponential
X = PY =
(
1 1
−1 1
) ( 1
2 t
2 − 4t + c1
− 110 t− 41100 + c2e10t
)
=
( 1
2 t
2 − 4110 t− 41100 + c1 + c2e10t
− 12 t2 + 3910 t− 41100 − c1 + c2e10t
)
= c1
(
1
−1
)
+ c2
(
1
1
)
e10t +
1
2
(
1
−1
)
t2 +
1
10
(−41
39
)
t− 41
100
(
1
1
)
38. λ1 = −1, λ2 = 1, K1 =
(
1
−1
)
, K2 =
(
1
1
)
, P =
(
1 1
−1 1
)
, P−1 =
1
2
(
1 −1
1 1
)
, P−1F =
(
2− 4e−2t
2 + 4e−2t
)
;
Y′ =
(−1 0
0 1
)
Y +
(
2− 4e−2t
2 + 4e−2t
)
y1 = 2 + 4e−2t + c1e−t, y2 = −2− 43e
−2t + c2et
X = PY =
(
1 1
−1 1
) (
2 + 4e−2t + c1e−t
−2− 43e−2t + c2et
)
=
( 8
3e
−2t + c1e−t + c2et
−4− 163 e−2t − c1e−t + c2et
)
= c1
(
1
−1
)
e−t + c2
(
1
1
)
et +
8
3
(
1
−2
)
e−2t +
(
0
−4
)
EXERCISES 10.5
Matrix Exponential
1. For A =
(
1 0
0 2
)
we have
A2 =
(
1 0
0 2
) (
1 0
0 2
)
=
(
1 0
0 4
)
,
A3 = AA2 =
(
1 0
0 2
) (
1 0
0 4
)
=
(
1 0
0 8
)
,
A4 = AA3 =
(
1 0
0 2
) (
1 0
0 8
)
=
(
1 0
0 16
)
,
and so on. In general
Ak =
(
1 0
0 2k
)
for k = 1, 2, 3, . . . .
Thus
eAt = I +
A
1!
t +
A2
2!
t2 +
A3
3!
t3 + · · ·
=
(
1 0
0 1
)
+
1
1!
(
1 0
0 2
)
t +
1
2!
(
1 0
0 4
)
t2 +
1
3!
(
1 0
0 8
)
t3 + · · ·
=
 1 + t +
t2
2!
+
t3
3!
+ · · · 0
0 1 + t +
(2t)2
2!
+
(2t)3
3!
+ · · ·
 = ( et 0
0 e2t
)
589
10.5 Matrix Exponential
and
e−At =
(
e−t 0
0 e−2t
)
.
2. For A =
(
0 1
1 0
)
we have
A2 =
(
0 1
1 0
) (
0 1
1 0
)
=
(
1 0
0 1
)
= I
A3 = AA2 =
(
0 1
1 0
)
I =
(
0 1
1 0
)
= A
A4 = (A2)2 = I
A5 = AA4 = AI = A,
and so on. In general,
Ak =
{
A, k = 1, 3, 5, . . .
I, k = 2, 4, 6, . . . .
Thus
eAt = I +
A
1!
t +
A2
2!
t2 +
A3
3!
t3 + · · ·
= I + At +
1
2!
It2 +
1
3!
At3 + · · ·
= I
(
1 +
1
2!
t2 +
1
4!
t4 + · · ·
)
+ A
(
t +
1
3!
t3 +
1
5!
t5 + · · ·
)
= I cosh t + A sinh t =
(
cosh t sinh t
sinh t cosh t
)
and
e−At =
(
cosh(−t) sinh(−t)
sinh(−t) cosh(−t)
)
=
(
cosh t − sinh t
− sinh t cosh t
)
.
3. For
A =
 1 1 11 1 1
−2 −2 −2

we have
A2 =
 1 1 11 1 1
−2 −2 −2

 1 1 11 1 1
−2 −2 −2
 =
 0 0 00 0 0
0 0 0
 .
Thus, A3 = A4 = A5 = · · · = 0 and
eAt = I + At =
 1 0 00 1 0
0 0 1
 +
 t t tt t t
−2t −2t −2t
 =
 t + 1 t tt t + 1 t
−2t −2t −2t + 1
 .
590
10.5 Matrix Exponential
4. For
A =
 0 0 03 0 0
5 1 0

we have
A2 =
 0 0 03 0 0
5 1 0

 0 0 03 0 0
5 1 0
 =
 0 0 00 0 0
3 0 0

A3 = AA2 =
 0 0 03 0 0
5 1 0

 0 0 00 0 0
3 0 0
 =
 0 0 00 00
0 0 0
 .
Thus, A4 = A5 = A6 = · · · = 0 and
eAt = I + At +
1
2
A2t2
=
 1 0 00 1 0
0 0 1
 +
 0 0 03t 0 0
5t t 0
 +
 0 0 00 0 0
3
2 t
2 0 0
 =
 1 0 03t 1 0
3
2 t
2 + 5t t 1
 .
5. Using the result of Problem 1,
X =
(
et 0
0 e2t
) (
c1
c2
)
= c1
(
et
0
)
+ c2
(
0
et
)
.
6. Using the result of Problem 2,
X =
(
cosh t sinh t
sinh t cosh t
) (
c1
c2
)
= c1
(
cosh t
sinh t
)
+ c2
(
sinh t
cosh t
)
.
7. Using the result of Problem 3,
X =
 t + 1 t tt t + 1 t
−2t −2t −2t + 1

 c1c2
c3
 = c1
 t + 1t
−2t
 + c2
 tt + 1
−2t
 + c3
 tt
−2t + 1
 .
8. Using the result of Problem 4,
X =
 1 0 03t 1 0
3
2 t
2 + 5t t 1

 c1c2
c3
 = c1
 13t
3
2 t
2 + 5t
 + c2
 01
t
 + c3
 00
1
 .
9. To solve
X′ =
(
1 0
0 2
)
X +
(
3
−1
)
591
10.5 Matrix Exponential
we identify t0 = 0, F(t) =
(
3
−1
)
, and use the results of Problem 1 and equation (6) in the text.
X(t) = eAtC + eAt
∫ t
t0
e−AsF(s) ds
=
(
et 0
0 e2t
) (
c1
c2
)
+
(
et 0
0 e2t
) ∫ t
0
(
e−s 0
0 e−2s
) (
3
−1
)
ds
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) ∫ t
0
(
3e−s
−e−2s
)
ds
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) (−3e−s
1
2e
−2s
) ∣∣∣∣t
0
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) (−3e−t + 3
1
2e
−2t − 12
)
=
(
c1e
t
c2e
2t
)
+
(−3 + 3et
1
2 − 12e2t
)
= c3
(
1
0
)
et + c4
(
0
1
)
e2t +
( −3
1
2
)
.
10. To solve
X′ =
(
1 0
0 2
)
X +
(
t
e4t
)
we identify t0 = 0, F(t) =
(
t
e4t
)
, and use the results of Problem 1 and equation (6) in the text.
X(t) = eAtC + eAt
∫ t
t0
e−AsF(s) ds
=
(
et 0
0 e2t
) (
c1
c2
)
+
(
et 0
0 e2t
) ∫ t
0
(
e−s 0
0 e−2s
) (
s
e4s
)
ds
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) ∫ t
0
(
se−s
e2s
)
ds
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) (−se−s − e−s
1
2e
2s
) ∣∣∣∣t
0
=
(
c1e
t
c2e
2t
)
+
(
et 0
0 e2t
) (−te−t − e−t + 1
1
2e
2t − 12
)
=
(
c1e
t
c2e
2t
)
+
(−t− 1 + et
1
2e
4t − 12e2t
)
= c3
(
1
0
)
et + c4
(
0
1
)
e2t +
(−t− 1
1
2e
4t
)
.
11. To solve
X′ =
(
0 1
1 0
)
X +
(
1
1
)
592
10.5 Matrix Exponential
we identify t0 = 0, F(t) =
(
1
1
)
, and use the results of Problem 2 and equation (6) in the text.
X(t) = eAtC + eAt
∫ t
t0
e−AsF(s) ds
=
(
cosh t sinh t
sinh t cosh t
) (
c1
c2
)
+
(
cosh t sinh t
sinh t cosh t
) ∫ t
0
(
cosh s − sinh s
− sinh s cosh s
) (
1
1
)
ds
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) ∫ t
0
(
cosh s− sinh s
− sinh s + cosh s
)
ds
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) (
sinh s− cosh s
− cosh s + sinh s
) ∣∣∣∣t
0
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) (
sinh t− cosh t + 1
− cosh t + sinh t + 1
)
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
sinh2 t− cosh2 t + cosh t + sinh t
sinh2 t− cosh2 t + sinh t + cosh t
)
= c1
(
cosh t
sinh t
)
+ c2
(
sinh t
cosh t
)
+
(
cosh t
sinh t
)
+
(
sinh t
cosh t
)
−
(
1
1
)
= c3
(
cosh t
sinh t
)
+ c4
(
sinh t
cosh t
)
−
(
1
1
)
.
12. To solve
X′ =
(
0 1
1 0
)
X +
(
cosh t
sinh t
)
we identify t0 = 0, F(t) =
(
cosh t
sinh t
)
, and use the results of Problem 2 and equation (6) in the text.
X(t) = eAtC + eAt
∫ t
t0
e−AsF(s) ds
=
(
cosh t sinh t
sinh t cosh t
) (
c1
c2
)
+
(
cosh t sinh t
sinh t cosh t
) ∫ t
0
(
cosh s − sinh s
− sinh s cosh s
) (
cosh s
sinh s
)
ds
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) ∫ t
0
(
1
0
)
ds
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) (
s
0
) ∣∣∣∣t
0
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
cosh t sinh t
sinh t cosh t
) (
t
0
)
=
(
c1 cosh t + c2 sinh t
c1 sinh t + c2 cosh t
)
+
(
t cosh t
t sinh t
)
= c1
(
cosh t
sinh t
)
+ c2
(
sinh t
cosh t
)
+ t
(
cosh t
sinh t
)
.
13. We have
X(0) = c1
 10
0
 + c2
 01
0
 + c3
 00
1
 =
 c1c2
c3
 =
 1−4
6
 .
593
10.5 Matrix Exponential
Thus, the solution of the initial-value problem is
X =
 t + 1t
−2t
− 4
 tt + 1
−2t
 + 6
 tt
−2t + 1
 .
14. We have
X(0) = c3
(
1
0
)
+ c4
(
0
1
)
+
(−3
1
2
)
=
(
c3 − 3
c4 + 12
)
=
(
4
3
)
.
Thus, c3 = 7 and c4 = 52 , so
X = 7
(
1
0
)
et +
5
2
(
0
1
)
e2t +
(−3
1
2
)
.
15. From sI−A =
(
s− 4 −3
4 s + 4
)
we find
(sI−A)−1 =

3/2
s− 2 −
1/2
s + 2
3/4
s− 2 −
3/4
s + 2
−1
s− 2 +
1
s + 2
−1/2
s− 2 +
3/2
s + 2

and
eAt =
(
3
2e
2t − 12e−2t 34e2t − 34e−2t
−e2t + e−2t − 12e2t + 32e−2t
)
.
The general solution of the system is then
X = eAtC =
(
3
2e
2t − 12e−2t 34e2t − 34e−2t
−e2t + e−2t − 12e2t + 32e−2t
)(
c1
c2
)
= c1
( 3
2
−1
)
e2t + c1
(− 12
1
)
e−2t + c2
( 3
4
− 12
)
e2t + c2
(− 34
3
2
)
e−2t
=
(1
2
c1 +
1
4
c2
) ( 3
−2
)
e2t +
(
−1
2
c1 − 34c2
) ( 1
−2
)
e−2t
= c3
(
3
−2
)
e2t + c4
(
1
−2
)
e−2t.
16. From sI−A =
(
s− 4 2
−1 s− 1
)
we find
(sI−A)−1 =

2
s− 3 −
1
s− 2 −
2
s− 3 +
2
s− 2
1
s− 3 −
1
s− 2
−1
s− 3 +
2
s− 2

and
eAt =
(
2e3t − e2t −2e3t + 2e2t
e3t − e2t −e3t + 2e2t
)
.
The general solution of the system is then
594
10.5 Matrix Exponential
X = eAtC =
(
2e3t − e2t −2e3t + 2e2t
e3t − e2t −e3t + 2e2t
) (
c1
c2
)
= c1
(
2
1
)
e3t + c1
(−1
−1
)
e2t + c2
(−2
−1
)
e3t + c2
(
2
2
)
e2t
= (c1 − c2)
(
2
1
)
e3t + (−c1 + 2c2)
(
1
1
)
e2t
= c3
(
2
1
)
e3t + c4
(
1
1
)
e2t.
17. From sI−A =
(
s− 5 9
−1 s + 1
)
we find
(sI−A)−1 =

1
s− 2 +
3
(s− 2)2 −
9
(s− 2)2
1
(s− 2)2
1
s− 2 −
3
(s− 2)2

and
eAt =
(
e2t + 3te2t −9te2t
te2t e2t − 3te2t
)
.
The general solution of the system is then
X = eAtC =
(
e2t + 3te2t −9te2t
te2t e2t − 3te2t
) (
c1
c2
)
= c1
(
1
0
)
e2t + c1
(
3
1
)
te2t + c2
(
0
1
)
e2t + c2
(−9
−3
)
te2t
= c1
(
1 + 3t
t
)
e2t + c2
( −9t
1− 3t
)
e2t.
18. From sI−A =
(
s −1
2 s + 2
)
we find
(sI−A)−1 =

s + 1 + 1
(s + 1)2 + 1
1
(s + 1)2 + 1
−2
(s + 1)2 + 1
s + 1− 1
(s + 1)2 + 1

and
eAt =
(
e−t cos t + e−t sin t e−t sin t
−2e−t sin t e−t cos t− e−t sin t
)
.
The general solution of the system is then
X = eAtC =
(
e−t cos t + e−t sin t e−t sin t
−2e−t sin t e−t cos t− e−t sin t
) (
c1
c2
)
= c1
(
1
0
)
e−t cos t + c1
(
1
−2
)
e−t sin t + c2
(
0
1
)
e−t cos t + c2
(
1
−1
)
e−t sin t
595
10.5 Matrix Exponential
= c1
(
cos t + sin t
−2 sin t
)
e−t + c2(
sin t
cos t− sin t
)
e−t.
19. The eigenvalues are λ1 = 1 and λ2 = 6. This leads to the system
et = b0 + b1
e6t = b0 + 6b1,
which has the solution b0 = 65e
t − 15e6t and b1 = − 15et + 15e6t. Then
eAt = b0I + b1A =
(
4
5e
t + 15e
6t 2
5e
t − 25e6t
2
5e
t − 25e6t 15et + 45e6t
)
.
The general solution of the system is then
X = eAtC =
(
4
5e
t + 15e
6t 2
5e
t − 25e6t
2
5e
t − 25e6t 15et + 45e6t
)(
c1
c2
)
= c1
( 4
5
2
5
)
et + c1
( 1
5
− 25
)
e6t + c2
( 2
5
1
5
)
et + c2
(− 25
4
5
)
e6t
=
(2
5
c1 +
1
5
c2
) ( 2
1
)
et +
(1
5
c1 − 25c2
) ( 1
−2
)
e6t
= c3
(
2
1
)
et + c4
(
1
−2
)
e6t.
20. The eigenvalues are λ1 = 2 and λ2 = 3. This leads to the system
e2t = b0 + 2b1
e3t = b0 + 3b1,
which has the solution b0 = 3e2t − 2e3t and b1 = −e2t + e3t. Then
eAt = b0I + b1A =
(
2e2t − e3t −2e2t + 2e3t
e2t − e3t −e2t + 2e3t
)
.
The general solution of the system is then
X = eAtC =
(
2e2t − e3t −2e2t + 2e3t
e2t − e3t −e2t + 2e3t
) (
c1
c2
)
= c1
(
2
1
)
e2t + c1
(−1
−1
)
e3t + c2
(−2
−1
)
e2t + c2
(
2
2
)
e3t
= (c1 − c2)
(
2
1
)
e2t + (−c1 + 2c2)
(
1
1
)
e3t
= c3
(
2
1
)
e2t + c4
(
1
1
)
e3t.
596
10.5 Matrix Exponential
21. The eigenvalues are λ1 = −1 and λ2 = 3. This leads to the system
e−t = b0 − b1
e3t = b0 + 3b1,
which has the solution b0 = 34e
−t + 14e
3t and b1 = − 14e−t + 14e3t. Then
eAt = b0I + b1A =
(
e3t −2e−t + 2e3t
0 e−t
)
.
The general solution of the system is then
X = eAtC =
(
e3t −2e−t + 2e3t
0 e−t
)(
c1
c2
)
= c1
(
1
0
)
e3t + c2
(−2
1
)
e−t + c2
(
2
0
)
e3t
= c2
(−2
1
)
e−t + (c1 + 2c2)
(
1
0
)
e3t
= c3
(−2
1
)
e−t + c4
(
1
0
)
e3t.
22. The eigenvalues are λ1 = 14 and λ2 =
1
2 . This leads to the system
et/4 = b0 +
1
4
b1
et/2 = b0 +
1
2
b1,
which has the solution b0 = 2et/4 + et/2 and b1 = −4et/4 + 4et/2. Then
eAt = b0I + b1A =
(−2et/4 + 3et/2 6et/4 − 6et/2
−et/4 + et/2 3et/4 − 2et/2
)
.
The general solution of the system is then
X = eAtC =
(−2et/4 + 3et/2 6et/4 − 6et/2
−et/4 + et/2 3et/4 − 2et/2
) (
c1
c2
)
= c1
(−2
−1
)
et/4 + c1
(
3
1
)
et/2 + c2
(
6
3
)
et/4 + c2
(−6
−2
)
et/2
= (−c1 + 3c2)
(
2
1
)
et/4 + (c1 − 2c2)
(
3
1
)
et/2
= c3
(
2
1
)
et/4 + c4
(
3
1
)
et/2.
23. From equation (3) in the text we have eDt = I + tD +
t2
2!
D2 +
t3
3!
D3 + · · · so that
PeDtP−1 = PP−1 + t(PDP−1) +
t2
2!
(PD2P−1) +
t3
3!
(PD3P−1) + · · · .
597
10.5 Matrix Exponential
But PP−1 + I, PDP−1 = A and PDnP−1 = An (see Problem 37, Exercises 8.12). Thus,
PeDtP−1 = I + tA +
t2
2!
A2 +
t3
3!
A3 + · · · = eAt.
24. From equation (3) in the text
eDt =

1 0 · · · 0
0 1 · · · 0
...
...
. . .
...
0 0 · · · 1
 +

λ1 0 · · · 0
0 λ2 · · · 0
...
...
. . .
...
0 0 · · · λn
 + 12! t2

λ21 0 · · · 0
0 λ22 · · · 0
...
...
. . .
...
0 0 · · · λ2n

+
1
3!
t3

λ31 0 · · · 0
0 λ32 · · · 0
...
...
. . .
...
0 0 · · · λ3n
 + · · ·
=

1 + λ1t + 12! (λ1t)
2 + · · · 0 · · · 0
0 1 + λ2t + 12! (λ2t)
2 + · · · · · · 0
...
...
. . .
...
0 0 · · · 1 + λnt + 12! (λnt)2 + · · ·

=

eλ1t 0 · · · 0
0 eλ2t · · · 0
...
...
. . .
...
0 0 · · · eλnt

25. From Problems 23 and 24 and equation (1) in the text
X = eAtC = PeDtP−1C =
(
e3t e5t
e3t 3e5t
) (
e3t 0
0 e5t
) ( 3
2e
−3t − 12e−3t
− 12e−5t 12e−5t
)(
c1
c2
)
=
(
3
2e
3t − 12e5t − 12e3t + 12e5t
3
2e
3t − 32e5t − 12e3t + 32e5t
)(
c1
c2
)
.
26. From Problems 23 and 24 and equation (1) in the text
X = eAtC = PeDtP−1C =
(−et e3t
et e3t
) (
et 0
0 e3t
) (− 12e−t 12e−t
1
2e
3t 1
2e
−3t
)(
c1
c2
)
=
(
1
2e
t + 12e
9t − 12et + 12e3t
− 12et + 12e9t 12et + 12e3t
)(
c1
c2
)
.
27. (a) The following commands can be used in Mathematica:
A={{4, 2},{3, 3}};
c={c1, c2};
m=MatrixExp[A t];
sol=Expand[m.c]
Collect[sol, {c1, c2}]//MatrixForm
598
10.5 Matrix Exponential
The output gives
x(t) = c1
(
2
5
et +
3
5
e6t
)
+ c2
(
−2
5
et +
2
5
e6t
)
y(t) = c1
(
−3
5
et +
3
5
e6t
)
+ c2
(
3
5
et +
2
5
e6t
)
.
The eigenvalues are 1 and 6 with corresponding eigenvectors(−2
3
)
and
(
1
1
)
,
so the solution of the system is
X(t) = b1
(−2
3
)
et + b2
(
1
1
)
e6t
or
x(t) = −2b1et + b2e6t
y(t) = 3b1et + b2e6t.
If we replace b1 with − 15c1 + 15c2 and b2 with 35c1 + 25c2, we obtain the solution found using the matrix
exponential.
(b) x(t) = c1e−2t cos t− (c1 + c2)e−2t sin t
y(t) = c2e−2t cos t + (2c1 + c2)e−2t sin t
28. x(t) = c1(3e−2t − 2e−t) + c3(−6e−2t + 6e−t)
y(t) = c2(4e−2t − 3e−t) + c4(4e−2t − 4e−t)
z(t) = c1(e−2t − e−t) + c3(−2e−2t + 3e−t)
w(t) = c2(−3e−2t + 3e−t) + c4(−3e−2t + 4e−t)
29. If det(sI−A) = 0, then s is an eigenvalue of A. Thus sI−A has an inverse if s is not an eigenvalue of A. For
the purposes of the discussion in this section, we take s to be larger than the largest eigenvalue of A. Under
this condition sI−A has an inverse.
30. Since A3 = 0, A is nilpotent. Since
eAt = I + At + A2
t2
2!
+ · · ·+ Ak t
k
k!
+ · · · ,
if A is nilpotent and Am = 0, then Ak = 0 for k ≥ m and
eAt = I + At + A2
t2
2!
+ · · ·+ Am−1 t
m−1
(m− 1)! .
In this problem A3 = 0, so
eAt = I + At + A2
t2
2
=
 1 0 00 1 0
0 0 1
 +
−1 1 1−1 0 1
−1 1 1
 t +
−1 0 10 0 0
−1 0 1
 t22
=
 1− t− t
2/2 t t + t2/2
−t 1 t
−t− t2/2 t 1 + t + t2/2

and the solution of X′ = AX is
X(t) = eAtC = eAt
 c1c2
c3
 =
 c1(1− t− t
2/2) + c2t + c3(t + t2/2)
−c1t + c2 + c3t
c1(−t− t2/2) + c2t + c3(1 + t + t2/2)
 .
599
10.5 Matrix Exponential
CHAPTER 10 REVIEW EXERCISES
CHAPTER 10 REVIEW EXERCISES
1. If X = k
(
4
5
)
, then X′ = 0 and
k
(
1 4
2 −1
) (
4
5
)
−
(
8
1
)
= k
(
24
3
)
−
(
8
1
)
=
(
0
0
)
.
We see that k = 13 .
2. Solving for c1 and c2 we find c1 = − 34 and c2 = 14 .
3. Since  4 6 61 3 2
−1 −4 −3

 31
−1
 =
 124
−4
 = 4
 31
−1
 ,
we see that λ = 4 is an eigenvalue with eigenvector K3. The corresponding solution is X3 = K3e4t.
4. The other eigenvalue is λ2 = 1− 2i with corresponding eigenvector K2 =
(
1
−i
)
. The general solution is
X(t) = c1
(
cos 2t
− sin 2t
)
et + c2
(
sin 2t
cos 2t
)
et.
5. We have det(A− λI) = (λ− 1)2 = 0 and K =
(
1
−1
)
. A solution to (A− λI)P = K is P =
(
0
1
)
so that
X = c1
(
1
−1
)
et + c2
[(
1
−1
)
tet +
(
0
1
)
et
]
.
6. We have det(A− λI) = (λ + 6)(λ + 2) = 0 so that
X = c1
(
1
−1
)
e−6t + c2
(
1
1
)
e−2t.
7. We have det(A− λI) = λ2 − 2λ + 5 = 0. For λ = 1 + 2i we obtain K1 =
(
1
i
)
and
X1 =
(
1
i
)
e(1+2i)t =
(
cos 2t
− sin 2t
)
et + i
(
sin 2t
cos 2t
)
et.
Then
X = c1
(
cos 2t
− sin 2t
)
et + c2
(
sin 2t
cos 2t
)
et.
8. We have det(A− λI) = λ2 − 2λ + 2 = 0. For λ = 1 + i we obtain K1 =
(
3− i
2
)
and
X1 =
(
3− i
2
)
e(1+i)t =
(
3 cos t + sin t
2 cos t
)
et + i
(− cos t + 3 sin t
2 sin t
)
et.
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CHAPTER 10 REVIEW EXERCISES
Then
X = c1
(
3 cos t + sint
2 cos t
)
et + c2
(− cos t + 3 sin t
2 sin t
)
et.
9. We have det(A− λI) = −(λ− 2)(λ− 4)(λ + 3) = 0 so that
X = c1
−23
1
 e2t + c2
 01
1
 e4t + c3
 712
−16
 e−3t.
10. We have det(A−λI) = −(λ+2)(λ2−2λ+3) = 0. The eigenvalues are λ1 = −2, λ2 = 1+
√
2i, and λ2 = 1−
√
2i,
with eigenvectors
K1 =
−75
4
 , K2 =
 11
2
√
2 i
1
 , and K3 =
 1− 12√2 i
1
 .
Thus
X = c1
−75
4
 e−2t + c2

 10
1
 cos√2t−
 012√2
0
 sin√2t
 et
+ c3

 012√2
0
 cos√2t +
 10
1
 sin√2t
 et
= c1
−75
4
 e−2t + c2
 cos
√
2t
− 12
√
2 sin
√
2t
cos
√
2t
 et + c3
 sin
√
2t
1
2
√
2 cos
√
2t
sin
√
2t
 et.
11. We have
Xc = c1
(
1
0
)
e2t + c2
(
4
1
)
e4t.
Then
Φ =
(
e2t 4e4t
0 e4t
)
, Φ−1 =
(
e−2t −4e−2t
0 e−4t
)
,
and
U =
∫
Φ−1F dt =
∫ (
2e−2t − 64te−2t
16te−4t
)
dt =
(
15e−2t + 32te−2t
−e−4t − 4te−4t
)
,
so that
Xp = ΦU =
(
11 + 16t
−1− 4t
)
.
12. We have
Xc = c1
(
2 cos t
− sin t
)
et + c2
(
2 sin t
cos t
)
et.
Then
Φ =
(
2 cos t 2 sin t
− sin t cos t
)
et, Φ−1 =
(
1
2 cos t − sin t
1
2 sin t cos t
)
e−t,
and
U =
∫
Φ−1F dt =
∫ (
cos t− sec t
sin t
)
dt =
(
sin t− ln | sec t + tan t|
− cos t
)
,
601
CHAPTER 10 REVIEW EXERCISES
so that
Xp = ΦU =
( −2 cos t ln | sec t + tan t|
−1 + sin t ln | sec t + tan t|
)
et.
13. We have
Xc = c1
(
cos t + sin t
2 cos t
)
+ c2
(
sin t− cos t
2 sin t
)
.
Then
Φ =
(
cos t + sin t sin t− cos t
2 cos t 2 sin t
)
, Φ−1 =
(
sin t 12 cos t− 12 sin t
− cos t 12 cos t + 12 sin t
)
,
and
U =
∫
Φ−1F dt =
∫ ( 1
2 sin t− 12 cos t + 12 csc t
− 12 sin t− 12 cos t + 12 csc t
)
dt
=
(− 12 cos t− 12 sin t + 12 ln | csc t− cot t|
1
2 cos t− 12 sin t + 12 ln | csc t− cot t|
)
,
so that
Xp = ΦU =
(−1
−1
)
+
(
sin t
sin t + cos t
)
ln | csc t− cot t|.
14. We have
Xc = c1
(
1
−1
)
e2t + c2
[(
1
−1
)
te2t +
(
1
0
)
e2t
]
.
Then
Φ =
(
e2t te2t + e2t
−e2t −te2t
)
, Φ−1 =
(−te−2t −te−2t − e−2t
e−2t e−2t
)
,
and
U =
∫
Φ−1F dt =
∫ (
t− 1
−1
)
dt =
( 1
2 t
2 − t
−t
)
,
so that
Xp = ΦU =
(− 12
1
2
)
t2e2t +
(−2
1
)
te2t.
15. (a) Letting
K =
 k1k2
k3

we note that (A− 2I)K = 0 implies that 3k1 + 3k2 + 3k3 = 0, so k1 = −(k2 + k3). Choosing k2 = 0, k3 = 1
and then k2 = 1, k3 = 0 we get
K1 =
−10
1
 and K2 =
−11
0
 ,
respectively. Thus,
X1 =
−10
1
 e2t and X2 =
−11
0
 e2t
602
CHAPTER 10 REVIEW EXERCISES
are two solutions.
(b) From det(A− λI) = λ2(3− λ) = 0 we see that λ1 = 3, and 0 is an eigenvalue of multiplicity two. Letting
K =
 k1k2
k3
 ,
as in part (a), we note that (A − 0I)K = AK = 0 implies that k1 + k2 + k3 = 0, so k1 = −(k2 + k3).
Choosing k2 = 0, k3 = 1, and then k2 = 1, k3 = 0 we get
K2 =
−10
1
 and K3 =
−11
0
 ,
respectively. Since the eigenvector corresponding to λ1 = 3 is
K1 =
 11
1
 ,
the general solution of the system is
X = c1
 11
1
 e3t + c2
−10
1
 + c3
−11
0
 .
16. For X =
(
c1
c2
)
et we have X′ = X = IX.
603