MA 262: Linear Algebra and Differential Equations Fall 2011, Purdue

Homework 14

1 Problem 22, Page 561

Let A be a 2 × 2 nondefective matrix. If all eigenvalues of A have negative real part, prove thatevery solution to x′ = Ax satisfies

limt→∞

x(t) = 0.

Proof. First of all, we know the fact that if a < 0, then eat → 0 as t → ∞.If A has two real eigenvalues, λ1 < 0, λ2 < 0, then

x(t) = c1eλ1tv1 + c2e

λ2tv2,

limt→∞

x(t) = 0.

1.2. If A has two complex eigenvalues, λ = α ± iβ, where α < 0. Actually, we just need oneeigenvalue λ = α+ iβ with corresponding eigenvector v = r + is, then

x1(t) = eαt(cos(bt)r − sin(bt)s),

x1(t) = eαt(sin(bt)r + cos(bt)s).

Similarly,

limt→∞

x1(t) = 0, limt→∞

x2(t) = 0,

⇒ limt→∞

x(t) = limt→∞

(c1x1(t) + c2x2(t)) = 0.

2 Problem 8, Page 577

Find the general solution to x′ = Ax+ b, where

A =

−1 −2 22 4 −10 0 3

, b =

−e3t

4e3t

3e3t

. (2.1)

Solution:

Step I: Solve the homogeneous system x′ = Ax.

1 Copy right reserved by Yingwei Wang

MA 262: Linear Algebra and Differential Equations Fall 2011, Purdue

It is easy to find that the eigenvalues and eigenvectors of A are λ1 = 0, v1 = (−2, 1, 0)T ; λ2 = 3,v2 = (1, 0, 2)T and v3 = (0, 1, 1)T .

The fundamental solution to the homogeneous system x′ = Ax is

X =

−2 e3t 01 0 e3t

0 2e3t e3t

. (2.2)

Step II: Find a particular solution to nonhomogeneous system x′ = Ax+ b. Let xp = Xu, whereu satisfies

Xu′ = b, (2.3)

⇒

−2 e3t 01 0 e3t

0 2e3t e3t

u′

1

u′

2

u′

3

. =

−e3t

4e3t

3e3t

, (2.4)

⇒

u′

1= 1

3e3t,

u′

2= −

1

3,

u′

3= 11

3.

(2.5)

⇒

u1 = 1

9e3t,

u2 = −1

3t,

u3 = 11

3t.

(2.6)

Now we find a particular solution

xp = Xu = e3t

−2

9−

1

3t

1

9+ 11

3t

3t

.

Step III: Give the general solution as

x = Xc+ xp, (2.7)

=

−2 e3t 01 0 e3t

0 2e3t e3t

c1c2c3

+ e3t

−2

9−

1

3t

1

9+ 11

3t

3t

. (2.8)

2 Copy right reserved by Yingwei Wang