1550078481-Ordinary_Differential_Equations__Roberts_

420 Ordinary Differential Equations

EXAMPLE 1 Determining the Type of Stability and

Sketching a Phase-Plane Portrait

Determine the type of stability of the origin and sketch a phase-plane

portrait for the system

(5)

SOLUTION

x' = -4x + y

y' = 2x - 5y.

The eigenvalues and associated eigenvectors of the matrix

A= (-4 1)

2 -5

are A1 = -3, >-2 = -6,

(Check this by using EIGEN or your computer software to find the eigenvalues

of A and their associated eigenvectors.) So the general solution of (5) is

(6)

The vector v 1 is a vector with its "tail" at the origin and its "head" at the

point (1, 1) in the xy-plane as shown in Figure 10.4. The graph of the vector

equation u = kv 1 , where k is a parameter, is the line m 1 , which contains the

vector v 1. Likewise, v2 is the vector with its "tail" at the origin and its "head"

at ( -1, 2) and the graph of the vector equation w = kv2 is the line m 2 , which

contains the vector v 2. If a particle starts at any point on the line m 1 in the

first quadrant, then c2 = 0 and c 1 > 0, since x > 0 in the first quadrant. Since

c2 = 0, the particle remains on the line m 1 (see the general solution of (5)-

equation (6)). And since .A 1 = -3 < 0, the particle moves toward the origin

along the line m 1 as t ---+ oo. If a particle starts at a point on the line m 1 in

the third quadrant, then c2 = 0 and c 1 < 0 , since x < 0 in the third quadrant.

Because c2 = 0, the particle remains on the line m 1 and since >. 1 = -3 < o;
the particle moves toward the origin as t ---+ oo. By a similar argument, any

particle which starts on the line m2 remains on m2, because c 1 = 0, and

proceeds toward the origin as t ---+ oo, because >. 2 = -6 < 0. Suppose a