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334 CHAPTER 10 Systems of Linear Differential Equations

y
10
y

5

x
–10 –5 0 5 10
x
–5

FIGURE 10.19 Proper node in Case 3(a).
FIGURE 10.18 Phase portrait in Example
10.21.
y
W + 2E
W + E
W
W − E
E
x

−W + 2E

FIGURE 10.20 W and E in Case 3(b).

This means that the trajectories in this case are half-lines from the origin. If λ> 0, these move
away from the origin as t increases, and if λ< 0, they move toward the origin. The origin in this
case is called a proper node. Figure 10.19 illustrates this for trajectories moving away from the
origin.

Case 3(b): A Does Not Have Two Linearly Independent Eigenvectors

In this case the general solution has the form
λt
λt
X(t) =[c 1 W + c 2 E]e + c 1 Ete ,
where E is an eigenvector and W is determined by the procedure outlined in Section 10.2.2.
To visualize the trajectories, begin with arrows from the origin representing W and E.Using
these we can draw vectors c 1 W + c 2 E, which may have various orientations relative to W and
E, depending on the signs and magnitudes of the constants. Some possibilities are shown in
Figure 10.20. For given c 1 and c 2 , c 1 W + c 2 E + c 1 Et sweeps out a straight line L as t varies. For
λt
agiven t, X(t) is e times this vector (see Figure 10.21). If λ is negative, this vector shrinks to
zero length as t →∞ and X(t) sweeps out a curve that approaches the origin tangent to E.If λ
is positive, reverse the orientation on this trajectory.

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October 14, 2010 20:32 THM/NEIL Page-334 27410_10_ch10_p295-342