Page 356 - Advanced engineering mathematics

336 CHAPTER 10 Systems of Linear Differential Equations

y

5

x
–30 –20 –10 0 10

–5

–10

FIGURE 10.22 Improper node in Example 10.22.

EXAMPLE 10.23
Let

−1 −2
A = .
4 3
Eigenvalues are 1 ± 2i,so α = 1 and β = 2 in the discussion. An eigenvector for the eigenvalue
1 + 2i is U + iV, where

−1 1
U = and V = .
2 0
The general solution is
t
X(t) =c 1 e [Ucos(2t) − Vsin(2t)]
t
+ c 2 e [usin(t) + Vcos(2t)].
Figure 10.23 is a phase portrait of this system, showing trajectories spiraling out from the origin
as t increases. The origin is a spiral source.

y

20

10

x
–15 –10 –5 0 5 10 15

–10

–20

–30

FIGURE 10.23 Spiral source in Example 10.23.

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October 14, 2010 20:32 THM/NEIL Page-336 27410_10_ch10_p295-342