Unformatted text preview: Systems of ﬁrstorder differential equations 155
Figure 4.11. Figure 4.12. nonhomogeneous autonomous differential equations
˙
x = −2x − y + 9
˙
y = −y + x + 3
˙
The equilibrium lines in the phase plane can readily be found by setting x = 0 and
˙
y = 0. Thus
˙
x=0
˙
y=0 implying
implying y = − 2x + 9
y=x+3 which can be solved to give a ﬁxed point, an equilibrium point, namely (x∗ , y∗ ) =
(2, 5). The solution equations for this system for initial condition, x0 = 2 and
y0 = 2 are
√
√
x(t) = 2 + 2 3 sin( 3t/2)e−(3t/2)
√
√
√
y(t) = 5 − (3 cos( 3t/2)) − 3 sin( 3t/2)e−(3t/2)
The equilibrium lines along with the trajectory are illustrated in ﬁgure 4.11.
The analysis of this example is the same as for examples 4.4 and 4.5. In this
case the ﬁxed point is at (x∗ , y∗ ) = (2, 5). The vectors of force are illustrated in
ﬁgure 4.12 by the arrows. What is apparent from this ﬁgure is that the system is
globally stable, and the dynamic forces are sending the system towards the ﬁxed
point in a counterclockwise motion. As we illustrated in ﬁgure 4.11, if the initial
point is (x0 , y0 ) = (2, 2), then the system begins in quadrant III and tends to the
ﬁxed point over time in a counterclockwise direction, passing ﬁrst into quadrant ...
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 Fall '11
 Dr.Gwartney
 Economics

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