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where use tomcat pdf-417 2d barcode printing toadd pdf 417 with java bar code Cle-4t [. + C2 e2t =I 0,. =I o. Note that [~m] ~ AUTONOMOUS SYSTEMS for t a large negative number. Also,. for t a large positive jdk barcode pdf417 number. To draw a more accurate phase plane diagram of (3.24), (3.

25), we consider the nullclines. The x nullcline is x -x + 3y = 0 or y = "3" and the y nullcline is. 3x - y = 0 y = 3x. The nullclines are draw n in Figure 11 as dashed lines. Note that any orbit crossing the nullcline y = is vertical at the crossing point and any orbit crossing the nullcline y = 3x is horizontal at the crossing point..

= 3x. -Ix Y -3. ----------~r_~~~--~----------. y= -x FIGURE 11. Saddle point. Finally, we consider the complex cases. Consider the system ax + by, -bx + ay,. (3.26) (3.27) 0.

where b =1= O. The characteristic equation for this system is ,,\2 _. 2a"\ + (a 2 + b2 ) =. and so the eigenvalues are the complex numbers a ilbl. Let x, y be a solution of (3.26), (3.

27), and let r(t), O(t) be the polar coordinates of x(t), y( t); that is, x(t) = r(t) cos O(t), y(t) = r(t) sin O(t). Since. 3.3. PHASE PLANE DIAGRAMS FOR LINEAR SYSTEMS we get that 2r(t)r"(t). Hence 2x(t)x"(t). + 2y(t)y"(t).. r"(t). x(t)x"(t) + y(t)y"(t) r(t) x(t)[ax(t) + by(t)] + y(t)[-bx(t) r(t) a[x 2(t) + y2(t)] r(t) ar(t).. + ay(t)]. Again assum e x, y is a solution of (3.26), (3.27) and consider tanO(t) Differentiating both sides, we get.

= x(t)". y(t). sec 2 0(t)e"(t). x(t)y"(t) - x"(t)y(t) x 2 (t) x(t)[-bx(t) + ay(t)]- [ax(t) x 2 (t) -b x 2(t) + y2(t) r2(t) cos 2 (t). b cos 2 (t) .. + by(t)]y(t). Therefore,. O"(t) = -b. Hence if x, PDF417 for Java y is a solution of (3.26), (3.27) and r(t), O(t) are the polar coordinates of x(t), y(t), then r, 0 solves the system.

(3.28) (3.29).

Case 5: ),1 PDF-417 2d barcode for Java = i(3, ),2 = -i(3, where (3 plane diagram for the system x". > 0 (center). Draw the phase (3.30) (3.31).

= by,. -bx,. where b i= O. In this case the system (3.30), (3.

31) is obtained from the system (3.26), (3.27) by setting a = O.

From before the eigenvalues are ),1 = ilbl, ),2 = -ilbl, so we are in the center case. If x, y is a solution of. 3. AUTONOMOUS SYSTEMS o solves the system (3.26), (3. applet pdf417 2d barcode 27) and r(t), O(t) are the polar coordinates of x(t), y(t), then r,.

r" 0". = 0,. (3.32) (3.33).

Equation (3 jvm PDF-417 2d barcode .32) implies that orbits stay the same distance from the origin, and equation (3.33) implies that the object has constant angular velocity -b.

In Figure 12 we draw the phase plane diagram when b < 0 (if b > 0 the arrows are pointed in the opposite direction.) In this case the equilibrium point (0,0) is called a center..

FIGURE 12. A center. Case 6: ).1 barcode pdf417 for Java = a + i(3, ).2 = a - i(3, where a Draw the phase plane diagram for the system x" y".

0, (3 > 0 (spiral point).. (3.34) (3.35).

ax+ by, -bx + ay,. where a =1= j2ee PDF 417 0 and b =1= O. From before the eigenvalues are ).1 = a + ilbl, ).

2 = a - ilbl, so we are in the spiral point case. If x, y is a solution of (3.26), (3.

27) and r(t), O(t) are the polar coordinates of x(t), y(t), then r, 0 solves the system (3.28), (3.29).

Equation (3.28) implies that if an object is not at the origin and a < 0, then as time increases it is getting closer to the origin. Equation (3.

29) implies that if an object is not at the origin, then its angular velocity is -b. In Figure 13 we draw the phase plane diagram when a < 0 and b > o. In this case if a < 0 the origin is called a stable.

STABILITY OF NONLINEAR SYSTEMS spiral poin jar barcode pdf417 t. If a > 0, then the object spirals away from the origin and we say the origin is an unstable spiral point..

FIGURE 13. A stable spiral point. Now that we have considered all the cases, let"s look at an example.

EXaIIlple 3.13 Draw the phase plane diagram for the system. 4x - 5y,.
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