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SSM-Ch04 - Chap 4 Sec 4.1 Systems of ODEs Phase Plane...

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Chap . 4 Systems of ODEs . Phase Plane . Qualitative Methods Sec . 4 . 1 Systems of ODEs as Models Example 2 . Spend time on Fig. 79 on p. 134 until you feel that you fully understand the difference between (a) (the usual representation in calculus) and (b), because trajectories will play an important role throughout this chapter. Try to understand the reasons for the following. The trajectory starts at the origin. It reaches its highest point where I 2 has a maximum (before t u003d 1). It has a vertical tangent where I 1 has a maximum, shortly after t u003d 1. As t increases from there to t u003d 5, the trajectory goes downward until it almost reaches the I 1 -axis at 3; this point is a limit as t u2192 u221e . In terms of t the trajectory goes up faster than it comes down. Problem Set 4 . 1 . Page 135 7 . Electrical network . The problem amounts to the determination of the two arbitrary constants in a general solution of a system of two ODEs in two unknown functions I 1 and I 2 , representing the currents in an electrical network shown in Fig. 78 in Sec. 4.1. You will see that this is quite similar to the corresponding task for a single second-order ODE. That solution is given by (6), in components I 1 ue0a2 t ue0a3 u003d 2 c 1 e u2212 2 t u002b c 2 e u2212 0.8 t u002b 3, I 2 ue0a2 t ue0a3 u003d c 1 e u2212 2 t u002b 0.8 c 2 e u2212 0.8 t . Setting t u003d 0 and using the given initial conditions I 1 ue0a2 0 ue0a3 u003d 0, I 2 ue0a2 0 ue0a3 u003d u2212 3 gives I 1 ue0a2 0 ue0a3 u003d 2 c 1 u002b c 2 u002b 3 u003d 0 I 2 ue0a2 0 ue0a3 u003d c 1 u002b 0.8 c 2 u003d u2212 3. (a) (b) From (a) you have c 2 u003d u2212 3 u2212 2 c 1 . From this and (b) you obtain c 1 u002b 0.8 ue0a2 u2212 3 u2212 2 c 1 ue0a3 u003d u2212 0.6 c 1 u2212 2.4 u003d u2212 3, hence c 1 u003d 1. Also c 2 u003d u2212 3 u2212 2 c 1 u003d u2212 3 u2212 2 u003d u2212 5. This yields the answer I 1 ue0a2 t ue0a3 u003d 2 e u2212 2 t u2212 5 e u2212 0.8 t u002b 3 I 2 ue0a2 t ue0a3 u003d e u2212 2 t u002b 0.8 ue0a2 u2212 5 ue0a3 e u2212 0.8 t u003d e u2212 2 t u2212 4 e u2212 0.8 t . You see that the limits are 3 and 0, respectively. Can you see this directly from Fig. 78 for physical reasons? 11 . Conversion of single ODEs to a system is an important process, which always follows the pattern shown in formulas (9) and (10) of Sec. 4.1. The present equation y u2032u2032 u2212 4 y u003d 0 can be readily solved. A general solution is y u003d c 1 e 2 t u002b c 2 e u2212 2 t . The point of the problem is not to explain a (complicated) solution method for a simple problem, but to explain the relation between systems and single ODEs and their solutions. In the present case the formulas (9) and (10) give y 1 u003d y , y 2 u003d y u2032 and y 1 u2032 u003d y 2 y 2 u2032 u003d 4 y 1 (because the given equation can be written y u2032u2032 u003d 4 y , hence y 1 u2032u2032 u003d 4 y 1 , but y 1 u2032u2032 u003d y 2 u2032 ). In matrix form (as in Example 3 of the text) this is y u2032 u003d Ay u003d 0 1 4 0 y . The characteristic equation is det ue0a2 A u2212 u03bb I ue0a3 u003d u2212 u03bb 1 4 u2212 u03bb u003d u03bb 2 u2212 4 u003d 0.
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Chap. 4 Systems of ODEs. Phase Plane. Qualitative Methods 31 The eigenvalues are u03bb 1 u003d 2 and u03bb 2 u003d u2212 2. For u03bb 1 you obtain an eigenvector from (13) in Sec. 4.0 with u03bb u003d u03bb 1 , that is, ue0a2 A u2212 u03bb 1 I ue0a3 x u003d u2212 2 1 4 u2212 2 x 1 x 2 u003d u2212 2 x
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