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Unformatted text preview: Solutions to HW7 (due 3/26, 3/27) Page 275 15:4321 1 2 3 442 2 4 x y To find the critical points, solve 2 x + y + 3 = 0 3 x 2 y 4 = 0 to get x = 2 and y = 1. There is only one equilibrium, and comparing to figure 5.12, it is a saddle (an unstable). 16:4321 1 2 3 442 2 4 x y To find the critical points, solve 5 x + 2 y = 0 x 4 y = 0 to get x = 0 and y = 0. There is only one equilibrium, and comparing to figure 5.12, it is a stable node. 1 17:4321 1 2 3 442 2 4 x y To find the critical points, solve 2 x + 13 y = 0 x 2 y = 0 to get x = 0 and y = 0. There is only one equilibrium, and comparing to figure 5.12, it is a center (stable). 18: x ' = x (7  x  2 y) y ' = y (5  x  y) 2 2 4 6 82 2 4 6 x y To find the critical points, solve x (7 x 2 y ) = 0 y (5 x y ) = 0. In the first equation we see that x = 0 or 7 x 2 y = 0. If x = 0, then the second equation becomes y (5 y ) = 0 so that y = 0 or y = 5. Looking at the second equation in the same way, we see that either y = 0 of 5 x y = 0. If y = 0 then the first equation gives that x = 0 or x = 7. If both x and y are nonzero, then 7 x 2 y = 0 and 5 x y = 0 so that x = 2 and y = 3. In all there are four critical points, (0,0), (0,5), (7,0) and (3,2). Comparing to figure 5.12 (0,0) is an unstable node, (0,5) is a stable node, (7,0) is a stable node and (3,2) is a saddle (unstable)....
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This note was uploaded on 04/10/2008 for the course PHYS 141 taught by Professor Dienes during the Spring '08 term at Arizona.
 Spring '08
 Dienes

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