**Unformatted text preview: **h thousands per day. b. We have y = 1 10 y (1-y 9 )-1 5 = 1 90 y (9-y )-1 5 = 1 90 (-y 2 +9 y-18) =-1 90 ( y-3)( y-6). The zeroes of-1 90 ( y-3)( y-6) are at y = 3 and y = 6, so these are the stable equilibria. The graph of-1 90 ( y-3)( y-6) is a parabola which is negative for y < 3, positive for 3 < y < 6 and negative for y > 6. Hence y = 3 is an unstable equilibrium, while y = 6 is a stable equlibrium. c. If the initial population of the lake is 7500 ﬁsh, then y (0) = 7 . 5 since y measures thousands of ﬁsh. The ﬁsh population will decrease and eventually approach 6000 (the stable equilibrium point y = 6.) If the initial population of the lake is 2500 ﬁsh, then y (0) = 2 . 5. Since-1 90 ( y-3)( y-6) is negative for y < 3, y ( t ) will continue to decrease, and eventually the lake will be completely farmed out. 1...

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- Spring '07
- lee
- 2 pts, Stability theory, Logistic function, initial population, 1 - k