**Unformatted text preview: **y = 1 10 y (1-y 6 )-2 15 = 1 60 y (6-y )-2 15 = 1 60 (-y 2 +6 y-8) =-1 60 ( y-2)( y-4). The zeroes of-1 60 ( y-2)( y-4) are at y = 2 and y = 6, so these are the equilibrium points. The graph of-1 60 ( y-2)( y-4) is a parabola which is negative for y < 2, positive for 2 < y < 4 and negative for y > 4. Hence y = 2 is an unstable equilibrium, while y = 4 is a stable equlibrium. (The phase-line diagram points left for y < 2, right for 2 < y < 4 and left for y > 4.) c. If the initial population of the lake is 1500 ﬁsh, then y (0) = 1 . 5 since y measures thousands of ﬁsh. Since-1 60 ( y-2)( y-4) is negative for y < 2, the ﬁsh population will decrease and eventually disappear; the lake will be completely farmed out. If the initial population of the lake is 3000 ﬁsh, then y (0) = 3. Since-1 60 ( y-2)( y-4) is positive for 2 < y < 4, y ( t ) will increase and eventually approach y = 6, or 4000 ﬁsh. 1...

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- Spring '07
- staff
- Math