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Unformatted text preview: y = 1 10 y (1y 6 )2 15 = 1 60 y (6y )2 15 = 1 60 (y 2 +6 y8) =1 60 ( y2)( y4). The zeroes of1 60 ( y2)( y4) are at y = 2 and y = 6, so these are the equilibrium points. The graph of1 60 ( y2)( y4) 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 phaseline 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 sh, then y (0) = 1 . 5 since y measures thousands of sh. Since1 60 ( y2)( y4) is negative for y < 2, the sh population will decrease and eventually disappear; the lake will be completely farmed out. If the initial population of the lake is 3000 sh, then y (0) = 3. Since1 60 ( y2)( y4) is positive for 2 < y < 4, y ( t ) will increase and eventually approach y = 6, or 4000 sh. 1...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Math

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