22BsolnsHW3

22BsolnsHW3 - Math 22B Solutions Homework 3 Spring 2008...

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Unformatted text preview: Math 22B Solutions Homework 3 Spring 2008 Section 2.5 4. dy dt = e y- 1, for-∞ < y < ∞ . Solution See attached figure 1 for graphs. To find the critical points, we solve the equation dy dt = 0 to get: e y- 1 = 0 ⇔ e y = 1 ⇔ y = ln 1 ⇔ y = 0 So the only critical point is at y = 0, and since f (0) = 1 > 0, the equilibrium soln. φ ( t ) = 0 is unstable. 13. dy dt = y 2 (1- y ) 2 Solution See attached figure 2 for the graphs. To find the critical values we again solve dy dt = 0 to get: y 2 (1- y ) 2 = 0 ⇔ y = 0 or y = 1 Now, since f ( y ) = y 2 (1- y ) 2 = y 4- 2 y 3 + y 2 , we see that f ( y ) = 4 y 3- 6 y 2 +2 y . So then f (0) = 0 and f (1) = 0, so both equilibrium solns. are semistable. 20. (a) dy dt = r (1- y k ) y- Ey Solution E must be smaller than r or else the population would deplete completely. Plug in y 1 = 0: = r (1- k )0- E (0) = 0 Plug in y 2 = K (1- E r ) > 0: = r (1- y k ) y- Ey = r (1- (1- E r )) K (1- E r )- EK (1- E r ) 1 = + EK (1- E r )- EK (1- E r ) (b) Solution...
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22BsolnsHW3 - Math 22B Solutions Homework 3 Spring 2008...

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