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HE1ss

# HE1ss - Hour Exam I There were two versions of this exam...

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Unformatted text preview: Hour Exam I There were two versions of this exam, and in each version there were two or- derings of the questions. So the simplest way to give solutions is just to list the problems with no order. The questions from your exam are in here somewhere. A 1 . Determine which equilibrium solutions to the equation dy dt = y 3 (1+ y ) 2 (2- y ) are asymptotically stable and which are unstable. Then make a rough sketch showing the behavior of the solutions to this equation. Solution. The equilibrium solutions here are y ≡ - 1 , y ≡ 0, and y ≡ 2, and dy dx ( x ) < 0 for x > 2, dy dx ( x ) > 0 for 0 < x < 2, dy dx ( x ) < 0 for- 1 < x < 0, and dy dx ( x ) < 0 for x <- 1. That makes y ≡ 2 asymptotically stable, and y ≡ 0 and y ≡ - 1 unstable. [Cannot do sketches here, but they were important.] A 2 . Determine which equilibrium solutions to the equation dy dt = y (1- y ) 2 ( y- 2) 3 are asymptotically stable and which are unstable. Then make a rough sketch showing the behavior of the solutions to this equation. Solution. The equilibrium solutions here are y ≡ , y ≡ 1, and y ≡ 2, and dy dx ( x ) > for x > 2, , dy dx ( x ) < 0 for 1 < x < 2, dy dx ( x ) < 0 for 0 < x < 1 and dy dx ( x ) > 0 for x < 0. That makes y ≡ 0 asymptotically stable, and y ≡ 1 and y ≡ 2 unstable. [Cannot do sketches here, but they were important.] B 1 . Suppose that P ( t ) is the population (in millions) at time t (in years) of species that is being “harvested” at a rate of h million per year. If the growth rate of the population without harvesting is r , then P ( t ) satisfies dP dt = rP- h. Suppose that r = 0 . 4 and h = 0 . 5. If the population at time t = 0 is 1 (million), at what time will the population reach zero? Leave your answer in logarithms. Solution. Since this problem gives you the equation, just start with dP dt = (0 . 4) P- (0 . 5) and this implies P ( t ) = 5 4 + Ce (0 . 4) t . Since P (0) = 1, this gives P ( t ) = 5 / 4- (1 / 4)exp( . 4 t ). So solving for P ( T ) = 0 gets you e (0 . 4) T = 5 and T = 5 2 ln5 . An alternate way to do this is to treat dP dt = rP- h. as a separable equation. That gives Z dP P- h/r = Z rdt and ln | P- h/r | = rt + C. substituting the values for h and r , and using P (0) = 1 to determine C , you get ln | P- 5 / 4 | = 0 . 4 t + ln1 / 4 . 2 So if P ( T ) = 0, then T = (1 / . 4)(ln(5 / 4)- ln(1 / 4)) = (5 / 2)ln5 ....
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HE1ss - Hour Exam I There were two versions of this exam...

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