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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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This note was uploaded on 05/01/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
 Spring '07
 staff
 Differential Equations, Equations

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