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Unformatted text preview: Mathematics 33B  Practice Midterm 1 Official exam to be administered Oct. 24th, 11:00am NAME (please print legibly): Your University ID Number: Your Discussion Section and TA: Signature: QUESTION VALUE SCORE 1 10 2 10 3 10 4 10 5 10 TOTAL 50 1 1. (10 points) 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. We have dy dx ( x ) < 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. Thus, y ≡ 2 asymptotically stable, and y ≡ 0 and y ≡  1 unstable. [Cannot do sketches here, but they are important.] 2 2. (10 points) 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 ln 5 . 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 + ln 1 / 4 . So if P ( T ) = 0, then T = (1 / . 4)(ln(5 / 4) ln(1 / 4)) = (5 / 2) ln 5 . 3 3. (10 points) (a) Find the solution to dy dx = 1 x (2 y 1) that satisfies y (1) = 2, AND find the domain of existence for this solution. Solution. Since this equation has no equilibrium solutions, you can begin with Z (2 y 1) dy = Z 1 x dx and y 2 y = ln  x  + C. Substituting y (1) = 2, you have 4 2 = ln 1 + C . So C = 2, and y 2 y ln  x   2 = 0 and y = 1 ± p 1 + 4 ln  x  + 8 2 You need the + sign to get y (1) = 2. So y ( x ) = 1 + p 9 + 4 ln  x  2 Since negative numbers have imaginary square roots, you need 9 + 4 ln  x  ≥ 0. So we need e 9 / 4 ≤  x  < ∞ . However, when the solution is defined on two intervals, you have to choose the one containing the initial value as the domain of existence. So the domain of existence is e 9 / 4 < x < ∞ and y ( x ) = (1 + √ 9 + 4 ln x ) / 2. b) What happens at the endpoint(s) of the domain of existence? Why doesn’t the existence theorem say that you should be able to continue the solution beyond them?...
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This note was uploaded on 02/08/2010 for the course MATH 33B taught by Professor Staff during the Winter '07 term at UCLA.
 Winter '07
 staff
 Math

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