It is easy to see that the population doubles when 120 60 e 14384 t so 0 14384

# It is easy to see that the population doubles when

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It is easy to see that the population doubles when 120 = 60 e 0 . 14384 t , so 0 . 14384 t d = ln(2) or the doubling time is t d = ln(2) r = 4 . 819 weeks . b. We begin by separating variables, so the general solution satisfies: Z dP P = Z ( a - b t ) dt or ln( P ( t )) = a t - bt 2 2 + C or P ( t ) = e C e a t - bt 2 2 . Since the initial value is P (0) = 60, it follows that e C = 60. Thus, P ( t ) = 60 e a t - bt 2 2 .

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We now use the data at t = 2 and 4 weeks. It follows from the solution above that 80 = 60 e 2 a - 2 b 90 = 60 e 4 a - 8 b . We rearrange the terms and take logarithms of both sides to get 2 a - 2 b = ln 4 3 4 a - 8 b = ln 3 2 . We solve these equations simultaneously to obtain 2 b = ln 4 3 - 1 2 ln 3 2 , so b = 0 . 042475. But a = b + 1 2 ln(4 / 3) or a = 0 . 1863. It follows that the solution is P ( t ) = 60 e 0 . 1863 t - 0 . 021237 t 2 . The population reaches a maximum when the derivative is zero, which occurs when t max = a b = 4 . 3865, so the maximum population is P ( t max ) = 90 . 286. 15. a. For the differential equation dy dt = t (2 - y ), the Euler formula is given by y n +1 = y n + h ( t n (2 - y n )) = y n + 0 . 25 ( t n (2 - y n )) . For this problem, y 0 = 4, we can use the Euler’s formula to create the following table: t 0 = 0 y 0 = 4 t 1 = 0 . 25 y 1 = y 0 + 0 . 25 ( t 0 (2 - y 0 )) = 4 + 0 . 25(0)(2 - 4) = 4 t 2 = 0 . 5 y 2 = y 1 + 0 . 25 ( t 1 (2 - y 1 )) = 4 + 0 . 25(0 . 25)(2 - 4) = 3 . 875 t 3 = 0 . 75 y 3 = y 2 + 0 . 25 ( t 2 (2 - y 2 )) = 3 . 875 + 0 . 25(0 . 5)(2 - 3 . 875) = 3 . 6406 t 4 = 1 . 0 y 4 = y 3 + 0 . 25 ( t 3 (2 - y 3 )) = 3 . 6406 + 0 . 25(0 . 75)(2 - 3 . 6406) = 3 . 3330 Thus, the approximate the solution at t = 1 is y 4 y (1) = 3 . 3330 . b. The differential equation is a separable differential equation. If we write the differential equation dy dt = - t ( y - 2), then we have the following integrals: Z dy y - 2 = - Z t dt ln | y - 2 | = - t 2 2 + C y ( t ) - 2 = e - t 2 / 2+ C = Ae - t 2 / 2 y ( t ) = 2 + Ae - t 2 / 2
With the initial condition, we find that A = 2. Thus, the solution to the initial value problem is y ( t ) = 2 + 2 e - t 2 / 2 . It follows that y (1) = 2 + 2 e - 0 . 5 3 . 21306. The error between the actual and Euler’s solution is 100 ( y 4 - y (1)) y (1) = 100 (3 . 3330 - 3 . 21306) 3 . 21306 = 3 . 73% . 16. a. For the differential equation, dR dt = - 0 . 05 R +0 . 2 e - 0 . 01 t with R (0) = 10 and h = 1, the Euler’s formula is R n +1 = R n + h ( - 0 . 05 R n + 0 . 2 e - 0 . 01 t n ) = R n - 0 . 05 R n + 0 . 2 e - 0 . 01 t n . Iterating this, we create a table t 0 = 0 R 0 = 10 t 1 = 1 R 1 = R 0 - 0 . 05 R 0 + 0 . 2 e - 0 . 01 t 0 = 10 - 0 . 5 + 0 . 2 = 9 . 7 t 2 = 2 R 2 = R 1 - 0 . 05 R 1 + 0 . 2 e - 0 . 01 t 1 = 9 . 7 - 0 . 485 + 0 . 198 = 9 . 413 t 3 = 3 R 3 = R 2 - 0 . 05 R 2 + 0 . 2 e - 0 . 01 t 2 = 9 . 413 - 0 . 471 + 0 . 096 = 9 . 138 Thus, the approximate the solution at t = 3 is R 3 R (3) = 9 . 138 . b. The DE is linear and can be written dR dt + 0 . 05 R = 0 . 2 e - 0 . 01 t with μ ( t ) = e 0 . 05 t . It follows that it can be written d dt e 0 . 05 t R ( t ) = 0 . 2 e 0 . 04 t or e 0 . 05 t R ( t ) = 0 . 2 Z e 0 . 04 t dt = 5 e 0 . 04 t + C. Thus, R ( t ) = 5 e - 0 . 01 t + Ce - 0 . 05 t , or R (0) = 10 = 5 + C. The solution is R ( t ) = 5 e - 0 . 05 t + 5 e - 0 . 01 t . The correct solution at t = 3 is R (3) = 9 . 15576. The percent error between the correct solution and the Euler solution 100 R 3 - R (3) R (3) = 100 9 . 138 - 9 . 15576 9 . 15576 = - 0 . 190% . 17. (Allee effect) Consider the DE given by the model: dP dt = P 9 - 0 . 01( P - 70) 2 = A ( P ) . The equilibria of this population model satisfy P ( 9 - 0 . 01( P - 70) 2 ) = 0 . Thus, P e = 0, 40, and 100. From the phase portrait below, it is easy to see that the equilibria P e = 0 and 100 are stable, while P e = 40 is unstable. The carrying capacity for this population is P e = 100, and the critical threshold number of animals required to avoid extinction is P e = 40.

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-20 0 20 40 60 80 100 120 -1000 -800 -600 -400 -200 0 200 400 600 800 1000 < > < > > P P ( 9 - 0 .
• Fall '08
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