Suppose that a certain population obeys the logistic equation dy/dt = ry[1 - (y/

Question

Suppose that a certain population obeys the logistic equation dy/dt = ry[1 - (y/K)]. If y0 = K/3, find the time at which the initial population has doubled. Find the value of corresponding to r = 0.025 per year. If y0/K = , find the time T at which y(T)/K = . where 0

Explanation / Answer

A similar problem was answered here dy/dt = (r/K)*y(K-y) dy/ [ y*(K-y)] = (r/K)*dt 1/[y(K-y)] = A/y + B/(K-y) 1 = A*(K-y) + By = (B - A)y + A*K => AK = 1 => A = 1/K B-A = 0 => B = 1/K => dy/ [ y*(K-y)] = (r/K)dt = (1/K)dy/y + (1/K)dy/(K-y) = (1/K)dy/y - (1/K)dy/(y-K) dy/y - dy/(y-K) = (r)dt ln(y) - ln(y0) - ln(y-K) + ln(y0 - K) = r*t ln(y/y-K) - ln(y0/y0-K) = r*t y/(y-K) = [y0/(y0 - K)]*exp(rt) invert 1 - K/y = (y0-K)exp(-rt)/y0 K/y = 1 - (y0-K)exp(-rt)/y0 y = K*{1 - (y0-K)exp(-rt)/y0}-1 a) y0 = K/3 y = K*{1 - (y0-K)exp(-rt)/y0}-1 2*[K/3] = K*{1 - (K/3-K)exp(-rt)/K/3}-1} 2K/3 = K*{1 + 2*exp(-r*t)}-1 2/3 = {1 + 2*exp(-r*t)-1 3/2 = 1 + 2*exp(-r*t) 1/4 = exp(-r*t) t = ln(1/4) / -r = -ln(1/4) / 0.025 = 55.4517744 ~ 55 years (b) If y0 / K = a, find the time T at which y (T) /K = ß, where 0 8 as a -> 0 or as ß -> 1. Find thevalue of T for r = 0.025 per year, a = 0.1, and ß =0.9 y = K*{1 - (y0-K)exp(-rt)/y0}-1 y/K = {1 - (y0-K)exp(-rt)/y0}-1 y/K = {1 - (1-K/y0)exp(-rt)}-1 ß = {1 - (1 - 1/a)*exp(-rt)}-1 1/ß = 1 - (1 - 1/a)*exp(-rt) 1 - 1/ß = (1-1/a)*exp(-rt) (ß-1)/ß = (a-1)/a *exp(-rt) [a*(ß-1)]/[ß*(a-1)] = exp(-rt) t = ln [ß*(a-1) / a*(ß-1)]/r t = ln [ß*(1-a) / a*(1-ß)]/r Note as a-> 0 , (1-a)/a =>8 ln(8) => 8 a = 0.1, ß = 0.9 t = ln[ 0.9*(1- 0.1) / 0.1*(1- 0.9)]/ 0.025 = 175.777966 yr ~ 176yrs

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