Question

(Problem 2:) (1) What is the general solution of the logistic equation, dN (a – bN)N? dt (2) Given that No = 3, a = ..001, calculate the carrying capacity. (3) Sketch the solution as t+00. (4) Calculate the doubling time tlog. Compare this doubling time with the doubling time texp of the exponential growth model: ON = aN with No = 3 and a = = .05. .05, b

Answer 1

Problem-2 1) dN = N(a - b) dt Let us write for convenience Na-bN)= =-{ N K 1 Where a=r=rate of growth and b=; K Where K denotes the Carrying Capacity dN N Hence 1 N dt K This model can be solved as follows. dN =rdt (1 - N/ KN Decompose the left hand side into partial fractions. dN IN = rdt + K-N N dN dN Integrate both the sides + : Ad dt K-N N to yield – log|K-N | + log |N= rt+C N After Arranging log = rt+ K-N So we have N K - N = C'ert Let us assume that when t=0, P(t)=P(0) N(0) Kert Then We can write, n(t) K + N(0)(ert – 1) N(0) e at b a + N(0)( ^-1.) b N(0) a e at at a+b N(O)(e -1) 2) Here, It is given that No = 3, a = .05, b = .001, The Carrying capacity K= a = 50 b N(0) a е at 3) For No = 3, a = .05, b= .001, N(t) at a+b N(0(“-1) or, N(t) .15e 054 .05+.003e05 Hence, lim lim -15e 054 N(1) .15 -1)) .003 50 056 t> t>0.05+.003 e' N(D) -50 -15e.ost .05+.003e" 056 -1)) t 0 50 100 150 200 4) .15e" Ost .15e Ost = 6 N(t) .05+.0036e954 – 1)) N(t) =65 .05+.003(ese – 1)) ht=15.181 Now for exponential growth an = aN with No = 3 and a = .05. N(t) = 3e05 N(t)=65e95 =6 In 2 204 = 250= = 13.863 Hence to = 13.863 -05