Explanation Given: The population community levels off at 150. That is carrying capacity K = 150 , the initial population growth P ( 0 ) = 50 and the final population increases 600 in first month.so P ( 1 ) = 60 . Result used: The differential form of a logistic equation is P ′ ( t ) = r P ( 1 − P K ) , where r Natural growth, carrying capacity K and population P . Calculation: Use the above result mentioned to write the logistic differential equation with K = 150 . That is, P ′ ( t ) = r P ( 1 − P 150 ) Rewrite the above logistic equation as follows, P ′ ( t ) P ( 1 − P 2000 ) = r Integrate on both sides of the above expression with respect to t as follows, ∫ P ′ ( t ) P ( 1 − P 150 ) d t = ∫ r d t ∫ ( P ′ ( t ) P ( t ) + P ′ ( t ) 150 − P ( t ) ) d t = ∫ r d t Take P ( t ) = z . P ′ ( t ) = d z d t P ′ ( t ) d t = d z Then, ∫ d z z + ∫ P ′ ( t ) 150 − P ( t ) d t = ∫ r d t ln z − ln ( 150 − P ( t ) ) = r t + C ln ( z 150 − P ( t ) ) = r t + C Substitute the value of z = P ( t ) in the above equation. That is, ln ( P ( t ) 150 − P ( t ) ) = r t + C Where C is constant. Take exponential on both sides of the above solution. That is, e ln ( P ( t ) 150 − P ( t ) ) = e r t + C P ( t ) 150 − P ( t ) = c 1 e r t Reciprocate the above solution as follows, 150 − P ( t ) P ( t ) = A e − r t The general solution of the logistic equation is 150 − P ( t ) P ( t ) = A e − r t . The initial growth of population is P ( 0 ) = 50 . Substitute the initial value P ( 0 ) = 50 in the above general equation

Single Variable Calculus Format: Unbound (saleable)

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ISBN: 9780134765761

Author: Briggs, William L.^cochran, Lyle^gillett, Bernard^

Publisher: Prentice Hall

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Chapter 9.5, Problem 18E

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To find: The logistic function of the given population problem and graph the solution for population function.

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