Explanation Given: Initial conditions are given as N ( 0 ) = 500 and N ( 1 ) = 1000 . Calculation: Consider the non-linear differential equation that corresponds to the given data. d N d t = N ( a − b N ) The solution of the non-linear differential equation is, N ( t ) = a N 0 b N 0 + ( a − b N 0 ) e − a t (1) Substitute 0 for t and 500 for N 0 in Equation (1). N ( t ) = a ( 500 ) b ( 500 ) + ( a − b ( 500 ) ) e − a ( 0 ) = 500 a 500 b + ( a − 500 b ) e 0 = 500 a 500 b + ( a − 500 b ) (2) Calculate the value of N ( t ) as t → ∞ from Equation (1). N ( t ) = lim t → ∞ a N 0 b N 0 + ( a − b N 0 ) e − a t = lim t → ∞ a N 0 b N 0 + ( a − b N 0 ) ( 0 ) = a N 0 b N 0 = a b Thus, lim t → ∞ N ( t ) = a b . The limiting number of people in the community is 50 , 000 . Thus, a b = 50 , 000 b = a 50 , 000 (3) It is given that N ( 1 ) = 1000 . Substitute 1 for t , a 50 , 000 for b and 1000 for N in Equation (2)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.8, Problem 2E

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The value of N(t) if the limiting number of people in the community is 50,000.

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Chapter 2.8, Problem 1E

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The value of N ( t ) if the limiting number of people in the community is 50 , 000 .

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Chapter 2 Solutions