Explanation Given: The differential Gompertz equation. is M ′ ( t ) = − r M ln ( M K ) with its initial point M ( 0 ) = M 0 and the value of r = 0.6 , K = 5500 and M 0 = 20 . Calculation: The differential Gompertz equation is M ′ ( t ) = − r M ln ( M K ) . The values are r = 0.6 , K = 5500 and M 0 = 20 . That means, the Gompertz equation is M ′ ( t ) = − 0.6 M ln ( M 5500 ) . Rewrite the above differential equation as follows, M ′ ( t ) M ln ( M 5500 ) = − 0.6 Integrate on both sides of the above expression with respect to t as follows, ∫ M ′ ( t ) M ln ( M 5500 ) d t = ∫ − 0.6 d t To solve the above integral, consider the following assumption. That is, Take ln ( M 500 ) = z Then, d ln ( M 5500 ) d t = d z d t 5500 M × d ( M 5500 ) d t = d z d t 5500 M × 1 5500 × d M d t = d z d t ( M ′ ( t ) M ) d t = d z Substitute these values in above integration ∫ M ′ ( t ) M ln ( M 5500 ) d t = ∫ − 0.6 d t . That is, ∫ d z z = ∫ − 0.6 d t ln z = − 0.6 t + C Substitute the value of z = ln ( M 5500 ) in the above solution. ln | ln ( M 5500 ) | = − 0.6 t + C Where C is constant. Take exponential on both sides of the above solution as follows, e ln | ln ( M 5500 ) | = e − 0

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter D1.5, Problem 22E

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Chapter D1.5, Problem 21E

Chapter D1.5, Problem 23E

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Calculus: Early Transcendentals (2nd Edition)

Ch. D1.1 - Prob. 1E Ch. D1.1 - Prob. 2E Ch. D1.1 - Prob. 3E Ch. D1.1 - If the general solution of a differential equation...Ch. D1.1 - Does the function y(t) = 2t satisfy the...Ch. D1.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...

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