Explanation Given: The differential Gompertz equation. is M ′ ( t ) = − r M ln ( M K ) with its initial point M ( 0 ) = M 0 . The value of r = 0.1 , K = 500 and M 0 = 50 . Result used: The differential Gompertz equation. is M ′ ( t ) = − r M ln ( M K ) with its initial point M ( 0 ) = M 0 . Calculation: The differential Gompertz equation is M ′ ( t ) = − r M ln ( M K ) and the value of r = 0.1 , K = 500 and M 0 = 50 . That means, the Gompertz equation is M ′ ( t ) = − 0.1 M ln ( M 500 ) . Rewrite the above differential equation as follows, M ′ ( t ) M ln ( M 500 ) = − 0.1 Integrate on both sides of the above expression with respect to t as follows, ∫ M ′ ( t ) M ln ( M 500 ) d t = ∫ − 0.1 d t Take ln ( M 500 ) = z . Then, d ln ( M 500 ) d t = d z d t 500 M × d ( M 500 ) d t = d z d t 500 M × 1 500 × 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 500 ) d t = ∫ − 0.1 d t as follows, ∫ d z z = ∫ − 0.1 d t ln z = − 0.1 t + C Substitute the value of z = ln ( M 500 ) in the above solution. That is, ln | ln ( M 500 ) | = − 0.1 t + C Where C is constant

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ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter D1.5, Problem 20E

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Chapter D1.5, Problem 19E

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The Gompertz equation for the given value and graph its solution.

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