a. Explanation Given: The general solution is ln | P 300 − P | = 0.1 t + C . Proof: Given that the general solution is ln | P 300 − P | = 0.1 t + C . Substitute t = 0 in the solution, ln | P 300 − P | = 0.1 t + C ln | P ( 0 ) 300 − P ( 0 ) | = 0.1 ( 0 ) + C ln | P ( 0 ) 300 − P ( 0 ) | b. To determine To show: The solution is P = 300 1 + 5 e − 0.1 t .

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 7.9, Problem 68E

To determine

To show: The initial condition P(0)=50 implies that C=−ln5.

To determine

To show: The solution is P=3001+5e−0.1t.

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Chapter 7.9, Problem 67E

Chapter 7.9, Problem 69E

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The initial condition P ( 0 ) = 50 implies that C = − ln 5 .

Solution Summary: The author explains that the initial condition P(0)=50 implies that C=-mathrmln5.