Explanation Given: The given differential equation is d P d t = P ( 10 − 1 − 10 − 7 P ) and the initial condition is P ( 0 ) = 5000 . Calculation: Consider the given differential equation. d P d t = P ( 10 − 1 − 10 − 7 P ) d P P ( 10 − 1 − 10 − 7 P ) = d t (1) Use partial fractions to decompose the left hand side of the Equation (1). 1 P ( 10 − 1 − 10 − 7 P ) = A P + B ( 10 − 1 − 10 − 7 P ) (2) Simplify the above equation. 1 P ( 10 − 1 − 10 − 7 P ) = A ( 10 − 1 − 10 − 7 P ) + B P P ( 10 − 1 − 10 − 7 P ) 1 = A ( 10 − 1 − 10 − 7 P ) + B P (3) Compare the coefficients of P on both sides of Equation (3). 0 = − A ( 10 − 7 ) + B B = A ( 10 − 7 ) (4) Compare the coefficients of constant terms on both sides of Equation (3). 1 = A ( 10 − 1 ) A = 1 10 − 1 = 10 Substitute 10 for A in Equation (4). B = ( 10 ) ( 10 − 7 ) = 10 − 6 Substitute 10 for A and 10 − 6 for B in Equation (2). 1 P ( 10 − 1 − 10 − 7 P ) = 10 P + 10 − 6 ( 10 − 1 − 10 − 7 P ) = 10 P + 10 − 6 10 − 7 ( 10 6 − P ) = 10 P − 10 P − 10 6 Substitute 10 P − 10 P − 10 6 for 1 P ( 10 − 1 − 10 − 7 P ) in Equation (1). [ 10 P − 10 P − 10 6 ] d P = d t Integrate both sides of the above equation. ∫ [ 10 P − 10 P − 10 6 ] d P = ∫ d t 10 ∫ 1 P d P − 10 ∫ 1 P − 10 6 d P = ∫ d t 10 ln | P | − 10 ln | P − 10 6 | = t + c 1 10 ln | P P − 10 6 | = t + c 1 Simplify the above equation. 10 ln | P P − 10 6 | = t + C 1 n | P P − 10 6 | = t + C 1 10 P P − 10 6 = e t + C 1 10 P P − 10 6 = C 2 e t 10 Simplify the above equation

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 2.8, Problem 3E

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The limiting value of the population and the time when the population will be equal to one-half of the limiting value.

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Chapter 2 Solutions

Advanced Engineering Mathematics

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The limiting value of the population and the time when the population will be equal to one-half of the limiting value.

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The limiting value of the population and the time when the population will be equal to one-half of the limiting value.

Chapter 2 Solutions