Program Explanation Given information: The pollutant concentration is c ( t ) = 10 ( 1 + cos t ) L / m 3 and it varies between 0 and 20, average depth of water is 2 meters the liquid pollutant flows at the rate of 2 × 10 5 m 3 per month. The period of oscillation is about 6 1 4 months. Explanation: The mount of solute x ( t ) satisfies the below differential equation. d x d t = r i c i − r o V x ....... (1) Here, r i , c i , r o are constants and V is the volume of the solution. The value of r i in millions of liters per second is calculated as, r i = 200 × 10 3 m 3 / month = 0.2 × 10 6 m 3 / month = 1 5 million liters The rate of inflow and outflow is same, so r o = 1 5 and the value of volume in millions of cubic meters is 2 . Substitute 1 5 , r i and 1 5 for r o , 10 ( 1 + cos t ) for c i and 2 for V in equation (1). d x d t = 1 5 ⋅ 10 ( 1 + cos t ) − 1 5 2 x d x d t = 2 ( 1 + cos t ) − 1 10 x d x d t + 1 10 x = 2 ( 1 + cos t ) The above equation is a linear first-order equation and its solution is calculated as, x e 1 10 ∫ d t = 2 ∫ ( 1 + cos t ) e 1 10 ∫ d t d t x e t 10 = 2 ∫ ( 1 + cos t ) e t 10 d t = 2 ∫ e t 10 d t + 2 ∫ cos t ( e t 10 ) d t Simplify the above equation. x e t 10 = 20 e t 10 + 2 ∫ cos t ( e t 10 ) d t + C ....... (2) Consider the integral ∫ cos t ( e t 10 ) d t as I and evaluate. I = ∫ cos t ( e t 10 ) d t The integral I can be solved with the integration by parts and then consider the below substitution. u = cos t ⇒ d u = − sin t d t d v = e t 10 d t ⇒ v = 10 e t 10 The integral I can be written as, I = 10 e t 10 cos t + 10 ∫ e t 10 sin t d t Consider the integral ∫ sin t ( e t 10 ) d t as I 1 and evaluate. I 1 = ∫ sin t ( e t 10 ) d t The integral I can be solved with the integration by parts and then consider the below substitution. u = sin t ⇒ d u = cos t d t d v = e t 10 d t ⇒ v = 10 e t 10 The integral I 1 can be written as, I 1 = 10 e t 10 sin t − 10 ∫ e t 10 cos t d t = 10 e t 10 sin t − 10 I The integration I can be written as, I = 10 e t 10 cos t + 10 I 1 = 10 e t 10 cos t + 10 ( 10 e t 10 sin t − 10 I ) = 10 e t 10 cos t + 100 e t 10 sin t − 100 I 101 I = 10 e t 10 cos t + 100 e t 10 sin t Simplify the above equation

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 1.5, Problem 46P

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Program Description: Purpose of problem is to calculate the amount of pollutant x(t) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10L/m3 .

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Chapter 1.5, Problem 45P

Chapter 1.6, Problem 1P

Chapter 1 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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Prob. 10P Ch. 1.3 - Prob. 11P Ch. 1.3 - Prob. 12P Ch. 1.3 - Prob. 13P Ch. 1.3 - Prob. 14P Ch. 1.3 - Prob. 15P Ch. 1.3 - Prob. 16P Ch. 1.3 - Prob. 17P Ch. 1.3 - Prob. 18P Ch. 1.3 - Prob. 19P Ch. 1.3 - Prob. 20P Ch. 1.3 - Prob. 21P Ch. 1.3 - Prob. 22P Ch. 1.3 - Prob. 23P Ch. 1.3 - Prob. 24P Ch. 1.3 - Prob. 25P Ch. 1.3 - Prob. 26P Ch. 1.3 - Prob. 27P Ch. 1.3 - Prob. 28P Ch. 1.3 - Verify that if c is a constant, then the function...Ch. 1.3 - Prob. 30P Ch. 1.3 - Prob. 31P Ch. 1.3 - Prob. 32P Ch. 1.3 - Prob. 33P Ch. 1.3 - (a) Use the direction field of Problem 5 to...Ch. 1.3 - Prob. 35P Ch. 1.4 - Prob. 1P Ch. 1.4 - Prob. 2P Ch. 1.4 - Prob. 3P Ch. 1.4 - Prob. 4P Ch. 1.4 - Prob. 5P Ch. 1.4 - Prob. 6P Ch. 1.4 - Prob. 7P Ch. 1.4 - Prob. 8P Ch. 1.4 - Prob. 9P Ch. 1.4 - Prob. 10P Ch. 1.4 - Prob. 11P Ch. 1.4 - Prob. 12P Ch. 1.4 - Prob. 13P Ch. 1.4 - Prob. 14P Ch. 1.4 - Prob. 15P Ch. 1.4 - Prob. 16P Ch. 1.4 - Prob. 17P Ch. 1.4 - Prob. 18P Ch. 1.4 - Prob. 19P Ch. 1.4 - Prob. 20P Ch. 1.4 - Prob. 21P Ch. 1.4 - 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Show that the substitution...Ch. 1.6 - Prob. 57P Ch. 1.6 - Prob. 58P Ch. 1.6 - Solve the differential equation dydx=xy1x+y+3 by...Ch. 1.6 - Prob. 60P Ch. 1.6 - Prob. 61P Ch. 1.6 - Prob. 62P Ch. 1.6 - Prob. 63P Ch. 1.6 - Prob. 64P Ch. 1.6 - Prob. 65P Ch. 1.6 - Prob. 66P Ch. 1.6 - Prob. 67P Ch. 1.6 - Prob. 68P Ch. 1.6 - Prob. 69P Ch. 1.6 - As in the text discussion, suppose that an...Ch. 1.6 - Prob. 71P Ch. 1.6 - Prob. 72P Ch. 1 - Prob. 1RP Ch. 1 - Prob. 2RP Ch. 1 - Prob. 3RP Ch. 1 - Prob. 4RP Ch. 1 - Prob. 5RP Ch. 1 - Prob. 6RP Ch. 1 - Prob. 7RP Ch. 1 - Prob. 8RP Ch. 1 - Prob. 9RP Ch. 1 - Prob. 10RP Ch. 1 - Prob. 11RP Ch. 1 - Prob. 12RP Ch. 1 - Prob. 13RP Ch. 1 - Prob. 14RP Ch. 1 - Prob. 15RP Ch. 1 - Prob. 16RP Ch. 1 - Prob. 17RP Ch. 1 - Prob. 18RP Ch. 1 - Prob. 19RP Ch. 1 - Prob. 20RP Ch. 1 - Prob. 21RP Ch. 1 - Prob. 22RP Ch. 1 - Prob. 23RP Ch. 1 - Prob. 24RP Ch. 1 - Prob. 25RP Ch. 1 - Prob. 26RP Ch. 1 - Prob. 27RP Ch. 1 - Prob. 28RP Ch. 1 - Prob. 29RP Ch. 1 - Prob. 30RP Ch. 1 - Prob. 31RP Ch. 1 - Prob. 32RP Ch. 1 - Prob. 33RP Ch. 1 - Prob. 34RP Ch. 1 - Prob. 35RP Ch. 1 - Prob. 36RP

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Program Description: Purpose of problem is to calculate the amount of pollutant x ( t ) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10 L / m 3 .

Solution Summary: The author calculates the amount of pollutant x(t) and shows that the graph is common with the steadily rising curve.

Concept explainers

Program Description: Purpose of problem is to calculate the amount of pollutant x(t) and show that the graph is common with the steadily rising curve and the time taken pollutant to reach 10L/m3 .

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Chapter 1 Solutions