a. Explanation Given: Consider the chemical rate equation y ′ ( t ) = − k y ( t ) (1) And y ′ ( t ) = − k y 2 ( t ) (2) Where y ( t ) is the concentration of the compound for t ≥ 0 and k > 0 , is a constant that determines the speed of the reaction and n is a positive integer. The initial concentration of the compound is y ( 0 ) = y 0 > 0 .and k = 0.3 . Calculation: To find the equilibrium solution, set the given differential equation equal to zero. The first chemical equation becomes, y ′ ( t ) = 0 − k y ( t ) = 0 < b. To determine To find: The direction field which approaches its equilibrium solution faster.

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter D1.2, Problem 48E

To determine

To find: The equilibrium solution in both cases and sketch the direction field at k=0.3 .

To determine

To find: The direction field which approaches its equilibrium solution faster.

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Chapter D1.2, Problem 47E

Chapter D1.2, Problem 49E

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Chapter D1 Solutions

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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The equilibrium solution in both cases and sketch the direction field at k = 0.3 .

Solution Summary: The author explains how to find the equilibrium solution in both cases and sketch the direction field at k=0.3.

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To find: The equilibrium solution in both cases and sketch the direction field at k=0.3 .

To find: The direction field which approaches its equilibrium solution faster.

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Chapter D1 Solutions