Explanation Calculation: Differentiate the function y = 2 ( 1 + c e − 6 t ) with respect to t . y = 2 ( 1 + c e − 6 t ) y = 2 ( 1 + c e − 6 t ) − 1 d y d t = − 2 ( 1 + c e − 6 t ) − 2 ( − 6 c e − 6 t ) d y d t = 12 c e − 6 t ( 1 + c e − 6 t ) 2 Thus, the first derivative of y is d y d t = 12 c e − 6 t ( 1 + c e − 6 t ) 2 . Substitute d y d t = 12 c e − 6 t ( 1 + c e − 6 t ) 2 and y = 2 ( 1 + c e − 6 t ) in differential equation d y d t = 3 y ( 2 − y ) to check whether the solution satisfies it or not. 12 c e − 6 t ( 1 + c e − 6 t ) 2 = 3 2 ( 1 + c e − 6 t ) ( 2 − 2 ( 1 + c e − 6 t ) ) 12 c e − 6 t ( 1 + c e − 6 t ) 2 = 12 ( 1 + c e − 6 t ) ( 1 + c e − 6 t <

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 5.3, Problem 64E

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To show: The function y=2(1+ce−6t) is a solution of the differential equation dydt=3y(2−y) where c is any real number and find the particular solution that passes through the point (0,1).

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Chapter 5.3, Problem 63E

Chapter 5.3, Problem 65E

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Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

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The function y = 2 ( 1 + c e − 6 t ) is a solution of the differential equation d y d t = 3 y ( 2 − y ) where c is any real number and find the particular solution that passes through the point ( 0 , 1 ) .

Solution Summary: The author explains that the function y=2(1+ce-6t) is a solution of the differential equation.

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To show: The function y=2(1+ce−6t) is a solution of the differential equation dydt=3y(2−y) where c is any real number and find the particular solution that passes through the point (0,1).

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