Explanation Theorem used: Separation of Variables: The solution of the differential equation, f ( y ) y ′ = g ( x ) is given implicitly by the equation, ∫ f ( y ) d y = ∫ g ( x ) d x . Calculation: The given differential equation y ′ 1 ( 2 − y ) 2 = e x is of the form f ( y ) y ′ = g ( x ) where, f ( y ) = 1 ( 2 − y ) 2 and g ( x ) = e x . By the above mentioned theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ 1 ( 2 − y ) 2 d y = ∫ e x d x That is, ∫ 1 ( 2 − y ) 2 d y = ∫ e x d x (1) Substitute u = 2 − y and d y = − d u in ∫ 1 ( 2 − y ) 2 d y and solve integral as follows. ∫ 1 ( 2 − y ) 2 d y = ∫ 1 u 2 ( − d u ) = − ∫ 1 u 2 d u = − ( u − 2 + 1 − 2 + 1 ) = 1 u = 1 2 − y That is, ∫ 1 ( 2 − y ) 2 d y = 1 2 − y

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Differential Equations

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Chapter 9.2, Problem 39E

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To find: The general solution for the differential equation y′=(2−y)2ex and its particular solution satisfying the initial condition y(0)=1.

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Chapter 9.2, Problem 38E

Chapter 9.2, Problem 40E

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Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

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The general solution for the differential equation y ′ = ( 2 − y ) 2 e x and its particular solution satisfying the initial condition y ( 0 ) = 1 .

Solution Summary: The author explains the general solution of the differential equation yprime, which is given implicitly by the equation, and its particular solution satisfying the initial condition.

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To find: The general solution for the differential equation y′=(2−y)2ex and its particular solution satisfying the initial condition y(0)=1.

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Chapter 9 Solutions