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 y ′ = x e x 2 is in the form of f ( y ) y ′ = g ( x ) where, f ( y ) = y and g ( x ) = x e x 2 . By the above theorem, the solution is given by, ∫ f ( y ) d y = ∫ g ( x ) d x ∫ y d y = ∫ x e x 2 d x That is, ∫ y d y = ∫ x e x 2 d x (1) Obtain the value of the integral ∫ x e x 2 d x by substitution method. Substitute u = x 2 and x d x = d u 2 in ∫ x e x 2 d x . ∫ x e x 2 d x = ∫ e u ( d u 2 ) = 1 2 ∫ e u d u = 1 2 e u = 1 2 e x 2 + C Thus, ∫ x e x 2 d x = 1 2 e x 2 + C . The equation (1) is simplified as follows

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 48E

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To find: The particular solution of the differential equation y′=xex2y satisfying the initial condition. y(0)=2.

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Chapter 9.2, Problem 47E

Chapter 9.2, Problem 49E

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

Ch. 9.1 - Show that y=Cx+1 is the general solution of the...Ch. 9.1 - Prob. 2MP Ch. 9.1 - Prob. 3MP Ch. 9.1 - Prob. 4MP Ch. 9.1 - Prob. 5MP Ch. 9.1 - Prob. 1ED Ch. 9.1 - Prob. 2ED Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E

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The particular solution of the differential equation y ′ = x e x 2 y satisfying the initial condition. y ( 0 ) = 2 .

Solution Summary: The author explains the differential equation yprime =xe2y satisfying the initial condition. The solution is given implicitly by the equation.

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To find: The particular solution of the differential equation y′=xex2y satisfying the initial condition. y(0)=2.

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