Explanation Theorem used: “Let the functions M , N , M y , and N x , be continuous in the rectangular region R : α < x < β , γ < y < δ . Then M ( x , y ) + N ( x , y ) y ′ = 0 is an exact differential equation in R if and only if M y ( x , y ) = N x ( x , y ) at each point of R ”. Calculation: The given equation is ( 9 x 2 + y − 1 ) − ( 4 y − x ) y ′ = 0 . Here, it is observed that M = 9 x 2 + y − 1 and N = x − 4 y . Differentiate M = 9 x 2 + y − 1 with respect to y . M y = d d y ( 9 x 2 + y − 1 ) = 1 Differentiate N = x − 4 y with respect to x . N x = d d x ( x − 4 y ) = 1 By the theorem stated above, the given equation is exact. There is a function ψ ( x , y ) such that ψ x ( x , y ) = 9 x 2 + y − 1 and ψ y ( x , y ) = x − 4 y . Integrate the equation ψ x ( x , y ) = 9 x 2 + y − 1 with respect to x . ψ ( x , y ) = ∫ ( 9 x 2 + y − 1 ) d x = 9 ( x 3 3 ) + y ( x ) + h ( y ) = 3 x 3 + x y − x + h ( y ) Differentiate ψ ( x , y ) = 3 x 3 + x y − x + h ( y ) with respect to y . ψ y ( x , y ) = x + h ′ ( y ) solve the equation ψ y ( x , y ) = N as follows. x + h ′ ( y ) = x − 4 y h ' ( y ) = − 4 y It is implies that h ( y ) = − 2 y 2 . That is, ψ ( x , y ) = 3 x 2 + x y − x − 2 y 2

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Question

Chapter 2.6, Problem 10P

To determine

To solve: The initial problem (9x2+y−1)−(4y−x)y′=0, y(1)=0 and determine at least approximately where the solution is valid.

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Chapter 2.6, Problem 9P

Chapter 2.6, Problem 11P

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Prob. 3P Ch. 2.1 - Prob. 4P Ch. 2.1 - Prob. 5P Ch. 2.1 - Prob. 6P Ch. 2.1 - Prob. 7P Ch. 2.1 - Prob. 8P Ch. 2.1 - Prob. 9P Ch. 2.1 - Prob. 10P

Ch. 2.1 - Prob. 11P Ch. 2.1 - Prob. 12P Ch. 2.1 - Prob. 13P Ch. 2.1 - Prob. 14P Ch. 2.1 - Prob. 15P Ch. 2.1 - Prob. 16P Ch. 2.1 - Prob. 17P Ch. 2.1 - Prob. 18P Ch. 2.1 - Prob. 19P Ch. 2.1 - Prob. 20P Ch. 2.1 - Prob. 21P Ch. 2.1 - Prob. 22P Ch. 2.1 - Prob. 23P Ch. 2.1 - Prob. 24P Ch. 2.1 - Prob. 25P Ch. 2.1 - Prob. 26P Ch. 2.1 - Prob. 27P Ch. 2.1 - Variation of Parameters. Consider the following...Ch. 2.1 - Prob. 29P Ch. 2.1 - In each of Problems 29 and 30, use the method of...Ch. 2.2 - Prob. 1P Ch. 2.2 - Prob. 2P Ch. 2.2 - Prob. 3P Ch. 2.2 - Prob. 4P Ch. 2.2 - Prob. 5P Ch. 2.2 - Prob. 6P Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - Prob. 9P Ch. 2.2 - Prob. 10P Ch. 2.2 - Prob. 11P Ch. 2.2 - Prob. 12P Ch. 2.2 - Prob. 13P Ch. 2.2 - Prob. 14P Ch. 2.2 - Prob. 15P Ch. 2.2 - Prob. 16P Ch. 2.2 - Prob. 17P Ch. 2.2 - Prob. 18P Ch. 2.2 - Prob. 19P Ch. 2.2 - Prob. 20P Ch. 2.2 - Prob. 21P Ch. 2.2 - Prob. 22P Ch. 2.2 - Prob. 23P Ch. 2.2 - Prob. 24P Ch. 2.2 - Prob. 25P Ch. 2.2 - Prob. 26P Ch. 2.2 - Prob. 27P Ch. 2.2 - Prob. 28P Ch. 2.2 - Prob. 29P Ch. 2.2 - Prob. 30P Ch. 2.2 - Prob. 31P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.3 - Prob. 10P Ch. 2.3 - Prob. 11P Ch. 2.3 - Prob. 12P Ch. 2.3 - Prob. 13P Ch. 2.3 - Prob. 14P Ch. 2.3 - Prob. 15P Ch. 2.3 - Prob. 16P Ch. 2.3 - Assume that the conditions are as in Problem 16...Ch. 2.3 - Prob. 18P Ch. 2.3 - Prob. 19P Ch. 2.3 - Prob. 20P Ch. 2.3 - Prob. 21P Ch. 2.3 - Prob. 22P Ch. 2.3 - Prob. 23P Ch. 2.3 - Prob. 24P Ch. 2.4 - In each of Problems 1 through 6, determine...Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.4 - Prob. 10P Ch. 2.4 - Prob. 11P Ch. 2.4 - Prob. 12P Ch. 2.4 - Prob. 13P Ch. 2.4 - Prob. 14P Ch. 2.4 - Prob. 15P Ch. 2.4 - Prob. 16P Ch. 2.4 - Prob. 17P Ch. 2.4 - Prob. 18P Ch. 2.4 - Prob. 19P Ch. 2.4 - Prob. 20P Ch. 2.4 - Prob. 21P Ch. 2.4 - Prob. 22P Ch. 2.4 - Prob. 23P Ch. 2.4 - Prob. 24P Ch. 2.4 - Prob. 25P Ch. 2.4 - Prob. 26P Ch. 2.4 - Prob. 27P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.5 - Prob. 12P Ch. 2.5 - Prob. 13P Ch. 2.5 - Prob. 14P Ch. 2.5 - Prob. 15P Ch. 2.5 - Prob. 16P Ch. 2.5 - Prob. 17P Ch. 2.5 - Prob. 18P Ch. 2.5 - Prob. 19P Ch. 2.5 - Prob. 20P Ch. 2.5 - Prob. 21P Ch. 2.5 - Prob. 22P Ch. 2.5 - Prob. 23P Ch. 2.5 - Prob. 24P Ch. 2.5 - Prob. 25P Ch. 2.5 - Prob. 26P Ch. 2.5 - Prob. 27P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2.6 - Prob. 5P Ch. 2.6 - Prob. 6P Ch. 2.6 - Prob. 7P Ch. 2.6 - Prob. 8P Ch. 2.6 - Prob. 9P Ch. 2.6 - Prob. 10P Ch. 2.6 - Prob. 11P Ch. 2.6 - Prob. 12P Ch. 2.6 - Prob. 13P Ch. 2.6 - Prob. 14P Ch. 2.6 - Prob. 15P Ch. 2.6 - Prob. 16P Ch. 2.6 - Prob. 17P Ch. 2.6 - Prob. 18P Ch. 2.6 - Prob. 19P Ch. 2.6 - Prob. 20P Ch. 2.6 - Prob. 21P Ch. 2.6 - Prob. 22P Ch. 2.7 - Prob. 1P Ch. 2.7 - Prob. 2P Ch. 2.7 - Prob. 3P Ch. 2.7 - Prob. 4P Ch. 2.7 - Prob. 5P Ch. 2.7 - Prob. 6P Ch. 2.7 - Prob. 7P Ch. 2.7 - Prob. 8P Ch. 2.7 - Prob. 9P Ch. 2.7 - Prob. 10P Ch. 2.7 - Prob. 11P Ch. 2.7 - Prob. 12P Ch. 2.7 - Prob. 14P Ch. 2.7 - Prob. 15P Ch. 2.7 - Prob. 16P Ch. 2.7 - Prob. 17P Ch. 2.8 - Prob. 1P Ch. 2.8 - Prob. 2P Ch. 2.8 - Prob. 3P Ch. 2.8 - Prob. 4P Ch. 2.8 - Prob. 5P Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - Prob. 8P Ch. 2.8 - Prob. 9P Ch. 2.8 - Prob. 10P Ch. 2.8 - Prob. 11P Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - Prob. 14P Ch. 2.8 - Prob. 15P Ch. 2.8 - Prob. 16P Ch. 2.8 - Prob. 17P Ch. 2.8 - Prob. 18P Ch. 2.9 - Prob. 1P Ch. 2.9 - Prob. 2P Ch. 2.9 - Prob. 3P Ch. 2.9 - Prob. 4P Ch. 2.9 - Prob. 5P Ch. 2.9 - Prob. 6P Ch. 2.9 - Prob. 7P Ch. 2.9 - Prob. 8P Ch. 2.9 - Prob. 9P Ch. 2.9 - Prob. 10P Ch. 2 - Prob. 1MP Ch. 2 - Prob. 2MP Ch. 2 - Prob. 3MP Ch. 2 - Prob. 4MP Ch. 2 - Prob. 5MP Ch. 2 - Prob. 6MP Ch. 2 - Prob. 7MP Ch. 2 - Prob. 8MP Ch. 2 - Prob. 9MP Ch. 2 - Prob. 10MP Ch. 2 - Prob. 11MP Ch. 2 - Prob. 12MP Ch. 2 - Prob. 13MP Ch. 2 - Prob. 14MP Ch. 2 - Prob. 15MP Ch. 2 - Prob. 16MP Ch. 2 - Prob. 17MP Ch. 2 - Prob. 18MP Ch. 2 - Prob. 19MP Ch. 2 - Prob. 20MP Ch. 2 - Prob. 21MP Ch. 2 - Prob. 22MP Ch. 2 - Prob. 23MP Ch. 2 - Prob. 24MP Ch. 2 - Prob. 25MP Ch. 2 - Prob. 28MP Ch. 2 - Prob. 29MP Ch. 2 - Prob. 31MP Ch. 2 - Prob. 32MP Ch. 2 - Prob. 33MP Ch. 2 - Prob. 34MP Ch. 2 - Prob. 35MP Ch. 2 - Prob. 36MP Ch. 2 - Prob. 37MP

Knowledge Booster

The initial problem ( 9 x 2 + y − 1 ) − ( 4 y − x ) y ′ = 0 , y ( 1 ) = 0 and determine at least approximately where the solution is valid.

To solve: The initial problem (9x2+y−1)−(4y−x)y′=0, y(1)=0 and determine at least approximately where the solution is valid.

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Chapter 2 Solutions