Program Explanation Given information: The differential equation is, x d y d x − 4 x 2 y + 2 y ln y = 0 Concept used: The linear first-order equation is, d y d x + P ( x ) y = Q ( x ) ....... (1) The solution of above equation is, y ( x ) e ∫ P ( x ) d x = ∫ ( Q ( x ) e ∫ P ( x ) d x ) d x + C Calculation: Consider v = ln y and differentiate with respect to x as d v d x = d d x ( ln y ) = 1 y d y d x d y d x = y d v d x The equation v = ln y can be written as e v = e ln y y = e v Substitute e v for y and e v d v d x for d y d x in the given differential equation as x e v d v d x − 4 x 2 e v + 2 e v v = 0 e v ( x d v d x − 4 x 2 + 2 v ) = 0 x d v d x − 4 x 2 + 2 v = 0 d v d x − 4 x + 2 x v = 0 Simplify the above equation as d v d x + 2 x v = 4 x

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 1.6, Problem 58P

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To calculate: The solution of the differential equation xdydx−4x2y+2ylny=0 .

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Chapter 1.6, Problem 57P

Chapter 1.6, Problem 59P

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Chapter 1 Solutions

EBK DIFFERENTIAL EQUATIONS

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The solution of the differential equation x d y d x − 4 x 2 y + 2 y ln y = 0 .

Solution Summary: The author explains that the solution of the differential equation xdy dx v=mathrmlny can be written as

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To calculate: The solution of the differential equation xdydx−4x2y+2ylny=0 .

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Chapter 1 Solutions