Explanation Given: The differential equation is x y ″ + ( 1 − cos x ) y ′ + x 2 y = 0 . Concept used: The linear differential equation is given as follows: y ″ + P ( x ) y ' + Q ( x ) y = 0 At point x = x 0 is an ordinary point if both P ( x ) and Q ( x ) are both analytic at x 0 . If a point in not ordinary, it is a singular point. A singular point x = x 0 is said to be regular singular point of the above differential equation if the function p ( x ) = ( x − x 0 ) P ( x ) and q ( x ) = ( x − x 0 ) 2 Q ( x ) are both analytic at x 0 . A singular point that is not regular is said to be an irregular singular point of the equation. Maclaurin series of cos x is represented as follows: cos x = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + … . Maclaurin series of e x is represented as follows: e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + … Calculation: The differential equation is given as follows: x y ″ + ( 1 − cos x ) y ′ + x 2 y = 0 Divide the above differential by x

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 6, Problem 19RE

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To check: The differential equation xy″+(1−cosx)y′+x2y=0 at the point x=0 is an ordinary point, singular point or an irregular point.

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Chapter 6, Problem 18RE

Chapter 6, Problem 20RE

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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To check: The differential equation x y ″ + ( 1 − cos x ) y ′ + x 2 y = 0 at the point x = 0 is an ordinary point, singular point or an irregular point.

Solution Summary: The author explains that the differential equation xy′′+(1-mathrmcos

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To check: The differential equation xy″+(1−cosx)y′+x2y=0 at the point x=0 is an ordinary point, singular point or an irregular point.

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Chapter 6 Solutions