Explanation Given: The differential equation is ( e x − 1 − x ) y ″ + x y = 0 . Procedure used: Consider the differential equation as shown below. y ″ + P ( x ) y ' + Q ( x ) y = 0 The 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, cos x = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + … Maclaurin series of e x is, e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + … Calculation: The linear differential equation is given as follows: ( e x − 1 − x ) y ″ + x y = 0 Divide the above differential by e x − 1 − x . y ″ + ( x e x − 1 − x ) y = 0 Where P ( x ) = 0 , Q ( x ) = x e x − 1 − x . Substitute e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + … in Q ( 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 20RE

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

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Chapter 6, Problem 19RE

Chapter 6, Problem 21RE

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

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

Solution Summary: The author evaluates whether the differential equation (ex-1-x-)y′′+xy=0 is an ordinary point, singular point or irregular point.

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

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Chapter 6 Solutions