Explanation The given differential equation is 2 y ″ − 3 y ′ − 2 y = 1 . The corresponding homogeneous differential equation is 2 y ″ − 3 y ′ − 2 y = 0 . Denote the differentiation by D y = d y d x and simplify. 2 y ″ − 3 y ′ − 2 y = 0 2 d 2 y d x 2 − 3 d y d x − 2 y = 0 2 D 2 y − 3 D y − 2 y = 0 ( 2 D 2 − 3 D − 2 ) y = 0 Therefore, the non-homogeneous differential equation 2 y ″ − 3 y ′ − 2 y = 1 can be written in the form of L ( y ) = g ( x ) as ( 2 D 2 − 3 D − 2 ) y = 1

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.5, Problem 4E

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To write: The differential equation 2y″−3y′−2y=1 in the form of L(y)=g(x), where L is a linear differential operator with constant coefficients, if possible, factor L.

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Chapter 4.5, Problem 3E

Chapter 4.5, Problem 5E

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The differential equation 2 y ″ − 3 y ′ − 2 y = 1 in the form of L ( y ) = g ( x ) , where L is a linear differential operator with constant coefficients, if possible, factor L .

Solution Summary: The author explains how to write the differential equation 2y′′-3yer -2y=1, where L is a linear differential operator with constant coefficients, if possible.

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To write: The differential equation 2y″−3y′−2y=1 in the form of L(y)=g(x), where L is a linear differential operator with constant coefficients, if possible, factor L.

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Chapter 4 Solutions