Explanation Procedure used: Consider a linear homogenous n th order equation with constant real coefficient as, a y n + b y n − 1 + ⋯ + c y ′ + d y = 0 . The corresponding characteristic equation is, a r n + b r n − 1 + ⋯ + c r + d = 0 . Let the roots of the equation be r 1 , r 2 , … , r n . (a) Real and distinct roots: If the roots are real and distinct, then n linearly independent solutions are, y 1 = e r 1 t , y 2 = e r 2 t ,…, y n = e r n t . (b) Repeated roots: If the root r = r 1 of the characteristics equation has multiplicity k , that is ( r − r 1 ) k is a factor of the characteristic equation, then k linearly independent solutions are, y 1 = e r 1 t , y 2 = t e r 1 t ,…, y k = t k − 1 e r 1 t . (c) Roots are complex conjugate: If the roots are r 1 = α + β i and r 2 = α − β i with multiplicity k , then 2 k linearly independent solutions associated with r 1 , 2 = α ± β i are, y 1 = e α t cos β t , y 2 = e α t sin β t , y 3 = t e α t cos β t ,…, y 2 k − 1 = t k − 1 e α t cos β t , y 2 k = t k − 1 e α t sin β t . Calculation: Consider the given Cauchy-Euler equation. x 3 y ‴ + 22 x 2 y ″ + 124 x y ′ + 140 y = 0 (1) Let us assume y = x r . So, y ′ = r x r − 1 y ″ = r ( r − 1 ) x r − 2 y ‴ = r ( r − 1 ) ( r − 2 ) x r − 3 Substitute x r for y , r x r − 1 for y ′ , r ( r − 1 ) x r − 2 for y ″ and r ( r − 1 ) ( r − 2 ) x r − 3 for y ‴ in the equation (1)

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

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Chapter 4.7, Problem 13E

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The general solution of the Cauchy-Euler equation x3y‴+22x2y″+124xy′+140y=0.

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Chapter 4.7, Problem 12E

Chapter 4.7, Problem 14E

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Chapter 4 Solutions

Introductory Differential Equations

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The general solution of the Cauchy-Euler equation x 3 y ‴ + 22 x 2 y ″ + 124 x y ′ + 140 y = 0 .

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The general solution of the Cauchy-Euler equation x3y‴+22x2y″+124xy′+140y=0.

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Chapter 4 Solutions