Explanation Procedure used: 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 Substitution of these derivatives into a higher order Cauchy-Euler equation yields a second order linear differential equation with constant coefficients. Consider a linear homogenous n th order equation with constant real coefficient. 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 . The, the different conditions are, (a) 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) Roots are repeated 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 a t cos β t , y 2 = e a t sin β t , y 3 = t e a t cos β t ,………………….., y 2 k − 1 = t k − 1 e a t cos β t , y 2 k = t k − 1 e a t sin β t . Calculation: Consider the Cauchy-Euler equation. x 3 y ‴ + 3 x 2 y ″ − 2 x y ′ − 2 y = 0 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 above equation

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Question

Chapter 4.7, Problem 18E

To determine

The general solution of the Cauchy-Euler equation x3y‴+3x2y″−2xy′−2y=0.

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Chapter 4.7, Problem 17E

Chapter 4.7, Problem 19E

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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 ‴ + 3 x 2 y ″ − 2 x y ′ − 2 y = 0 .

The general solution of the Cauchy-Euler equation x3y‴+3x2y″−2xy′−2y=0.

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Chapter 4 Solutions