Explanation Procedure used: If x = e t then, t = ln x . So, d t d x = 1 x So, d y d x = d y d t d t d x = 1 x d y d t Differentiate the above equation. d 2 y d x 2 = d d x ( 1 x d y d t ) = 1 x d d t ( d y d t ) d t d x + d d x ( 1 x ) d y d t = 1 x 2 d 2 y d t 2 − 1 x 2 d y d t = 1 x 2 ( d 2 y d t 2 − d y d t ) Substitution of these derivatives into a second 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. 3 x 2 y ″ − 4 x y ′ + 2 y = 0 3 x 2 d 2 y d x 2 − 4 x d y d x + 2 y = 0 Substitute 1 x d y d t for d y d x and 1 x 2 ( d 2 y d t 2 − d y d t ) for d 2 y d x 2 in the above equation

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 31E

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The solution to the initial value problem 3x2y″−4xy′+2y=0 with the initial conditions y(1)=2 and y′(1)=1.

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Chapter 4.7, Problem 30E

Chapter 4.7, Problem 32E

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Chapter 4 Solutions

Introductory Differential Equations

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The solution to the initial value problem 3 x 2 y ″ − 4 x y ′ + 2 y = 0 with the initial conditions y ( 1 ) = 2 and y ′ ( 1 ) = 1 .

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The solution to the initial value problem 3x2y″−4xy′+2y=0 with the initial conditions y(1)=2 and y′(1)=1.

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Chapter 4 Solutions