Explanation Definition used: If S = { y 1 , y 2 } is the fundamental set of solutions of the homogenous equation y ″ + p ( t ) y ′ + q ( t ) y = 0 of the differential equation y ″ + p ( t ) y ′ + q ( t ) y = f ( t ) then, u ′ 1 = − y 2 f ( t ) W ( S ) and u ′ 2 = y 1 f ( t ) W ( S ) The particular solution is given by, y p = u 1 y 1 + u 2 y 2 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 2 y ″ − 5 x y ′ + 9 y = x 3 The corresponding homogenous equation is, x 2 y ″ − 5 x y ′ + 9 y = 0 x 2 d 2 y d x 2 − 5 x d y d x + 9 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. x 2 ⋅ 1 x 2 ( d 2 y d t 2 − d y d t ) − 5 x ⋅ 1 x d y d t + 9 y = 0 d 2 y d t 2 − d y d t − 5 d y d t + 9 y = 0 d 2 y d t 2 − 6 d y d t + 9 y = 0 The corresponding characteristic equation is, r 2 − 6 r + 9 = 0 ( r − 3 ) ( r − 3 ) = 0 r = 3 , 3 Therefore, the general solution of the Cauchy Euler equation x 2 y ″ − 5 x y ′ + 9 y = 0 is, y h ( t ) = c 1 e ( 3 ) t + c 2 t e ( 3 ) t = c 1 e 3 t + c 2 t e 3 t Substitute t = ln x in the above equation. y h ( x ) = c 1 e 3 ln x + c 2 t e 3 ln x = c 1 x 3 + c 2 ( ln x ) x 3 The fundamental set of solutions of the equation x 2 y ″ − 5 x y ′ + 9 y = 0 is given by, S = { y 1 , y 2 } = { x 3 , x 3 ln x } Hence, calculate the value of Wronskian, W ( S ) = | x 3 x 3 ln x d d t ( x 3 ) d d t ( x 3 ln x ) | = | x 3 x 3 ln x 3 x 2 x 2 ( 3 ln x + 1 ) | = x 5 ( 3 ln x + 1 ) − 3 x 5 ln x = x 5 Convert the Cauchy-Euler equation to its standard form

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Question

Chapter 4.7, Problem 22E

To determine

The general solution of the non homogenous Cauchy-Euler equation x2y″−5xy′+9y=x3.

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Chapter 4.7, Problem 21E

Chapter 4.7, Problem 23E

Chapter 4 Solutions

Introductory Differential Equations

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