Explanation Procedure used: 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 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 . The forcing function if f i ( t ) = t m is, F = { 1 , t , t 2 , ⋯ , t m } The forcing function if f i ( t ) = t m e α t is, F = { e α t , e α t t , e α t t 2 , ⋯ , e α t t m } The forcing function if f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t is, F = { e α t cos β t , e α t t cos β t , e α t t 2 cos β t , ⋯ , e α t t m cos β t , e α t sin β t , e α t t sin β t , e α t t 2 sin β t , ⋯ , e α t t m sin β t } The general solution if roots are imaginary in form of r 1 , 2 = α ± β i is, y ( t ) = e a t ( c 1 cos β t + c 2 sin β t ) The particular solution can be obtained by taking the linear combination of all functions in the associated sets. If any repeated function present then it should be appear only once in the particular solution. The general solution of the differential equation is y ( x ) = y p + y h Where, y p is particular solution and y h is complementary function. 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 . The forcing function if f i ( t ) = t m is, F = { 1 , t , t 2 , ... , t m } The forcing function if f i ( t ) = t m e α t is, F = { e α t , e α t t , e α t t 2 , ... , e α t t m } The forcing function if f i ( t ) = t m e α t cos β t or f i ( t ) = t m e α t sin β t is, F = { e α t cos β t , e α t t cos β t , e α t t 2 cos β t ,

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Question

Chapter 4.7, Problem 30E

To determine

The general solution of the non homogenous Cauchy-Euler equation x3y‴+16x2y″+70xy′+80y=x−13.

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Chapter 4.7, Problem 29E

Chapter 4.7, Problem 31E

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