Explanation Given: The given boundary value problem is: x 2 y ″ ( x ) + 3 x y ′ ( x ) + 2 y ( x ) = h ( x ) ; y ′ ( 1 ) = 0 , y ′ ( e π ) = 0 Procedure used: Let L be the differential operator given by the L [ y ] = A 2 y ″ + A 1 y ′ + A 0 y . The formal adjoint of L is the differential operator L + define by L + [ y ] = ( A 2 y ) ′′ − ( A 1 y ) ′ + A 0 y provided A 2 ′′ , A 1 ′ and A 0 are continuous on [ a , b ] . If the roots of auxiliary equation having complex conjugate α ± β i . Then, The general solution is: y ( x ) = c 1 e α x cos β x + c 2 e α x sin β x Find the constant with the help of boundary conditions. Then, apply the Fredholm alternate theorem to determine the condition for solution. Fredholm alternate theorem: Let B is the set of linear boundary conditions. The nonhomogeneous boundary value problem: L [ y ] ( x ) = h ( x ) , a < x < b , B [ y ] = 0 has a solution if and only if ∫ a b h ( x ) z ( x ) d x = 0 Calculation: Consider the given boundary value problem: x 2 y ″ ( x ) + 3 x y ′ ( x ) + 2 y ( x ) = 0 Here A 0 = 2 , A 1 = 3 x and A 2 = x 2 . Substitute in the formal adjoint formula: L + [ y ] = ( x 2 y ) ′′ − ( 3 x y ) ′ + 2 y = [ x 2 y ′ + 2 x y ] ′ − [ 3 x y ′ + 3 y ] + 2 y = [ x 2 y ″ + 2 x y ′ + 2 x y ′ + 2 y ] − 3 x y ′ − 3 y + 2 y = x 2 y ″ + x y ′ + y Let y = x r and substitute in the above equation The auxiliary equation becomes:

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 11.4, Problem 25E

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The condition on h for the given nonhomogeneous boundary value problem.

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The condition on h for the given nonhomogeneous boundary value problem.

Solution Summary: The author explains the condition on h for the given nonhomogeneous boundary value problem. Let L be the differential operator given by Lleft[yright]=A_2y

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The condition on h for the given nonhomogeneous boundary value problem.

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