Explanation Given: The given boundary value problem is: y ″ + y = h ( x ) , y ( 0 ) = 0 , y ( 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 proof: Consider the given boundary value problem: y ″ + y = h ( x ) , y ( 0 ) = 0 , y ( π ) = 0 , Here A 0 = 1 , A 1 = 0 and A 2 = 1

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 29E

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To prove:

The given boundary value problem has a unique solutions for each h that is continuous on [0,π] and satisfies ∫0πh(x)sinxdx=0.

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Chapter 11.4, Problem 28E

Chapter 11.5, Problem 1E

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Fundamentals of Differential Equations and Boundary Value Problems

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To prove: The given boundary value problem has a unique solutions for each h that is continuous on [ 0 , π ] and satisfies ∫ 0 π h ( x ) sin x d x = 0 .

Solution Summary: The author explains that the given boundary value problem has unique solutions for each h that are continuous on [0,pi ] and satisfies displaystyle

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To prove:

The given boundary value problem has a unique solutions for each h that is continuous on [0,π] and satisfies ∫0πh(x)sinxdx=0.

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Chapter 11 Solutions

To prove: The given boundary value problem has a unique solutions for each h that is continuous on [ 0 , π ] and satisfies ∫ 0 π h ( x ) sin x d x = 0 .

Solution Summary: The author explains that the given boundary value problem has unique solutions for each h that are continuous on [0,pi ] and satisfies displaystyle

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To prove: The given boundary value problem has a unique solutions for each h that is continuous on [0,π] and satisfies ∫0πh(x)sinxdx=0.

Want to see the full answer?

Chapter 11 Solutions

To prove:

The given boundary value problem has a unique solutions for each h that is continuous on [0,π] and satisfies ∫0πh(x)sinxdx=0.