Explanation Given: The boundary-value problem is, y ″ + λ y = 0 , y ( 0 ) = 0 , y ( π 4 ) = 0 . Procedure used: The general solution is obtained by considering the different cases for the parameter λ as zero, strictly negative and strictly positive. Apply the boundary conditions to find the eigenvalues and corresponding eigenfunctions. Calculation: When λ = 0 , the equation becomes y ″ = 0 , for which the general solution is given by y = c 1 x + c 2 . Substitute x = 0 and y ( 0 ) = 0 in above the general solution. 0 = 0 + c 2 c 2 = 0 Substitute x = π 4 and y ( 0 ) = 0 in the above general solution. 0 = c 1 ( π 4 ) + c 2 c 1 = 0 Therefore, the solution is y ( x ) = 0 . When λ < 0 , assume that λ = − α 2 , where without loss of generality, the number α is positive. Then, the auxiliary equation is m 2 − α 2 = 0 with roots − α and α , for which the general solution is given by y = c 1 cosh α x + c 2 sinh α x . Substitute x = 0 and y ( 0 ) = 0 in above general solution. 0 = c 1 + 0 c 1 = 0 Substitute x = π 4 and y ( 0 ) = 0 in the above general solution. 0 = 0 + c 2 c 2 = 0 Therefore, the solution is y ( x ) = 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.9, Problem 12E

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Chapter 3 Solutions

Advanced Engineering Mathematics

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