Explanation Given: The boundary-value problem is, y ″ + λ y = 0 , y ′ ( 0 ) = 0 , y ′ ( π ) = 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 . Differentiate the above general solution with respect to x . y ′ = c 1 . (1) Substitute x = 0 and y ′ ( 0 ) = 0 in the equation (1). c 1 = 0 For the non-trivial solution, y = c 2 . Therefore, λ = 0 is an eigenvalue and corresponding eigenvalue is y = 1 . 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 . Differentiate the above general solution with respect to x . y ′ = c 1 α sinh α x + c 2 α cosh α x . (2) Substitute x = 0 and y ′ ( 0 ) = 0 in the equation (2). c 2 = 0 Substitute x = π and y ′ ( π ) = 0 in the equation (2). α c 1 sinh α π = 0 Here α ≠ 0, sinh α π ≠ 0 , so that c 1 = 0 . Therefore, the solution is y ( x ) = 0 . When λ > 0 , write λ = α 2 , where α > 0 without loss of generality

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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The eigenvalues and eigenfunctions for the given boundary-value problem.

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