Explanation Given: The boundary-value problem is y ″ + λ y = 0 , y ( − π ) = 0 , y ( π ) = 0 . Calculation: In the given boundary value problem when λ = 0 the equation becomes y ″ = 0 . The general solution of which is y = c 1 x + c 2 . Substitute x = π and y ( π ) = 0 in above the general solution is, 0 = − c 1 π + c 2 (1) Substitute x = π and y ( π ) = 0 in above the general solution is, 0 = c 1 π + c 2 (2) Substitute the value of c 2 from equation (2) in equation (1), 0 = − c 1 π + ( − c 1 π ) c 1 = 0 Substitute the value of c 1 in equation (1), c 2 = 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 which the general solution is given by y = c 1 cosh α x + c 2 sinh α x . Substitute x = π and y ( π ) = 0 in above the general solution, 0 = c 1 cosh α π + c 2 sinh α π (3) Substitute x = − π and y ( − π ) = 0 in above the general solution, 0 = c 1 cosh α π − c 2 sinh α π (4) Substitute the value of c 2 from equation (4) in equation (3), c 1 = 0 Substitute the value of c 1 in equation (4), c 2 = 0 Therefore, the solution is y ( x ) = 0

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

Solution Summary: The author explains the eigenvalues and enumerates the equations for the given boundary-value problem.

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