Explanation Given: For given boundary value problem boundary conditions are y ( 1 ) = 0 and y ′ ( e ) = 0 . Calculation: The given boundary value problem is, x 2 y ″ + x y ′ + λ y = 0 (1) There are three cases to find the Eigenvalues, Case 1: In case 1, λ = 0 Substitute λ = 0 in equation (1), x 2 y ″ + x y ′ = 0 The general solution is expressed as, y ( x ) = c 1 + c 2 ln x (2) Differentiating both sides of the equation (2) with respect to x , y ′ ( x ) = 0 + c 2 x = c 2 x At boundary condition y ( 1 ) = 0 and y ′ ( e ) = 0 , At y ′ ( e ) = 0 , y ′ ( e ) = c 2 ( e ) 0 = c 2 ( e ) c 2 = 0 Substitute c 2 = 0 in equation (2), y ( x ) = c 1 At y ( 1 ) = 0 , c 1 = 0 Substitute the value of c 1 and c 2 in equation (2), y ( x ) = 0 For eigenvalue λ = 0 the eigenfunction is y ( x ) = 0 . Case 2: In case 2, λ < 0 Let λ = − α 2 , where α is a positive integer. Substitute λ = − α 2 in equation (1), x 2 y ″ + x y ′ − α 2 y = 0 The general solution is expressed as, y ( x ) = c 1 x − α + c 2 x α (3) Differentiating both sides of the equation (3) with respect to x , y ′ ( x ) = − α c 1 x − α − 1 + α c 2 x α − 1 At boundary condition y ( 1 ) = 0 and y ′ ( e ) = 0 , At y ′ ( e ) = 0 , y ′ ( e ) = − α c 1 e − α − 1 + α c 2 e α − 1 0 = − α c 1 e − α − 1 + α c 2 e α − 1 α c 1 e − α − 1 = α c 2 e α − 1 c 1 = c 2 e 2 α Substitute c 1 = c 2 e 2 α in equation (3), y ( x ) = c 2 e 2 α x − α + c 2 x α = c 2 ( e 2 α x − α + x α ) At y ( 1 ) = 0 , c 2 = 0 Therefore, the values are c 1 = 0 and c 2 = 0 Substitute the value of c 1 = 0 and c 2 = 0 in equation (3), y ( x ) = 0 For Eigenvalue λ < 0 the Eigenfunction is y ( x ) = 0

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