Explanation Given: For given boundary value problem boundary conditions are y ( 0 ) = 0 and y ( 1 ) + y ′ ( 1 ) = 0 . Calculation: The given boundary value problem is, y ″ + λ y = 0 . (1) There are three cases to find the Eigen values, Case 1: When λ = 0 , substitute λ = 0 in equation (1), y ″ = 0 The general solution is expressed as, y ( x ) = c 1 + c 2 x . (2) Differentiating both sides of the equation (2) with respect to x , y ′ ( x ) = c 2 At boundary condition y ( 0 ) = 0 and y ( 1 ) + y ′ ( 1 ) = 0 , y ( 0 ) = 0 y ( 0 ) = c 1 + c 2 ( 0 ) c 1 = 0 At y ( 1 ) + y ′ ( 1 ) = 0 , the value of c 2 is, 2 c 2 = 0 c 2 = 0 Substitute the value of c 1 and c 2 in equation (2), y ( x ) = 0 For Eigen value λ = 0 the Eigen function is y ( x ) = 0 . Case 2: When λ < 0 Let λ = − α 2 , where α is a positive integer. Substitute λ = − α 2 in equation (1), y ″ − α 2 y = 0 The general solution is expressed as, y ( x ) = c 1 cosh α x + c 2 sinh α x . (3) Differentiating both sides of the equation (3) with respect to x , y ′ ( x ) = α c 1 sinh α x + α c 2 cosh α x At boundary condition y ( 0 ) = 0 and y ( 1 ) + y ′ ( 1 ) = 0 , At y ( 0 ) = 0 , y ( 0 ) = c 1 cosh 0 + c 2 sinh 0 0 = c 1 + 0 c 1 = 0 Substitute c 1 = 0 in equation (3), y ( x ) = c 2 sinh α x Differentiate both sides of the above equation with respect to x as follows. y ′ ( x ) = α c 2 cosh α x y ( 1 ) + y ′ ( 1 ) = 0 c 2 sinh α x + α c 2 cosh α x = 0 c 2 ( sinh α x + α cosh α x ) = 0 Since α > 0 , ( sinh α x + α cosh α x ) ≠ 0 , Then c 2 = 0

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 5.2, Problem 38E

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To prove: The Eigen value is λn=αn2 and the Eigen function is yn(x)=sinαnx for the given boundary value problem.

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Chapter 5.2, Problem 37E

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Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

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To prove: The Eigen value is λ n = α n 2 and the Eigen function is y n ( x ) = sin α n x for the given boundary value problem.

Solution Summary: The author explains the Eigen value and Eigen function for the given boundary value problem.

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To prove: The Eigen value is λn=αn2 and the Eigen function is yn(x)=sinαnx for the given boundary value problem.

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