Explanation Given: The boundary-value problem is y ″ + 2 y ′ + ( λ + 1 ) y = 0 , y ( 0 ) = 0 , y ( 5 ) = 0 . Calculation: The auxiliary equation corresponding to the differential equation y ″ + 2 y ′ + ( λ + 1 ) y = 0 , y ( 0 ) = 0 , y ( 5 ) = 0 is as follows: m 2 + 2 m + ( λ + 1 ) = 0 The value of m is calculated as follows: m 2 + 2 m + ( λ + 1 ) = 0 m = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( λ + 1 ) 2 ( 1 ) m = − 1 ± − λ In the given boundary value problem when λ = 0 , the general solution is obtained as follows: y = c 1 e − x + c 2 x e − x ( 1 ) Substitute x = 0 and y ( 0 ) = 0 in equation (1). 0 = c 1 e − 0 + c 2 ⋅ 0 ⋅ e − 0 c 1 = 0 Substitute x = 5 and y ( 5 ) = 0 in equation (1). 0 = 0 ⋅ e − 5 + c 2 ⋅ 5 ⋅ e − 5 c 2 = 0 Therefore, the solution is obtained as y ( x ) = 0 . When λ < 0 , assume that λ = − α 2 , without loss of generality, the number α is positive. Now the general solution is given as follows: y = e − x ( c 1 cosh α x + c 2 sinh α x ) ( 2 ) Substitute x = 0 and y ( 0 ) = 0 in equation (2). 0 = e − 0 ( c 1 cosh α ⋅ 0 + c 2 sinh α ⋅ 0 ) c 1 = 0 Substitute x = 5 and y ( 5 ) = 0 in equation (2)

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 5.2, Problem 15E

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The eigen values and eigen functions for the boundary-value problem y″+2y′+(λ+1)y=0, y(0)=0, y(5)=0.

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Chapter 5.2, Problem 14E

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The eigen values and eigen functions for the boundary-value problem y ″ + 2 y ′ + ( λ + 1 ) y = 0 , y ( 0 ) = 0 , y ( 5 ) = 0 .

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The eigen values and eigen functions for the boundary-value problem y″+2y′+(λ+1)y=0, y(0)=0, y(5)=0.

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