Explanation Procedure used: Particular solution to a linear DE by greens function: 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. Further, substitute the initial values to find the coefficients. 2. If y 1 ( x ) and y 2 ( x ) are two fundamental solution of the differential equation then particular solution is given by y p = u 1 y 1 ( x ) + u 2 y 2 ( x ) . 3. Use y p = ∫ x 0 x G ( x , t ) f ( t ) d t to find the particular solution. 4. Obtain the final solution by y ( x ) = y h + y p . Calculation: The given initial-value problem is y ″ + 3 y ′ + 2 y = sin e x , y ( 0 ) = − 1 , y ′ ( 0 ) = 0 . Consider the differential equation y ″ + 3 y ′ + 2 y = sin e x . The associated homogeneous differential equation is y ″ + 3 y ′ + 2 y = 0 . The auxiliary equation to y ″ + 3 y ′ + 2 y = 0 is m 2 + 3 m + 2 = 0 . Calculate the solution to the auxiliary equation, m 2 + 3 m + 2 = 0 ( m + 1 ) ( m + 2 ) = 0 m + 1 = 0 or m + 2 = 0 Hence, the roots of the auxiliary equation are m 1 = − 1 and m 2 = − 2 . Thus, the two solutions to the DE are y 1 = e − x and y 2 = e − 2 x . Hence, the solution of homogeneous DE y ″ + 3 y ′ + 2 y = 0 is y h ( x ) = c 1 e − x + c 2 e − 2 x . Consider the boundary values y ( 0 ) = − 1 , y ′ ( 0 ) = 0 , Substitute y ( 0 ) = − 1 into y h = c 1 e − x + c 2 e − 2 x , − 1 = c 1 + c 2 c 2 = − 1 − c 1 The first differential of y h = c 1 e − x + c 2 e − 2 x is, y ′ h = − c 1 e − x − 2 c 2 e − 2 x Substitute y ′ ( 0 ) = 0 into y ′ h = − c 1 e − x − 2 c 2 e − 2 x , 0 = − c 1 − 2 c 2 c 1 = − 2 c 2 c 2 = − 1 + 2 c 2 c 2 = 1 Hence, c 1 = − 2 . Thus, the values of coefficients are c 1 = − 2 and c 2 = 1 . Thus, the solution to the homogeneous DE y ″ + 3 y ′ + 2 y = 0 is, y h = − 2 e − x + e − 2 x . The particular solution to the nonhomogeneous differential equation is given by y p = ∫ x 0 x G ( x , t ) f ( t ) d t . Now, to determine the Wronskian, use variable transformation, [ y 1 ( x ) = − 2 e − x defines y 1 ( t ) = − 2 e − t , y 2 ( x ) = e − 2 x defines y 2 ( t ) = e − 2 t and f ( x ) = sin e x defines f ( t ) = sin e t ] W ( t ) = | − 2 e − t e − 2 t 2 e − t − 2 e − 2 t | = 4 e − 3 t − 2 e − 3 t = 2 e − 3 t Thus, for any t , W ( t ) ≠ 0 , determine the greens function G ( x , t ) , G ( x , t ) = y 1 ( t ) y 2 ( x ) − y 1 ( x ) y 2 ( t ) W ( t ) = − 2 e − t e − 2 x − ( − 2 e − x ) ( e − 2 t ) 2 e − 3 t = − e − 2 x + 2 t + e t − x = − e 2 ( t − x ) + e t − x Substitute the green function into y p = ∫ x 0 x G ( x , t ) f ( t ) d t and the initial limit is x 0 = 0

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3.10, Problem 25E

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The solution of the initial-value problem y″+3y′+2y=sinex,y(0)=−1,y′(0)=0.

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Chapter 3.10, Problem 24E

Chapter 3.10, Problem 26E

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Chapter 3 Solutions

Advanced Engineering Mathematics

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 42E Ch. 3.2 - Prob. 1E Ch. 3.2 - Prob. 2E Ch. 3.2 - Prob. 3E Ch. 3.2 - Prob. 4E Ch. 3.2 - Prob. 5E Ch. 3.2 - Prob. 6E Ch. 3.2 - Prob. 7E Ch. 3.2 - Prob. 8E Ch. 3.2 - Prob. 9E Ch. 3.2 - Prob. 10E Ch. 3.2 - Prob. 11E Ch. 3.2 - Prob. 12E Ch. 3.2 - Prob. 13E Ch. 3.2 - Prob. 14E Ch. 3.2 - Prob. 15E Ch. 3.2 - Prob. 16E Ch. 3.2 - Prob. 17E Ch. 3.2 - Prob. 18E Ch. 3.2 - Prob. 19E Ch. 3.2 - Prob. 20E Ch. 3.2 - 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The solution of the initial-value problem y ″ + 3 y ′ + 2 y = sin e x , y ( 0 ) = − 1 , y ′ ( 0 ) = 0 .

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The solution of the initial-value problem y″+3y′+2y=sinex,y(0)=−1,y′(0)=0.

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Chapter 3 Solutions