Explanation Procedure used: Particular solution to a linear DE by Green’s function: 1. Consider the differential equation and determine the complementary function to the homogeneous differential equation. 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. Calculation: Consider the differential equation y ″ − 2 y ′ + 2 y = f ( x ) . The associate homogeneous differential equation to y ″ − 2 y ′ + 2 y = f ( x ) is y ″ − 2 y ′ + 2 y = 0 . The auxiliary equation to y ″ − 2 y ′ + 2 y = 0 is m 2 − 2 m + 2 = 0 . Thus, calculate the solution to the auxiliary equation with help of m = − b ± b 2 − 4 a c 2 a , m = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 2 ) ( 1 ) 2 ( 1 ) = 2 ± ( 4 − 8 ) 2 = 1 ± i Hence, the roots of the auxiliary equation are m 1 = 1 + i and m 2 = 1 − i . Thus, the two solutions to the DE are y 1 = e x cos x and y 2 = e x sin 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. Thus, y 1 ( x ) = e x cos x defines y 1 ( t ) = e t cos t and y 2 ( x ) = e x sin x defines y 2 ( t ) = e t sin t

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 3.10, Problem 6E

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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

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The particular solution of the differential equation y ″ − 2 y ′ + 2 y = f ( x ) in the integral form.

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Chapter 3 Solutions