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 boundary 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. Find greens function from G ( x , t ) = y 1 ( t ) y 2 ( x ) W t . 4. Obtain the final solution by y ( x ) = y h + y p . Calculation: Consider the differential equation y ″ − 2 y ′ + 2 y = e x . The associated homogeneous differential equation is y ″ − 2 y ′ + 2 y = 0 . The auxiliary equation to the homogeneous DE y ″ − 2 y ′ + 2 y = 0 is m 2 − 2 m + 2 = 0 . Find the roots of m 2 − 2 m + 2 = 0 . Use, m = − ( b ) ± b 2 − 4 a c 2 a . Thus, m 1 = − ( − 2 ) ± 4 − 4 ( 2 ) 2 ( 1 ) = 2 ± 2 i 2 = 1 + i Thus, the roots of the auxiliary equation are m 1 = 1 + i and m 2 = 1 − i Hence, the solution of homogeneous DE y ″ − 2 y ′ + 2 y = 0 is y h = e x ( c 1 cos x + c 2 sin x ) . Consider the boundary values y ( 0 ) = 0 , y ( π 2 ) = 0 . Substitute y ( 0 ) = 0 into y h = e x ( c 1 cos x + c 2 sin x ) . Thus, 0 = c 1 Thus, choose c 2 = 1 . Hence, y 1 = e x sin x . Substitute y ( π 2 ) = 0 . 0 = e π 2 ( c 2 ) Choose, c 1 = 1 . Thus, y 2 = e x cos 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, determine the Wronskian. Use, variable transformation

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4.8, Problem 41E

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The solution of the given boundary value problem y″−2y′+2y=ex,y(0)=0,y(π2)=0.

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Differential Equations with Boundary-Value Problems (MindTap Course List)

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The solution of the given boundary value problem y ″ − 2 y ′ + 2 y = e x , y ( 0 ) = 0 , y ( π 2 ) = 0 .

Solution Summary: The author describes the solution of a linear DE by greens function.

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The solution of the given boundary value problem y″−2y′+2y=ex,y(0)=0,y(π2)=0.

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