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 x 2 y ″ − 2 x y ′ + 2 y = x ln x , y ( 1 ) = 1 , y ′ ( 1 ) = 0 . Consider the differential equation x 2 y ″ − 2 x y ′ + 2 y = x ln x . The associated homogeneous differential equation is x 2 y ″ − 2 x y ′ + 2 y = 0 . Consider that y = x m is a solution of homogeneous DE. Substitute y = x m into x 2 y ″ − 2 x y ′ + 2 y = 0 , x 2 m ( m − 1 ) x m − 2 − 2 x m x m − 1 + 2 x m = 0 x 2 m ( m − 1 ) x m x − 2 − 2 x m x m x − 1 + 2 x m = 0 x m ( m 2 − m ) − 2 m x m + 2 x m = 0 Since for any m , x m ≠ 0 . Calculate the roots of m 2 − 3 m + 2 = 0 as, m 2 − 3 m + 2 = 0 m 2 − m − 2 m + 2 = 0 m ( m − 1 ) − 2 ( m − 1 ) = 0 ( m − 1 ) ( 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 = x and y 2 = x 2 . Hence, the solution of homogeneous DE x 2 y ″ − 2 x y ′ + 2 y = 0 is y h = c 1 x + c 2 x 2 . Consider the boundary values y ( 1 ) = 1 , y ′ ( 1 ) = 0 , Substitute y ( 1 ) = 1 into y h = c 1 e − x + c 2 e − 2 x , 1 = c 1 + c 2 c 1 = 1 − c 2 The first differential of y h = c 1 e − x + c 2 e − 2 x is, y ′ h = c 1 + 2 c 2 x Substitute y ′ ( 1 ) = 0 into y h = c 1 x + c 2 x 2 , 0 = c 1 + 2 c 2 0 = 1 − c 2 + 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 x 2 y ″ − 2 x y ′ + 2 y = 0 is, y h ( x ) = 2 x − x 2 . Divide the DE x 2 y ″ − 2 x y ′ + 2 y = x by x 2 .to obtain a standard DE, y ″ − 2 1 x y ′ + 2 x 2 y = ln x x The particular solution to the nonhomogeneous differential equation is given by y p = ∫ x 0 x G ( x , t ) f ( t ) d t

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

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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

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Solution Summary: The author explains how to solve a linear DE by greens function.

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