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 ′ + y = f ( x ) . The associate homogeneous differential equation to y ″ + 2 y ′ + y = f ( x ) is y ″ + 2 y ′ + y = 0 . The auxiliary equation to y ″ + 2 y ′ + y = 0 is m 2 + 2 m + 1 = 0 . Thus, calculate the solution to the auxiliary equation as, y ″ + 2 y ′ + y = 0 m 2 + 2 m + 1 = 0 ( m + 1 ) 2 = 0 Hence, the roots of the auxiliary equation are m 1 = − 1 and m 2 = − 1 . Thus, the two solutions to the DE are y 1 = e − x and y 2 = x e − 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 defines y 1 ( t ) = e − t and y 2 ( x ) = x e − x defines y 2 ( t ) = t e − 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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Chapter 4.8, Problem 3E

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

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Chapter 4.8, Problem 2E

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

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

Solution Summary: The author explains the particular solution to a linear DE by Green's function. The auxiliary equation is .

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

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