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. 4. The general solution to a nonhomogeneous differential equation is given by y ( x ) = y h + y p , where, y h is the solution to the homogeneous equation associate to the differential equation L ( x ) with nonhomogeneous boundary conditions and y p is the solution to the nonhomogeneous differential equation with homogeneous boundary values. Calculation: Consider the differential equation y ″ − 16 y = x e − 2 x . The associate homogeneous differential equation to y ″ − 16 y = x e − 2 x is y ″ − 16 y = 0 . The auxiliary equation to y ″ − 16 y = 0 is m 2 − 16 = 0 . Thus, calculate the solution to the auxiliary equation as, m 2 − 16 = 0 m 2 = 16 m = ± 4 Hence, the roots of the auxiliary equation are m 1 = 4 and m 2 = − 4 . Thus, the two solutions to the DE are y 1 = e 4 x and y 2 = e − 4 x . Thus, the solution to homogeneous DE associate to y ″ − 16 y = f ( x ) is y h = c 1 e 4 x + c 2 e − 4 x . Since, there are no boundary values or initial values, leave the coefficients unknown. 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 4 x defines y 1 ( t ) = e 4 t , y 2 ( x ) = e − 4 x defines y 2 ( t ) = e − 4 t and f ( x ) = x e − 2 x defines f ( t ) = t e − 2 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 7E

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The general solution to the differential equation y″−16y=xe−2x in the integral form of particular solution.

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Chapter 4.8, Problem 6E

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

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The general solution to the differential equation y ″ − 16 y = x e − 2 x in the integral form of particular solution.

Solution Summary: The author explains the general solution to the differential equation y′′-16y=xe-2x in the integral form of particular solution.

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The general solution to the differential equation y″−16y=xe−2x in the integral form of particular solution.

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