Explanation Procedure used: Solution of the differential equation with undetermined coefficients-Annihilator Approach: The Annihilator Approach represents the differential equation L ( y ) = g ( x ) , where, the function g ( x ) may consist any or the entire finite sum, product of multiple polynomials, sine and cosine functions and exponential functions as well. Steps: 1. Look for the complementary solution y c for the homogeneous equation L ( y ) = 0 . 2. Operate both sides of the equation L ( y ) = g ( x ) by the differential operator which annihilates the function g ( x ) . 3. Solve for a homogeneous differential equation by looking for the general solution of the differential equation obtained L 1 L ( y ) = 0 . 4. Solve for the particular solution of the non-homogeneous differential equation L ( y ) = g ( x ) by eliminating the duplicated term in the complementary solution of L 1 L ( y ) = 0 . 5. Determine the undetermined coefficients by substituting the particular solution into non-homogeneous differential equation L ( y ) = g ( x ) . 6. Write the general solution from complementary and particular solution in the form y = y c + y p . Calculation: Consider the differential equation y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 . Find the roots of auxiliary equation of homogeneous DE y ‴ − 3 y ″ + 3 y ′ − y = 0 associate to y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 . Therefore, y ‴ − 3 y ″ + 3 y ′ − y = 0 m 3 − 3 m 2 + 3 m − 1 = 0 ( m − 1 ) 3 = 0 [ Since ( m − 1 ) 2 = m 3 − 3 m 2 + 3 m − 1 ] Thus, the solutions of the auxiliary equation are m 1 = 1 , m 2 = 1 and m 3 = 1 . Thus, the complimentary function is y c ( x ) = c 1 e x + c 2 x e x + c 3 x 2 e x . The term − x + 16 can be annihilated by the differential operator D and the operator ( D − 1 ) annihilates e x . D 2 ( D − 1 ) ( e x − x + 16 ) = 0 Apply D 2 ( D − 1 ) to the given differential equation. D 2 ( D − 1 ) ( y ‴ − 3 y ″ + 3 y ′ − y ) = 0 D 2 ( D − 1 ) ( D 3 − 3 D 2 + 3 D − 1 ) ( y ) = 0 The auxiliary equation is m 2 ( m − 1 ) ( m 3 − 3 m 2 + 3 m − 1 ) = 0 . Thus, the roots of the auxiliary equation are m 1 = 1 = m 2 = m 3 = m 4 , m 5 = 0 and m 6 = 0 . Thus, the solution of the DE is y ( x ) = c 1 e x + c 2 x e x + c 3 x 2 e x + c 4 x 3 e x + c 5 + c 6 x . Since the terms c 1 e x , c 2 x e x and c 3 x 2 e x are repeated. Therefore, the particular solution is y p ( x ) = A x 3 e x + B + C x , Differentiate the above particular solution with respect to x

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 4.5, Problem 61E

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To solve: The differential equation y‴−3y″+3y′−y=ex−x+16 by undetermined coefficients.

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Chapter 4.5, Problem 60E

Chapter 4.5, Problem 62E

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

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The differential equation y ‴ − 3 y ″ + 3 y ′ − y = e x − x + 16 by undetermined coefficients.

Solution Summary: The author explains the Annihilator Approach, which solves the differential equation with undetermined coefficients.

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To solve: The differential equation y‴−3y″+3y′−y=ex−x+16 by undetermined coefficients.

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