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 ( 4 ) − 4 y ″ = 5 x 2 − e 2 x . Find the roots of auxiliary equation of homogeneous DE y ( 4 ) − 4 y ″ = 0 associate to y ( 4 ) − 4 y ″ = 5 x 2 − e 2 x . Therefore, y ( 4 ) − 4 y ″ = 0 m 4 − 4 m 2 = 0 m 2 ( m 2 − 4 ) = 0 The solutions of the auxiliary equation are m 1 = − 2 , m 2 = 2 , m 3 = 0 and m 4 = 0 . Thus, the complimentary function is y c ( x ) = c 1 e − 2 x + c 2 e 2 x + c 3 + c 4 x . The exponential term e 2 x will be annihilated by ( D − 2 ) and the algebraic term 5 x 2 will be annihilated by D 3 . Therefore, D 3 ( D − 2 ) ( 5 x 2 − e 2 x ) = 0 Apply D 3 ( D − 2 ) to the given differential equation. D 3 ( D − 2 ) ( y ( 4 ) − 4 y ″ ) = 0 D 3 ( D − 2 ) ( D 4 − 4 D 2 ) ( y ) = 0 D 5 ( D − 2 ) 2 ( D + 2 ) ( y ) = 0 The auxiliary equation is m 5 ( m − 2 ) 2 ( m + 2 ) = 0 . Thus, the roots of the auxiliary equation are m 1 = − 2 , m 2 = 2 , m 3 = 0 , m 4 = 0 , m 5 = 2 and m 6 = m 7 = m 8 = 0 . Thus, the solution of the DE is y ( x ) = c 1 e − 2 x + c 2 e 2 x + c 3 + c 4 x + c 5 x e 2 x + c 6 x 2 + c 7 x 3 + c 8 x 4 . Since the terms c 1 e − 2 x , c 2 e 2 x , c 3 and c 4 x are repeated, the particular solution is y p ( x ) = A x e 2 x + B x 2 + C x 3 + D x 4 . Therefore, Differentiate y p ( x ) = A x e 2 x + B x 2 + C x 3 + D x 4 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 64E

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To solve: The differential equation $y(4)−4y″=5x2−e2x$ by undetermined coefficients.

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Chapter 4.5, Problem 63E

Chapter 4.5, Problem 65E

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

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The differential equation y ( 4 ) − 4 y ″ = 5 x 2 − e 2 x 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 $y(4)−4y″=5x2−e2x$ by undetermined coefficients.

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