Explanation Procedure used: General solution of a differential equation by undermined coefficients: 1. Consider the given differential equation a 1 y ( n ) + ... + a n − 1 y ′ + a n y = g ( x ) and determine the homogeneous differential equation associated with the DE. 2. Calculate the roots of auxiliary equation of homogeneous equation. 3. Determine g ( x ) . 4. Assume the particular solution to the differential equation in form of an equation. 5. Substitute, the particular solution into the differential equation. 6. Compare the coefficients and determine the general solution. Calculation: The given differential equation is, y ″ + 2 y ′ − 24 y = 16 − ( x + 2 ) e 4 x . Determine the associate homogenous equation of differential equation y ″ + 2 y ′ − 24 y = 16 − ( x + 2 ) e 4 x . That is, y ″ + 2 y ′ − 24 y = 0 . The auxiliary equation corresponding to the homogenous equation, y ″ + 2 y ′ − 24 y = 0 is m 2 + 2 m − 24 = 0 . Obtain the roots of m 2 + 2 m − 24 = 0 as, m 2 + 2 m − 24 = 0 m 2 + 6 m − 4 m − 24 = 0 m ( m + 6 ) − 4 ( m + 6 ) = 0 ( m + 6 ) ( m − 4 ) = 0 m 1 = 4 and m 2 − 6 Therefore, the roots of the auxiliary equation are m 1 = 4 and m 2 − 6 . Hence, the complementary function is y c = c 1 e 4 x + c 2 e − 6 x . Compare the differential equation y ″ + 2 y ′ − 24 y = 16 − ( x + 2 ) e 4 x with y ″ + 2 y ′ − 24 y = g ( x ) . Hence, g ( x ) = 16 − ( x + 2 ) e 4 x . Assume that a particular solution of the given differential equation is also in the form of a quadratic function y p = A + ( B x 2 + C x ) e 4 x . Compute the first and the second derivative of the above equation. y ′ p = [ 4 ( ( B x 2 + C x ) e 4 x ) + ( 2 B x + C ) e 4 x ] y ″ p = 4 { 4 ( ( B x 2 + C x ) e 4 x + ( 2 B x + C ) e 4 x ) } + { ( 2 B x + C ) e 4 x + 2 B e 4 x } y ″ p = [ ( 2 B − A ) ( e 2 x cos x + 2 e 2 x sin x ) + ( 2 A + B ) ( − e 2 x sin x + 2 e 2 x cos x ) ] y ″ p = 16 ( B x 2 + C x ) e 4 x + 8 ( 2 B x + C ) e 4 x + 2 B e 4 x Substitute these values in the differential equation, y ″ p + 2 y ′ p − 24 y = 16 − ( x + 2 ) e 4 x

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 3.4, Problem 20E

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To solve: The differential equation y″+2y′−24y=16−(x+2)e4x by undetermined coefficients.

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Chapter 3.4, Problem 19E

Chapter 3.4, Problem 21E

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The differential equation y ″ + 2 y ′ − 24 y = 16 − ( x + 2 ) e 4 x by undetermined coefficients.

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To solve: The differential equation y″+2y′−24y=16−(x+2)e4x by undetermined coefficients.

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