Explanation Procedure used: General solution to a differential equation by method of undermined coefficients. 1 Consider the DE and determine the homogeneous DE L ( x ) = 0 associate to DE. 2 Determine auxiliary equation and find the roots. 3 From the roots of auxiliary equation find the complementary function y c . 4 Use the method of undermined coefficients to determine the particular solution y p . 5 Substitute, the particular solution y p into DE and find the constants. 6 Find the general solution to the DE from y = y c + y p . Calculation: Consider the differential equation y ″ − 3 y ′ + 5 y = 4 x 3 − x . The homogeneous DE equation associate to y ″ − 3 y ′ + 5 y = 4 x 3 − x is y ″ − 3 y ′ + 5 y = 0 . The auxiliary equation to y ″ − 3 y ′ + 5 y = 0 is m 2 − 3 m + 5 = 0 . Find the roots of auxiliary equation using m = − b ± b 2 − 4 a c 2 ( a ) . m = − ( − 3 ) ± ( − 3 ) 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) = 3 ± 11 i 2 = 3 2 ± 11 2 i Hence, the roots of the auxiliary equation are m 1 = 3 2 + 11 2 i and m 2 = 3 2 − 11 2 i . Since the roots are conjugate complex numbers, the complementary function is y c = e 3 2 x ( c 1 cos 11 2 x + c 2 sin 11 2 x ) . Consider the particular solution in the form of a polynomial function, y p = A x 3 + B x 2 + C x + D . Differentiate y p with respect to x . y p = A x 3 + B x 2 + C x + D y ′ p = 3 A x 2 + 2 B x + C y ″ p = 6 A x + 2 B Substitute y p , y ′ p and y ″ p into the fundamental DE, 6 A x + 2 B − 3 ( 3 A x 2 + 2 B x + C ) + 5 ( A x 3 + B x 2 + C x + D ) = 4 x 3 − 2 x 5 A x 3 + ( − 9 A + 5 B ) x 2 + ( 6 A − 6 B + 5 C ) x + ( 2 B − 3 C + 5 D ) = 4 x 3 − 2 x Compare the coefficients of x 3

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 4, Problem 23RE

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To find: The general solution of the differential equation y″−3y′+5y=4x3−2x.

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Chapter 4, Problem 22RE

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A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - In Problems 14 the given family of functions is...Ch. 4.1 - Given that y=c1+c2x2 is a two-parameter family of...Ch. 4.1 - Find two members of the family of solutions in...Ch. 4.1 - Given that x(t) = c1 cos t + c2 sin t is the...Ch. 4.1 - Use the general solution of x + 2x = 0 given in...Ch. 4.1 - In Problems 9 and 10 find an interval centered...Ch. 4.1 - In Problems 9 and 10 find an interval centered...

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The general solution of the differential equation y ″ − 3 y ′ + 5 y = 4 x 3 − 2 x .

Solution Summary: The author explains how to find the general solution to a differential equation by method of undermined coefficients.

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To find: The general solution of the differential equation y″−3y′+5y=4x3−2x.

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