Explanation Given: The initial value problem is, y ‴ + 2 y ″ − 9 y ′ − 18 y = − 18 x 2 − 18 x + 22 y ( 0 ) = − 2 , y ′ ( 0 ) = − 8 , y ″ ( 0 ) = − 12 Approach: 1) Use elimination method for solving algebraic systems. 2) Use method of undetermined coefficients to find particular solution of nonhomogeneous ordinary differential equation. Calculation: The corresponding homogeneous equation is y ‴ + 2 y ″ − 9 y ′ − 18 y = 0 ( 1 ) The auxiliary equation of Equation ( 1 ) is, r 3 + 2 r 2 − 9 r − 18 = 0 ( r + 2 ) ( r 2 − 9 ) = 0 ( r + 2 ) ( r − 3 ) ( r + 3 ) = 0 ( 2 ) Roots of Equation ( 2 ) are r = − 3 , 3 , − 2 . So, the complimentary solution of Equation ( 1 ) is y c ( x ) = C 1 e − 2 x + C 2 e − 3 x + C 3 e 3 x . The particular solution regarding the expression − 18 x 2 − 18 x + 22 is y p ( x ) = C 4 x 2 + C 5 x + C 6 ( 3 ) Differentiate Equation ( 3 ) with respect to x . y p ′ ( x ) = 2 C 4 x + C 5 ( 4 ) Differentiate Equation ( 4 ) with respect to x . y p ′′ ( x ) = 2 C 4 ( 5 ) Differentiate Equation ( 5 ) with respect to x . y p ′′ ′ ( x ) = 0 ( 6 ) Substitute C 4 x 2 + C 5 x + C 6 for y p ( x ) , 2 C 4 x + C 5 for y p ′ ( x ) , 2 C 4 for y p ′′ ( x ) , and 0 for y p ′′ ( x ) in given differential equation. 4 C 4 − 9 ( 2 C 4 x + C 5 ) − 18 ( C 4 x 2 + C 5 x + C 6 ) = − 18 x 2 − 18 x + 22 − 18 C 4 x 2 − 18 ( C 4 + C 5 ) x + ( 4 C 4 − 9 C 5 − 18 C 6 ) = − 18 x 2 − 18 x + 22 ( 7 ) Compare the coefficients of powers of x , x 2 and constant term in Equation ( 7 )

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 6.3, Problem 31E

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Solution Summary: The author explains the method of elimination for solving algebraic systems, and the auxiliary equation of Equation (1).

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To find: The solution to initial value problem.

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Chapter 6 Solutions

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The solution to initial value problem.