Explanation The given differential equation is y ″ − y ′ − 2 y = − 2 t + 4 t 2 . Consider the homogeneous equation y ″ − y ′ − 2 y = 0 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 − r − 2 = 0 . The roots of the characteristic equation are as follows. r 2 − r − 2 = 0 r 1 , 2 = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 2 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = 1 ± 1 + 8 2 r 1 , 2 = 1 ± 9 2 Simplify further, r 1 , 2 = 1 ± 3 2 r 1 , 2 = 1 + 3 2 , 1 − 3 2 r 1 , 2 = 2 , − 1 r 1 = − 1 and r 2 = 2 Note that, the general equation is y c ( t ) = c 1 e r 1 t + c 2 e r 2 t where r 1 and r 2 are the distinct root of the characteristic equation. Substitute r 1 = − 1 and r 2 = 2 in y c ( t ) = c 1 e r 1 t + c 2 e r 2 t . y c ( t ) = c 1 e − t + c 2 e 2 t Thus, the complementary solution is y c ( t ) = c 1 e − t + c 2 e 2 t . Let Y = A t 2 + B t + C ( ∵ g ( t ) = − 2 t + 4 t 2 and g ( t ) is polynomial ) . Obtain the first derivative of Y = A t 2 + B t + C as follows. Y ′ = d d t ( A t 2 + B t + C ) = A ( 2 t ) + B = 2 A t + B Obtain the second derivative of Y = A t 2 + B t + C as follows. Y ″ = d d t ( Y ′ ) = d d t ( 2 A t + B ) = 2 A + 0 = 2 A Obtain the particular solution as follows

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 3.5, Problem 2P

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The general solution of the given differential equation.

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Chapter 3 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 3.1 - Prob. 1P Ch. 3.1 - Prob. 2P Ch. 3.1 - Prob. 3P Ch. 3.1 - Prob. 4P Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 7P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P

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