Explanation The given differential equation is y ″ + y ′ − 6 y = 12 e 3 t + 12 e − 2 t . Consider the homogeneous equation y ″ + y ′ − 6 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 − 6 = 0 . The roots of the characteristic equation are as follows. r 2 + r − 6 = 0 r 1 , 2 = − 1 ± ( 1 ) 2 − 4 ( 1 ) ( − 6 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 1 ± 1 + 24 2 r 1 , 2 = − 1 ± 25 2 Simplify further, r 1 , 2 = − 1 ± 5 2 r 1 , 2 = − 1 + 5 2 , − 1 − 5 2 r 1 , 2 = 2 , − 3 r 1 = − 3 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 = − 3 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 − 3 t + c 2 e 2 t Thus, the complementary solution is y c ( t ) = c 1 e − 3 t + c 2 e 2 t . Let Y = A e 3 t + B e − 2 t ( ∵ g ( t ) = 12 e 3 t + 12 e − 2 t ) . Obtain the first derivative of Y = A e 3 t + B e − 2 t as follows. Y ′ = d d t ( A e 3 t + B e − 2 t ) = A ( 3 e 3 t ) + B ( − 2 e − 2 t ) = 3 A e 3 t − 2 B e − 2 t Obtain the second derivative of Y = A e 3 t + B e − 2 t as follows. Y ″ = d d t ( Y ′ ) = d d t ( 3 A e 3 t − 2 B e − 2 t ) = 3 A ( 3 e 3 t ) − 2 B ( − 2 e − 2 t ) = 9 A e 3 t + 4 B e − 2 t Obtain the particular solution as follows

Elementary Differential Equations And Boundary Value Problems, 11th Edition

11th Edition

ISBN: 9781119375753

Author: William E. Boyce, Richard C. DiPrima, Douglas B. Meade

Publisher: WILEY

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Chapter 3.5, Problem 3P

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Chapter 3 Solutions

Elementary Differential Equations And Boundary Value Problems, 11th Edition

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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