To determine The general solution of the given differential equation. Explanation The given differential equation is y ″ + 2 y ′ + y = 2 e − t . Consider the homogeneous equation y ″ + 2 y ′ + 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 + 2 r + 1 = 0 . The roots of the characteristic equation are as follows. r 2 + 2 r + 1 = 0 r 1 , 2 = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 2 ± 4 − 4 2 r 1 , 2 = − 2 ± 0 2 Simplify further, r 1 , 2 = − 2 2 r 1 , 2 = − 1 r = − 1 Note that, the general equation is y c ( t ) = c 1 e r t + c 2 t e r t where r is the double root of the characteristic equation. Substitute r = − 1 in y c ( t ) = c 1 e r t + c 2 t e r t . y c ( t ) = c 1 e − t + c 2 t e − t Thus, the complementary solution is y c ( t ) = c 1 e − t + c 2 t e − t . Let Y = A t 2 e − t ( ∵ g ( t ) = 2 e − t ) . Obtain the first derivative of Y = A t 2 e − t as follows. Y ′ = d d t ( A t 2 e − t ) = A [ t 2 ( − e − t ) + e − t ( 2 t ) ] = − A t 2 e − t + 2 A t e − t Obtain the second derivative of Y = A t 2 e − t as follows

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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