Explanation The given differential equation is 2 y ″ + 3 y ′ + y = t 2 + 3 sin t . Consider the homogeneous equation 2 y ″ + 3 y ′ + y = 0 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: 2 r 2 + 3 r + 1 = 0 . The roots of the characteristic equation are as follows. 2 r 2 + 3 r + 1 = 0 r 1 , 2 = − 3 ± ( 3 ) 2 − 4 ( 2 ) ( 1 ) 2 ( 2 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 3 ± 9 − 8 4 r 1 , 2 = − 3 ± 1 4 Simplify further, r 1 , 2 = − 3 ± 1 4 r 1 , 2 = − 3 + 1 4 , − 3 − 1 4 r 1 , 2 = − 1 2 , − 1 r 1 = − 1 and r 2 = − 1 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 = − 1 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 − 1 2 t Thus, the complementary solution is y c ( t ) = c 1 e − t + c 2 e − 1 2 t . Let Y 1 = A + B t + C t 2 , Y 2 = D cos t + E sin t ( ∵ g ( t ) = g 1 ( t ) + g 2 ( t ) here g 1 ( t ) = t 2 , g 2 ( t ) = 3 sin t ) . Solve for Y 1 = A + B t + C t 2 : Obtain the first derivative of Y 1 = A + B t + C t 2 as follows. Y 1 ′ = d d t ( A + B t + C t 2 ) = 0 + B + 2 C t = B + 2 C t Obtain the second derivative of Y 1 = A + B t + C t 2 as follows. Y 1 ′ ′ = d d t ( Y 1 ′ ) = d d t ( B + 2 C t ) = 0 + 2 C = 2 C Obtain the particular solution as follows. Substitute Y 1 ′ = B + 2 C t , Y 1 ′ ′ = 2 C and Y 1 = A + B t + C t 2 in 2 y ″ + 3 y ′ + y = t 2 . 2 ( 2 C ) + 3 ( B + 2 C t ) + A + B t + C t 2 = t 2 4 C + 3 B + 6 C t + A + B t + C t 2 = t 2 ( A + 4 C + 3 B ) + ( 6 C + B ) t + C t 2 = t 2 Comparing coefficient on both sides, A + 4 C + 3 B = 0 B + 6 C = 0 C = 1 Substitute C = 1 in B + 6 C = 0 . B + 6 ( 1 ) = 0 B = − 6 Substitute B = − 6 , C = 1 in B + 6 C = 0 . A + 4 ( 1 ) + 3 ( − 6 ) = 0 A + 4 − 18 = 0 A = 14 Substitute A = 14 , B = − 6 , C = 1 in Y 1 = A + B t + C t 2

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Differential Equations

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Chapter 3.5, Problem 9P

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The general solution of the given differential equation.

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