Explanation The given differential equation is y ″ + 9 y = t 2 e 3 t + 6 . Consider the homogeneous equation y ″ + 9 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 + 9 = 0 . The roots of the characteristic equation are as follows. r 2 + 9 = 0 r 1 , 2 = − ( 0 ) ± ( 0 ) 2 − 4 ( 1 ) ( 9 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = 0 ± 0 − 36 2 r 1 , 2 = ± − 36 2 Simplify further, r 1 , 2 = ± i 6 2 r 1 , 2 = 6 i 2 , − 6 i 2 r 1 , 2 = 3 i , − 3 i ( ∵ r = a + b i ) Here, roots are complex. Note that, the general equation is y c ( t ) = c 1 e a t cos b t + c 2 e a t sin b t where a is the real part of the root and b is the imaginary part of the roots. Substitute a = 0 and b = 3 in y c ( t ) = c 1 e a t cos b t + c 2 e a t sin b t . y c ( t ) = c 1 e 0 t cos ( 3 t ) + c 2 e 0 t sin ( 3 t ) y c ( t ) = c 1 cos ( 3 t ) + c 2 sin ( 3 t ) ( ∵ e 0 t = 1 ) Therefore, the complementary solution of the given differential equation is y c ( t ) = c 1 cos ( 3 t ) + c 2 sin ( 3 t ) . Let Y 1 = A , Y 2 = A e 3 t + B t e 3 t + C t 2 e 3 t ( ∵ g ( t ) = g 1 ( t ) + g 2 ( t ) where g 1 ( t ) = 6 , g 2 ( t ) = t t e 3 t ) . Solve for Y 1 = A : Obtain the first derivative of Y 1 = A as follows. Y 1 ′ = d d t ( A ) = 0 Obtain the second derivative of Y 1 = A as follows. Y 1 ′ ′ = d d t ( Y 1 ′ ) = d d t ( 0 ) = 0 Obtain the particular solution as follows. Substitute Y 1 ′ = 0 , Y 1 ′ ′ = 0 and Y = A in y ″ + 9 y = 6 . ( 0 ) + 9 A = 6 A = 6 9 A = 2 3 Substitute A = 2 3 in Y = A . Y 1 = 2 3 Solve for Y 2 = A e 3 t + B t e 3 t + C t 2 e 3 t : Obtain the first derivative of Y 2 = A e 3 t + B t e 3 t + C t 2 e 3 t as follows. Y ′ 2 = d d t ( A e 3 t + B t e 3 t + C t 2 e 3 t ) = A d d t ( e 3 t ) + B d d t ( t e 3 t ) + C d d t ( t 2 e 3 t ) = A ( 3 e 3 t ) + B [ t ( 3 e 3 t ) + e 3 t ( 1 ) ] + C [ t 2 ( 3 e 3 t ) + e 3 t ( 2 t ) ] = 3 A e 3 t + 3 B t e 3 t + B e 3 t + 3 C t 2 e 3 t + 2 C t e 3 t Obtain the second derivative of Y 2 = A e 3 t + B t e 3 t + C t 2 e 3 t as follows. Y 2 ′ ′ = d d t ( Y ′ 2 ) = d d t ( 3 A e 3 t + 3 B t e 3 t + B e 3 t + 3 B t 2 e 3 t + 2 B t e 3 t ) = 3 A d d t ( e 3 t ) + 3 B d d t ( t e 3 t ) + B d d t ( e 3 t ) + 3 C d d t ( t 2 e 3 t ) + 2 C d d t ( t e 3 t ) = 3 A ( 3 e 3 t ) + 3 B ( t ( 3 e 3 t ) + e 3 t ) + B ( 3 e 3 t ) + 3 C ( t 2 ( 3 e 3 t ) + e 3 t ( 2 t ) ) + 2 C ( t ( 3 e 3 t ) + e 3 t ) Simplify further, Y 2 ′ ′ = 9 A e 3 t + 9 B t e 3 t + 3 B e 3 t + 3 B e 3 t + 9 C t 2 e 3 t + 6 C t e 3 t + 6 C t e 3 t + 2 C e 3 t = 9 A e 3 t + 9 B t e 3 t + 6 B e 3 t + 9 C t 2 e 3 t + 12 C t e 3 t + 2 C e 3 t Obtain the particular solution as follows

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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