Explanation The given differential equation is y ″ + y ′ − 2 y = 2 t and the initial condition are y ( 0 ) = 0 and y ′ ( 0 ) = 1 . 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 = 1 , − 2 r 1 = − 2 and r 2 = 1 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 = − 2 and r 2 = 1 in y c ( t ) = c 1 e r 1 t + c 2 e r 2 t . y c ( t ) = c 1 e − 2 t + c 2 e t Thus, the complementary solution is y c ( t ) = c 1 e − 2 t + c 2 e t . Let Y = A t + B ( ∵ g ( t ) = 2 t and g ( t ) is polynomial ) . Obtain the first derivative of Y = A t + B as follows. Y ′ = d d t ( A t + B ) = A ( 1 ) + 0 = A Obtain the second derivative of Y = A t + B as follows. Y ″ = d d t ( Y ′ ) = d d t ( A ) = 0 Obtain the particular solution as follows. Substitute Y ′ = A , Y ″ = 0 and Y = A t + B in y ″ + y ′ − 2 y = 2 t . 0 + A − 2 ( A t + B ) = 2 t A − 2 A t − 2 B = 2 t ( A − 2 B ) − 2 A t = 2 t Compare the coefficient on both sides, − 2 A = 2 ⇒ A = − 1 A − 2 B = 0 Substitute A = − 1 in A − 2 B = 0 . − 1 − 2 B = 0 − 2 B = 1 B = − 1 2 Substitute A = − 1 and B = − 1 2 in Y = A t + B . Y = − t − 1 2 Obtain the general solution of the given differential equation as follows. Substitute Y = − t − 1 2 and y c ( t ) = c 1 e − 2 t + c 2 e t in y ( t ) = y c ( t ) + Y

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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