Explanation The given differential equation with initial condition is y ″ + 4 y ′ + 4 y = 0 , y ( − 1 ) = 2 , y ′ ( − 1 ) = 1 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + 4 r + 4 = 0 . The roots of the characteristic equation are as follows. r 2 + 4 r + 4 = 0 r 1 , 2 = − 4 ± ( 4 ) 2 − 4 ( 1 ) ( 4 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 4 ± 16 − 16 2 r 1 , 2 = − 4 ± 0 2 Simplify further, r 1 , 2 = − 4 2 r 1 , 2 = − 2 r = − 2 Note that, the general equation is y ( t ) = c 1 e r t + c 2 t e r t where r is the double root of the characteristic equation. Substitute − 2 for r in y ( t ) = c 1 e r t + c 2 t e r t . y ( t ) = c 1 e − 2 t + c 2 t e − 2 t Therefore, the general solution of the given differential equation is y ( t ) = c 1 e − 2 t + c 2 t e − 2 t _ . Obtain the derivative of the general solution as follows. y ′ = d d t ( c 1 e − 2 t + c 2 t e − 2 t ) = c 1 ( − 2 e − 2 t ) + c 2 d d t ( t e − 2 t ) = − 2 c 1 e − 2 t + c 2 [ t d d t ( e − 2 t ) + e − 2 t d d t ( t ) ] = − 2 c 1 e − 2 t + c 2 [ t ( − 2 e − 2 t ) + e − 2 t ] = − 2 c 1 e − 2 t − 2 c 2 t e − 2 t + c 2 e − 2 t Thus, the derivative of the general solution is y ′ ( t ) = − 2 c 1 e − 2 t − 2 c 2 t e − 2 t + c 2 e − 2 t . Initial conditions: Substitute –1 for t and 2 for y ( t ) in y ( t ) = c 1 e − 2 t + c 2 t e − 2 t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Elementary Differential Equations

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