Explanation The given differential equation with initial condition is 9 y ″ + 6 y ′ + 82 y = 0 , y ( 0 ) = − 1 , y ′ ( 0 ) = 2 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: 9 r 2 + 6 r + 82 = 0 . The roots of the characteristic equation is as follows. 9 r 2 + 6 r + 82 = 0 r 1 , 2 = − 6 ± ( 6 ) 2 − 4 ( 9 ) ( 82 ) 2 ( 9 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 6 ± 36 − 2952 18 r 1 , 2 = − 6 ± − 2916 18 Simplify further, r 1 , 2 = − 6 ± 54 i 18 r 1 , 2 = − 6 18 ± 54 i 18 r = − 1 3 ± 3 i ( ∵ r = a + b i ) Here, roots are complex. Note that, the general equation is y ( 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 = − 1 3 and b = 3 in y ( t ) = c 1 e a t cos b t + c 2 e a t sin b t . y ( t ) = c 1 e − 1 3 t cos ( 3 t ) + c 2 e − 1 3 t sin ( 3 t ) Therefore, the general solution of the given differential equation is y ( t ) = c 1 e − 1 3 t cos ( 3 t ) + c 2 e − 1 3 t sin ( 3 t ) . Obtain the derivative of the general solution as follows. y ′ = d d t ( c 1 e − 1 3 t cos ( 3 t ) + c 2 e − 1 3 t sin ( 3 t ) ) = c 1 d d t ( e − 1 3 t cos ( 3 t ) ) + c 2 d d t ( e − 1 3 t sin ( 3 t ) ) = c 1 [ e − 1 3 t d d t ( cos ( 3 t ) ) + cos ( 3 t ) d d t ( e − 1 3 t ) ] + c 2 [ e − 1 3 t d d t ( sin ( 3 t ) ) + sin ( 3 t ) d d t ( e − 1 3 t ) ] = c 1 [ e − 1 3 t ( − 3 sin ( 3 t ) ) + cos ( 3 t ) ( − 1 3 e − 1 3 t ) ] + c 2 [ e − 1 3 t ( 3 cos ( 3 t ) ) + sin ( 3 t ) ( − 1 3 e − 1 3 t ) ] = c 1 [ − 3 e − 1 3 t ( sin ( 3 t ) ) − 1 3 cos ( 3 t ) ( e − 1 3 t ) ] + c 2 [ 3 e − 1 3 t ( cos ( 3 t ) ) − 1 3 sin ( 3 t ) ( e − 1 3 t ) ] Thus, the derivative of the general solution is y ′ ( t ) = c 1 [ − 3 e − 1 3 t ( sin ( 3 t ) ) − 1 3 cos ( 3 t ) ( e − 1 3 t ) ] + c 2 [ 3 e − 1 3 t ( cos ( 3 t ) ) − 1 3 sin ( 3 t ) ( e − 1 3 t ) ]

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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