To determine The general solution of the given differential equation. Explanation The given differential equation is u ″ + ω 0 2 u = cos ω t , ω 2 ≠ ω 0 2 . Consider the homogeneous equation u ″ + ω 0 2 u = 0 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + ω 0 2 = 0 . The roots of the characteristic equation are as follows. r 2 + ω 0 2 = 0 r 1 , 2 = 0 ± ( 0 ) 2 − 4 ( 1 ) ( ω 0 2 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = ± 0 − 4 ω 0 2 2 r 1 , 2 = ± − 4 ω 0 2 2 Simplify further, r 1 , 2 = ± i 2 ω 0 2 2 r 1 , 2 = ± ω 0 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 = ω 0 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 ( ω 0 t ) + c 2 e 0 t sin ( ω 0 t ) = c 1 cos ( ω 0 t ) + c 2 sin ( ω 0 t ) Therefore, the complementary solution of the given differential equation is y c ( t ) = c 1 cos ( ω 0 t ) + c 2 sin ( ω 0 t ) . Let U = A cos ( ω t ) + B sin ( ω t ) ( ∵ g ( t ) = cos ω t ) . Obtain the first derivative of U = A cos ( ω t ) + B sin ( ω t ) as follows. U ′ = d d t ( A cos ( ω t ) + B sin ( ω t ) ) = A ( − ω sin ω t ) + B ( ω cos ω t ) = − A ω sin ω t + ω B cos ω t Obtain the second derivative of U = A cos ( ω t ) + B sin ( ω t ) as follows. U ″ = d d t ( U ′ ) = d d t ( − A ω sin ω t + ω B cos ω t ) = − ω A ( ω cos ω t ) + ω B ( − ω sin ω t ) = − ω 2 A cos ω t − ω 2 B sin ω t Obtain the particular solution as follows. Substitute U ′ = − A ω sin ω t + ω B cos ω t , U ″ = − ω 2 A cos ω t − ω 2 B sin ω t and U = A cos ( ω t ) + B sin ( ω t ) in u ″ + ω 0 2 u = cos ω t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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