To determine The solution of the given IVP. Explanation Given: The given IVP is ( D 3 − 4 D ) y = 10 cos x + 5 sin x , y ( 0 ) = 3 , y ′ ( 0 ) = − 2 , y ″ ( 0 ) = − 1 . Procedure used: Let y ( n ) + a n − 1 ( x ) y ( n − 1 ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = r ( x ) be the n th order homogeneous ODE with constant coefficients, where y ( n ) = d n y d x n is the first term and r ( x ) ≠ 0 . A general solution of the above equation on an open interval I of the x -axis is of the form r ( x ) ≠ 0 Obtain the characteristic equation by substituting y = e λ x in the ODE as follows. y ( x ) = y h ( x ) + y p ( x ) Here y h ( x ) is a general solution of the corresponding homogeneous ODE. And y p ( x ) is any solution of the given equation on I containing no arbitrary constants. Calculation: Consider the given IVP, ( D 3 − 4 D ) y = 10 cos x + 5 sin x , y ( 0 ) = 3 , y ′ ( 0 ) = − 2 , y ″ ( 0 ) = − 1 y ′ ′ ′ − 4 y ′ = 10 cos x + 5 sin x , y ( 0 ) = 3 , y ′ ( 0 ) = − 2 , y ″ ( 0 ) = − 1 Consider the corresponding homogeneous equation of the given ODE, y ′ ′ ′ − 4 y ′ = 0 Obtain the characteristic equation as follows. λ 3 − 4 λ = 0 Compute the roots of the characteristic equation as follows. λ 3 − 4 λ = 0 λ ( λ 2 − 4 ) = 0 λ ( λ − 2 ) ( λ + 2 ) = 0 λ 1 = 0 , λ 2 = − 2 , λ 3 = 2 The homogeneous solution of the given ODE is given as follows. y h = c 1 e 0 + c 2 e − 2 x + c 3 e 2 x = c 1 + c 2 e − 2 x + c 3 e 2 x Let the particular solution is of the form y p = A cos x + B sin x . Obtain the first three derivatives of the above equation as follows. y ′ p = − A sin x + B cos x y ″ p = − A cos x − B sin x y ′ ′ ′ p = A sin x − B cos x Substitute the above derivatives in the given ODE as follows. y ′ ′ ′ − 4 y ′ = 10 cos x + 5 sin x A sin x − B cos x − 4 ( − A sin x + B cos x ) = 10 cos x + 5 sin x cos x ( − 5 B ) + sin x ( 5 A ) = 10 cos x + 5 sin x − 5 B cos x + 5 A sin x = 10 cos x + 5 sin x Equate like terms on both sides as follows

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Author: Erwin Kreyszig

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