To determine The solution of the given IVP. Explanation Given: The given IVP is y i v + 5 y ″ + 4 y = 90 sin 4 x , y ( 0 ) = 1 , y ′ ( 0 ) = 2 , y ″ ( 0 ) = − 1 , y ′ ′ ′ ( 0 ) = − 32 . 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, The given IVP is y i v + 5 y ″ + 4 y = 90 sin 4 x , y ( 0 ) = 1 , y ′ ( 0 ) = 2 , y ″ ( 0 ) = − 1 , y ′ ′ ′ ( 0 ) = − 32 . Consider the corresponding homogeneous equation of the given ODE, y i v + 5 y ″ + 4 y = 0 Obtain the characteristic equation as follows. λ 4 + 5 λ 2 + 4 = 0 Compute the roots of the characteristic equation as follows. λ 2 = − 5 ± 5 2 − 16 2 λ 2 = − 5 ± 9 2 λ 2 = − 5 ± 3 2 λ 2 = − 4 , − 1 Simplify further as follows. λ 1 = − i , λ 2 = i , λ 3 = − 2 i , λ 4 = 2 i The homogeneous solution of the given ODE is given as follows. y h = c 1 cos x + c 2 sin x + c 3 cos 2 x + c 4 sin 2 x Let the particular solution is of the form y p = A cos 4 x + B sin 4 x . Obtain the first three derivatives of the above equation as follows. y ′ p = − 4 A sin 4 x + 4 B cos 4 x y ″ p = − 16 A cos x − 16 B sin 4 x y ′ ′ ′ p = 64 A sin 4 x − 64 B cos 4 x y p i v = 256 A cos 4 x + 256 B sin 4 x Substitute the above derivatives in the given ODE as follows. 256 A cos 4 x + 256 B sin 4 x + 5 ( − 16 A cos x − 16 B sin 4 x ) + 4 ( A cos 4 x + B sin 4 x ) = 90 sin 4 x cos 4 x ( 256 A − 80 A + 4 A ) + sin 4 x ( 256 B − 80 B + 4 B ) = 90 sin 4 x 180 A cos 4 x + 180 B sin 4 x = 90 sin 4 x Equate like terms on both sides as follows. 180 A = 0 , 180 B = 90 A = 0 , B = 1 2 The particular solution of the given ODE is given as follows

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