The solution of the given IVP and graph the solution. The solution of the given IVP is y = 1 2 e 5 x + 1 2 e − 5 x − cos 4 x _ . Given: The given IVP is y i v − 9 y ″ − 400 y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 41 , y ′ ′ ′ ( 0 ) = 0 . 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, y i v − 9 y ″ − 400 y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 41 , y ′ ′ ′ ( 0 ) = 0 Consider the homogeneous equation of the given ODE, y i v − 9 y ″ − 400 y = 0 Obtain the characteristic equation as follows. λ 4 − 9 λ 2 − 400 = 0 Compute the roots of the characteristic equation as follows. λ 4 − 9 λ 2 − 400 = 0 ( λ 2 − 25 ) ( λ 2 + 16 ) = 0 λ 2 = 25 or λ 2 = − 16 λ 1 = 5 , λ 2 = − 5 , λ 3 = − 4 i , λ 4 = 4 i The solution of the given ODE is given as follows. y = c 1 e 5 x + c 2 e − 5 x + e 0 x ( c 3 cos 4 x + c 4 sin 4 x ) = c 1 e 5 x + c 2 e − 5 x + c 3 cos 4 x + c 4 sin 4 x Obtain the first 3 derivatives of the general solution obtained as follows. y ′ = 5 c 1 e 5 x − 5 c 2 e − 5 x − 4 c 3 sin 4 x + 4 c 4 cos 4 x y ″ = 25 c 1 e 5 x + 25 c 2 e − 5 x − 16 c 3 cos 4 x − 16 c 4 sin 4 x y ′ ′ ′ = 125 c 1 e 5 x − 125 c 2 e − 5 x + 64 c 3 sin 4 x − 64 c 4 cos 4 x Apply the given initial values on the general solution and its derivatives obtain as follows. y ( 0 ) = c 1 e 0 + c 2 e 0 + c 3 cos 0 + c 4 sin 0 = 0 y ′ ( 0 ) = 5 c 1 e 0 − 5 c 2 e 0 − 4 c 3 sin 0 + 4 c 4 cos 0 = 0 y ″ ( 0 ) = 25 c 1 e 0 + 25 c 2 e 0 − 16 c 3 cos 0 − 16 c 4 sin 0 = 41 y ′ ′ ′ ( 0 ) = 125 c 1 e 0 − 125 c 2 e 0 + 64 c 3 sin 0 − 64 c 4 cos 0 = 0 Simplify further as follows. c 1 + c 2 + c 3 = 0 5 c 1 − 5 c 2 + 4 c 4 = 0 25 c 1 + 25 c 2 − 16 c 3 = 41 125 c 1 − 125 c 2 − 64 c 4 = 0 Simplify further as follows. c 1 + c 2 + c 3 = 0 5 c 1 − 5 c 2 + 4 c 4 = 0 25 c 1 + 25 c 2 − 16 c 3 − 25 ( c 1 + c 2 + c 3 ) = 41 − 25 ( 0 ) 125 c 1 − 125 c 2 − 64 c 4 − 25 ( 5 c 1 − 5 c 2 + 4 c 4 ) = 0 − 25 ( 0 ) Simplify further as follows. c 1 + c 2 + c 3 = 0 5 c 1 − 5 c 2 + 4 c 4 = 0 − 41 c 3 = 41 − 164 c 4 = 0 Simplify further as follows. c 1 + c 2 + c 3 = 0 5 c 1 − 5 c 2 + 4 c 4 = 0 c 3 = − 1 c 4 = 0 Simplify further as follows. c 1 + c 2 = 1 5 c 1 − 5 c 2 = 0 c 3 = − 1 c 4 = 0 Simplify further as follows. c 1 + c 2 = 1 5 c 1 − 5 c 2 + 5 ( c 1 + c 2 ) = 0 + 5 ( 1 ) c 3 = − 1 c 4 = 0 Simplify further as follows. c 1 + c 2 = 1 10 c 1 = 5 c 3 = − 1 c 4 = 0 Simplify further as follows. c 2 = 1 2 c 1 = 1 2 c 3 = − 1 c 4 = 0 Substitute the values of the constants in the general solution as follows. y = 1 2 e 5 x + 1 2 e − 5 x − cos 4 x Thus, the solution of the given ODE is y = 1 2 e 5 x + 1 2 e − 5 x − cos 4 x _ . Plot the graph of the solution as follows. Observe from the above figure that the solution has a global minimum at the origin.

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