The solution of the given IVP and graph the solution. The solution of the given IVP is y = e x ( 1 8 cos x + 7 8 sin x ) + e − x ( 3 8 cos x − 17 8 sin x ) _ b. Given: The given IVP is y i v + 4 y = 0 , y ( 0 ) = 1 2 , y ′ ( 0 ) = − 3 2 , y ″ ( 0 ) = 5 2 , y ′ ′ ′ ( 0 ) = − 7 2 . 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 + 4 y = 0 , y ( 0 ) = 1 2 , y ′ ( 0 ) = − 3 2 , y ″ ( 0 ) = 5 2 , y ′ ′ ′ ( 0 ) = − 7 2 Consider the homogeneous equation of the given ODE, y i v + 4 y = 0 Obtain the characteristic equation as follows. λ 4 + 4 = 0 Compute the roots of the characteristic equation as follows. λ 4 + 4 λ 2 + 4 − 4 λ 2 = 0 ( λ 2 + 2 ) 2 − ( 2 λ ) 2 = 0 ( λ 2 + 2 + 2 λ ) ( λ 2 + 2 − 2 λ ) = 0 λ = − 2 ± 2 2 − 4 ( 1 ) ( 2 ) 2 or λ = 2 ± ( − 2 ) 2 − 4 ( 1 ) ( 2 ) 2 λ = − 2 ± 2 i 2 or λ = 2 ± 2 i 2 λ = − 1 ± i or λ = 1 ± i The solution of the given ODE is given as follows. y = e x ( c 1 cos x + c 2 sin x ) + e − x ( c 3 cos x + c 4 sin x ) Obtain the first 3 derivatives of the general solution obtained as follows. y ′ = [ e x ( c 1 cos x + c 2 sin x ) − e − x ( c 3 cos x + c 4 sin x ) + e x ( − c 1 sin x + c 2 cos x ) + e − x ( − c 3 sin x + c 4 cos x ) ] = e x ( cos x ( c 1 + c 2 ) + sin x ( c 2 − c 1 ) ) + e − x ( cos x ( c 4 − c 3 ) + sin x ( − c 4 − c 3 ) ) y ″ = [ e x ( cos x ( c 1 + c 2 ) + sin x ( c 2 − c 1 ) ) − e − x ( cos x ( c 4 − c 3 ) + sin x ( − c 4 − c 3 ) ) + e x ( − sin x ( c 1 + c 2 ) + cos x ( c 2 − c 1 ) ) + e − x ( − sin x ( c 4 − c 3 ) + cos x ( − c 4 − c 3 ) ) ] = e x ( 2 c 2 cos x − 2 c 1 sin x ) − e − x ( − 2 c 3 cos x − 2 c 4 sin x ) y ′ ′ ′ = [ e x ( 2 c 2 cos x − 2 c 1 sin x ) + e − x ( − 2 c 3 cos x − 2 c 4 sin x ) + e x ( − 2 c 2 sin x − 2 c 1 cos x ) − e − x ( 2 c 3 sin x − 2 c 4 cos x ) ] = e x [ cos x ( 2 c 2 − 2 c 1 ) + sin x ( − 2 c 1 − 2 c 2 ) ] + e − x [ cos x ( − 2 c 3 + 2 c 4 ) + sin x ( 2 c 4 − 2 c 3 ) ] Apply the given initial values on the general solution and its derivatives obtain as follows. y ( 0 ) = e 0 ( c 1 cos 0 + c 2 sin 0 ) + e 0 ( c 3 cos 0 + c 4 sin 0 ) = 1 2 y ′ ( 0 ) = e 0 ( cos 0 ( c 1 + c 2 ) + sin 0 ( c 2 − c 1 ) ) + e 0 ( cos 0 ( c 4 − c 3 ) + sin 0 ( − c 4 − c 3 ) ) = − 3 2 y ″ ( 0 ) = e 0 ( 2 c 2 cos x − 2 c 1 sin 0 ) − e 0 ( − 2 c 3 cos 0 − 2 c 4 sin 0 ) = 5 2 y ′ ′ ′ ( 0 ) = ( e 0 [ cos x ( 2 c 2 − 2 c 1 ) + sin 0 ( − 2 c 1 − 2 c 2 ) ] + e 0 [ cos 0 ( − 2 c 3 + 2 c 4 ) + sin 0 ( 2 c 4 − 2 c 3 ) ] ) = − 7 2 Simplify further as follows. c 1 + c 3 = 1 2 c 1 + c 2 + c 4 − c 3 = − 3 2 2 c 2 + 2 c 3 = 5 2 2 c 2 − 2 c 1 − 2 c 3 + 2 c 4 = − 7 2 Simplify further as follows. c 1 = 1 2 − c 3 c 1 + c 2 + c 4 − c 3 + ( c 1 + c 3 ) = − 3 2 + 1 2 c 2 = 5 4 − c 3 2 c 2 − 2 c 1 − 2 c 3 + 2 c 4 + ( 2 c 2 + 2 c 3 ) = − 7 2 + 5 2 Simplify further as follows. c 1 = 1 2 − c 3 2 ( 1 2 − c 3 ) + 5 4 − c 3 + c 4 = − 1 c 2 = 5 4 − c 3 4 ( 5 4 − c 3 ) − 2 ( 1 2 − c 3 ) + 2 c 4 = − 1 Simplify further as follows. c 1 = 1 2 − c 3 − 3 c 3 + c 4 = − 13 4 c 2 = 5 4 − c 3 − 2 c 3 + 2 c 4 = − 5 Simplify further as follows. c 1 = 1 2 − c 3 − 3 c 3 + c 4 = − 13 4 c 2 = 5 4 − c 3 − 2 c 3 + 2 c 4 − 2 ( − 3 c 3 + c 4 ) = − 5 − 2 ( − 13 4 ) Simplify further as follows. c 1 = 1 2 − c 3 − 3 c 3 + c 4 = − 13 4 c 2 = 5 4 − c 3 4 c 3 = 3 2 c 3 = 3 8 Simplify further as follows. c 1 = 1 2 − 3 8 = 1 8 c 4 = − 13 4 + 9 8 = − 17 8 c 2 = 5 4 − 3 8 = 7 8 c 3 = 3 8 Substitute the values of the constants in the general solution as follows. y = e x ( 1 8 cos x + 7 8 sin x ) + e − x ( 3 8 cos x − 17 8 sin x ) Thus, the solution of the given ODE is y = e x ( 1 8 cos x + 7 8 sin x ) + e − x ( 3 8 cos x − 17 8 sin x ) _ . Plot the graph of the solution as follows. Observe from the above figure that the solution has a local minimum near the origin.

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