The solution of the given IVP and graph the solution. The solution of the given IVP is y = 4 e − x + 5 e − 0.5 x cos ( 3 x ) _ . Given: The given IVP is 4 y ′ ′ ′ + 8 y ″ + 41 y ′ + 37 y = 0 , y ( 0 ) = 9 , y ′ ( 0 ) = − 6.5 , y ″ ( 0 ) = − 39.75 . 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, 4 y ′ ′ ′ + 8 y ″ + 41 y ′ + 37 y = 0 , y ( 0 ) = 9 , y ′ ( 0 ) = − 6.5 , y ″ ( 0 ) = − 39.75 Consider the homogeneous equation of the given ODE, 4 y ′ ′ ′ + 8 y ″ + 41 y ′ + 37 y = 0 Obtain the characteristic equation as follows. 4 λ 3 + 8 λ 2 + 41 λ + 37 = 0 Compute the roots of the characteristic equation as follows. ( λ + 1 ) ( 4 λ 2 + 4 λ + 37 ) = 0 λ 1 = − 1 or 4 λ 2 + 4 λ + 37 = 0 λ 1 = − 1 or λ = − 4 ± 4 2 − 4 ( 4 ) ( 37 ) 2 ( 4 ) λ 1 = − 1 or λ = − 4 ± 4 − 36 8 λ 1 = − 1 or λ = − 1 2 ± 3 i The solution of the given ODE is given as follows. y = c 1 e − x + c 2 e − 0.5 x cos ( 3 x ) + c 3 e − 0.5 x sin ( 3 x ) Obtain the first 2 derivatives of the general solution obtained as follows. y = c 1 e − x + c 2 e − 0.5 x cos ( 3 x ) + c 3 e − 0.5 x sin ( 3 x ) y ′ = − c 1 e − x − 0.5 c 2 e − 0.5 x cos ( 3 x ) − 3 c 2 e − 0.5 x sin ( 3 x ) − 0.5 c 3 e − 0.5 x sin ( 3 x ) + 3 c 3 e − 0.5 x cos ( 3 x ) = − c 1 e − x + e − 0.5 x ( − 0.5 c 2 cos ( 3 x ) − 3 c 2 sin ( 3 x ) − 0.5 c 3 sin ( 3 x ) + 3 c 3 cos ( 3 x ) ) y ″ = [ c 1 e − x − 0.5 e − 0.5 x ( − 0.5 c 2 cos ( 3 x ) − 3 c 2 sin ( 3 x ) − 0.5 c 3 sin ( 3 x ) + 3 c 3 cos ( 3 x ) ) + e − 0.5 x ( 1.5 c 2 sin ( 3 x ) − 9 c 2 cos ( 3 x ) − 1.5 c 3 cos ( 3 x ) − 9 c 3 sin ( 3 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 cos ( 0 ) + c 3 e 0 sin ( 0 ) = 9 y ′ ( 0 ) = − c 1 e 0 + e 0 ( − 0.5 c 2 cos ( 0 ) − 3 c 2 sin ( 0 ) − 0.5 c 3 sin ( 0 ) + 3 c 3 cos ( 0 ) ) = − 6.5 y ″ ( 0 ) = [ c 1 e 0 − 0.5 e 0 ( − 0.5 c 2 cos ( 0 ) − 3 c 2 sin ( 0 ) − 0.5 c 3 sin ( 0 ) + 3 c 3 cos ( 0 ) ) + e 0 ( 1.5 c 2 sin ( 0 ) − 9 c 2 cos ( 0 ) − 1.5 c 3 cos ( 0 ) − 9 c 3 sin ( 0 ) ) ] = − 39.75 Simplify further as follows. c 1 + c 2 = 9 − c 1 − 0.5 c 2 + 3 c 3 = − 6.5 [ c 1 − 0.5 ( − 0.5 c 2 + 3 c 3 ) + − 9 c 2 − 1.5 c 3 ] = − 39.75 Simplify further as follows. c 1 + c 2 = 9 − c 1 − 0.5 c 2 + 3 c 3 = − 6.5 c 1 − 8.75 c 2 − 3 c 3 = − 39.75 Simplify further as follows. c 1 + c 2 = 9 − c 1 − 0.5 c 2 + 3 c 3 = − 6.5 c 1 − 8.75 c 2 − 3 c 3 + ( − c 1 − 0.5 c 2 + 3 c 3 ) = − 39.75 + ( − 6.5 ) Simplify further as follows. c 1 + c 2 = 9 − c 1 − 0.5 c 2 + 3 c 3 = − 6.5 − 9.25 c 2 = − 46.25 c 2 = 5 Simplify further as follows. c 1 = 4 c 3 = − 6.5 + 4 + 2.5 3 = 0 c 2 = 5 Substitute the values of the constants in the general solution as follows. y = 4 e − x + 5 e − 0.5 x cos ( 3 x ) Thus, the solution of the given ODE is y = 4 e − x + 5 e − 0.5 x cos ( 3 x ) _ . Plot the graph of the solution as follows. Observe from the above figure that the solution is diverging to infinity as x tends to infinity.

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