To determine The general solution of the given ODE. Explanation Given: The given ODE is ( D 4 + 10 D 2 + 9 I ) y = 6.5 sinh 2 x . 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 ODE, ( D 4 + 10 D 2 + 9 I ) y = 6.5 sinh 2 x y i v + 10 y ″ + 9 y = 6.5 ( e 2 x − e − 2 x 2 ) y i v + 10 y ″ + 9 y = 3.25 ( e 2 x − e − 2 x ) Consider the corresponding homogeneous equation of the given ODE, y i v + 10 y ″ + 9 y = 0 Obtain the characteristic equation as follows. λ 4 + 10 λ 2 + 9 = 0 Compute the roots of the characteristic equation as follows. λ 4 + 10 λ 2 + 9 = 0 ( λ 2 ) 2 + 10 λ 2 + 9 = 0 λ 2 = − 10 ± 10 2 − 36 2 = − 5 ± 4 λ 2 = − 9 , − 1 λ 1 = − i , λ 2 = i , λ 3 = − 3 i , λ 4 = 3 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 3 x + c 4 sin 3 x Let the particular solution is of the form y p = A e 2 x + B e − 2 x

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Chapter 3.3, Problem 3P

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