To determine The solution of the given IVP. Explanation Given: The given IVP is y ′ ′ ′ + 5 y ″ + 24 y ′ + 20 y = x , y ( 0 ) = 1.94 , y ′ ( 0 ) = − 3.95 , y ″ ( 0 ) = − 24 . 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 ′ ′ ′ + 5 y ″ + 24 y ′ + 20 y = x , y ( 0 ) = 1.94 , y ′ ( 0 ) = − 3.95 , y ″ ( 0 ) = − 24 Consider the corresponding homogeneous equation of the given ODE, y ′ ′ ′ + 5 y ″ + 24 y ′ + 20 y = 0 Obtain the characteristic equation as follows. λ 3 + 5 λ 2 + 24 λ + 20 = 0 Compute the roots of the characteristic equation as follows. ( λ + 1 ) ( λ 2 + 4 λ + 20 ) = 0 λ 1 = − 1 , λ = − 4 ± 4 2 − 4 ( 1 ) ( 20 ) 2 λ 1 = − 1 , λ = − 4 ± 16 − 80 2 λ 1 = − 1 , λ = − 4 ± 8 i 2 λ 1 = − 1 , λ 2 = − 2 − 4 i , λ 3 = − 2 + 4 i The homogeneous solution of the given ODE is given as follows. y h = c 1 e − x + c 2 e − 2 x cos 4 x + c 3 e − 2 x sin 4 x Let the particular solution is of the form y p = A x + B . Obtain the first three derivatives of the above equation as follows. y ′ p = A y ″ p = 0 y ′ ′ ′ p = 0 Substitute the above derivatives in the given ODE as follows. 0 + 5 ( 0 ) + 24 A + 20 ( A x + B ) = x 20 A x + ( 24 A + 20 B ) = x Equate like terms on both sides as follows. 20 A = 1 and 24 A + 20 B = 0 A = 1 20 and B = − 24 20 ( 1 20 ) A = 0.05 and B = − 0.06 The particular solution of the given ODE is given as follows. y p = 0.05 x − 0.06 The general solution of the given ODE is as follows, y = y h + y p = c 1 e − x + c 2 e − 2 x cos 4 x + c 3 e − 2 x sin 4 x + 0.05 x − 0.06 Obtain the first 2 derivatives of the general solution obtained as follows. y = c 1 e − x + c 2 e − 2 x cos 4 x + c 3 e − 2 x sin 4 x + 0.05 x − 0.06 y ′ = − c 1 e − x − 2 c 2 e − 2 x cos 4 x + c 2 e − 2 x ( − 4 sin 4 x ) − 2 c 3 e − 2 x sin 4 x + c 3 e − 2 x ( 4 cos 4 x ) + 0.05 = − c 1 e − x + cos 4 x ( 4 c 3 e − 2 x − 2 c 2 e − 2 x ) + sin 4 x ( − 4 c 2 e − 2 x − 2 c 3 e − 2 x ) + 0

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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