To determine The solution of the given IVP. Explanation Given: The given IVP is ( D 3 − 2 D 2 − 9 D + 18 I ) y = e 2 x , y ( 0 ) = 4.5 , y ′ ( 0 ) = 8.8 , y ″ ( 0 ) = 17.2 . 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, ( D 3 − 2 D 2 − 9 D + 18 I ) y = e 2 x , y ( 0 ) = 4.5 , y ′ ( 0 ) = 8.8 , y ″ ( 0 ) = 17.2 y ′ ′ ′ − 2 y ″ − 9 y ′ + 18 y = e 2 x , y ( 0 ) = 4.5 , y ′ ( 0 ) = 8.8 , y ″ ( 0 ) = 17.2 Consider the corresponding homogeneous equation of the given ODE, y ′ ′ ′ − 2 y ″ − 9 y ′ + 18 y = 0 Obtain the characteristic equation as follows. λ 3 − 2 λ 2 − 9 λ + 18 = 0 Compute the roots of the characteristic equation as follows. ( λ − 2 ) ( λ 2 − 9 ) = 0 ( λ − 2 ) ( λ + 3 ) ( λ − 3 ) = 0 λ 1 = 2 , λ 2 = − 3 , λ 3 = 3 The homogeneous solution of the given ODE is given as follows. y h = c 1 e 2 x + c 2 e − 3 x + c 3 e 3 x Let the particular solution is of the form y p = A e 2 x . Observe that it is already in the homogeneous solution, multiply by x . y p = A x e 2 x Obtain the first three derivatives of the above equation as follows. y ′ p = 2 A x e 2 x + A e 2 x y ″ p = 4 A x e 2 x + 2 A e 2 x + 2 A e 2 x = 4 A x e 2 x + 4 A e 2 x y ′ ′ ′ p = 8 A x e 2 x + 4 A e 2 x + 8 A e 2 x = 8 A x e 2 x + 12 A e 2 x Substitute the above derivatives in the given ODE as follows. y ′ ′ ′ − 2 y ″ − 9 y ′ + 18 y = e 2 x 8 A x e 2 x + 12 A e 2 x − 2 ( 4 A x e 2 x + 4 A e 2 x ) − 9 ( 2 A x e 2 x + A e 2 x ) + 18 ( A x e 2 x ) = e 2 x A e 2 x ( 12 + 8 x − 8 − 8 x − 9 − 18 x + 18 x ) = e 2 x − 5 A e 2 x = e 2 x Equate like terms on both sides as follows. − 5 A = 1 A = − 1 5 The particular solution of the given ODE is given as follows. y p = − 1 5 x e 2 x The general solution of the given ODE is as follows, y = y h + y p = c 1 e 2 x + c 2 e − 3 x + c 3 e 3 x − 1 5 x e 2 x Obtain the first 2 derivatives of the general solution obtained as follows

10th Edition

ISBN: 9781119664697

Author: Kreyszig

Publisher: WILEY

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Chapter 3.3, Problem 12P

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The solution of the given IVP.

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ADVANCED ENGINEERING MATHEMATICS

Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - 1–6 BASES: TYPICAL EXAMPLES To get a feel for...Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P Ch. 3.1 - Prob. 11P

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