Explanation The given differential equations is, x 3 y ‴ − 6 x 2 y ″ + 17 x y ′ − 17 y = 0 with initial conditions are y ( 1 ) = − 2 , y ′ ( 1 ) = 0 , y ″ ( 1 ) = 0 . Let y = x r . Obtain the first, second and third derivative of y as follows: y ′ = r x r − 1 y ″ = r ( r − 1 ) x r − 2 y ‴ = r ( r − 1 ) ( r − 2 ) x r − 3 Substitute y ′ = r x r − 1 , y ″ = r ( r − 1 ) x r − 2 and y ‴ = r ( r − 1 ) ( r − 2 ) x r − 3 in x 3 y ‴ − 6 x 2 y ″ + 17 x y ′ − 17 y = 0 as follows: x 3 ( r ( r − 1 ) ( r − 2 ) x r − 3 ) − 6 x 2 ( r ( r − 1 ) x r − 2 ) + 17 x ( r x r − 1 ) − 17 x r = 0 r ( r − 1 ) ( r − 2 ) x r − 6 r ( r − 1 ) x r + 17 r x r − 17 x r = 0 x r ( r 3 − 2 r 2 − r 2 + 2 r − 6 r 2 + 6 r + 17 r − 17 ) = 0 x r ( r 3 − 9 r 2 + 25 r − 17 ) = 0 Note that, x r ≠ 0 . That is, r 3 − 9 r 2 + 25 r − 17 = 0 ( r − 1 ) ( r 2 − 8 r + 17 ) = 0 Hence r = 1 , r = 4 ± i . The general solution of the equation is y = C 1 x + C 2 x 4 cos ( ln ( x ) ) + C 3 x 4 sin ( ln ( x ) ) . Substitute y ( 1 ) = − 2 in y = C 1 x + C 2 x 4 cos ( ln ( x ) ) + C 3 x 4 sin ( ln ( x ) )

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

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Chapter 4.7, Problem 38E

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To solve: The initial value problem x3y‴−6x2y″+17xy′−17y=0.

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Chapter 4.7, Problem 37E

Chapter 4.7, Problem 39E

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Chapter 4 Solutions

Introductory Differential Equations

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The initial value problem x 3 y ‴ − 6 x 2 y ″ + 17 x y ′ − 17 y = 0 .

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To solve: The initial value problem x3y‴−6x2y″+17xy′−17y=0.

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Chapter 4 Solutions