Explanation The given initial value problem is y ‴ − 12 y ″ + 27 y ′ + 40 y = 0 , y ( 0 ) = − 3 , y ′ ( 0 ) = − 6 , y ″ ( 0 ) = − 12 . The auxiliary equation associated with y ‴ − 12 y ″ + 27 y ′ + 40 y = 0 is r 3 − 12 r 2 + 27 r + 40 = 0 . When r = − 1 , r 3 − 12 r 2 + 27 r + 40 = ( − 1 ) 3 − 12 ( − 1 ) 2 + 27 ( − 1 ) + 40 = − 1 − 12 − 27 + 40 = − 40 + 40 = 0 Thus, r + 1 is a factor. By the polynomial division, obtain the other factor as follows: r + 1 r 2 − 13 r + 40 r 3 − 12 r 2 + 27 r + 40 r 3 + r 2 _ − 13 r 2 + 27 r − 13 r 2 − 13 r _ 40 r + 40 40 r + 40 _ 0 Thus, the equation r 3 − 12 r 2 + 27 r + 40 = 0 becomes ( r + 1 ) ( r 2 − 13 r + 40 ) = 0 . Factorize the polynomial r 2 − 13 r + 40 = 0 as follows: r 2 − 5 r − 8 r + 40 = 0 r ( r − 5 ) − 8 ( r − 5 ) = 0 ( r − 8 ) ( r − 5 ) = 0 r = 5 , 8 That is, the roots r = − 1 , r = 5 and r = 8 are distinct and real. The solutions will be e − 1 t , e 5 t and e 8 t . Thus, the general solution will be, y ( t ) = c 1 e − t + c 2 e 5 t + c 3 e 8 t . Obtain the particular solution as follows: Apply the initial condition y ( 0 ) = − 3 in y ( t ) = c 1 e − t + c 2 e 5 t + c 3 e 8 t , y ( 0 ) = c 1 e ( 0 ) + c 2 e 5 ( 0 ) + c 3 e 8 ( 0 ) − 3 = c 1 ( 1 ) + c 2 ( 1 ) + c 3 ( 1 ) c 1 + c 2 + c 3 = − 3 … ( 1 ) Differentiate y ( t ) = c 1 e − t + c 2 e 5 t + c 3 e 8 t with respect to t as, y ′ ( t ) = − c 1 e − t + 5 c 2 e 5 t + 8 c 3 e 8 t . Apply the initial condition y ′ ( 0 ) = − 6 in y ′ ( t ) = − c 1 e − t + 5 c 2 e 5 t + 8 c 3 e 8 t , y ′ ( 0 ) = − c 1 e ( 0 ) + 5 c 2 e 5 ( 0 ) + 8 c 3 e 8 ( 0 ) − 6 = − c 1 ( 1 ) + 5 c 2 ( 1 ) + 8 c 3 ( 1 ) − c 1 + 5 c 2 + 8 c 3 = − 6 … ( 2 ) Differentiate y ′ ( t ) = − c 1 e − t + 5 c 2 e 5 t + 8 c 3 e 8 t with respect to t as, y ″ ( t ) = c 1 e − t + 25 c 2 e 5 t + 64 c 3 e 8 t

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4, Problem 33RP

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The particular solution to the initial value problem y‴−12y″+27y′+40y=0,y(0)=−3,y′(0)=−6,y″(0)=−12.

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Chapter 4, Problem 32RP

Chapter 4, Problem 34RP

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Fundamentals of Differential Equations (9th Edition)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

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The particular solution to the initial value problem y ‴ − 12 y ″ + 27 y ′ + 40 y = 0 , y ( 0 ) = − 3 , y ′ ( 0 ) = − 6 , y ″ ( 0 ) = − 12 .

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The particular solution to the initial value problem y‴−12y″+27y′+40y=0,y(0)=−3,y′(0)=−6,y″(0)=−12.

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