Explanation Given: The differential equation is, y ( i v ) + 4 y ‴ − 3 y ″ − 64 y ′ − 208 y = 0 Approach: Consider the differential equation given by, P ( D ) y = 0 Let r 1 , r 2 , … , r k be the distinct roots of the auxiliary equation, so that, P ( r ) = ( r − r 1 ) m 1 ( r − r 2 ) m 2 ⋯ ( r − r k ) m k Here m i denotes the multiplicity of the root r = r i . a. If r i is real, then the functions e r 1 x , x e r 2 x , … , x m i − 1 e r i x are linearly independent solutions to the differential equation. b. If r j is complex, say r j = a + i b , then the functions e a x cos b x , c e a x cos b x , … , x m j − 1 e a x cos b x e a x sin b x , c e a x sin b x , … , x m j − 1 e a x sin b x Corresponding to the conjugate roots r = a ± i b are linearly independent solutions to the given differential equation. c. The n real-valued solutions y 1 , y 2 , … , y n for roots r 1 , r 2 , … , r k are linearly independent on any interval. The general solution is given as, y ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) + ⋯ + c n y n ( x ) Calculation: The differential equation is, y ( i v ) + 4 y ‴ − 3 y ″ − 64 y ′ − 208 y = 0 … ( 1 ) The auxiliary equation for equation ( 1 ) is given as, t 4 + 4 t 3 − 3 t 2 − 64 t − 208 = 0 … ( 2 ) Substitute 4 for t in the equation ( 2 ) as, ( 4 ) 4 + 4 ( 4 ) 3 − 3 ( 4 ) 2 − 64 ( 4 ) − 208 = 0 256 + 256 − 48 − 256 − 208 = 0 0 = 0 Therefore, ( t − 4 ) is the factor for the equation ( 2 ) . Divide equation ( 2 ) by ( t − 4 ) to get the equation, t 3 + 8 t 2 + 29 t + 52 = 0 … ( 3 ) Substitute − 4 for t in the equation ( 3 ) as,

Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

4th Edition

ISBN: 9780321985811

Author: Stephen W. Goode, Scott A. Annin

Publisher: Pearson (edition 4)

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Chapter 8.2, Problem 49P

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The general solution to the differential equation y(iv)+4y‴−3y″−64y′−208y=0.

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Chapter 8.2, Problem 48P

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Differential Equations And Linear Algebra, Books A La Carte Edition (4th Edition)

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The general solution to the differential equation y ( i v ) + 4 y ‴ − 3 y ″ − 64 y ′ − 208 y = 0 .

Solution Summary: The author explains that the general solution to the differential equation is, y(iv)+4

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The general solution to the differential equation y(iv)+4y‴−3y″−64y′−208y=0.

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