Explanation The given differential equation is, a y ″ + b y ′ + c y = 0 , where b 2 − 4 a c < 0 . Given that, the characteristic equation of the differential equation has complex roots λ ± i μ . The characteristic equation is, a r 2 + b r + c = 0 and the complex roots are, λ ± i μ . That is, − b ± b 2 − 4 a c 2 a = r − b ± b 2 − 4 a c 2 a = λ ± i μ Therefore, the values of λ and μ are, λ = − b 2 a μ = | b 2 − 4 a c | 2 a The given functions are, u ( t ) = e λ t cos μ t and v ( t ) = e λ t sin μ t . The derivatives of the above functions are, u ′ ( t ) = λ e λ t cos μ t − μ e λ t sin μ t v ′ ( t ) = λ e λ t sin μ t + μ e λ t cos μ t The second derivative of the function u is, u ″ ( t ) = λ 2 e λ t cos μ t − μ λ e λ t sin μ t − μ λ e λ t sin μ t − μ 2 e λ t cos μ t = ( λ 2 − μ 2 ) e λ t cos μ t − 2 μ λ e λ t sin μ t The second derivative of the function v is, v ″ ( t ) = λ 2 e λ t sin μ t + μ λ e λ t cos μ t + μ λ e λ t cos μ t − μ 2 e λ t sin μ t = ( λ 2 − μ 2 ) e λ t sin μ t + 2 μ λ e λ t cos μ t Consider the function u . Substitute u , u ′ and u ″ in the equation, a y ″ + b y ′ + c y = 0 as follows. a y ″ + b y ′ + c y = 0 a u ″ + b u ′ + c u = 0 a ( ( λ 2 − μ 2 ) e λ t cos μ t − 2 μ λ e λ t sin μ t ) + b ( λ e λ t cos μ t − μ e λ t sin μ t ) + c ( e λ t cos μ t ) = 0 ( a ( λ 2 − μ 2 ) + b λ + c ) e λ t cos μ t + ( − 2 a μ λ − b μ ) e λ t sin μ t = 0 Now, substitute λ and μ in the above equation

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Question

Chapter 3.3, Problem 23P

To determine

To show: The functions, u(t)=eλtcosμt and v(t)=eλtsinμt are the solutions of the differential equation, ay″+by′+cy=0 which has characteristic equation with complex roots, λ±iμ.

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Chapter 3.3, Problem 22P

Chapter 3.3, Problem 24P

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Chapter 3 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

Ch. 3.1 - Prob. 1P Ch. 3.1 - Prob. 2P Ch. 3.1 - Prob. 3P Ch. 3.1 - Prob. 4P Ch. 3.1 - Prob. 5P Ch. 3.1 - Prob. 6P Ch. 3.1 - Prob. 7P Ch. 3.1 - Prob. 8P Ch. 3.1 - Prob. 9P Ch. 3.1 - Prob. 10P

Ch. 3.1 - Prob. 11P Ch. 3.1 - Prob. 12P Ch. 3.1 - Prob. 13P Ch. 3.1 - Prob. 14P Ch. 3.1 - Prob. 15P Ch. 3.1 - Prob. 16P Ch. 3.1 - Prob. 17P Ch. 3.1 - Prob. 18P Ch. 3.1 - Prob. 20P Ch. 3.2 - Prob. 1P Ch. 3.2 - Prob. 2P Ch. 3.2 - Prob. 3P Ch. 3.2 - Prob. 4P Ch. 3.2 - Prob. 5P Ch. 3.2 - Prob. 6P Ch. 3.2 - Prob. 7P Ch. 3.2 - Prob. 8P Ch. 3.2 - Prob. 9P Ch. 3.2 - Prob. 10P Ch. 3.2 - Prob. 11P Ch. 3.2 - Prob. 12P Ch. 3.2 - Prob. 13P Ch. 3.2 - Prob. 14P Ch. 3.2 - Prob. 15P Ch. 3.2 - Prob. 16P Ch. 3.2 - Prob. 17P Ch. 3.2 - Prob. 18P Ch. 3.2 - Prob. 19P Ch. 3.2 - Prob. 20P Ch. 3.2 - Prob. 21P Ch. 3.2 - Prob. 22P Ch. 3.2 - Prob. 23P Ch. 3.2 - Prob. 24P Ch. 3.2 - Prob. 25P Ch. 3.2 - Prob. 26P Ch. 3.2 - Prob. 27P Ch. 3.2 - Prob. 28P Ch. 3.2 - Prob. 29P Ch. 3.2 - Prob. 30P Ch. 3.2 - Prob. 31P Ch. 3.2 - Prob. 32P Ch. 3.2 - Prob. 33P Ch. 3.2 - Prob. 34P Ch. 3.2 - Prob. 35P Ch. 3.2 - Prob. 36P Ch. 3.2 - Prob. 37P Ch. 3.2 - Prob. 38P Ch. 3.3 - Prob. 1P Ch. 3.3 - Prob. 2P Ch. 3.3 - Prob. 3P Ch. 3.3 - Prob. 4P Ch. 3.3 - Prob. 5P Ch. 3.3 - Prob. 6P Ch. 3.3 - Prob. 7P Ch. 3.3 - Prob. 8P Ch. 3.3 - Prob. 9P Ch. 3.3 - Prob. 10P Ch. 3.3 - Prob. 11P Ch. 3.3 - Prob. 12P Ch. 3.3 - Prob. 13P Ch. 3.3 - Prob. 14P Ch. 3.3 - Prob. 15P Ch. 3.3 - Prob. 16P Ch. 3.3 - Prob. 17P Ch. 3.3 - Prob. 18P Ch. 3.3 - Prob. 19P Ch. 3.3 - Prob. 20P Ch. 3.3 - Prob. 21P Ch. 3.3 - Prob. 22P Ch. 3.3 - Prob. 23P Ch. 3.3 - Prob. 24P Ch. 3.3 - Prob. 25P Ch. 3.3 - Prob. 26P Ch. 3.3 - Prob. 27P Ch. 3.3 - Prob. 28P Ch. 3.3 - Prob. 29P Ch. 3.3 - Prob. 30P Ch. 3.3 - Prob. 31P Ch. 3.3 - Prob. 32P Ch. 3.3 - Prob. 33P Ch. 3.3 - Prob. 34P Ch. 3.3 - Prob. 35P Ch. 3.3 - Prob. 36P Ch. 3.4 - Prob. 1P Ch. 3.4 - Prob. 2P Ch. 3.4 - Prob. 3P Ch. 3.4 - Prob. 4P Ch. 3.4 - Prob. 5P Ch. 3.4 - Prob. 6P Ch. 3.4 - Prob. 7P Ch. 3.4 - Prob. 8P Ch. 3.4 - Prob. 9P Ch. 3.4 - Prob. 10P Ch. 3.4 - Prob. 11P Ch. 3.4 - Prob. 12P Ch. 3.4 - Prob. 13P Ch. 3.4 - Prob. 14P Ch. 3.4 - Prob. 15P Ch. 3.4 - Prob. 16P Ch. 3.4 - Prob. 17P Ch. 3.4 - Prob. 18P Ch. 3.4 - Prob. 19P Ch. 3.4 - Prob. 20P Ch. 3.4 - Prob. 21P Ch. 3.4 - Prob. 22P Ch. 3.4 - Prob. 23P Ch. 3.4 - Prob. 24P Ch. 3.4 - Prob. 25P Ch. 3.4 - Prob. 26P Ch. 3.4 - Prob. 27P Ch. 3.4 - Prob. 28P Ch. 3.4 - Prob. 29P Ch. 3.4 - Prob. 30P Ch. 3.4 - Prob. 31P Ch. 3.4 - Prob. 32P Ch. 3.4 - Prob. 33P Ch. 3.4 - Prob. 34P Ch. 3.5 - Prob. 1P Ch. 3.5 - Prob. 2P Ch. 3.5 - Prob. 3P Ch. 3.5 - Prob. 4P Ch. 3.5 - Prob. 5P Ch. 3.5 - Prob. 6P Ch. 3.5 - Prob. 7P Ch. 3.5 - Prob. 8P Ch. 3.5 - Prob. 9P Ch. 3.5 - Prob. 10P Ch. 3.5 - Prob. 11P Ch. 3.5 - Prob. 12P Ch. 3.5 - Prob. 13P Ch. 3.5 - Prob. 14P Ch. 3.5 - Prob. 15P Ch. 3.5 - Prob. 22P Ch. 3.5 - Prob. 23P Ch. 3.5 - Prob. 24P Ch. 3.5 - Prob. 25P Ch. 3.5 - Prob. 26P Ch. 3.5 - Prob. 27P Ch. 3.5 - Prob. 28P Ch. 3.5 - Prob. 29P Ch. 3.5 - Prob. 30P Ch. 3.6 - Prob. 1P Ch. 3.6 - Prob. 2P Ch. 3.6 - Prob. 3P Ch. 3.6 - Prob. 4P Ch. 3.6 - Prob. 5P Ch. 3.6 - Prob. 6P Ch. 3.6 - Prob. 7P Ch. 3.6 - Prob. 8P Ch. 3.6 - Prob. 9P Ch. 3.6 - Prob. 10P Ch. 3.6 - Prob. 11P Ch. 3.6 - Prob. 12P Ch. 3.6 - Prob. 13P Ch. 3.6 - Prob. 14P Ch. 3.6 - Prob. 15P Ch. 3.6 - Prob. 16P Ch. 3.6 - Prob. 17P Ch. 3.6 - Prob. 18P Ch. 3.6 - Prob. 19P Ch. 3.6 - Prob. 20P Ch. 3.6 - Prob. 21P Ch. 3.6 - Prob. 22P Ch. 3.6 - Prob. 23P Ch. 3.6 - Prob. 24P Ch. 3.6 - Prob. 25P Ch. 3.6 - Prob. 26P Ch. 3.7 - Prob. 1P Ch. 3.7 - Prob. 2P Ch. 3.7 - Prob. 3P Ch. 3.7 - Prob. 4P Ch. 3.7 - Prob. 5P Ch. 3.7 - Prob. 6P Ch. 3.7 - Prob. 7P Ch. 3.7 - Prob. 8P Ch. 3.7 - Prob. 9P Ch. 3.7 - Prob. 10P Ch. 3.7 - Prob. 11P Ch. 3.7 - Prob. 12P Ch. 3.7 - Prob. 13P Ch. 3.7 - Prob. 14P Ch. 3.7 - Prob. 15P Ch. 3.7 - Prob. 16P Ch. 3.7 - Prob. 17P Ch. 3.7 - Prob. 18P Ch. 3.7 - Prob. 19P Ch. 3.7 - Prob. 20P Ch. 3.7 - Prob. 21P Ch. 3.7 - Prob. 22P Ch. 3.7 - Prob. 23P Ch. 3.7 - Prob. 24P Ch. 3.8 - Prob. 1P Ch. 3.8 - Prob. 2P Ch. 3.8 - Prob. 3P Ch. 3.8 - Prob. 4P Ch. 3.8 - Prob. 5P Ch. 3.8 - Prob. 6P Ch. 3.8 - Prob. 7P Ch. 3.8 - Prob. 9P Ch. 3.8 - Prob. 11P Ch. 3.8 - Prob. 14P

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The functions, u ( t ) = e λ t cos μ t and v ( t ) = e λ t sin μ t are the solutions of the differential equation, a y ″ + b y ′ + c y = 0 which has characteristic equation with complex roots, λ ± i μ .

To show: The functions, u(t)=eλtcosμt and v(t)=eλtsinμt are the solutions of the differential equation, ay″+by′+cy=0 which has characteristic equation with complex roots, λ±iμ.

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Chapter 3 Solutions