Explanation Given information: The linear differential equation y ″ + 4 y ′ + 5 y = 0 . The initial conditions y ( 0 ) = 2 and y ′ ( 0 ) = − 7 . Formula used: If m 1 = α + i β and m 2 = α − i β are complex zeros of the characteristics equation, then the general solution is y = C 1 e α x cos β x + C 2 e α x sin β x . Calculation: The characteristic equation for the differential equation y ″ + 4 y ′ + 5 y = 0 is m 2 + 4 m + 5 = 0 . The zeros of m 2 + 4 m + 5 = 0 can be calculated as: m 2 + 4 m + 5 = 0 m = − 4 ± ( 4 ) 2 − 4 × 5 2 m = − 4 ± − 4 2 m = − 4 ± 2 i 2 m = − 2 ± i Thus, α = − 2 and β = 1 As the characteristic equation hascomplex zeros the general solution is y = C 1 e − 2 x cos x + C 2 e − 2 x sin x

11th Edition

ISBN: 9781337604789

Author: Larson

Publisher: Cengage

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Chapter 16, Problem 30RE

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To calculate: The particular solution of the linear differential equation y″+4y′+5y=0 that satisfies the initial condition y(0)=2 and y′(0)=−7.

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Chapter 16, Problem 29RE

Chapter 16, Problem 31RE

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The particular solution of the linear differential equation y ″ + 4 y ′ + 5 y = 0 that satisfies the initial condition y ( 0 ) = 2 and y ′ ( 0 ) = − 7 .

Solution Summary: The author calculates the particular solution of the linear differential equation y′′+4

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To calculate: The particular solution of the linear differential equation y″+4y′+5y=0 that satisfies the initial condition y(0)=2 and y′(0)=−7.

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Chapter 16 Solutions