Explanation Given information: The given differential equation is d 2 q d t 2 + 4 d q d t + 8 q = 3 sin 4 t and assume that q ( 0 ) = 0 and q ' ( 0 ) = 0 . Concept Used: If y p is the particular solution of the differential equation y ' ' + a y ' + b y = F ( x ) and y h is the general solution, then y = y h + y p is the general solution of the non homogeneous equation. Calculation: Consider the given differential equation is d 2 q d t 2 + 4 d q d t + 8 q = 3 sin 4 t . The characteristics equation is m 2 + 4 m + 8 = 0 . First, we have to find the solution of the characteristics equation m 2 + 4 m + 8 = 0 . m 2 + 4 m + 8 = 0 ( m + 2 ) 2 + 4 = 0 ( m + 2 ) 2 = − 4 m + 2 = ± 2 i m = − 2 ± 2 i m = − 2 + 2 i o r − 2 − 2 i Therefore, q h = C 1 e − 2 t sin 2 t + C 2 e − 2 t cos 2 t Let q p be generalized form of 3 sin 4 t . q p = A cos 4 t + B sin 4 t q p ' = − 4 A sin 4 t + 4 B cos 4 t q p ' ' = − 16 A cos 4 t − 16 B sin 4 t Substituting in the original equation yields q ' ' + 4 q ' + 8 = 3 sin 4 t − 16 A cos 4 t − 16 B sin 4 t − 16 A sin 4 t + 16 B cos 4 t + 8 A cos 4 t + 8 B sin 4 t = 3 sin 4 t − 8 A cos 4 t + 16 B cos 4 t − 16 A sin 4 t − 8 B sin 4 t = 3 sin 4 t ( − 8 A + 16 B ) cos 4 t − ( 16 A + 8 B ) sin 4 t = 3 sin 4 t Equating the coefficient of like terms yields − 16 A − 8 B = 3 − 8 A + 16 B = 0 − 8 A = − 16 B A = 2 B So, A = − 3 20 a n d B = − 3 40 Therefore, q p = − 3 20 cos 4 t − 3 40

Student Solutions Manual For Larson/edwards? Multivariable Calculus, 11th

11th Edition

ISBN: 9781337275392

Author: Larson, Ron; Edwards, Bruce H.

Publisher: Cengage Learning

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Chapter 16, Problem 55RE

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To calculate: The charge q as a function of time t for the electrical circuit describedfor the differential equation is d2qdt2+4dqdt+8q=3sin4t.

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Chapter 16, Problem 54RE

Chapter 16, Problem 56RE

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Chapter 16 Solutions

Student Solutions Manual For Larson/edwards? Multivariable Calculus, 11th

Ch. 16.1 - Exactness What does it mean for the...Ch. 16.1 - Integrating Factor When is it beneficial to use an...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Testing for Exactness In Exercises 3-6, determine...Ch. 16.1 - Prob. 5E Ch. 16.1 - Prob. 6E Ch. 16.1 - Prob. 7E Ch. 16.1 - Solving an Exact Differential Equation In...Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The charge q as a function of time t for the electrical circuit describedfor the differential equation is d 2 q d t 2 + 4 d q d t + 8 q = 3 sin 4 t .

Solution Summary: The author calculates the charge q as a function of time t for the electrical circuit.

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To calculate: The charge q as a function of time t for the electrical circuit describedfor the differential equation is d2qdt2+4dqdt+8q=3sin4t.

Want to see the full answer?

Chapter 16 Solutions