(a) Explanation The given equation with initial conditions is, 3 u ″ − u ′ + 2 u = 0 , u ( 0 ) = 2 and u ′ ( 0 ) = 0 . Obtain the general solution of the differential equation as follows. The characteristic equation of the given differential equation is, 3 r 2 − r + 2 = 0 . The discriminant of the characteristic equation is, D = b 2 − 4 a c = ( − 1 ) 2 − 4 ( 3 ) ( 2 ) = − 23 Since, D < 0 , the roots of the equation are conjugate complex numbers. The roots of the equation, 3 r 2 − r + 2 = 0 is computed as follows. 3 r 2 − r + 2 = 0 r = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 3 ) ( 2 ) 2 ( 3 ) r = 1 ± − 23 6 r = 1 6 ± 23 6 i r 1 = 1 6 + 23 6 i and r 2 = 1 6 − 23 6 i Therefore, the value of the real numbers are, λ = 1 6 and μ = 23 6 and the corresponding expressions for y are, u 1 ( t ) = e ( 1 6 + 23 6 i ) t u 2 ( t ) = e ( 1 6 − 23 6 i ) t Hence, the general solution is obtained as follows. u ( t ) = c 1 e t 6 cos ( 23 6 t ) + c 2 e t 6 sin ( 23 6 t ) The initial conditions are given as, u ( 0 ) = 2 and u ′ ( 0 ) = 0 . The value of the constant c 1 is obtained on applying the initial condition in y . c 1 e 0 cos ( 0 ) + c 2 e 0 sin ( 0 ) = 2 c 1 = 2 Thus, the equation u becomes, u ( t ) = 2 e t 6 cos ( 23 6 t ) + c 2 e t 6 sin ( 23 6 t ) (b) To determine The solution for which, the first time. | u ( t ) | = 10 is true.

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 16P

(a)

To determine

The solution of the initial value problem, 3u″−u′+2u=0, u(0)=2 and u′(0)=0.

(b)

To determine

The solution for which, the first time. |u(t)|=10 is true.

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Chapter 3.3, Problem 15P

Chapter 3.3, Problem 17P

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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

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The solution of the initial value problem, 3 u ″ − u ′ + 2 u = 0 , u ( 0 ) = 2 and u ′ ( 0 ) = 0 .

The solution of the initial value problem, 3u″−u′+2u=0, u(0)=2 and u′(0)=0.

The solution for which, the first time. |u(t)|=10 is true.

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Chapter 3 Solutions