Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 17 - Second-Order Differential Equations - Review - Exercises - Page 1222: 20


$x(t)=0.6 e^{-4t} \sin 4t$

Work Step by Step

Need to plug the given values such as: $2\dfrac{d^2 x}{dt^2}+16\dfrac{dx}{dt}+64 x=0$ The auxiliary equation is: $2r^2+16r+64=0 \implies r= -4 \pm 4i$ Given: $x'=2.4$ Here, we have $x= e^{-4t}(c_1 \cos 4t +c_2 \sin 4t)$ Set $t=0$, then we have $x(0)= e^{0}(c_1 \cos 0 +c_2 \sin 0)$ Now, $c_1=0; x'(t) = -e^{-4t}(c_1 \cos (4t) +c_2 \sin (4t))+e^{-4t}(-4c_1 \sin (4t) +4c_2 \cos (4t))$ When $t=0$, then we have $x'(0) = -e^{0}((0) \times \cos 0 +c_2 \times \sin 0)+e^{0}(-4c_1 \sin 0 +4c_2 \cos 0)$ Then, we get $c_2=0.6$ Answer: $x(t)=e^{-4t}[(0) \times \cos (4t) +(0.6) \times \sin (4t)]$ Hence, $x(t)=0.6 (e^{-4t}) \sin (4t)$
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