Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Review - Exercises - Page 1181: 19


$Q(t)=-0.02e^{-10t}[\cos 10t + \sin 10t]+0.03$

Work Step by Step

After plugging the data in, we get $\dfrac{d^2 Q}{dt^2}+20\dfrac{dQ}{dt}+200Q=6$ The auxiliary equation is: $r^2+20r+200=0$ $\implies r= -10 \pm 10i$ Here,we have $Q_c= e^{-10t}(\sin 10t +\cos 10t)$ The particular solution is: $Q_p=A$ and $ Q'_p=0; Q''_p=0$ $A=\dfrac{3}{100}$ (Simplify) The particular solution is: $Q_p=\dfrac{3}{100}$ Now, $Q(t)=e^{-10t}(c_1\sin 10t +c_2\cos 10t)+\dfrac{3}{100}$ Consider at $t=0$ $Q(0)=e^{-10t}(c_1\sin 10t +c_2\cos 10t)+\dfrac{3}{100} \implies \dfrac{1}{100}=e^{0}(c_1\sin 0 +c_2\cos 0)+\dfrac{1}{100}$ This yields: $c_1=-\dfrac{1}{50}=-0.02 ; c_2=-\dfrac{1}{50}=-0.02$ Hence, $Q(t)=-0.02e^{-10t}[\cos 10t + \sin 10t]+0.03$
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