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 1221: 19


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

Work Step by Step

Need to plug the given values such as: $\dfrac{d^2 Q}{dt^2}+20\dfrac{dQ}{dt}+200Q=6$ The auxiliary equation is: $r^2+20r+200=0 \implies r= -10 \pm 10i$ Here, we have $Q_c= e^{-10t}(\sin 10t +\cos 10t)$ and $Q_p=A$ and $Q'_p=0; Q''_p=0$ On solving, we get $A=\dfrac{3}{100}$ Then, we have $Q_p=\dfrac{3}{100}$ $Q(t)=e^{-10t}(c_1\sin (10t) +c_2\cos (10t))+\dfrac{3}{100}$ ...(1) Set $t=0$ in the above equation (1) $Q(0)=\dfrac{1}{100}=e^{0}(c_1\sin 0 +c_2\cos 0)+\dfrac{1}{100}$ we get $c_2=-\dfrac{1}{50}$ and $c_1=-\dfrac{1}{50}$ Hence, we have $Q(t)=-\dfrac{1}{50}e^{-10t}[\cos (10t) + \sin (10t))+\dfrac{3}{100}$ Re-arrange as : $Q(t)=-0.02e^{-10t}[\cos (10t) + \sin (10t)]+0.03$
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