Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 35


$$y=t- \frac{\pi}{3} \cos 3t -\frac{\pi}{3} $$

Work Step by Step

Given$$ \frac{d y}{d t}=1+\pi\sin (3t) , \quad y(\pi)=\pi $$ Since :\begin{align*} y&= \int (1+\pi\sin (3t) ) dt\\ &=t- \frac{\pi}{3} \cos 3t+c \end{align*} at $t=\pi,\ \ \ y= \pi $, then $c= -\pi/3$, hence $$y=t- \frac{\pi}{3} \cos 3t -\frac{\pi}{3} $$
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