## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 36

#### Answer

$$y=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) - \frac{\pi}{8}$$

#### Work Step by Step

Given$$\frac{d y}{d t}=\cos (3\pi t)+\sin (4\pi t) , \quad y(1/3)= 0$$ Since :\begin{align*} y&= \int (\cos (3\pi t)+\sin (4\pi t) ) dt\\ &=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) +c \end{align*} at $t=\frac{1}{3},\ \ \ y= 0$, then $c= -\pi/8$, hence: $$y=\frac{1}{3 \pi }\sin (3\pi t)-\frac{1}{4\pi }\cos (4\pi t) - \frac{\pi}{8}$$

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