Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 253: 54



Work Step by Step

$\frac{dy}{dt}=(4t+3)^{-2};$ $y(1)=0$ $dy=(4t+3)^{-2}dt$ $\int dy=\int(4t+3)^{-2}dt$ $y+C=\frac{1}{4}\int4(4t+3)^{-2}dt$ $y=\frac{1}{4}(-(4t+3)^{-1})+C$ Substitute in initial condition to solve for $C$ $0=\frac{1}{4}(-\frac{1}{7})+C$ $C=\frac{1}{28}$ $y=-\frac{1}{4(4t+3)}+\frac{1}{28}$
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