University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Practice Exercises - Page 421: 33


$2 \ln (\sqrt y+1)=\ln |x|+\ln (4)$

Work Step by Step

Separate the differentials and integrate. $\int \dfrac{1}{\sqrt y(\sqrt y+1)} dy=\int \dfrac{dx}{x}$ Re-write as: $\int \dfrac{2 dy}{2 \sqrt y(\sqrt y+1)} =\int \dfrac{dx}{x}$ This implies that $2 \ln |(\sqrt y+1)|=\ln |x|+c$ Applying the initial conditions $y(1)=1$, we get Thus, $2 \ln |(\sqrt (1)+1)|=\ln |1|+c\implies c=2 \ln (2)=\ln (4)$ Hence, $2 \ln (\sqrt y+1)=\ln |x|+\ln (4)$
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