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: 32


$\ln |\ln (y)|=\tan^{-1} x+\ln (2)$

Work Step by Step

Separate the differentials and integrate. $\int (\ln y)^{-1} (\dfrac{1}{y}) dy=\int \dfrac{dx}{1+x^2}$ This implies that $\ln |\ln (y)|=\tan^{-1} x+c$ Applying the initial conditions $y(0)=e^2$, we get Thus, $\ln |\ln (e^2)|=\tan^{-1} (0)+c\implies c=\ln (2)$ Hence, $\ln |\ln (y)|=\tan^{-1} x+\ln (2)$
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