University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 5


$y'+y=\dfrac{2}{1+4e^{2x}}$ and $\dfrac{\pi}{2}$

Work Step by Step

Apply chain rule of differentiation. Thus, $y'=-e^{-x} \tan^{-1} (2e^x) + e^{-x} (2e^x)[\dfrac{1}{1+(2e^x)^2}]$ or, $y'=-e^{-x} \tan^{-1} (2e^x) + \dfrac{2}{1+4e^{2x}}=-y+\dfrac{2}{1+4e^{2x}}$ or, $y'+y=\dfrac{2}{1+4e^{2x}}$ This is the required differential equation. Apply the initial conditions. Since we have $y=-e^{-x} \tan^{-1} (2e^x) $ Thus, $y(-\ln 2)=-e^{-(-\ln 2)} \tan^{-1} (2e^{(-\ln 2)}) =2 (\dfrac{\pi}{4})=\dfrac{\pi}{2}$
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