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: 22


$\ln (1+e^y)=e^x+x+c$

Work Step by Step

Given: $\dfrac{dy}{dx}=e^{x-y}+e^{x}+e^{-y}+1$ Re-arrange the given equation and integrate as follows:. Then, $\dfrac{dy}{dx}=(e^{x}+1)(e^{-y}+1)$ Then $ \int (e^{x}+1) dx=\int \dfrac{1}{e^{-y}+1} dy \implies \int (e^x+1) dx=\int \dfrac{e^y}{1+e^{y}} dy $ ....(1) Let us take the help of $u$ substitution. plug $u=e^y \implies du=e^y dy$ Equation (1) becomes: $ \int (e^{x}+1) dx=\int \dfrac{du}{1+u}$ $e^x+x+c=\ln (1+u)$ Hence, $\ln (1+e^y)=e^x+x+c$
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