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: 15


$2e^{\sqrt x}+e^{-y}=c$

Work Step by Step

Given: $\sqrt x\dfrac{dy}{dx}=e^{y+\sqrt x}$ Re-arrange the given equation and integrate as follows:. Then $\int \dfrac{e^{\sqrt x}dx}{\sqrt x}= \int \dfrac{dy}{e^{y}} \implies 2 \int \dfrac{e^{\sqrt x}dx}{2 \sqrt x}= \int e^{-y} dy$....(1) Let us take the help of $u$ substitution. plug $u=\sqrt x \implies \dfrac{dy}{2 \sqrt x}=du$ Equation (1) becomes: $ 2 \int e^{u}=-e^{-y}+c$ $\implies 2e^{u}=-e^{-y}+c$ Hence, $2e^{\sqrt x}+e^{-y}=c$
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