Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 15

Answer

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

Work Step by Step

$\sqrt x\frac{dy}{dx}$ = $e^{(y+\sqrt x)}$ $e^{-y}dy$ = $\frac{e^{\sqrt x}}{\sqrt x}dx$ $\int(e^{-y})dy$ = $\int\frac{e^{\sqrt x}}{\sqrt x}dx$ $u$ = $\sqrt x$ $du$ = $\frac{1}{2\sqrt x}dx$ $2du$ = $\frac{1}{\sqrt x}dx$ $\int(e^{-y})dy$ = $2\int(e^{u})du$ $-e^{-y}$ = $2e^{u}+C$ $-e^{-y}$ = $2e^{\sqrt x}+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.