Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises: 4

Answer

$y = -ln(x^2/2 + C)$

Work Step by Step

$y' +xe^y = 0$ can be rewritten as : $dy/dx + xe^y = 0$ Now let's separate the y on one side and the x on the other side: $dy/dx = -xe^y$ $e^{-y}dy = -xdx$ **It is to be noted that $1/e^y = e^{-y}$** Now let's integrate both sides: $\int e^{-y}dy =\int -xdx$ $-e^{-y} = -x^2/2 + C$ $e^{-y} = x^2/2 + C_2$ To isolate y, we need to add ln() to both sides: $ln(e^{-y}) = ln(x^2/2 + C_2)$ $-y = ln(x^2/2 + C_2)$ $y = -ln(x^2/2 + C_2)$
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