Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 9 - Differential Equations - 9.3 Separable Equations - 9.3 Exercises - Page 645: 4

Answer

$\displaystyle y=-\ln|\frac{x^2}{2}+C|$

Work Step by Step

$\displaystyle y'+xe^y=0$ $\displaystyle y'=\frac{dy}{dx}=-xe^y$ Seperation of variables: $\displaystyle e^{-y}dy=-xdx$ Integrate both sides $\displaystyle \int e^{-y}dy=\int -x$ $dx$ $\displaystyle -e^{-y}=-\frac{x^2}{2}+C$ $\displaystyle e^{-y}=\frac{x^2}{2}+C$ $\displaystyle -y=\ln|\frac{x^2}{2}+C|$ $\displaystyle y=-\ln|\frac{x^2}{2}+C|$ The reason why the answer has $\displaystyle \frac{x^2}{2}+C$ and not $\displaystyle \frac{x^2}{2}-C$ when $-1$ is multiplied on both sides is because $C$, the constant of integration, has not taken on a definite value and therefore remains the same.
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