Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Review - Exercises - Page 657: 12

Answer

$y=\ln \left(\frac{e}{1-ex^3}\right)$

Work Step by Step

Solve for $y$: $\frac{dy}{dx}=3x^2e^{y}$ (Separate the variables) $\frac{1}{e^{y}}dy=3x^2dx$ $-e^{-y}dy=-3x^2dx$ (Integrate) $\int -e^{-y}dy=\int -3x^2 dx$ $e^{-y}=-x^3+C$ Substitute the initial value $y(0)=1$: $e^{-1}=-0^3+C$ $e^{-1}=0+C$ $C=e^{-1}$ Now we have: $e^{-y}=-x^3+e^{-1}$ $\frac{1}{e^y}=\frac{1-ex^3}{e}$ $e^{y}=\frac{e}{1-ex^3}$ Taking the logarithm, $y=\ln (\frac{e}{1-ex^3})$ Here is the graph of this solution.
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