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: 6

Answer

$\displaystyle \frac{u^2}{2}+\frac{u^5}{5}=\frac{t^3}{3}-\frac{1}{t}+C$

Work Step by Step

$\displaystyle \frac{du}{dt}=\frac{1+t^4}{ut^2+u^4t^2}=\frac{1+t^4}{t^2(u+u^4)}$ $\displaystyle \int(u+u^4)du=\int\frac{1+t^4}{t^2}dt=\int(\frac{1}{t^2}+t^2)dt$ $\displaystyle \frac{u^2}{2}+\frac{u^5}{5}=\frac{t^3}{3}-\frac{1}{t}+C$
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