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: 6

Answer

The answer is in an implicit form : $-t^{-1} + t^3/3 + C $

Work Step by Step

$du/dt = (1 + t^4)/(ut^2 + u^4t^2) $ This can be rewritten as: $du/dt = (1 + t^4)/((t^2)(u + u^4)) $ Now it will be easier to separate u on one side and t on the other side: $(u + u^4) du = (1 + t^4)/(t^2) dt $ This can be simplified to : $(u + u^4) du = (t^{-2} + t^2) dt $ Now let's integrate both sides: $\int(u + u^4) du = \int(t^{-2} + t^2) dt $ $u^2/2 + u^5/5 = -t^{-1} + t^3/3 + C $
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