University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 4



Work Step by Step

We have: $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$ Apply product rule: Thus, we have: $y'=(-\dfrac{1}{2})[\dfrac{4x^3}{(\sqrt{1+x^4})^3}] \int^{x}_1\sqrt{1+t^4}dt+\dfrac{1}{\sqrt{1+x^4}} (\sqrt{1+x^4})$ or, $y'=(\dfrac{-2x^3}{1+x^4})[(\dfrac{1}{\sqrt{1+x^4}})\int^{x}_1\sqrt{1+t^4}dt]+1$ or, $y'+(\dfrac{2x^3}{1+x^4})y=1$ This is the required differential equation.
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