Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 400: 4


See the below result.

Work Step by Step

Given: $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$ Use product rule. This implies: $y'=(-\dfrac{1}{2})[\dfrac{4x^3}{(\sqrt{1+x^4})^3}] \int^{x}_1\sqrt{1+t^4}dt+\dfrac{1}{\sqrt{1+x^4}} (\sqrt{1+x^4})$ and $y'=(\dfrac{-2x^3}{1+x^4})[(\dfrac{1}{\sqrt{1+x^4}})\int^{x}_1\sqrt{1+t^4}dt]+1$ Thus, $y'+(\dfrac{2x^3}{1+x^4})y=1$ Hence, this is our required differential equation.
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