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: 3



Work Step by Step

since, we have $y=(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt$ Apply product rule. Thus, we have $y'=(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}$ Then $x^2 y'=x^2[(-\dfrac{1}{x^2}) \int^{x}_1\dfrac{e^t}{t}dt+\dfrac{1}{x} {\dfrac{e^x}{x}}]=-\int^{x}_1\dfrac{e^t}{t}dt+e^x$ or, $=-x[(\dfrac{1}{x})\int^{x}_1\dfrac{e^t}{t}dt]+e^x$ or, $=-xy+e^x$
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