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: 8


$x^2y'=xy-y^2$ and $y(e)=e$

Work Step by Step

Apply the quotient rule of differentiation: Thus, $y'=\dfrac{ \ln x -x(\dfrac{1}{x})}{(\ln x)^2}=\dfrac{1}{\ln x} -\dfrac{1}{(\ln x)^2}$ Since, we have $y=\dfrac{ x}{\ln x}$ Therefore, $x^2y'=\dfrac{x^2}{\ln x} -\dfrac{x^2}{(\ln x)^2}$ or, $x^2y'=xy-y^2$ Apply the initial conditions: $y(e)=\dfrac{e}{\ln e}=e$
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