Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.3 Exercises: 13

Answer

$y=ke^{\frac{\ln^2{x}}{2}}$

Work Step by Step

$y\ln{x}-xy'=0$ $y\ln{x}=xy'$ $\frac{\ln{x}}{x}=\frac{y'}{y}$ $\int\frac{dy}{y}=\int\frac{\ln{x}}{x}dx$ $\ln{y}=\frac{\ln^2{x}}{2}+C$ $e^{\ln{y}}=y=e^{\frac{\ln^2{x}}{2}+C}=e^{\frac{\ln^2{x}}{2}}e^C$ Let $k=e^C$ then $y=ke^{\frac{\ln^2{x}}{2}}$
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