Answer
$\bf True$.
Work Step by Step
$\bf True$.
$f(x)=y=\frac{lnx}{x}$
$y'=\frac{1-lnx}{x^{2}}$
Now, $x^{2}y'+xy=x^{2}\frac{1-lnx}{x^{2}}+x\frac{lnx}{x}$
$=1-lnx+lnx$
$=1$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 9 - Differential Equations - Review - True-False Quiz - Page 674: 2
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