Answer
$y^{2}$=C-8cosx
Work Step by Step
yy'=4sinx
$y\frac{dy}{dx}$+4sinx
$\int$ydy+$\int$4sinxdx
$\frac{y^{2}}{2}$=-4cosx+$C_{1}$
$y^{2}$=$C_1-8$cosx
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 6 - Differential Equations - 6.3 Exercises - Page 421: 9
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