Answer
$$yy'=-8\cos{(\pi x)}$$
$$y=\sqrt{-\frac{16}{\pi}\sin{(\pi x)}+C}$$
Work Step by Step
$yy'=-8\cos{(\pi x)}$
$\int y dy=\int -8\cos{(\pi x)}dx$
$\frac{y^2}{2}=-\frac{8}{\pi}\sin{(\pi x)}+C$
$y=\sqrt{-\frac{16}{\pi}\sin{(\pi x)}+C}$
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: 10
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