Answer
$y=\sqrt{\frac{7}{6}e^x+C}$
Work Step by Step
$12yy'-7e^x=0$
$12yy'=7e^x$
$\int 12ydy=\int7e^xdx$
$12\frac{y^2}{2}=7e^x+C$
$y^2=\frac{7}{6}e^x+C$
$y=\sqrt{\frac{7}{6}e^x+C}$
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: 14
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