Answer
$y^2=5e^{2x}+4$
Work Step by Step
$yy'-5e^{2x}=0$
$\int ydy=\int 5e^{2x}dx$
$\frac{y^2}{2}=\frac{5e^{2x}}{2}+C'$
Or, $y^2=5e^{2x}+C$
We are given that $y(0)=-3,$ so $(-3)^2=5e^0+C$
Or, $C=9-5=4$
Therefore, $y^2=5e^{2x}+4$
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 - Review Exercises - Page 432: 38
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