Calculus 10th Edition

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



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, $y^2=5e^{2x}+4$
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