Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.3 Exercises: 15



Work Step by Step

$yy'-2e^x=0\hspace{15mm}y(0)=3$ First we want the general solution. $yy'=2e^x$ $\int ydy=\int 2e^xdx$ $\frac{y^2}{2}=2e^x+C$ $y^2=4e^x+C$ $y=\sqrt{4e^x+C}$ Now applying our initial condition we have $3=\sqrt{4+C}$ $C=3^2-4=5$, and so our particular solution is $y=\sqrt{4e^x+5}$
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