Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.1 Exercises - Page 403: 2


$y=e^{-2x}$ is a solution

Work Step by Step

First take the derivative of y to find y' $y= e^{-2x}$ $y'= -2e^{-2x}$ Then substitute y and y' into the differential equation $ 3(-2e^{-2x} ) + 5(e^{-2x} )=-e^{-2x}$ $-6e^{-2x} +5e^{-2x} = -e^{-2x}$ $-e^{-2x} = -e^{-2x} $ The solution checks with the differential equation so $y=e^{-2x}$ is a solution.
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