Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.1 - Modeling with Differential Equations - 9.1 Exercises - Page 590: 1


$y=(2/3)e^x+e^{-2x}$ is a solution to the differential equation $y'+2y=2e^x$

Work Step by Step

First we must find the $y$ and $y'$ values for the equation. $y$ is given to us in the problem as the claim $y=(2/3)e^x+e^{-2x}$ Using simple derivation we can calculate $y'$ $d((2/3)e^x+e^{-2x})/dx = dy/dx$ $dy/dx = (2/3)e^x-2e^{-2x}$ We then take the $y$ and $y'$ expressions and substitute them into the original differential equation. $y'+2y=2e^x$ $((2/3)e^x-2e^{-2x})+2((2/3)e^x+e^{-2x})=2e^x$ $(2/3)e^x-2e^{-2x}+(4/3)e^x+2e^{-2x}=2e^x$ $(2/3)e^x+(4/3)e^x=2e^x$ $(6/3)e^x=2e^x$ $2e^x=2e^x$ Since $2e^x=2e^x$, $y=(2/3)e^x+e^{-2x}$ is a solution to the differential equation $y'+2y=2e^x$
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