## Calculus 8th Edition

Published by Cengage

# Chapter 9 - Differential Equations - 9.1 Modeling with Differential Equations - 9.1 Exercises: 1

#### Answer

$2e^x=2e^x$

#### Work Step by Step

given: $y=\frac{2}{3}e^x+e^{-2x}$ $y'=\frac{dy}{dx}$ $=\frac{2}{3}e^x-2e^{-2x}$ now we have expressions for both y and y', which we can plug into our differential equation. If this produces a true statement, then the given expression for y is a valid solution. $y'+2y=2e^x$ $\frac{2}{3}e^x-2e^{-2x}+2[\frac{2}{3}e^x+e^{-2x}]=2e^x$ $\frac{2}{3}e^x-2e^{-2x}+\frac{4}{3}e^x+2e^{-2x}=2e^x$ $2e^x=2e^x$

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