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.
$y'+2y=2e^x$
$\frac{2}{3}e^x-2e^{-2x}+2[\frac{2}{3}e^x+e^{-2x}]=2e^x$
$\frac{2}{3}e^x-2e^{-2x}+\frac{4}{3}e^x+2e^{-2x}=2e^x$
$2e^x=2e^x$
