Answer
$y=(2/3)e^x+e^{-2x}$ 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 $y'+2y=2e^x$
