Answer
$y = \frac{2}{3}{e^x} + {e^{ - 2x}}{\text{ is a solution}}$
Work Step by Step
$$\eqalign{
& y = \frac{2}{3}{e^x} + {e^{ - 2x}},{\text{ }}y' + 2y = 2{e^x} \cr
& {\text{Differentiate the given function}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{2}{3}{e^x}} \right] + \frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}\left( {{e^x}} \right) + \left( { - 2{e^{ - 2x}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}{e^x} - 2{e^{ - 2x}} \cr
& or \cr
& y' = \frac{2}{3}{e^x} - 2{e^{ - 2x}} \cr
& \cr
& {\text{Substitute }}y{\text{ and }}y'{\text{ into the given differential equation}} \cr
& \underbrace {y' + 2y = 2{e^x}}_ \downarrow \cr
& \left( {\frac{2}{3}{e^x} - 2{e^{ - 2x}}} \right) + 2\left( {\frac{2}{3}{e^x} + {e^{ - 2x}}} \right) = 2{e^x} \cr
& \frac{2}{3}{e^x} - 2{e^{ - 2x}} + \frac{4}{3}{e^x} + 2{e^{ - 2x}} = 2{e^x} \cr
& \frac{6}{3}{e^x} = 2{e^x} \cr
& 2{e^x} = 2{e^x} \cr
& {\text{The statement is true}}{\text{, then }}y = \frac{2}{3}{e^x} + {e^{ - 2x}}x{\text{ is a solution}} \cr
& {\text{of the given differential equation}}{\text{.}} \cr} $$