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