Answer
It is a solution of the differential equation.
Work Step by Step
Find the first and second derivatives.
$y=C_1 e^{-x} cos{x} + C_2 e^{-x} sin{x}$
$y= e^{-x}(C_1cos(x) +C_2 sin(x))$
$y'= -e^{-x} ( C_1cos(x) + C_2 sin(x) ) + e^{-x} (-C_1 sin(x) +C_2 cos(x))$
$y'= e^{-x} [(C_2 -C_1)cos(x) - (C_1+C_2)sin(x)]$
$y''= -e^{-x}[(C_2-C_1)cos(x) - (C_1+C_2) sin(x)] + e^{-x}[-(C_2 - C_1)sin(x) - (C_1 + C_2) cos(x)] $
$y''= 2e^{-x}(C_1 sin(x) - C_2cos(x))$
Plug back into the differential equation
$ 2e^{-x}((C_1 sin(x) - C_2cos(x)) + 2e^{-x} [(C_2 -C_1)cos(x) - (C_1+C_2)sin(x)] + 2e^{-x}(C_1cos(x) +C_2 sin(x)) =0 $
Simplify
$2e^{-x}(C_1sin(x) -C_2cos(x) +C_2cos(x) -C_1 cos(x) -C_1sin(x) -C_2sin(x) + C_1cos(x) +C_2sin(x))= 0$
All terms cancel out and
$0=0$
So, it is a solution to the differential equation.
