Answer
$y = e^{-t}[Acos(t)+Bsin(t)]$
Work Step by Step
$y'' + 2y' +2 = 0$ has characteristic polynomial $r^2 +2r +2 = 0$
Completing the square:
$(r+1)^2 = -2+1 \to r = -1\pm i$
Taking out the common term $e^{-t}$
$y = e^{-t}[C_1 e^{it} + C_2 e^{-it}]$
Applying Euler's formula
$y = e^{-t}[C_1(cos(t)+ isin(t)) + C_2(cos(-t)+isin(-t))] $
Cosine is even and sine is odd, so this may be rewritten as
$y = e^{-t}[C_1(cos(t)+ isin(t)) + C_2(cos(t)-isin(t))] $
Grouping real and imaginary terms yields
$e^{-t}[(C_1+C_2)cos(t) + i(C_1 -C_2)sin(t)]$
Let's rename these grouped constants $A,B$ respectively
$y = e^{-t}[Acos(t)+Bsin(t)]$
