Answer
$y= e^t[Acos(t) + Bsin(t)]$
Work Step by Step
y''-2y'+2y = 0 is linear second order homogeneous with characteristic equation $r^2 -2r +2 = 0$
This can be solved by completing the square
$r^2 -2r +1 = -2 +1$
$(r-1)^2 = -1$
$r = 1\pm i $
$y = C_1 e^{(1+i)t}+ C_2 e^{(1-i)t}$
Take out the common term $e^t$
$y = e^t[C_1e^{it}+C_2 e^{-it}]$
Apply 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 can be rewritten as
$y = e^t[C_1(cos(t)+isin(t))+C_2(cos(t)-isin(t))]$
Grouping real and imaginary terms yields
$e^t[(C_1 + C_2)cos(t)+ (C_1 - iC_2)sin(t)]$
These two combinations of constants yields two new constants; let's call them $A$ and $B$.
$y= e^t[Acos(t) + Bsin(t)]$
