Answer
$y = e^t[Acos(5t)+ Bsin(5t)]$
Work Step by Step
$y'' − 2y' + 6y = 0$ has characteristic equation $r^2-2r + 6 = 0$
This can be solved by completing the square
$r^2 - 2r+1 = -6+1$
$(r-1)^2 = -5$
$r = 1\pm5i$
Taking out the common factor of $e^t$ y can be written as
$y = e^t[C_1 e^{5it}+ C_2 e^{-5it}]$
$y = e^{t}[C_1 (cos(5t)+isin(5t))+C_2(cos(-5t)+isin(-5t))]$
Cosine is even and sine is odd so this may be rewritten
$y = e^{t}[C_1 (cos(5t)+isin(5t))+C_2(cos(5t)-isin(5t))]$
Grouping real and imaginary terms we have
$y = e^t[(C_1+C_2)cos(5t) + (C_1 -iC_2)sin(5t)]$
Let these two groupings of constants be $A,B$ respectively
$y = e^t[Acos(5t)+ Bsin(5t)]$
