Solving $y'' + 2y' + 2y = 0$: How to eliminate imaginary unit from solution?

Question

$$y'' + 2y' + 2y = 0$$

$\downarrow$ (write characteristic equation)

$\lambda^2 +2\lambda + 2 = 0$

$\downarrow$ (solve characteristic equation)

$\lambda = -1 \pm i$

$\downarrow$ (write general solution)

$y = Ae^{(-1 - i)t} + Be^{(-1 + i)t}$

$\downarrow$ (apply Euler's formula)

$y = Ae^{-t}(\cos(-t) + i\sin(-t)) + Be^{-t}(\cos(t) + i\sin(t))$

$\downarrow$ (perform minor algebra/trig rearrangement)

$y = (A + B)e^{-t}\cos(t) + (B - A)e^{-t}\sin(t)i$

Where do I go from here to eliminate the $i$? Plugging in the exponential formulas for sine and cosine leads back to the original general solution, with $i$'s remaining.

Answer 1

Both $(A+B)$ and $(B-A)i$ are arbitrary constants. Rename them $C_1$ and $C_2$ to get the general solution $$y=C_1e^{-t}\cos{(t)}+C_2e^{-t}\sin{(t)}$$

Answer 2

In real solutions the coefficients of complex conjugate terms are themselves complex conjugates. So if you render

$y=\color{blue}{(A+Bi)}e^{(-1+i)t}+\color{blue}{(A-Bi)}e^{(-1-i)t}$

and carry through as in the question, the factors of $i$ will cancel naturally.