I'm not sure if I am doing this right but I have the characteristic polynomial as $r^2 + 4r + 5$, which factors to give the roots $-2 +- i$.
I then get $y_1(x) = c_1 e^{(-2+i)x}, y_2(x) = c_2 e^{(-2 - i)x}$ which gives
$y(x) = y_1 + y_2 = c_1 e^{(-2+i)x} + c_2 e^{(-2-i)x}$
I am stuck on how to apply Euler's identity. Any help would be appreciated.
$c_1 e^{(2+i)x}+c_2e^{(2-i)x} = e^{2x} (c_1 e^{ix} + c_2 e^{-ix})$
$\cos x = \frac{1}{2}(e^{ix} + e^{-ix}),\sin x = \frac{1}{2i}(e^{ix} - e^{-ix})$
$c_1 e^{(2+i)x}+c_2e^{(2-i)x} = A e^{2x} \cos x + B e^{2x} \sin x$