My book states that:

$\displaystyle y'' +4y = 0$

has a general solution of:

$\displaystyle y(t) = c_1\cos {2t} + c_2\sin{2t}$

I dont see how they achieve this since the initial equation has a characteristic form of:

$\displaystyle r^2 + 4 = 0$

$\displaystyle r_{1,2} = \pm 2$

$\displaystyle \therefore $

$\displaystyle y(t) = c_1e^{2t} + c_2e^{-2t}$