Consider this IVP:
$$y''' + 3 y'' + 4y' + 12 y = 0$$
$y(0) = a, y'(0) = b, y''(0) = c $
where $a,b,$ and $c$ are real numbers
So let's start by finding the characteristic equations
$$r^3 + 3r^2 + 4r + 12 = 0$$ $$(r + 3) (r^2 + 4) = 0 $$ $$ r = -3, \pm 2i $$
So the general solution is
$$\implies y = c_1 e^{-3x} + c_2 \cos 2x + c_3 \sin 2x$$ $$ y' = -3c_1 e^{-3x} - 2c_2 \sin 2x + 2c_3 \cos2x $$ $$ y'' = 9c_1 e^{-3x} - 4c_2 \cos 2x - 4c_3 \sin 2x $$
Initial conditions
$$ c_1 + c_2 = a \\ - 3c_1 + 2c_3 = b \\ 9c_1 - 4c_2 = c$$
So $$ c_1 = \frac{1}{13} (4a + c) $$ $$ c_2 = \frac{1}{13} (9a - c) $$ $$ c_3 = \frac{1}{26} ( 12 a + 13 + 3c) $$
Therefore
$$ y = \frac{1}{13} \left [ (4a + c) e^{-3x} + (9a - c) \cos 2x + \frac{1}{2} (12a + 13b + 3c) \sin 2x \right ] $$
It should be easy for you to check the conditions directly, remembering that the cosine term drops out of the first derivative and the sine term drops out of the other two conditions.
I do get the same result as yours.