Why can we ignore the factor of $i$ in the general solution of 2nd Order linear, homogeneous ODE's with constant coefficients I have that the general solution is $$ e^{\alpha x} (A \cos\beta x +iB \sin\beta x) $$ for any ODE (as described in title) with complex roots $ \alpha \pm i \beta $ with $\alpha,\beta \ \in \ \mathbb{R} $.
I don't understand how ignoring the factor of $i$ gives the same result. 
edit: I should have clarified what I mean by "ignoring the factor of $i$". My lecturer's notes state that the correct general solution is $$ e^{\alpha x} (A \cos\beta x +B \sin\beta x) $$ with no factor of $i$ in the second term.

The general solution for the homogeneous part is $c_1e^{(\alpha+i\beta)x}+c_2e^{(\alpha-i\beta)x}$ because as you mentioned above, $\alpha\pm i\beta$ are the roots of characteristic equation.
Now we know that if $y_1(x)$ and $y_2(x)$ are two linearly independent solutions (i.e. the complete solution to homogeneous part be $c_1y_1+c_2y_2$), then $y_3 = ay_1(x)+by_2(x)$ and $y_4(x) = cy_1(x)+dy_2(x)$ are two linearly independent solutions, too if $\frac{a}{c}\neq \frac{b}{d}$ or equivalently $ad-bc \neq 0$, in other words another complete solution to homogeneous part which is equivalent to the last pair is $c_3y_3+c_4y_4$.
With the above knowledge, if we consider $y_1(x)=e^{(\alpha+i\beta)x}=e^{\alpha x}(\cos(\beta)+i\sin(\beta))$ and $y_2(x)=e^{(\alpha-i\beta)x}=e^{\alpha x}(\cos(\beta)-i\sin(\beta))$, then another pair of complete solution is $y_3(x) = \frac{1}{2}y_1(x)+\frac{1}{2}y_2(x)=e^{\alpha x}\cos(\beta x)$ and $y_4(x) = \frac{1}{2}y_1(x)-\frac{1}{2}y_2(x)=e^{\alpha x}\sin(\beta x)$ because $a=b=c=\frac{1}{2},d=\frac{-1}{2}$ and $ad-bc=\frac{-1}{2}\neq 0$. 
Therefore the complete solution can also be expressed as $c_3y_3+c_4y_4=c_3e^{\alpha x}\cos(\beta x)+c_4e^{\alpha x}\sin(\beta x)$.
Please notice that there is no difference in using $c_1y_1+c_2y_2$ or $c_3y_3+c_4y_4$. The only difference is when you substitute the initial conditions and you want to calculate the constant coefficients, then $c_1,c_2$ will become complex numbers and are harder to deal with but $c_3,c_4$ will be real numbers and easier to handle. Just try it on a simple equation and you get the idea.
