What is the difference between solutions to 2nd order homogeneous ODE? I’m studying Vibrations, and we have two forms to the 2nd order homogeneous ODE:
$$mx ̈+kx ̇=0$$
$$x(t)=C_1 e^{iw_n t}+C_2 e^{-iw_n t}$$
and
$$x(t)=A\cos(w_n t)+B\sin(w_n t)$$
Even though I can use either solution for a given problem (for simple problems), what is the difference between solutions to 2nd order homogeneous ODE? When to use one over the other? Are there situations when one form is more useful than the other?
 A: No difference. Write $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$
$$A\cos(wt)+B\sin(wt)=\frac{1}{2}A(e^{iwt}+e^{-iwt})+\frac{1}{2i}B(e^{iwt}-e^{-iwt})=\frac{1}{2}(A-iB)e^{iwt}+\frac{1}{2}(A+iB)e^{-iwt}$$
the two solutions are the same for $$C_1=\frac{1}{2}(A-iB)\\C_2=\frac{1}{2}(A+iB) $$
A: 
what is the difference between solutions to 2nd order homogeneous ODE?

The second form is a special case of the first form when $C_2 = C^*_1$ in which case $x(t)$ is a real valued function of time (I'm assuming here that both $A$ and $B$ are real)
Using Euler's formula, the first form can be written as
$$\left(C_1 + C_2\right)\cos(\omega_n t) + i\left(C_1 - C_2\right)\sin(\omega_n t)$$
where, in general, $C_1$ and $C_2$ are complex numbers.  Now, if $C_2 = C^*_1$ then this is just
$$2\Re\{C_2\}\cos(\omega_n t) + 2\Im\{C_2\}\sin(\omega_n t)$$
and so
$$A = 2\Re\{C_2\}$$
$$B = 2\Im\{C_2\}$$

When to use one over the other?

For $x(t)$ real, the second form is explicitly real.
