Understanding the equivalence of two general solutions of the harmonic oscillator differential equation The solution of the following differential equation
$$ -kx(t) = m \frac{d^{2}x}{dt^{2}}, $$
with $\omega = \sqrt{k/m}$, is
$$ x(t) = C_{1}e^{-i\omega t} + C_{2}e^{i\omega t}.$$
The real part of this,
$$ x(t) = x_0 e^{i(-\omega t + \phi)},$$
is the solution of my differential where $x_0$ is the amplitude of a harmonic oscillator and $\phi$ being a phase difference.
I have found another form of the general solution of this differential equation is
$$ x(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t).$$
With the following parameters $x_0 = \sqrt{C_{1}^{2} + C_{2}^{2}}$ ,
$\cos\phi = C_1/x_0$ and $\sin\phi=-C_2/x_0$ you can write
$$x(t) = x_0\cos(\omega t + \phi),$$
which is equivalent of the $\Re\{x_0e^{i(-\omega t + \phi)}\}$. But I don't understand how the general solution of the differential equation can be written in these two ways.
 A: You should not be using the same symbols $C_1$ and $C_2$ in the two solutions…
Write
$$
x(t)=D_1 e^{-i \omega t}+D_2 e^{i\omega t} \tag{1}
$$
and expand using Euler’s formula: $e^{i\omega t}=\cos(\omega t)+i\sin(\omega t)$.  Eq.(1) then takes the form
$$
x(t)=(D_1+D_2)\cos(\omega t) + i (D_1-D_2)\sin(\omega t) \tag{2}
$$
so you can now declare $C_1=D_1+D_2$ and $C_2=i (D_1-D_2)$.
Note that here we must switch the physics on: obviously the position is real
so $D_1+D_2$ must be real whereas $D_1-D_2$ must be pure imaginary.
This constrains the choices of $D_1$ to $D_2$.  This is fine because the solution of Eq.(1) should depend on 2 real parameters, whereas it would depend on 4 real parameters if $D_1$ and $D_2$ were arbitrary complex numbers.    In other words, $D_1$ and $D_2$ must be such hat
$D_1=D_2^*$ so that then $C_1$ is related to the real part of $D_1$ and $C_2$ to the complex part of $D_1$.
You now start with Eq.(2) as
$$
x(t)=C_1\cos(\omega t)+C_2\sin(\omega t)
$$
and then continue with $C=\sqrt{C_1^2+C_2^2}$, $C_1=C\cos\varphi$ and $C_2=C\sin\varphi$, which immediately gives
$$
x(t)=C \cos(\omega t+\varphi)
$$
