Understanding the equivalence of two general solutions of the harmonic oscillator differential equation

Question

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.

Answer 1

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 $$ x(t)=C \cos(\omega t+\varphi) $$