Given solutions from two ODE's:
$x(t)=\alpha \sin(\omega t+\phi)$
$y(t)=\alpha \omega \cos(\omega t+\phi)$
Which two ODEs correspond to each of these solutions?
My considerations:
Obviously, the oscillator equation must be used:
$\frac{dx}{dt^2}+\omega^2x=0$
or
$\frac{dy}{dt^2}+\omega^2y=0$
These equations provide the solution $\sin(\omega t)$. How to modify them to take into account the amplitude and phase of the sinusoidal signal?
Your O.D.E does allow for a phase expression. First of all, the general solution is, as you say:
$$x=A\sin(\omega t)+B\cos(\omega t)$$
But now consider this:
$$\alpha\sin(\omega t+\epsilon)=\alpha\sin(\omega t)\cos(\epsilon)+\alpha\cos(\omega t)\sin(\epsilon)$$
If we write that as equivalent to the first general solution, we have:
$$A\sin(\omega t)+B\cos(\omega t)\equiv\alpha\sin(\omega t)\cos(\epsilon)+\alpha\cos(\omega t)\sin(\epsilon)\\A=\alpha\cos(\epsilon)\\B=\alpha\sin(\epsilon)$$
And so it indeed does make sense to let the solution be written in phase-amplitude form.
Furthermore:
$$A^2+B^2=\alpha^2(\cos^2(\epsilon)+\sin^2(\epsilon))=\alpha^2\\\alpha=\sqrt{A^2+B^2}\\\frac{B}{A}=\frac{\alpha\sin(\epsilon)}{\alpha\cos(\epsilon)}=\tan(\epsilon)$$
So you can completely determine the phase-amplitude form from the initial conditions.