I'll work a little backwards, but arrive at a form for $x(t)$ and $y(t)$, which you can use for your simulation.Having those differential equation and making the switch to complex coordinate $z=x+iy$ you get the following diff. equation
$\ddot{z}+2i\omega\dot{z}\sin{\lambda}+\omega_{p}^2z=0$
with $\omega_{p}^2=\frac{g}{L}$. For this kind of diff. equation you take a solution of the following type
$z(t)=Z_{0}(t)e^{-i\omega\sin{\lambda}t}$. Inserting this into the eq. above you arrive at
$\ddot{Z}(t)+(\alpha^2+\omega_{p}^2)Z(t)=0$. Where $\alpha=\omega\sin{\lambda}$ and $\alpha^2$ is tiny when we compare it to $\omega_{p}^2$ so we can neglect it. So, this leave you with $\ddot{Z}(t)+\omega_{p}^2Z(t)=0$. And the solution for this has the following general expression
$Z(t)=Ae^{i\omega_{p}t}+Be^{-i\omega_{p}t}$
and the complete solution is now
$z(t)=e^{-i\alpha t}(Ae^{i\omega_{p}t}+Be^{-i\omega_{p}t})$
Here you can see that are two special cases which correspond to harmonic oscillations of the pendulum, when $A=B$ and $A=-B$. In the first care you find
$z(t)=2Ae^{-i\alpha t}\cos{\omega_{p}t}$ and in the second case
$z(t)=2ie^{-i\alpha t}\sin{\omega_{p}t}$
The first solution corresponds to the initial condition $z(t=0)=2A$ and the second to $z(t=0)=0$. In these solutions, you can see that the exponential factor is due to the Coriolis force. To get rid of these exponential you apply Euler formula. After doing this, to find the "real" trajectories, you take the real and imaginary part of $z$. Hence, for the first solution you find
$Re(z)=x(t)=2A\cos(\alpha t)\cos(\omega_{p}t)$
$Im(z)=y(t)=-2A\sin(\alpha t)\cos(\omega_{p}t)$
You can do the same for the last solution. Having these, you can simply find $\dot{x}(t)$ and $\dot{y}(t)$. Hope it helps.
You seem to be saying that friction couldn't speed it up, because nothing else is moving that fast. Well, how fast is it moving?
We can imagine the gyroscope axis parallel to the z axis, and the casing to be aligned such that the x axis goes through it. If the casing is tipped slightly, the gyroscope resists that turning and one side of the shaft has firm contact with the side of the groove.
Now, I cannot tell easily from the photo the size of the shaft portion within the groove, but it appears to have a radius to me no larger than about 4mm. If we can imagine that the rotor has a speed of 4000rpm, what would that translate to at the edge of the shaft contact point?
$$v = {\omega r}$$
$$v = {4000rpm \times 2mm}$$
$$v = {418.9 radians/s \times 0.2cm}$$
$$v = 83cm/s$$
What's the diameter of the groove? Maybe 3.5 inches? If so the circumference would be about 27cm. So if you could spin the thing in way that made the axis tumble more than 3 times a second, then the motion of the axis would be such that it is carrying the edge of the shaft past the groove faster than rotation is carrying it backward. Friction would serve to increase the rotation, not decrease it.
You'll have to let me know if these assumptions are correct. I appear to have made guesses about the size of the shaft in the groove, the speed it can rotate, the diameter of the groove, and the rate of "tumbling" that the rotor does when you are accelerating it.
Best Answer
First order of business is to find where the heck on Earth you are. First, $\omega = 360 \sin(\phi)/day$, where $\omega$ is 216.528 degrees; $\phi$ is the latitude of your position. North of the equator is positive, south negative. This gives you a band to follow around the earth horizontally, positions where could possibly be. You can further narrow your position down because a Foucault pendulum can be used to find the acceleration of gravity at its position. Once you figure this out, you can go to NASA websites and check out when this location has its next or last total solar eclipse.