A tunnel inside the Earth (but not an ordinary tunnel) I decided to dig a tunnel inside the Earth. In equatorial plane. It should be designed in such a way that it follows the Coriolis effect. That means if, say a stone is dropped from rest into the tunnel, then it travels in the tunnel without touching the walls of the tunnel until it reaches the tunnel's lip on the other side somewhere near the surface of the Earth. So the question itself is an equation of the curve of the tunnel. I suspect that there is no nice closed form for the curve. But who knows? Nevertheless, it would be good to know how deep the tunnel goes and where it rises again to the surface and the travel time.    
You could say that it would be impossible to do this on Earth, but you actually could do this on the Moon.
 A: I am not sure whether you meant initially at rest relative to the universe, or to the surface of the Earth. Here are the answers to both versions:

Universe
Let the latitude be $\theta_0$. In the non-rotating reference frame (of the universe), the motion of the stone is in simple harmonic motion. So $r(t) = r_0 \sin (\omega t)$, where $2\pi / \omega$ is the time it takes to get to the center of the earth. The latitude is constant. And the longitude $\phi_0$ (in the non-rotating reference frame) is also constant. But the earth spins with constant angular velocity $\Omega_0 = 2\pi$ radians per 24 hours. So in the rotating coordinate $\phi'(t) = \phi_0 - \Omega t$. 
So measured relative to the surface of the earth in spherical coordinates you have the parametric description
$$ r(t) = r_0\sin(\omega t) $$
where $r_0$ is the radius of the Earth. (Note: this assumes that the Earth is perfectly spherical and of uniform density inside, which is obviously not quite physical. Of course, digging a tunnel like that is also not quite physical...) 
$$ \theta(t) = \theta_0, \phi(t) = \phi_0 - \Omega t $$
Finding $\omega$ from the assumption that Earth is perfectly spherical and of uniform density is left as an exercise to the reader. 
A second interesting exercise is to find the conditions on the density of Earth and on the rate of revolution that allows for the stone to travel in a closed orbit (again, this is much much simpler when considered in the non-rotating reference frame...)

Surface of the earth
Again we work in the non-rotating reference frame. We have again, via the constant density assumption, (and assuming that the mass of stone is 1) that the potential energy is $P = \alpha r^2$. The kinetic energy is $2K = \dot{r}^2 + r^2\dot{\theta}^2 + (r\cos\theta)^2\dot{\phi}^2$. The conservation of angular momentum means that angular component of the velocity is of size $L / r$, where $L$ can be computed from the rate of rotation of the Earth. So the conserved energy is 
$$ E = \dot{r}^2 + \frac{L^2}{r^2} + \alpha r^2 $$
This gives an ODE for $r$. Similarly using a tilted coordinate system you can solve for the angles using the conservation of angular momentum by integrating an ODE and plugging in the solution for $r$. 
For the maximum depth, however, you don't need to explicitly solve the ODE: The energy is known initially: $E_0 = 0+ \frac{L^2}{r_0^2} + \alpha r_0^2$, where $L$ depends only on the rate of revolution for Earth, and $\alpha$ on the mass of Earth (assuming uniform density). $r_0$ is the radius of Earth. At the maximum depth, $\dot{r}$ is again 0. So you are down to finding the "other" positive root of the quartic polynomial 
$$ E_0 r^2 = L^2 + \alpha r^4 $$
which you can solve explicitly using the quadratic formula
$$ r^2 = \frac{E_0 \pm \sqrt{ E_0^2 - 4 \alpha L^2}}{2\alpha } $$
where the + solution is the radius of the earth, and the - solution is the depth. 
Plugging in physical numbers: at the initial drop, $\dot{\theta} = 0$ and $\dot{\phi} = \Omega = \frac{2\pi}{86400} s^{-1}$. The radius of Earth we take to be $6.38 \times 10^6 m$, the mass $6\times 10^{24} kg$. So $\alpha = G M / r_0^3 = 1.5 \times 10^{-6} s^{-2}$. 
The conserved angular momentum is initially
$$ L_0^2 = r_0^4 \cos(\theta_0)^4 \Omega^2 = 8.8\times 10^{18}\cos\theta_0^4 m^4 s^{-2} $$
and the conserved Energy is initially
$$ E_0 = L_0^2 r_0^{-2} + \alpha r_0^2 = (\cos\theta_0)^4 2.15 \times 10^5 + 6.1\times 10^7 m^2 s^{-2} \sim 6.1\times 10^7 m^2 s^{-2} $$
(the initial angular momentum contribution is very small). 
Now $E_0^2 = 3.7 \times 10^{15}$ and $4\alpha L_0^2 = 5.28\times 10^{13}\cos(\theta_0)^4$, so the answer is that the maximum depth is very close to the center of the earth! Using the binomial expansion we get that 
$$ r^2 \sim \frac{L_0^2}{E_0}  \implies r \sim 3.9\times 10^5 \times \cos(\theta_0)^2 m$$
So if you start at the equator where $\theta_0 = 0$, the hole will get about 94% to the center of the Earth. 
A: Well...
In that measure you have to notice central gravitation field.
So define all the forces applied to the stone during the fall - Gravitation force $F_g$ as function of distance to Earth/Moon core, Centrifugal force $F_r$ as function of distance to axis and Coriolis Force $F_c$. Sum of them is mass times acceleration.
$$ ma=m\frac{d^2r}{dt^2}=F_g+F_c+F_r $$
Integrate twice and here's your result.
I'd rather use simulink or another program to solve it numerically...
Special conditions:
 North/south pole: No Coriolis effect, no rotation: The stone shall touch the opposite ground and return as linear harmonic oscillator.
