How to mathematically describe a planet's eccentric orbit?
The formula is Kepler's equation, but to understand it you need to know three values:

$M$ is the "Mean Anomaly". It increases linearly from 0 to 360 over the period of one orbit, measured from periapse to periapse. So if a planet has a period of 100 days, then the mean anomaly at day 0 is 0, at day 50 it 180degrees, at day 25 it is 90 degrees. This is only the position of the body that is in a circular orbit. It is the first stage of calculating the position of the body in an elliptical orbit.

$E$ is the "Eccentric Anomaly". This is also not the actual location of the planet. It is defined by $\cos\,E = x/a$ where $x$ is the x-coordinate of the of the planet, measured from the centre of the ellipse not the focus, and $a$ is the semi-major axis of the ellipse

$T$ is the "True anomaly", the angle between the actual position of the planet and the line through the major axis of the ellipse, measure at the focus of the ellipse (ie at the sun). 

These are related by the Kepler equation. $$M = E- e\sin E,$$where $e$ is the eccentricity of the ellipse. You can calculate M directly, then solve this equation (numerically) to find E. Providing the eccentricty isn't very close to one, Newton's method is effective at finding a solution to this equation.

Then the true anomaly is given as 
$$
T   = 2 \, \mathop{\mathrm{arg}}\left(\sqrt{1-e} \, \cos\frac{E}{2} , \sqrt{1+e}\sin\frac{E}{2}\right) 
$$
(where $\mathrm{arg}(x,y)$ is available as atan2(y,x) in many programming languages)

It is possible to use the Eccentric anomaly to calculate the coordinates of the planet directly using $\cos\,E=x/a$and $\sin\,E=y/b$ where a and b are the semi-major and -minor axes of the ellipse.

The word "anomaly" doesn't mean "false result", but "angular deviation (from periapse)" and has been in use since Kepler's time.