Why do heavier objects fall faster in air? We all know that in an idealised world all objects accelerate at the same rate when dropped regardless of their mass.  We also know that in reality (or more accurately, in air) a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc.  A loose explanation is that the increased mass of the lead feather somehow defeats the air resistance more effectively than the duck's feather.
Is there a more formal mathematical explanation for why one falls faster than the other?
 A: Short answer: air drag!
Gravity is acting in both feathers the most massive receives a stronger pull to down. Air drag is counter acting that movement and is proportional to the velocity (in a very complex way, references here and here)
That stronger pull helps to overcome the increasing drag opositing force. That's why the lead feather ill accelerate faster and reach a bigger terminal velocity.
The same principle is applied to race cars. Two cars, same shape, the one with the most potent engine can accelerate more and reach a greater max velocity.
Another Example: Skydivers usually dresses something to increase air drag and stands in a position to help the drag to lower the terminal velocity and increase the fall time. A stading up jumper ill fall a lot faster.
Edit
After some discussion on the buoyance effect I searched a while about a bird's feather density, a value not easy to get. I found this reference (it'a a .pdf document) about chiken's feather and contains a lot of considerations about density. After the lecture we can use a value of 0,89g/cm3 and that almost as dense as water. So any buoyance effect is negligible. If one still want's to discuss negligible forces we can pick also the gravity variation on altitude or the effect off relativistic physics over the acceleration of a body.
A: 
We also know that in reality a lead feather falls much faster than a
  duck's feather with exactly the same dimensions/structure etc

No, not in reality, in air.  In a vacuum, say, on the surface of the moon (as demonstrated here), they fall at the same rate.

Is there a more formal mathematical explanation for why one falls
  faster than the other?

If the two objects have the same shape, the drag force on the each object, as a function of speed $v$, is the same.
The total force accelerating the object downwards is the difference between the force of gravity and the drag force:
$$F_{net} = mg - f_d(v)$$
The acceleration of each object is thus
$$a = \frac{F_{net}}{m} = g - \frac{f_d(v)}{m}$$
Note that in the absence of drag, the acceleration is $g$.  With drag, however, the acceleration, at a given speed, is reduced by
$$\frac{f_d(v)}{m}$$
For the much more massive lead feather, this term is much smaller than for the duck's feather.
A: Linear momentum!
I think the easiest way to gasp the concept is to think at atomic level about momentum of the objects and the air's atoms that cause friction.
Linear momentum is equal to $m\times v$ so heavier object have higher momentum. Imagine an atom that comes from the opposite direction of the falling object and collide with the object.

For making the things easier we suppose an inelastic one-dimensional collision. Basically when the atoms collide with the object the object lose part of its momentum, mass is constant so this cause the object to slow down.
Imagine two objects with different mass but same velocity, the object with greater mass (e.g. a metal sphere )will lose only a little bit of its linear momentum hence it will keep going with a velocity near to the initial one, in the other hand the object with smaller mass (e.g. a ballon) and hance momentum will lose a great portion of its linear momentum and so it will slow down considerably!
A: A good approximation of the drag force for an object falling through the atmosphere is $-cv^2$, with $c$ a constant independent of mass. Thus, $$m \dot{v} = mg - cv^2$$ is the equation of motion with initial condition $v(0)=0$. We write $$ t = m \int_0 ^{v(t)} \frac{d v}{mg -cv^2}$$ and the final result is $$ v (t) = \sqrt{\frac{mg}{c}} \tanh \left( t \sqrt{ \frac{gc}{m}} \right),$$ which is a function increasing as $m$ increases for $t$  constant, therefore heavier objects fall faster than lighter ones in presence of drag due to air. The terminal speed is $$\lim_{t\to \infty} v(t) = \sqrt{\frac{mg}{c}}.$$ For a person in free fall with drag, the terminal speed is about 50 m/s.
The previous analysis depends on the fact that the cross-section of the falling object remains constant, which is often far from true and alters the result significantly, since, for example, a feather curves when falling while a feather of the same shape made of metal will not curve and will be heavier, making the difference in falling speed more pronounced. Indeed, a plot with varying $c$, which is $\propto A^{-1}$ with $A$ the cross-section, indicates that the effect of the cross-section on the speed is much more important than that of the different masses. Also, we assume that wind currents and turbulence are negligible, another assumption that may change the result in real conditions significantly.
Edit:
This analysis, as anticipated, may fail spectacularly if one takes into consideration that an asymmetric object in general rotates in a chaotic fashion if it exceeds certain threshold angle, which can be said to depend on the density, cf. this article
A: Whether the lead feather falls faster or the ducks feather falls faster, depends on the direction of external force per unit area acting on them, mass per unit area of each of them.   
Probability of lead feather falling faster is greater than the ducks feather, because of greater probability of less downward external force on both.   
Downward force should not be neglected considering atmospheric pressure, air pressure can hold up a $10m$ high column of water (Torricelli's law). Quite a tall glass! Click here to see the video.$_1$  

Credits: $_1$ Modern's ABC of Physics-2012 Edition-Page No.695. 
