Oneway,
While I understand the logic behind your assessment, there are several things to consider...
First, the force due to drag is not linear with relative airspeed. In fact, it is proportional to the square of the relative airspeed.
Second, the velocity gained or lost whilst in a given wind layer is Integral[ drag force dt ]. If the time changes, then the amount of velocity gained in a layer will change. This factor only washes out if we're at terminal velocity.
Now, consider three wind layers, one still (0 mph) 10-15k, one west (25 mph) 5-10k, one east (25 mph) 0-5k, and you are traveling due north. Assuming the bomb reaches terminal velocity when it hits the west-layer, the velocity gained is given by
dv/dt = - k (25 - v(t))^2
or in terms of position
dx^2/dt^2 = - k (25 - dx/dt)^2
from the wind. This is a second order nonlinear differential equation, which in general does not have a solution which you can write down nicely, let alone a solution which is linear.
The next layer will have another of the same type of equation. In real life, it is virtually impossible that these effects will "wash out" if you are heading due north since the force is proportional to the square of the velocity.
In this perfect scenario, there should be two drop headings which give a true overall trajectory i.e., you've picked out a heading so that these will wash out. HOWEVER this is due to an inherent symmetry in the system - you do not care about the whether a quantity is left or right wrt to the airplane so long as it cancels out, so with only two crosswinds, we have a symmetry to exploit. With no such symmetry, there should only be one solution.