Học sinh 16 tuổi chinh phục bài toán thế kỷ

Trích:
http://www.reddit.com/r/worldnews/co...dle_an/c4sxd91
The problem he solved is as follows:

Let (x(t),y(t)) be the position of a particle at time t. Let g be the acceleration due to gravity and c the constant of friction. Solve the differential equation:

(x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )

subject to the constraint that (x''(t),y''(t)+g) is always opposite in direction to (x'(t),y'(t)).

Finding the general solution to this differential equation will find the general solution for the path of a particle which has drag proportional to the square of the velocity (and opposite in direction). Here's an explanation how this differential equation encodes the motion of such a particle:

The square of the velocity is:
x'(t)2 + y'(t)2

The total acceleraton is:
( x''(t)2 + y''(t)2 )1/2

The acceleration due to gravity is g in the negative y direction.
Thus the drag (acceleration due only to friction) is:
( x''(t)2 + (y''(t)+g)2 )1/2

Thus path of such a particle satisfies the differential equation:
( x''(t)2 + (y''(t)+g)2 )1/2 = c*(x'(t)2 + y'(t)2 )

Of course, we also require the direction of the drag (x''(t),y''(t)+g) to be opposite to the direction of the velocity (x'(t),y'(t)). Once we find the intial position and velocity of the particle, uniqueness theorems tell us its path is uniquely determined.

EDIT: I've verified Ray's solution with Maple. Here.
http://www.reddit.com/r/worldnews/co...dle_an/c4szejb

EDIT: Pelly has shown there is a very easy derivaton of it!
http://www.reddit.com/r/worldnews/co...dle_an/c4t03fl

Here's a forward solution (found by reverse-engineering the answer):
http://www.reddit.com/r/math/comment...lating/c4szzld
Consider a projectile moving in gravity with quadratic air resistance. The governing equations are

u' = -a * u * sqrt( u2 + v2 )
v' = -a * v * sqrt( u2 + v2 ) - g

where a is the coefficient of air resistance defined by |F| = ma|v|2 .

Cross-multiply and rearrange to find
a * sqrt( u2 + v2 ) * (uv'-vu') = gu'

Substitute v = su and separate variables:
a * sqrt( 1 + s2 ) * s' = g*u'/u3

Integrate both sides to get the answer:
g/u2 + a(v * sqrt( u2 + v2 )/u2 + arcsinh|v/u|) = cons

Knowing some German helps. When you go to this page and you click the picture, it enlarges it.

You can see some math symbols above his right shoulder. That may be all we can get until he puts the text of his presentation on the web somewhere.

Here's the full equation (also from the original site). It reads:
https://www.jugend-forscht.de/images...7_download.jpg