Here's summmary of what I'm reading through the interwebs:

**The original post is here**

"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 a drag proportional to the square of the velocity (and opposite in direction). Here's an explanation of 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 acceleration 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 [the preceding should probably read "the impedance (acceleration due to friction plus gravity) is"]:

$$\bigg( x''(t)2 + (y''(t)+g)2 \bigg)(1/2)$$

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."

**The original post is here**

"Here's a forward solution (found by reverse-engineering the answer):

Consider a projectile moving in gravity with quadratic air resistance. The governing equations are

$$u' = -a u \sqrt{ u^2 + v^2 }$$

$$v' = -a v \sqrt{ u^2 + v^2 } - g$$

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

Cross-multiply and rearrange to find

$$a \sqrt{ u^2 + v^2 } (uv'-vu') = gu'$$

Substitute $v = su$ and separate variables:

$$a \sqrt{ 1 + s^2} s' = g\frac {u'}{u^3}$$

Integrate both sides to get the answer:

$$\frac g {u^2} + a \left(\frac{v \sqrt{ u^2 + v^2 }}{u^2} + \sinh^{-1}\left|\frac v u\right| \right)= C"$$

"From what I can tell from the image the solution in the image is implicit and was derived by Parker at NCSU in 1977. It is still impressing for a 16-year-old. Here's the paper by Parker if anyone's is interested."

"Am I crazy... or is the equation not anywhere in the paper?"

"It isn't the same equation, but there are solutions using the logarithm definition of inverse hyperbolic functions. The solution isn't the same, but there's an implicit solution too. It's impressing because he found a way to solve the differential equation while Parker stopped at the hairy integral at the end of the "Exact Solutions" section. But yeah, my comment does look confusing, as the exact same equation isn't there."

"It was of little interest, thus nobody was really breaking their teeth on it.
But it still is amazing, especially at such a young age."

"Exactly. This solution is implicit, therefore it has little use in actual calculations as you would need to numerically solve it in order to use it, you might as well solve the differential equation numerically directly. Exact solutions similar to the one presented here have been known since 1977 in a paper I posted in another thread. Anyway, the trick used to solve the ODE is quite clever, especially for a 16-year-old."