Challenging differential equation for a physics problem( higher math needed)?

1/Your logic is right. Unfortunately, Eva messes it up completely by introducing a second derivative (!!??!) in place of the displacement itself. The second derivative is, as you correctly note,the acceleration which is the lhs of your final equation. 3/ I don't think it can be solved completely with elementary functions. However, you can get a first integral as follows: I rewrite the equation x"=g-k/(H-x) for brevity . Let's then multiply both sides by x' (the velocity) to get x'x" = gx'-k x'/(H-x) so that all three terms are time derivatives of easy to write functions. Integrating you get : 1/2x'^2 = gx +kLog(H-x) + C The integration constant C is easily found by observing that the initial velocity is 0, that is, when x=0, then x'=0. Hence: 0 = kLogH+C so that the equation for the velocity becomes: 1/2x'^2 = gx + kLog(1-x/H) Here you see that in the beginning (x close to 0) the velocity squared grows linearly with x, But later the log becomes strongly negative and the movement stops. The maximum displacement is given by x'=0, i.e. gx+kLog(1-x/H) = 0 which has, of course, the solution x=0 but also another one corresponding to x_max. This can only be solved numerically. You can find a graphical approximation by looking for the intersections of the two curves exp(-gx/k) and 1-x/H Unfortunately, I can't make a plot here. To go further, you could observe that the diffeq obtained is separable by taking the square root and writing it dx/sqrt(gx+kLog(1-x/H)) = dt and integrating and finally inverting. But that's very theoretical. I don't believe any simple primitive of the 1/sqrt(..) function exists and you would still have the problem to invert it. However, you might solve it numerically to get your results. 4/ Given that the whole story (elevator being set free, falling and being stopped by the air pressure) should be short, you might indeed think that the compression is adiabatic (no heat exchange during the process) rather than isothermal. In that case the air will indeed heat up. The pressure-volume relation becomes in this case pv^gamma = cte instead of pv=cte with gamma the ratio of constant pressure over constant volume heat capacities which is about 1.4 for air. Then you can follow the same course. Instead of having to integrate x'/(H-x) leading to a log, you'll have x'/(H-x)^gamma leading to a power law -(1-x/H)^(1-gamma) It might make the square root more manageable for analytical integration but I'm not sure and have no time to check.