Help Identifying ODE: x'' + k(x')^2 + c = 0

[sir_manning]

Hey

I need help identifying the differential equation: x'' + k(x')^2 + c = 0 . Can anyone point me in the right direction?

Thanks.

 

[Mute]

What do you mean by "identifying"? That differential equation could be a model equation for a lot of things! Did you want a specific name for the form or something?

In any event, if you set x' = v you get a first order ODE,

v' + kv^2 + c = 0.

One system that this equation describes is a falling object subject to wind resistance at high velocities, where c would be the acceleration due to gravity, g. (High velocities because at low speeds air resistance tends to go as v instead of v^2).

 

[HallsofIvy]

Mute's dead on. As far as "identifying" is concerned, it is a second order, non-linear equation. As Mute said, letting v= x' you get the first order, separable, differential equation v= -(kv²+ c) or
$$\frac{dv}{kv^2+ c}= -tdt$$

That's easily integrable but find x from x'= v may give you an integral that has no simple anti- derivative.

 

[sir_manning]

Thanks - it is in fact an equation for a falling object. V = x' and performing a change of variable gave me the answer I needed.