# Air resistance differential equation

Hello all, I want to say thank you in advance for any and all advice on my question. My classical mechanics textbook (Marion Thornton) has been taking me through motion for a particle with retarding forces.

The example it keeps giving is:

m dv/dt = -kmv

which can be solved for:
v = v0e-kt and
x = v0/k(1-e-kt)

But out of curiosity I tried using the actual drag force equation "1/2ρCAv2" instead of "kmv." But I can't figure out how to solve the differential:
##\ddot x ## + 1/2ρCA##\dot x ##2 = 0

How do you solve this thing? I'm stuck since it's not the standard
x'' + ax' + bx = 0

My solution yielded:

$$\int\frac{\mathrm{d}\dot x }{ \dot x^2} = \frac{1}{2m}\rho CA\int \mathrm{dt}$$

which just gives some weird thing:
$$t = e^{\frac{1}{2m}\rho CAx}$$
which can't be right.

Last edited:

Orodruin
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Having a quadratic air resistance term, your differential equation is no longer linear so you will need to apply other methods. As you have noted, the differential equation you get for ##v## is separable, i.e.,
$$\frac{dv}{v^2} = K\, dt.$$
However, your integration of this does not seem correct to me. Please show the details of your computation.

I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##

Chestermiller
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I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##
What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.

Thank you by the way Orodruin for mentioning that it's nonlinear. I'm doing some research on "homogeneous first-order nonlinear ordinary differential equations." I guess that's the kind of equation this is.

What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.
Would that make a big difference tho? I could always just set up the constant to be trivial.

Ray Vickson
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Would that make a big difference tho? I could always just set up the constant to be trivial.

Try it for yourself: with a constant of integration, and without it. Do you get mathematically equivalent ##x(t)## formulas?

Last edited:
• Chestermiller
Chestermiller
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Would that make a big difference tho? I could always just set up the constant to be trivial.
Ask yourself this: Do you think I would have responded the way I did if I didn't already know that it would alleviate your difficulty?