# Differential equation question

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#1
It's only a first order equation, but I'm stuck with the algebra.

The equation is: m(dv/dt) = mg - kv^2

=> (m/mg - kv^2)(dv/dt) = 1

=> (m/mg - kv^2) dv = dt

Integrating both sides gives: -(m/2kv)ln(mg - kv^2) = t + c

I want v(t), but the problem is the 2kv bit in the ln term. I can't see an easy way of re-arranging in terms of v :/. Have I done the integration wrong?
0
16 years ago
#2
(Original post by Nylex)
It's only a first order equation, but I'm stuck with the algebra.

The equation is: m(dv/dt) = mg - kv^2

=> (m/mg - kv^2)(dv/dt) = 1

=> (m/mg - kv^2) dv = dt

Integrating both sides gives: -(m/2kv)ln(mg - kv^2) = t + c

I want v(t), but the problem is the 2kv bit in the ln term. I can't see an easy way of re-arranging in terms of v :/. Have I done the integration wrong?
If you differentiate what you have it is not the same as the integrand. You cant use that method because the derivative of mg - kv^2 is -2kv and that is not present in the numerator of the integrand. It is possible to use a = v.dv/dx then use the fact that v = dx/dt
0
16 years ago
#3
Or to find v(t) you could try some variation of the v = tanhx substitution that takes into account the constants
0
16 years ago
#4
m dv/dt = mg - kv²

terminal velocity, u, (when acceleration is 0): u = root(mg/k) (substitute this in for simplicity)

=> dv/dt = g [1 - (v/u)²]

using substitution z = v/u: (for simplicity)

=> INT [u / (1-z²)] dz = g INT  dt

=> 1/2 ln [(1+z)/(1-z)] = -1/2 ln[(1-z)/(1+z)] = gt/u + C
(the change of sign because it is easier to work with e^-t than it is to work with e^t)

=> [(1-z)/(1+z)] = Ae^[-(2gt/u)] = F(t) for simplicity

at t=0, v=0 => z = 0; => A = 1

=> z = (1-F(t))/(1+F(t))

=> v = u [1-e^(-2gt/u)]/[1+e^(-2gt/u)]

if you want to, you can substitute u back in:

v = root(mg/k) [1-e^(-2 root(gk/m) t)]/[1+e^(-2 root(gk/m) t)]
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#5
Ah ****, thanx both of you.
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