# Differential equation/rational function

1. Jan 31, 2012

### nos

Hi all,

Three forces are acting on an object.
Gravity, mg
Lorentz force, av(a is a constant)
Air resistance, kv^2( k is a constant)
v is the velocity of the object

I figured that this differential equation must be the right one:

M(total)$\frac{dv}{dt}$= mg-av-kv^2

Where m and M(total) are different masses

$\frac{dv}{kv^2+av-mg}$ = -$\frac{dt}{M(total)}$

Integrate both sides:

-$\frac{2}{\sqrt{a^2+4kmg}}$ artanh($\frac{2kv+a}{\sqrt{a^2+4kmg}}$) = -$\frac{T}{M(total)}$ + C

So I want to find the expression for velocity v.
How do I begin? And should the integration constant even be there? When I put limits on the integral(v goes from 0 to v and t goes from 0 to t) then there is no constant of integration.
Thanks so much.

2. Jan 31, 2012

### nos

I'm actually starting to think that maybe the whole anti-derivative is wrong.

3. Jan 31, 2012

### nos

So I figured using the limits on the integral gives me:

arctanh($\frac{2kv+a}{\sqrt{a^2+4kmg}}$)-arctanh($\frac{0+a}{\sqrt{a^2+4kmg}}$)=$\frac{\sqrt{a^2+4kmg}}{2M(total)}T$

Taking the second term arctanh tot the other side and taking the tanh of both sides:

2kv+a = $\sqrt{a^ 2+4kmg}$ tanh($\frac{\sqrt{a^2+4kmg}}{2M(total)}T$
+arctanh($\frac{a}{\sqrt{a^2+4kmg}}$))

which will finally give v in terms of t:

v(t) = $\frac{-a}{2k}$+$\sqrt{a^ 2+4kmg}$ tanh($\frac{\sqrt{a^2+4kmg}}{2M(total)}T$
+arctanh($\frac{a}{\sqrt{a^2+4kmg}}$))/2k