Differential Equations

1. Feb 4, 2008

captainjack2000

Object mass m released from height h falls to ground under gravity. There is air resistance and a horizontal wind of velocity w so that if the velocity of object is v total force on it due to its passage through air is k(w-v) where k>0. Find an expression for time T object takes to reach ground?

Equation of motion: dv/dt = -g+ k/m(w-v)

could someone show me how to start this? Am I meant to take an integrating factor
I= e^(k/m) and integrate it wrt to t? If so how do I integrate it?????

2. Feb 4, 2008

HallsofIvy

Staff Emeritus
I am a bit concerned about the fact that the wind is horizontal but you are treating the force as if it were entirely vertical. Are you sure you are not supposed to separate into horizontal and vertical components? If you haven't been working with components of vector, that probably not what is intended so don't worry about it.
Your equation of motion is dv/dt= -g+ k/m(w- v)= -g+ (k/m) w- (k/m) v. If you write it as dv/dt+ (k/m)v= (k/m)w- g does it make more sense? Can you find an integrating factor, $\nu(t)$ that makes $\nu(dv/dt+ (k/m)v) = d(\nu v)/dt$?

I have no idea what you mean by "an integrating factor I= e^(k/m)". An integrating factor is a function of t! What formula do you have for the integrating factor of a linear equation?

3. Feb 4, 2008

captainjack2000

Sorry the integrating factor was meant to say I=exp(kt/m)
do I then write (dv/dt)exp(kt/m) + exp(kt/m)(k/m)v = (k/m)wexp(kt/m)- gexp(kt/m) ?
Sorry i'm still confused how to solve for v(t) since I don't know how to rearrange this into a separable equation?

4. Feb 4, 2008

HallsofIvy

Staff Emeritus
Well the whole point of an integrating factor is that
d(ekt/mv)/dt= ekt/mdv/dt+ (k/m)ekt/mv.

Your equation is the same as
d(ekt/m)/dt= [(k/m)w- g]ekt/m. Now integrate both sides with respect to t.