# Thread: Diff EQ Falling body

1. ## Diff EQ Falling body

Consider the falling body problem where:
m(dv/dt) = W - B - CV. W is weight, B is boyancy and CV is drag force. W,B,C are constants, V is dependant on t of course V(t).

Consider an additional force acting upon the object that depends only on time symbolized by n(t). Assume the body is dropped from rest.

I must set up an IVP for V(t), t > or = 0.
Im a little confused how to approuch this with the extra n(t).
It also doesnt explicity say if n(t) is + or -, but i assume its -, do you agree?

m(dv/dt) = W - B - CV - n(t)

How shall i separate these variables and attempt to solve?
Thanks to all in help.

Chris

2. Originally Posted by chrsr345
Consider the falling body problem where:
m(dv/dt) = W - B - CV. W is weight, B is boyancy and CV is drag force. W,B,C are constants, V is dependant on t of course V(t).

Consider an additional force acting upon the object that depends only on time symbolized by n(t). Assume the body is dropped from rest.

I must set up an IVP for V(t), t > or = 0.
Im a little confused how to approuch this with the extra n(t).
It also doesnt explicity say if n(t) is + or -, but i assume its -, do you agree? I would assume it to be positive. That's what I usually tend to do in situations like these...

m(dv/dt) = W - B - CV - n(t)

How shall i separate these variables and attempt to solve?
Thanks to all in help.

Chris
First divide the equation through by m to get:

$\frac{\,dV}{\,dt}=\frac{W}{m}-\frac{B}{m}-\frac{C}{m}V+\frac{n(t)}{m}$

This DE is linear:

$\frac{\,dV}{\,dt}+\frac{C}{m}V=\frac{W}{m}-\frac{B}{m}+\frac{n(t)}{m}$

Apply the method of the integrating factor, where $P(t)=\frac{C}{m}$

Can you try to take it from here?

--Chris