# Particle motion diff eq

1. Sep 8, 2005

### infraray

I'm given F=kvx. I need to find x(t), k is a positive constant, and the particle passes thru the origin with speed Vo at time t=0.

I start by rewriting the formula as such:

dv/dt=k*dx/dt*x

I am confused now though because I now have dv,dt, and dx. I assume I need to get everything in terms of dx and dt. Can I use seperation of variables or do I need another technique?

2. Sep 8, 2005

### quasar987

Is v the speed?

If so, you got F=kvx, which can be rewriten as

$$F = k\frac{dx}{dt}x$$

and F = ma, which can be rewritten as

$$F = m\frac{d^2x}{dt^2}$$

$$k\frac{dx}{dt}x = m\frac{d^2x}{dt^2}$$

3. Sep 8, 2005

### amcavoy

Now is this where you have trouble? $kx=mx^{2}$ has solutions x=0 and x=k/m. What does this tell you about the differential equation?

Last edited: Sep 8, 2005
4. Sep 8, 2005

### infraray

Ok, after some tinkering I was able to get:
$$k\frac{dx}{dt}x = m\frac{d^2x}{dt^2}$$
Since it doesn't appear to be able to be put into a 1st order equation I assume I am out of luck with seperation, or am I?

Sorry, but I'm not sure what x=0, x=k/m tells me about the equation. Are they linear dependent? The lighter the object the further it travels?

It's been over 2 years since diff eq and my head is a mess right now. I feel helpless. My teacher keeps saying this is easy, but when you've been away from it for a while, it is very hard. It's not like riding a bike that's for sure.

5. Sep 9, 2005

### HallsofIvy

Staff Emeritus
Try using the fact (chain rule) that
$$\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$$
to eliminate t from the equation.

$$m\frac{dv}{dt}= mv\frac{dv}{dx}= F= kxv$$