# Differential mechanics equation

1. Oct 3, 2006

### thenewbosco

here is the question and my solution until i have become stumped:

The force acting on a particle of mass m is given by : F=kvx where k is a positive constant. The particle passes through the origin with speed vo at t=0. Find x as a function of t.

what i have done is set up the following differential equation:

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

is this correct? and if so, how do i solve this type of differential equation?
i am not so strong at de's

2. Oct 3, 2006

### jpr0

$$\frac{d^2 x}{d t^2} - 2\mu x \frac{dx}{dt} = 0$$

(just relabelling constants, $\mu = k/2m$), Then you notice that

$$2\mu x \frac{dx}{dt} = \frac{d}{dt}\left(\mu x^2\right)$$

So the entire LHS of your equation can be written as... [something]

By the way, are you sure that this is the force you're given?

The solutions you get from this should be pretty non-trivially dependent on your initial conditions (as is the case with most non-linear ODEs), for example if you analyse your force, when x is positive, and v is positive, your force is positive, and so your particle will accelerate off to infinity. If x is positive, and v is negative, then the particle is accelerated towards the origin. If x is negative, and v is negative, the particle is accelerated toward the origin (slowing the particle), until it reaches a steady state when v=0, (since the force acting on the particle is zero). If x is negative, and v is positive, the particle is accelerated away from the origin along -x (deccelerating the particle). If the particles initial momentum is not enough to reach the origin, (or just enough to reach it) it will then come to a halt (again because v=0, and also F=0). If it has enough momentum to go beyond the origin, then again it will escape to infinity.

In what context did you come across this force?

Last edited: Oct 3, 2006
3. Oct 3, 2006

### thenewbosco

thanks for the help. however i am still not sure what to do after rewriting like this, is there anymore help someone can provide?

4. Oct 4, 2006