# Differential Equation- Help!

1. Apr 25, 2008

### SandraH

1. The problem statement, all variables and given/known data

We're required to analyse a particle moving in the potential U(x) = a/x^2 (a > 0). Setting F = U(x) and using the Newton equation F = ma, this gives rise to the DE:

d2x = a/m * (1/x^2)

dt^2

I can't for the life of me figure out what method to use or even what sort of DE this is! That 1/x^2 is blowing my mind.

Any clues would really help.

3. The attempt at a solution

I've tried treating this as a non-homogeneous constant-coefficient ODE of the form

ax'' + bx' + cx = f(t)

(setting f(t) = 1/ x(t)^2 )

I try to solve

d2x = x(t)^-2

dt^2

Ignoring constants for now.

Complementary solution (solution to the complementary equation:
ax"(t) + bx'(t) + cx(t) + d = 0)

-> x_c = at + b

Particular solution

-> xp = x(t)^-2 + x(t)^-1 + x(t) + c1 ????

Next I find the derivatives of the particular solution

x'p = -2x(t)^-3 * x'(t) -x(t)^-2 * x'(t) + x'(t)

And using the chain rule:

x''p = [ -2x(t)^-3 * x''(t) + x'(t) * 6 x(t)^-3 * x'(t)] + [-x(t)^-2 * x''(t) + x'(t)* 2x(t)^-3 * x'(t)] + x''(t)

At this point (right before finding constants) I seriously begin to doubt whether this is the method to use. Any hints?

I also tried using the modified Newton equation (t - t0 = +/- sqrt(m/2) int[1/sqrt(E-U(y)] dy... it's called 'reduction to quadrature') which yielded a very hard integral I couldn't do by substitution or parts

int ((E - a/y^2)^-1/2 ) dy

I'm yet to try partial fractions, but I seriously doubt whether that would work...

Thanks!

Sandra
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 25, 2008

### Hootenanny

Staff Emeritus
Welcome to PF Sandra,

Thanks for posting such a complete post detailing your efforts, it's refreshing change from students who just expect us to do their homework for them! Your problem is right at the beginning of your solution:
This is incorrect, the relationship between a conservative force F and the potential U is,

$$\underline{F} = -\nabla U$$

Which in one dimension simplifies to,

$$F = -\frac{d}{dx}U$$

3. Apr 27, 2008

### SandraH

ahaaa a problem right at the beginning. I feel sheepish..
Thanks Hootenanny

4. Apr 27, 2008

### Hootenanny

Staff Emeritus
Your welcome

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook