Analyzing a Particle in a Non-Homogeneous Differential Equation

[SandraH]

Homework Statement

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.

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

 

[Hootenanny]

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:

We're required to analyse a particle moving in the potential U(x) = a/x^2 (a > 0). Setting F = U(x)

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,

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

 

[SandraH]

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

 

[Hootenanny]

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

Your welcome