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Homework Help: Differential Equation- Help!

  1. Apr 25, 2008 #1
    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. jcsd
  3. Apr 25, 2008 #2

    Hootenanny

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    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,

    [tex]\underline{F} = -\nabla U[/tex]

    Which in one dimension simplifies to,

    [tex]F = -\frac{d}{dx}U[/tex]
     
  4. Apr 27, 2008 #3
    ahaaa a problem right at the beginning. I feel sheepish..
    Thanks Hootenanny
     
  5. Apr 27, 2008 #4

    Hootenanny

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    Your welcome :smile:
     
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