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X'' - 1/x^2 =0

  1. Sep 10, 2007 #1
    Can anybody solve this problem?
    I don't think this is something that we see in an elementary diff eq textbook.
    Thank you in advance.

    Yang
     
  2. jcsd
  3. Sep 10, 2007 #2
    Sorry. I should have put this into "diff eq."
     
  4. Sep 10, 2007 #3

    berkeman

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    I moved it to Homework Help, Calculus. Homework and coursework questions (even if not for homework, as long as they are homework-like) need to be posted in the Homework Help forums, and you need to show us your work before we can offer tutorial help.

    So how would you approach solving this type of equation?

    Welcome to the PF, BTW.
     
  5. Sep 10, 2007 #4
    Thanks. I was solving (which I thought is) a trivial diff eq.
    where there is some repeling force from an origin to an electron, say, at x.
    So, the diff eq. really boils down to

    x'' - 1/x^2 = 0

    which I wonder if it even has a closed form solution.
    Thank you all in advance.

    Yang
     
  6. Sep 10, 2007 #5

    dextercioby

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    HINT: Multiply the equation by x' and then try to write the whole LHS as a total derivative.
     
  7. Sep 10, 2007 #6
    Perhaps a little more hint:)

    Thank you, Kurt.
     
  8. Sep 10, 2007 #7

    dextercioby

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    I'm not Kurt. My name is Daniel. Kurt's a part of the name appearing in the signature.

    How about writing it like that ?

    [tex]x' x'' -\frac{x'}{x^{2}}=0 [/tex]
     
  9. Sep 10, 2007 #8
    Sorry, Daniel. I guess it's been too long since I left college. My brain is all rusty:(
     
  10. Sep 10, 2007 #9
    Perhaps [tex]\frac{d}{dt} (x')^2 = 2 x' x'' [/tex] helps?
     
  11. Sep 10, 2007 #10

    dextercioby

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    Yes, it does. Also [itex]\frac{d}{dt} \frac{1}{x'}=-\frac{x'}{x^{2}} [/itex].
     
  12. Sep 10, 2007 #11
    Thanks a lot. So, I have
    [tex] \frac{1}{2}(x')^2 - \frac{1}{x} = C [/tex] where C is a constant.
    I feel so hopeless. I still don't see how I can solve this:(
     
  13. Sep 10, 2007 #12

    dextercioby

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    Well, there's a plus before the 1/x. Check it out. Separate the derivative and then separate variables.
     
  14. Sep 10, 2007 #13
    Thank you again. This is not how I should approach it, isn't it?
    [tex] x' = 2\sqrt{C-1/x} [/tex]
     
  15. Sep 10, 2007 #14

    dextercioby

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    Okay, now, do you know how to separate variables ?
     
  16. Sep 10, 2007 #15
    Oh, I see. Isn't it
    [tex] \frac{dx}{2\sqrt{C-1/x}} = dt [/tex]
     
  17. Sep 10, 2007 #16
    I guess no (though I learned it once upon a time). Sorry.
     
  18. Sep 10, 2007 #17
    No, I don't know how to separate variables. Please help...
     
  19. Sep 10, 2007 #18

    dextercioby

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    Yes, now integrate in both terms.
     
  20. Sep 10, 2007 #19
    Thank you, Daniel.
     
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