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A simple ODE I think

  1. Nov 23, 2011 #1

    lavinia

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    dc/dx = x[itex]^{2}[/itex]e[itex]^{-xc}[/itex]
     
  2. jcsd
  3. Nov 23, 2011 #2
    ?? it doesnt seem too simple, first its non separable non linear, but it is only first order. not quite sure what the post is for
     
  4. Nov 23, 2011 #3
    Make the subst, [itex]u(x) \equiv x \, c(x)[/itex] and see what ODE you get for [itex]u(x)[/itex].
     
  5. Nov 23, 2011 #4

    lavinia

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    Hi Raymo I must apologise. The equationI wrote is wrong.

    The right equation is dc/dx = e[itex]^{-xc}[/itex]

    This solves for the y coordinate of a unit speed geodesic in the plane with a particular metric of constant negative curvature. The x coordinate is easy. Maybe a conformal change of coordinates would give an easier equation.
     
  6. Nov 24, 2011 #5
    This is not a correct hint for this equation.

    I think I found the way though. Solve your equation in terms of [itex]c[/itex]:

    [tex]
    \ln c' = - x \, c
    [/tex]

    [tex]
    c = -x \frac{1}{\ln p}, \ p \equiv c'(x)
    [/tex]

    Your equation becomes an implicit ODE of the D'Alemebert - Clauraut type. It is reduced to a linear ODE w.r.t. [itex]x = x(p)[/itex] after differentiating w.r.t. [itex]x[/itex].

    What differential equation do you get at this step? Can you perform the integral in a closed form using elementary functions?
     
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