A simple ODE I think

lavinia · Nov 23, 2011

dc/dx = x[itex]^{2}[/itex]e[itex]^{-xc}[/itex]

raymo39 · Nov 23, 2011

?? 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

Dickfore · Nov 23, 2011

Make the subst, [itex]u(x) \equiv x \, c(x)[/itex] and see what ODE you get for [itex]u(x)[/itex].

lavinia · Nov 23, 2011

raymo39 said: ↑

?? 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

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.

Dickfore · Nov 24, 2011

lavinia said: ↑

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.

Dickfore said: ↑

Make the subst, [itex]u(x) \equiv x \, c(x)[/itex] and see what ODE you get for [itex]u(x)[/itex].

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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A simple ODE I think

lavinia

raymo39

Dickfore

lavinia

Dickfore