Solving the D'Alembert-Claureaut Equation for a Unit Speed Geodesic in the Plane

[lavinia]

dc/dx = x$^{2}$e$^{-xc}$

 

[raymo39]

?? it doesn't seem too simple, first its non separable non linear, but it is only first order. not quite sure what the post is for

 

[Dickfore]

Make the subst, $u(x) \equiv x \, c(x)$ and see what ODE you get for $u(x)$.

 

[lavinia]

?? it doesn't 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$^{-xc}$

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]

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

The right equation is dc/dx = e$^{-xc}$

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.

Make the subst, $u(x) \equiv x \, c(x)$ and see what ODE you get for $u(x)$.

This is not a correct hint for this equation.

I think I found the way though. Solve your equation in terms of $c$:

$$\ln c' = - x \, c$$

$$c = -x \frac{1}{\ln p}, \ p \equiv c'(x)$$

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

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