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:

<br /> \ln c&#039; = - x \, c<br />

<br /> c = -x \frac{1}{\ln p}, \ p \equiv c&#039;(x)<br />

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?