# A simple ODE I think

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

raymo39
?? 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
Make the subst, $u(x) \equiv x \, c(x)$ and see what ODE you get for $u(x)$.

Gold Member
?? 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$^{-xc}$

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?