# A simple ODE I think

1. Nov 23, 2011

### lavinia

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

2. Nov 23, 2011

### 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

3. Nov 23, 2011

### Dickfore

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

4. Nov 23, 2011

### lavinia

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.

5. Nov 24, 2011

### Dickfore

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?