- #1

lavinia

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

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- Thread starter lavinia
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- #1

lavinia

Science Advisor

Gold Member

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

- #2

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- #3

- 2,967

- 5

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

- #4

lavinia

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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.

- #5

- 2,967

- 5

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.

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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