## non linear differential equation

I can't type in latex so in this post d^2a is the secpnd derivative of a, while (da)^2 is the square of the derivative.

This equation arose from the G_thetatheta compinent of the einstein tensor. Iwas solving tfor the shcwarzchildmetric where where the cosmos constant is nonzero.

the equation is:

e^2a(d^2a +2(da)^2 +2/r(da))= L where L is the cosmological constantand r is the radial coordinate. Umm how do we solve this?

 Quote by Terilien I can't type in latex so in this post d^2a is the secpnd derivative of a, while (da)^2 is the square of the derivative. This equation arose from the G_thetatheta compinent of the einstein tensor. Iwas solving tfor the shcwarzchildmetric where where the cosmos constant is nonzero. the equation is: e^2a(d^2a +2(da)^2 +2/r(da))= L where L is the cosmological constantand r is the radial coordinate. Umm how do we solve this?
$$e^{2a}(da'' +2da^2 + \frac{2da}{r}) = L$$

is that right?
 Yes it is. can you helpme? essentially I determined from G_tt and G_rr that b=-a just like with the ordinary metric. When I plugged it into the G_thetathata equation, it was still fairly ugly, and icouldnt find a solution by inspection. Non linears are annoying.

## non linear differential equation

Is it not

$$a^{\prime \prime} + 2 a^{\prime 2} + \frac{2a}{r} = L e^{-2a}$$

and, presumably the derivative is with respect to r, yes?
 yes it is. It may be that. It is possible that I've made a small mistake in my calculations of miswrote someon paper. How do we solve that? Can it be done by inspection? non linearity scares me. could someone help me solve it?