No idea here, but *[this](http://www.wolframalpha.com/input/?i=%28L^2%29%27%27+%2B+B%28L^2%29%27+%2B+2gL+%3D+A+)**?
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Hooray for the confusing notation mathematicians have adopted for derivatives, eh? Although things like "L^2" for the function x -> L(x) * L(x) don't exactly help either. (I can't help you, I've never had to solve differential equations -- lucky me.)
I see you've edited your post to clarify things. :-)
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I've never been so good at differential equations, but maybe you'd want to check Monge's technique. If I recall correctly, this technique is able to solve pretty much anything of low order. Though maybe that was for partial differential equations, can't remember well.
What I can say though is that you can't assume that L->Inf is t->Inf, as your L could be anything. Unless you were given special information about your equation, in general you can't assume that.
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Try this wolframalpha-link:
WolframAlpha
Check if it is the correct ODE. Solution seems to be ok, but not very beautiful. you can get c_1 with your boundary condition. Looks like L->0 for t->inf .
Or do you need the complete path to the solution?
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Try using Laplace transform. It changes differential equations into algebraic formulas and makes it easier to solve.
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..and I thought what better place than this very forum.
What I have is:
(L(t)^2)'' + B(L(t)^2)' + 2g*L(t) = A
Differentiating with respect to t!
and I'm trying to find L(t). The only boundary condition that is obvious to me is L(0) = 0. A, B, g constants > 0.
Do I need more boundary conditions to solve this? It's possible that L -> inf as t -> inf but I'm not certain.
Can this be rewritten as the first order ODE:
z2' = A - Bz2 - 2g*sqrt(z1)
where z1 = L^2 and z2 = (L^2)'?
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