Deleted

L(t) = 3

Half life 3 confirmed.

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+)**?

I'm not entirely sure what you mean, but do you just want to solve: (L^2)'' + B(L^2)' + 2g*L = A for L?
If so, it's not very hard and the answer will be : = (A-2) / 2(B+g), where B+g =/= 0 > g=-B.

will try to solve it tomorrow

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.

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?

maybe this?
L(t)= A/[2(1+B+G)]

The answer is 42

Try using Laplace transform. It changes differential equations into algebraic formulas and makes it easier to solve.

Noctis to Wormy, do you read me? Noctis to Wormy, are you there? Please respond!

These things evoke bad memories. My maths is not advanced enough to help you.