Need help solving a second order non-linear ordinary differential equation...

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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)'?

1 decade ago*

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19 Comments

mRevan

L(t) = 3

Half life 3 confirmed.

1 decade ago

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Wormy

Bonus: You just cured cancer.

Good job.

1 decade ago

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Ansatsunin

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+)**?

1 decade ago

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Wormy

Nope. I want to differentiate with respect to t, not L. Wolfram alpha is able to solve the ODE when the 2gL term is omitted but runs out of computation time otherwise.

1 decade ago

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92Arash

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.

1 decade ago

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Wormy

Differentiating with respect to t, not L, to find L(t).

1 decade ago

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Nudiustertian

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. :-)

1 decade ago

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hfxx

will try to solve it tomorrow

1 decade ago

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Phoenixio7

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.

1 decade ago

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Wormy

Seems to be for partial, but I'll look into it.

I was basing that assumption on the physical meaning of the ODE. It's still probably not the case though.

1 decade ago

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BadTenMan

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?

1 decade ago

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Wormy

Not the correct ODE, this is. Don't care about the steps, just need the solution.

1 decade ago

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SolSys

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

1 decade ago

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