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The following system of ODEs – is it recognized as distinct system, with meaningful background and uses?

$$\frac{dx}{dt} = - [x(t)]^2 - x(t)y(t)$$ $$\frac{dy}{dt} = - [y(t)]^2 - x(t)y(t)$$

This is probably not needed, but initial conditions: $x(t=0) = x_0, \space y(t=0) = y_0$

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    possible duplicate of [Using Octave to solve systems of two non-linear ODEs](http://math.stackexchange.com/questions/245435/using-octave-to-solve-systems-of-two-non-linear-odes)2012-11-30
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    Why are you asking the same question again? [here](http://math.stackexchange.com/questions/245435/using-octave-to-solve-systems-of-two-non-linear-odes) is your old question.2012-11-30
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    @MhenniBenghorbal: it is not the same question.. The other is – to be strict – not even about finding solution, but about using Octave2012-11-30

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It is a two-dimensional Lotka-Volterra equation. The most general LV-equation has applications in population dynamics, networks and chemical reactions.

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    All summands are negative – is this allowed in L-V equations?2012-12-01
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    @Drenet Yes. In fact it is very common.2012-12-01
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I don't recognise the ODE system, but you can obtain a solution analytically.

Hint: Let $u=1/x$ and $v=1/y$, as though each is a Bernoulli differential equation. You can then show that, if $x_0\neq0$ and $y_0\neq0$, a solution is given by $$x(t)=\frac{x_0}{1+\left(x_0+y_0\right)t},\\ y(t)=\frac{y_0}{1+\left(x_0+y_0\right)t}.$$

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    You don't need the conditions $x_0 \ne 0$ and $y_0 \ne 0$.2012-11-30
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    @RobertIsrael Perhaps, but for the derivation I used, this condition is required.2012-11-30
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    Here's a useful principle: once the solution has been found, it doesn't really matter how you found it. To check that it is a solution, just plug it in to the differential equation. If it works, it works.2012-11-30
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    How to go from your solution, to the following? http://chemistry.stackexchange.com/questions/500/what-is-probability-of-3-possible-products-between-two-chemical-species2012-11-30
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I doubt that it has a name, but you might notice that trajectories are rays from the origin (apart from the fixed points along $x+y=0$).