# Analytically solving a regular nonlinear 1st order quadratic ODE system

I just sincerely hope somebody can help me to analytically solve the following ODE system. $x(t)$, $y(t)$, $z(t)$ are $3$ functions of $t$, and $C_1, C_2, C_3, C_4, C_5, C_6$ are just constants.

$$\left\{ \begin{array}{l} x'(t)=[1-x(t)][C_1 y(t)+C_2 z(t)]\\ y'(t)=[1-y(t)][C_3 x(t)+C_4 z(t)]\\ z'(t)=[1-z(t)][C_5 x(t)+C_6 y(t)] \end{array} \right.$$

I appreciate a lot if you can offer me any closed-form solutions analytically, especially a general solution with $N$ functions. Or if you can point me to any related materials, I will also appreciate. : )

Furthermore, if this is not solvable, is there any possible way to make approximation functions for them...?

Thanks a lot!

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Thanks! ${}{}{}$ – joriki Aug 11 '12 at 13:14
M.SE is for timeless questions, not urgent ones. – Michael Greinecker Aug 11 '12 at 13:23
@MichaelGreinecker Give the guys a break – Pedro Tamaroff Aug 11 '12 at 13:33
Just so you know, saying "urgent" in your post actually disinclines some people from answering your questions. – J. M. Aug 11 '12 at 13:56
Dears, thanks a lot for your comments. I will modify my post : ) – Dobby Aug 12 '12 at 8:09

It appears your system is autonomous. It follows that the method of characteristics may help you solve it. Essentially, just solve for $dt$ then integrate. As $x'(t)=\frac{dx}{dt}$ and $y'(t) = \frac{dy}{dt}$ and $z'(t) = \frac{dz}{dt}$ you can see that $$dt = \frac{dx}{x'(t)} = \frac{dy}{y'(t)} = \frac{dy}{y'(t)}$$ You replace the expressions $x'(t),y'(t),z'(t)$ in the above with the expressions involving the unknown $x,y,z$ hence yielding the timeless system in $x,y,z$. Solutions should be level curves. These level curves give you the point-sets which the solutions to your give ODE will parametrize.
Since I'm not particularly interested in doing those integrations for your problem I'll illustrate this method with an easier problem: $$\frac{dx}{dt} = -y \qquad \frac{dy}{dt} =x$$ Hence, $$dt = \frac{dx}{-y} = \frac{dy}{x} \ \Rightarrow \ \ xdx = ydy \ \ \Rightarrow \boxed{x^2+y^2=R^2}$$ Of course this system is more easily solved by subsitution which reveals $x(t) = R\cos(t)$ and $y(t)=R\sin(t)$ which are clearly solutions to my given problem and parametrizations to the boxed level curve as I advertised.