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I have a system of equations with two equations containing the Lambert W function as follows,

$$\begin{cases} x = 1 - W_0(\frac{C_1 e^{y + 1}}{y + 1}) \\ y = 1 - W_0(\frac{C_2 e^{x + 1}}{x + 1}) \end{cases}$$

I have already solved the system numerically using a simple iterative method. A gentelman told me that the answer to the system is $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ where $k_1 = C_2 - C_1$ and $k_2 = C_2 - C_1$ but I have no clue how can one compute the answer.

  1. How can I get the answer to the system (i.e the steps)?
  2. While we have two curves in the system of equation, the intersection (i.e the answer of the system) must be a point not a curve. But $y = \frac{k_1 - k_2 x}{k_1 x - k_2}$ is a hyperbola. How can I describe it?
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Where do the $k_1$ and $k_2$ suddenly come from? –  Raskolnikov Mar 14 '11 at 21:37
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OK, for the solution to be right $k_1=C_1-C_2$ and $k_2=-(C_1+C_2)$. (Although some multiple would work as well.) My suggestion: invert the Lambert W function. –  Raskolnikov Mar 14 '11 at 21:44
    
Sorry, I fixed the question. –  Mohsen Mar 14 '11 at 21:47
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Thanks a lot. I solved the first part then. Since $W_0(f(y)) = 1 - x$, we can say $(1 - x)e^{1 - x} = f(y)$. We do the same thing with the other equation and the rest is straightforward. But I am still confused. Why I got a hyperbola as an answer instead of a point? Did we lost something? I mean a part of the information from the equations? –  Mohsen Mar 14 '11 at 23:13
    
Yes, you combine the two equations into one, thereby you still have one independent equation to use. So you could for example pick one of the original equations and keep it. –  Raskolnikov Mar 14 '11 at 23:16
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