Thanks for taking the time to check this. I have been struggling to
solve the following system of equations:

ma+ 
(ma^(ra/rb))*(1-(1/rb)-y)*(y^(1/rb))*((1-y)^(-1-(1/rb)))
=
(mb^(rb/ra))* (x^((1/ra)-1))*((1-x)^(-1-(1/ra)))*(1/(ra da))

and the symmetric equation

mb+ 
(mb^(rb/ra))*(1-(1/ra)-x)*(x^(1/ra))*((1-x)^(-1-(1/ra)))
=
(ma^(ra/rb))* (y^((1/rb)-1))*((1-y)^(-1-(1/rb)))*(1/(rb db))

where 0<x<0.5(1-(1/ra)), 0<y<0.5(1-(1/rb)) are the variables, 1< ra,
1< rb, 1< ma, 1< mb, 0<da<1, 0<db<1 are the parameters.
It comes up as the equilibrium condition to a game theoretic model, I
am trying to get an explicit solution, or at least get a complete
characterization of the solution. While it looks very messy, it
behaves quite nicely. The RHS of first of equation is monotonically
decreasing and the LHS of second equation is monotonically increasing
for 0<x<0.5(1-(1/ra)). The RHS of second equation is monotonically
decreasing and the LHS of first equation is monotonically increasing
for 0<y<0.5(1-(1/rb)).
Do you have any suggestions? Thank you for your time.

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Clarification of Question by baroqqque-ga on 19 Oct 2006 06:02 PDT

Quick correction, 0< ma, 0 < mb instead of 1< ma, 1< mb

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Clarification of Question by baroqqque-ga on 23 Oct 2006 03:49 PDT

1/(ra da) stands for 1/(ra*da)
The reason I put the tag differential equations is I suspected the
explicit solution might be one' probably not, though

I don't understand the notations:

1/(ra da) in the RHS of the first equation, and

1/(rb db) in the RHS of the second equation.

Although the Subject of the Question refers to differential equations,
I don't see how a differential expression, if that's what these are,
can be combined with the rest of the equations.

regards, mathtalk-ga

Perhaps 1/(ra*da) and 1/(rb*db) are equivalents, as da,db are
described as "parameters".  I think this means da,db are given scalar
quantities, as are ra,rb.

-- mathtalk-ga