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I want to find the solution of these equations:

x' = -3 x (-1 + x^2)/(3 (x^2) - 1) + Sqrt[3] λ x^2 (-2 + 3 x^2) y/(4*(3 (x^2) - 1))
y'=-Sqrt[3]/2  λ x y^2 + 3/2 y (1 + 1/4 x^2 (-2 + x^2) y^2)

λ is a constant parameter. I use this code to solve:

Solve[{X == 0, Y == 0}, {x, y}]

I know that must find 9 {x,y} but the solutions of this are not complete, they are Repetitious.

enter image description here

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  • $\begingroup$ With [Lambda] = 1/6; and Solve[{x' == 0, y' == 0}, {x, y}] i get a bunch of nice numeric results, so check your solve pls $\endgroup$ – user9660 Dec 18 '15 at 16:23
  • $\begingroup$ Are those primes derivatives, or just decoration on your variables? $\endgroup$ – J. M. will be back soon Dec 18 '15 at 16:24
  • $\begingroup$ @J.M. the "difference-equations" tag explains all :) $\endgroup$ – Dr. belisarius Dec 18 '15 at 16:25
  • $\begingroup$ @Dr. (titles, yecch): I don't see a recurrence anywhere in the post, so I'm tempted to remove that tag actually. ;) $\endgroup$ – J. M. will be back soon Dec 18 '15 at 16:27
  • $\begingroup$ @lou but i need to solution be by [Lambda] $\endgroup$ – milad Dec 18 '15 at 16:30
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Not nice, but works

x1 = -3 x (-1 + x^2)/(3 (x^2) - 1) + Sqrt[3] λ x^2 (-2 + 3 x^2) y/(4*(3 (x^2) - 1));
y1 = -Sqrt[3]/2 λ x y^2 + 3/2 y (1 + 1/4 x^2 (-2 + x^2) y^2);
r = Reduce[{x1 == 0, y1 == 0}, {x, y}, Backsubstitution -> True];

yv = Numerator /@ Take[Union@Flatten@ReplaceList[y, {ToRules@r}], -4]
yr = Thread[yv -> Array[yy, 4]]
yrr = Thread[yv/Sqrt[2] -> Array[yy, 4]/Sqrt[2]]
ypol = Times @@ (y - yv) // Expand

xv = Numerator /@ Take[Union@Flatten@ReplaceList[x, {ToRules@r}], -4];
xr = Thread[xv -> Array[xx, 4]]
xrr = Thread[1/xv -> 1/Array[xx, 4]];
xpol = Times @@ (x - xv) // Expand

Column@{r /. yr /. xr /. xrr /. yrr // ToRules}

Mathematica graphics

Where the xx[..] and yy[..] are specified by xr and yr and are the roots of xpol and ypol respectively.

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