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$$\dot{x}=x(1-x-\frac{3y}{4(x+1)})$$ $$\dot{y}=y(y-1)$$ One of the fixed points of this system is $(1,0)$ (Easily found through putting $\dot{x}$ and $\dot{y}$ equal to zero). This point is an asymptotically unstable fixed point. Establish the stability of the fixed points. This is established by forming the matrix of partial derivatives evaluated at the fixed points which gives the following: [ \begin{bmatrix} 1-2x-\frac{3}{4}\frac{y}{(x+1)^2} & -\frac{3x}{4(x+1)} \\ 0 & 2y-1 \end{bmatrix} = \begin{bmatrix} -1 & \frac{-3}{8} \\ 0 & -1 \end{bmatrix}

This matrix yield the eigenvalue $\lambda=-1$ with an algebraic multiplicity of 2 while it has just one eigenvector $v = (1,0)^T$. Which leads me to conclude that I am dealing with an asymptotically unstable point.

The question that is being asked at this point is to prove that for initial values $(x_0,y_0)^T$ for which counts $x_0>0$, $0<y_0<1)$ and $3y_0<4(1-x_0^2)$ how could I prove that for all these solutions the limit is $(1,0)$? I feel like there is something that I am missing. Any help would be much appreciated.

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