Slowing down a nonlinear differential system to compute its asymptotics How do we solve following system of differential equations.
$$x'(t)=- \frac{x}{2}+\frac{x}{2}[\lambda y-\frac{1}{2}(1+\lambda)x+1-x-y]$$ 
$$y'(t)= \frac{x}{2}[-\lambda y+\frac{1}{2}x]$$
$$x(0)= \frac{2 \lambda}{(1+\lambda)^2},y(0)= \frac{1}{(1+\lambda)^2}$$
In text book it given that the asymptotic value of $x(t)$ is 
$$x(t= \infty)= \frac{1-\lambda}{1+\lambda}$$
Can some one help me to solve this. Thanks.
 A: 
Once supplemented by some missing hypothesis and cleansed of some unfortunate misprints, the exercise is a nice conceptual question about the trajectories of a given system in two different time scales.

Let $X(t)=\begin{pmatrix}x(t)\\ y(t)\end{pmatrix}$ and $A=\begin{pmatrix}3+\lambda&2(1-\lambda)\\ -1&2\lambda\end{pmatrix}$ then the system of differential equations in the question can be rewritten as $$X'(t)=-\tfrac14x(t)\cdot A\cdot X(t).$$ Consider the solution $U(t)=\begin{pmatrix}u(t)\\ v(t)\end{pmatrix}$ of the differential system $$U'(t)=-A\cdot U(t),$$ with the same initial condition $U(0)=X(0)$, then $X(t)=U(\theta(t))$ where $\theta$ solves $$\theta'(t)=\tfrac14x(t),\qquad\theta(0)=0.$$ In other words, $X$ and $U$ follow the same path but in two different time scales. Furthermore, if $U(t)$ crosses the line $u=0$ for the first time at some point $(0,v_*)$, then $\theta'(t)=0$ at that time hence, due to the time change, $X(t)$ accumulates on this point in the sense that $X(t)\to(0,v_*)$ when $t\to+\infty$, in particular, $$y(t)\to y_\infty=v_*.$$
One sees that the change of time scale is quite drastic since it leads to $X$ staying confined to a strict subset of the whole path of $U$, and that the exercise should ask to show that $y(t)\to y_\infty$ (not that $x(t)\to y_\infty$ which is absurd since $x(t)\to0$, as explained above).
To go further, note that 
a missing hypothesis is that $\lambda$ is in $(0,1)$.
The eigenvalues of the matrix $A$ are $a=\lambda+1$, with eigenvector $V_a=\begin{pmatrix}1-\lambda\\ -1\end{pmatrix}$, and $b=2(\lambda+1)$, with eigenvector $V_b=\begin{pmatrix}2\\ -1\end{pmatrix}$,
hence, for every $t$, $$U(t)=c_ae_a(t)V_a+c_be_b(t)V_b,\qquad e_a(t)=\mathrm e^{-at},\qquad e_b(t)=\mathrm e^{-bt},$$ for some suitable constants $c_a$ and $c_b$. 
The initial condition $a^2X(0)=\begin{pmatrix}2\lambda\\ 1\end{pmatrix}$ implies that $a^2U(0)=-2V_a+V_b$, hence, for every $t$, $$a^2U(t)=-2e_a(t)V_a+e_b(t)V_b.$$
In particular, $$a^2u(t)=-2(1-\lambda)e_a(t)+2e_b(t)=2e_a(t)\cdot(e_a(t)-(1-\lambda)),$$ hence $u(t)=0$ for the first time when $$e_a(t)=1-\lambda.$$ Finally, note that, for every $t$, $$a^2v(t)=2e_a(t)-e_b(t)=e_a(t)\cdot(2-e_a(t)),$$ hence, when $u(t)=0$, $a^2v(t)=(1-\lambda)(1+\lambda)$, that is, $$v_*=y_\infty=\frac{1-\lambda}{1+\lambda}.$$

Supplementary material: Below is a plot of the curve $(U(t))_{t\geqslant0}$ starting from $U(0)=\frac1{(1+\lambda)^2}\begin{pmatrix}2\lambda\\ 1\end{pmatrix}$ for $\lambda=\frac12$. The curve $(X(t))_{t\geqslant0}$ starting from $X(0)=U(0)$ is the part of the curve on the right of the ordinate axis $x=0$. The point $(0,y_*)$ is $(0,\frac13)$.
$\qquad\qquad\qquad$
$$\texttt{ParametricPlot[{4s(2s-1)/9,4s(2-s)/9},{s,0,1}]}$$
For every $\lambda$, $U(t)$ and $X(t)$ stay on the part of parabola with equation $$(1+\lambda)(x+2y)^2=4(x+(1-\lambda)y).$$
The parabola when $\lambda=\frac12$:
$\qquad\qquad\qquad$
$$\texttt{graph[3(x+2y)^2=4(2x+y),{x,-.2,4},{y,-0.5,0.5}]}$$
Finally, one can generalize these results to every starting point, noting that the dynamics of $(u,v)$ is conveniently encoded by the variables $(w,z)$ defined by $$w=u+(1-\lambda)v,\qquad z=(u+2v)^2,$$ since $$w'=-2(1+\lambda)w,\qquad z'=-2(1+\lambda)z.$$
As a direct consequence, starting from some $(w_0,z_0)$, one stays on the $(w,z)$-parabola of equation $$w_0\,z(t)-z_0\,w(t)=0,$$ thus, for every starting point $(x_0,y_0)$, the solution $(x(t),y(t))$ stays on the parabola
$$
(x_0+(1-\lambda)y_0)\cdot(x+2y)^2-(x_0+2y_0)^2\cdot(x+(1-\lambda)y)=0,$$ and $(x(t),y(t))\to(0,y_\infty)$ when $t\to+\infty$, with
$$
y_\infty=\frac{(1-\lambda)\cdot(x_0+2y_0)^2}{4\cdot(x_0+(1-\lambda)y_0)}.$$
