Non linear second order ODE I really need help solving this : 
$$y_{xx}-\left(y^{3}-y\right)-\varepsilon\frac{1}{2}\left(1-y^{2}\right)=0
  $$
With boundary conditions :
$$ y(\pm \infty  )=-1 $$
I need to find a solution that is accurate up to $O(\epsilon)$
i all so know that
$$ y^{(k)}(\pm \infty) = 0\;\;\;\;\; for\; \; k>0 $$
Thanks alot! 
Please note that holding boundary condition is not easy and very important, 
the answer given below is a great effort but unfortunatly it is not a good answer.
any help would be great! 
 A: $$\frac{d^2v}{dx^2}-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$
This is an autonomous ODE. So, the change of function is :  $\frac{dv}{dx}=f(v)$
$\frac{d^2v}{dx^2}=\frac{df}{dv}\frac{dv}{dx}=f'f$
$$f'f-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$
$$\frac{1}{2}\frac{d(f^2)}{dv}=\left(v^{3}-v\right)+\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$
$$f^2=\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+c_1$$
$$\frac{dv}{dx}=\sqrt{\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+c_1}$$ 
Especially in case of $\varepsilon=0$ with conditions  $v(\pm\infty)=-1$ and $v'(\pm\infty)=0$ :
$\frac{1}{2}(-1)^4-(-1)^2+c_1=0$ hense $c_1=\frac{1}{2}$
$$x=\int \frac{dv}{\sqrt{\frac{1}{2}v^4-v^2+\varepsilon\left(v-\frac{1}{3}v^{3}\right)+\frac{1}{2}}}+c_2$$
The exact analytical result is a very complicated formula involving an elliptic integral of the first kind. Further approximate calculus can be done in considering a series expansion relatively to $\varepsilon$, at least up to $O(\varepsilon^2)$.
We search an approximate on the form :
$$x=F_0(v)+\varepsilon F_1(v)+O(\varepsilon^2)$$
$$F_0(v)=\int \frac{dv}{\sqrt{\frac{1}{2}v^4-v^2+\frac{1}{2}}}+c_2=-\sqrt{2}\tanh^{-1}(v)+c_2$$
$$F_1(v)=\int\frac{\sqrt{2}v(v^2-3)}{3(v^2-1)^3}dv=-\frac{v^2-2}{6(v^2-1)^2}+c_3$$
