How can I prove the uniqueness of an ODE: $ -v''(x)+\frac{v(x)}{(1+v^2(x))^2}+v(x)=1 $? The question has been answer here

Given the ODE
$$
-v''(x)+\frac{v(x)}{(1+v^2(x))^2}+v(x)=1
$$
satisfies the condition $x\in(0,1)$, $v(0)=v(1)=1$, and $v(\cdot)$ is symmetric with respect to $1/2$.

I am wondering that can we determine the solution $v$ uniquely such that the above condition satisfies? This ODE is not linear nor any standard form so I am not quiet sure about it...although the two boundary condition is given. Moreover, I know $v$ is non-negative and quasi-convex.
Typically, by quasi-convexity and symmetricity, we have $v$ is monotone decreasing in $(0,1/2)$ and monotone increasing in $(1/2,1)$.
However, I am still not so sure that the uniqueness...
 A: By Theorem 7.7 (described in "The Theory of Differential Equations" by Walter Kelley and Allan Peterson, page 313), this BVP has a unique solution.  
[Note that using this theorem, no appeal needs to be made to the: symmetry, non-negativity, and/or quasi-convex properties mentioned.]
In the language of the theorem quoted, we have:
$$v''(t)=f(t,v)\text{ where } f(t,v)=f(v)=\frac{v}{(1+v^2)^2}+v-1$$
Here $f(t,v)$ is continuous on $[0,1]\times \mathbb{R}$ and satisfies the (Lipschitz) condition:
$$|f(t,x) − f(t,y)| = |f(x) - f(y)| ≤ K|x − y|$$ where $K=2$.
[Reason:  $|f′(v)| ≤ 2$ for all $v$ in $\mathbb{R}$, so the MVT implies $|f(x) - f(y)|/|x - y| = |f′(\text{squiggle})|$ (for some squiggle in the interval $(x , y)$ ).]
Thus, by the theorem, as long as the two boundary conditions specified are close enough together (specifically, as long as $b-a < \frac{2 \sqrt{2}}{\sqrt{K}}$ ), the given BVP has a unique solution.
Here $a=0$ and $b=1$, so $b-a = 1 < 2 = \frac{2 \sqrt{2}}{\sqrt{K}}$, so we do have a unique solution.
