Equicontinuous set Let $\mathcal E$ be the set of all functions $u\in C^1([0,2])$ such that $u(x)\geq 0$ for every $x\in[0,2]$ and $|u'(x)+u^2(x)|<1$ for every $x\in [0,2]$.
Prove that the set $\mathcal F:=\{u_{|[1,2]}: u\in\mathcal E\}$ is an equicontinuous subset of $C^0([1,2]).$
The point I am stuck on is that i can't see how to combine the strange hypothesis imposed on every $u\in\mathcal E$, in particular i solved the two differential equations $$u'(x)=1-u^2(x),\qquad u'(x)=-1-u^2(x),$$ which result to be the extremal case of the condition given. In particular the two solutions are $$u_1(x)=\frac{ae^t-be^{-t}}{ae^t+be^{-t}},\qquad u_2(x)=\frac{a\cos(x)-b\sin(x)}{a\cos(x)+b\sin(x)}.$$ I feel however i'm not ong the right path so any help is appreciated. 
P.S. Those above are a big part of my efforts and thoughts on this problem so i hope they won't be completely useless :P 
Edit In the first case the derivative is $$u'_1(x)=\frac{2ab}{(ae^t+be^{-t})}\geq 0$$
while for the other function we have, for $x\in[0,2],$ $$u'_2(x)=-\frac{\sin(2x) ab}{(a\cos(x)+b\sin(x))^2}\leq 0.$$ Moreover $u_1(1)>u_2(1)$, since $$\frac{ae-b^{e-1}}{ae+be^{-1}}>\frac{a\cos(1)-b\sin(1)}{a\sin(1)+b\cos(1)}\Leftrightarrow (a^2e+be^{-1})(\sin(1)-\cos(1)),$$ and $\sin(1)>\cos(1).$ Now, all this bounds i've found are useful to solve the problem?
 A: Suppose $u \in \mathcal{E}$.  
It's enough to show $u(t) \le 3$ for all $t \in [1,2]$, since then we'll have $-10 \le u'(t) \le 1$, and any set of functions with uniformly bounded first derivatives is certainly equicontinuous.  We also know that $u' \le 1$ on $[0,2]$, and so by the mean value theorem it suffices to show that $u(1) \le 2$.  If $u(0) \le 1$ we are also done, so assume $u(0) > 1$.
Let $v$ be the solution of $v'(t) = 1 - v(t)^2$ with $v(0) = u(0) > 1$.  This is given by your formula for $u_1$ with, say, $b=1$ and some $a < -1$.  I claim $u(t) \le v(t)$.  This will complete the proof, since it is easy to check that $v(1) < 2$.  (We have $v(1) = \frac{ae-e^{-1}}{ae+e^{-1}}$, which is increasing in $a$; compute its value at $a=-1$.)
Set $w(t) = v(t) - u(t)$.  We have $w(0)=0$ and $w'(t) > u(t)^2 - v(t)^2$.  Suppose to the contrary there exists $s \in [0,1]$ such that $u(s) > v(s)$; let $s_0$ be the infimum of all such $s$.  Then necessarily $u(s_0) = v(s_0)$, so $w(s_0)=0$ and $w'(s_0) > u(s_0)^2 - v(s_0)^2 = 0$.  So for all small enough $\epsilon$, $w(s_0 + \epsilon) > 0$.  This contradicts our choice of $s_0$ as the infimum.  So in fact $u \le v$ and we are done.
