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?