Suppose for two given functions $f\_1,f\_2 \colon \mathbb{R}^2 \to \mathbb{R}$ there exist unique solutions $y\_1$ and $y\_2$ with the intersection of their intervals of existence $[0,\epsilon)$ to the integral equations $$y\_k(x)=\int\_{0}^{x} f\_k(t,y\_k(t)) dt$$. Moreover, suppose that $f\_1(x,y)\leq f\_2(x,y)$ (or strictly if that makes things easier), does it follow that $y\_1 \leq y\_2$ on $[0,\epsilon)$?


It's true if $f\_1$ and $f\_2$ are Lipschitz in $y$ and continuous in $x$ (i.e., satisfy the hypotheses of the standard uniqueness and existence theorem) or if $f\_1$ and $f\_2$ are decreasing in $y$ (i.e., satisfy the hypotheses of Peano's uniqueness theorem). The assumption of unique solutions is certainly necessary.



Does $f_1(x,y)<f_2(x,y)$ imply $y_1<y_2$ for solutions to the integral equation $y_k'=f_k(x,y_k)$?
In general the answer seems to be no: Let 
$$
f\_1(t,y)=\begin{cases} -1, &t\leq 0,\quad y\in \mathbb R\\\
 1,& t>0,\quad y>t/2,\\\
 -1,& t>0,\quad -t/2 \leq y \leq t/2,\\\
 -e^{-n^2y},& t\in [1/n,1/(n-1)),\quad y< -t/2,\quad n\geq 1,
 \end{cases}
$$
and 
let $f\_2(t,y)=-f\_1(t,-y)$ so that $f\_1(t,y)< f\_2(t,y)$ when $(t,y)\in \mathbb R^2$. 


For $x\geq 0$ we have the solutions $y\_1(x)=x$ and $y\_2(x)= -x$ and in order to show that e.g. $y\_1$ is unique one must show that $y\_1$ cannot be such that $-x/2 \leq y\_1(x)\leq x/2$ some $x>0$ and furthermore show that if $-\infty < y\_1(x)< -x/2$ when $1/n \leq x \leq 1/(n-1)$ for some $n>1$ then 
$$
1 > e^{n^2y(1/(n-1))}-e^{n^2y(1/n)}= \frac n{n-1} > 1.
$$


