True or false about existence of solution for an ODE Consider the ODE $x'=f(x,t)$ in $\mathbb{R}^{n}$ and $t\in\mathbb{R}$. Let $\gamma(t)$ a soluitino for the ODE such that $\gamma(0)=0$ and $\gamma(1)=x_{0}$. So, given a neighborhood $U_{x_{0}}$ of $x_{0}$, there is $\epsilon>0$ such that, if $g(x,t)$ satisfies $|g(x,t)-f(x,t)|<\epsilon$ for all $x$ and $t$, so the solution $\eta(t)$ of $x'=g(x,t)$ such that $\eta(0)=0$ satisfies $\eta(1)\in U_{x_{0}}$.
Is this statement true or false? Intuitively, seems true, but I don't know how to demonstrate.
 A: ** When $f(t)$ and $g(t)$ do not depend on $x$ this works:
$|\dot\eta(t) - \dot \gamma(t)| = |f(t)-g(t)|<\epsilon$, so integrate $-\epsilon < \dot \eta(t) - \dot \gamma(t)  < \epsilon$
and use that $\eta(0) = \gamma(0) = 0$ to get:
$|\eta(t) - \gamma(t) | < \epsilon t$.
Then set $t = 1$ to get that $\eta(1)$ is $\epsilon$-close to $x_0 = \gamma(1)$.
So, for a general neighborhood of $x_0$, it will depend on $t$ (how far you want to send the solutions) and how good your bound $\epsilon$ for $|f - g|$ is. In the worst case, an $O(\epsilon)$ perturbation can have an $O(1)$ effect on solutions within time scale $O(1/\epsilon)$ (but of course for certain examples things could be better)
**edit: thanks to LutzL and user539887's comments, we need to add conditions on how to compare $f(x,t)$ and $g(y,t)$ when $x\ne y$. For example, when $f$ has a Lipschitz condition $|f(x,t) - f(y,t)|\le M|x-y|$, here is a way we can work out some bounds:
Set $u(t) = \eta(t) - \gamma(t)$, then $u(0) = 0$ and
$|\dot u| = |g(\eta(t), t) - f(\gamma(t), t) + f(\eta(t), t) - f(\eta(t), t)|<\epsilon + M |u|$.
then compare directly with solutions of $\dot u_+ = \epsilon + M|u_+|, \dot u_- = -\epsilon - M|u_-|$ to get
$-\frac{\epsilon}{M}(e^{Mt} - 1) = u_-(t) \le u(t) \le u_+(t) = \frac{\epsilon}{M}(e^{Mt} - 1)$ for $t\ge 0$
Or $|u(t)|\le \frac{\epsilon}{M}(e^{Mt}-1)$ for $t\ge 0$.
