Let $f \in \mathcal{C}(\mathbb{R}^{n+1},\mathbb{R}^n)$, and $a,b \in \mathbb{R}^+$, such that $0<a<a+\epsilon \leq b$ (where $\epsilon >0$).
and we define the following ODEs : Assuming the existence of an absolutely continuous solution $x(.)$ defined on $[0,b]$. $$\begin{cases} \dot x(t)=f(t,x(t)) & t\in[0,b]\\ x(a)=x_a, \end{cases} $$ Now, we introduce a small perturbation, and I want to prove that $x_{\epsilon}$ is defined on [0,b] for small $\epsilon$ (i.e $\exists \delta >0 \,\, \epsilon < \delta \text{ such that }x_{\epsilon} \text{ defined on } [0,b] $). $$\begin{cases} \dot x_{\epsilon}(t)=f(t,x_{\epsilon}(t)) & t\in[0,b]\\ x_{\epsilon}(a+\epsilon)=(1+\epsilon)x_a, \end{cases} $$ We have the existence of a neighborhood of $(a+\epsilon)$ : $]a+\epsilon-\nu,a+\epsilon+\nu[$ for some $\nu>0$.
First of all, I tried to prove that $x_{\epsilon}$ is defined at $0$ : $\exists \delta >0 \,\, \epsilon < \delta \text{ such that }x_{\epsilon} \text{ defined at } 0$.
By contradiction, we have for all $\delta>0$, $\exists \epsilon<\delta \text{ such that } x_{\epsilon} \text{ is not defined at } 0$ i.e $\lim_{y\rightarrow 0}x_{\epsilon}(y)=\infty.$ So if we tend $\delta $ to $0$, we will get $\lim_{y\rightarrow 0}x(y)=\infty.$ which is impossible.