Initial value problem and grönwall's inequality I have the following problem:
Consider the Initial Value Problem $x'=f(t,x), \; x(t_0)=x_0 $, where $f:I\times \mathbb{R}^n \to \mathbb{R}^n $ is locally lipschitz with respect to $x$. Suppose that exists $c_1,c_2 \geq 0$ such that 
$$\| f(t,x) \| \leq c_1 \|x\| + c_2 , \; \forall (t,x)\in I \times \mathbb{R}^n $$
Prove that the maximal solution $\varphi(t)$ of the IVP is defined $\forall t\in I$. 
Hint: Use the Grönwall's inequality.
I have done this:
If $\| f(t,x) \| \leq c_1 \|x\| + c_2 , \; \forall (t,x)\in I \times \mathbb{R}^n $, in particular:
 $$\| f(t,\varphi(t)) \| =\|\varphi'(t) \|\leq c_1 \|\varphi(t)\| + c_2= c_1\left\|x_0+\int_{t_0}^tf(t,\varphi(s))ds \right\|+c_2 =c_2+c_1\|x_0\|+c_1 \int_{t_0}^t \left\| f(t,\varphi(s)) \right\|ds$$
Then, by the Grönwall's inequality,
$$\|\varphi'(t) \| \leq e^{c_1 |t-t_0|}\left( c_2+c_1 \|x_0 \| \right)$$
The professor said that we should prove firstly that $\varphi(t)$ is bounded, $\forall t \in I$. I have proved that $\varphi '(t) $ is bounded. It implies that $\varphi(t) $ is bounded? And how can I finish the problem?
Thank you very much 
 A: Ok, I have got this:
Using the LutzL's comment, $\| \varphi(t) - x_0 \| \leq \left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t-t_0) }-1 \right)  $, assuming $t\geq t_0$
So, I have solved this exercise by contradiction. If $I=(a,b)$, suppose that the maximal interval of definition of $\varphi(t)$ is $(t_-,t_+)$, with $t_+< b $. In the interval $[t_0, t_+) $ I have proved that 
$$\| \varphi(t) - x_0 \| \leq \left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t_+-t_0) }-1 \right)  $$
Then, $\varphi(t)\in \bar B(x_0, R)$, $R=\left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t_+-t_0) }-1 \right)$. Let $K$ be the compact $K=[t_0,t_+]\times \bar B(x_0,R)$. By the 'boundary approximation theorem' (I'm not sure what is the name of this theorem in english), $\exists t_K\in (t_-,t_+)$  such that $\forall t\geq t_K, \; (t,\varphi(t)) \notin K$. But $t\in [t_0,t_+]$, so $\varphi(t)\notin \bar B(x_0,R)$, which is a contradiction. Then $t_+=b$, and using the same argument, $t_-=a$. 
Is it correct?
Thanks!
