When doest it make sense to talk about globally well posedness for the ODE $y'=y$? 
My Naive Question: How to formulate the following question, rigorous and  then to find answer:"There exists a unique continuous solution to the ODE $y'=y$ for all times? Note that the solution is not bounded."

My attempt: Suppose $y:\mathbb R \to  \mathbb R$ be a function and 
$$\frac{dy(x)}{dx}= y(x),$$ with initial condition 
$$y(t_{0})=A \in C(\mathbb R), \ (t_{0}\in \mathbb R)$$ (where $A$ constant function). 
Then, by septation of variable, we have, $y(x)= e^{x+C}, \ (x\in \mathbb R),$ where $C=\log(A/e^{t_{0}}).$
Therefore,  for the given time,  solution belongs to $C(\mathbb R, C(\mathbb R)).$ (And the exponential function(solution depends on initial time $t_{0}$) is unbounded with respect to space variable $x$).
Can we say solution function is unbounded with respect to time variable? What is a standard meaning of global well posdeness for the OED? Doest it make sense of talking of global well poseness of the above ODE $y'=y$? If I want to prove the ODE $y'=y$ is globally well posedness then what should I prove?
[Please let me know if I have made any serious mistake; or any other suggestions to make everything here precise.]
Thanks,
 A: Well-posed problem is the problem that satisfy the conditions (the sketch)


*

*the solution exists in a certain class of functions

*the solution is unique in that class of functions

*the solution depends "in a continuous way" of the parameters of the problem (the notion of being "continuous way" to define in each case).
In the case of differential equation $y'=y$ it is reasonable to talk about the solutions of the Cauchy problem $y(t_0)=y_0$ in the space $C(\Bbb R)$. It easy to see that the solution exists in that class and is, indeed, unique (a side remark - the separation of variables can lead to way too many errors, use this method with caution) and writes $$y(t;y_0)=e^ty_0.$$
As you already noticed this function is unbounded, but we can say that for every compact $K\subset\Bbb R$ 
$$\lim_{p\to s}\sup_{t\in K}|y(t;p)-y(t;s)|=0,$$
i.e. on every compact the solutions for differetial initial data converge in the uniform norm as the initial data converge. This is "continouos dependence on the parameters of the problem". You can also generalise this continuous dependence on the Cauchy problem $y'=ay,\,y(t_0)=y_0$ with $a$ and $y_0$ being parameters. 
