Bound on differential equation using Picard's theorem I've shown that if I have a differential equation: $y' = f(x,y)$, $y(0) = 0$ where $f$ is continuously differentially and $|f| \le K$ and $|\frac{\partial f}{\partial y}| \le L$ in some compact set containing $(0,0)$. 
Then I have shown that: $|y| \le \frac{K}{L} (e^{L|x|}-1)$ - just by using Picard iterated integrals as $|f(x,y)| \le L|y| + K$ and so $|y| \le Kx + L\int\limits_0^x  |y|  dx$ and then solving the right hand side of that.
Now I am given a differential equation $y' = e^y$ with $y(0) = 0$ and so I can find the exact solution:
$$y = log(\frac{1}{1-x})$$
Now this seems to contradict the bound as $x \to 1$, I cannot figure out where I have gone wrong here?
Any help is much appreciated
 A: I don't have enough points to comments, but I think the key is "$f_y\leq L$ in some compact set containing $(0,0)$". So, probably everythink is valid only for $x$ s.t. $(x,y(x))$ is in this set and $y(x)$ may be large elsewhere.
A: The existence theorem only guarantees "local" solutions.  It just says there is some open interval $I$ around $x=0$ and a function on that interval satisfying the differential equation.  The interval of existence does not have to extend to the boundary of the set where f and its derivative are bounded.
To be more precise, suppose $f$ and $\partial_y f$ are bounded on $[a,b]\times[c,d] \subset R^2$. Then for some open interval $I \subset [a,b]$ containing $0$, there exists a solution $y$ such that $(x,y(x)) \in [a,b]\times[c,d]$, $y(0)=0$ and $y'(x)= f(x,y(x))$.  
Nothing guarantees that $1 \in I$.
Another example is $y' = y^2$ with the initial condition $y(0)=1$.  This looks harmless because $f(x,y)= y^2$, which is bounded and has a bounded derivative on any compact set.  However the solution is 
$$y(x) = \frac{1}{1-x}$$
which blows up at $x=1$.
