I.V.P. $y'=1+y^2$ $y(0)=0$ 
I.V.P. $y'=1+y^2$, $y(0)=0$.
a) find the maximum interval there exist an unique solution for the given I.V.P.
b) find the solution of the given  I.V.P.

My try:
a)
I should to find M,L and h.
Such that :
1) $M=\max |f(x,y)|$.
2) $L$ is Lipschitz constant:
$$|f(x,y_2)-f(x,y_1)|\leq L|y_2-y_1|.$$
3) $h=\min(a,{b\over M})$.
So
1) $M=|f(x,y)|=|1+y^2|\leq 1+|y|^2$
$M=1+b^2$
2) $L=|{\partial f\over \partial y}|=|2y|$
$\leq 2|y|=2b$
$L=2b$
3) $h=min(a,{b\over M})$
$=min(a,{b\over {1+b^2}})$
$b=1$
[by use the First Derivative Test]
So
$M=2$ , $L=2$ and $h={1\over 2}$.
True ?

For b)
$y'=1+y^2$
${dy\over dx}=1+y^2$
$dy=(1+y^2) dx$
${1\over {(1+y^2)}} dy= dx$
$\tan^{-1} (y)=x+c$
From condition
$x=0, y=0$
$\tan^{-1} (0)=0+c$
$c=0$
So
$\tan^{-1} (y)=x$
$y= \tan (x)$
True ?
Thanks.

True.
So how far can
$y = \tan(x)$
go until it blows up?
