Largest interval of solution of ODE I am currently self-learning a little bit of the theory of ordinary differential equations, and for now I am limiting for equations in which the Picard-Lindelöf theorem applies:
$y'=F(t,y)$ where $F$ is defined and continuous on some open set $U$ of $\mathbb{R}\times\mathbb{R}^n$ and Lipschitz uniformly continuous in $y$ with inital conditions $y(t_0)=y_0$
Given $a,b\in\mathbb{R}_+^*$ sufficiently small that the compact set $R=[t_0-a,t_0+a]\times\mathcal{D}(y_0,b) \subset U$, there is a solution in $I=[t_0-a',t_0+a']$ where $a'=\min(a,\frac{b}{M})$ and $M=\lVert{F}\rVert_{\infty,R}$ is the norm of $F$ in $R$ and $\mathcal{D}(y_0,b)$ is the disk with center $y_0$ and radius $b$


*

*An example
I considered the smooth function $F$ defined in $\mathbb{R}^2$ by:
$$F(t,y)=\cos^2t+y^2$$
with the initial condition $y(0)=0$


*

*Optimizing $a'$
This function is Lipschitz uniformly continuous in $y$ on any bounded open subset of $\mathbb{R}^2$.
Let $a,b\in \mathbb{R}_+^*$, we can show with some work that $a'=\min(a,\frac{b}{b^2+1})$, and by noticing that:$$\sup_{b\in\mathbb{R}_+^*}\frac{b}{b^2+1}=\frac{1}{2} \text{ which is attained on the point }b=1$$ we can prove that $a'=\frac{1}{2}$ works
Now such $a'$ is optimal in a sense that it is the largest one which can be extracted by the Picard-Lindelöf theorem, but I am suspecting that we can further get a larger value of $a'$, and as an evidence I used Mathematica which suggested that the largest "radius" is $a'\in [1.88,1.89]$



*

*My question
are there some good assumptions on $F$ that can give 


*

*some bounds on $a'$?

*or even better its value? 

 A: The differential equation can be solved in "closed form":
$$ y(t) = -{\frac {{\it MathieuCPrime} \left( 1/2,-1/4,t
 \right) }{{\it MathieuC} \left( 1/2,-1/4,t \right) }}
$$
The solution exists up until the first zero of the denominator, which 
is approximately $1.888380846$.
A: Knowing that the solution exists for $|t|\le a=\frac12$ with values $|y(t)|\le b=1$, you can find an upper bound for the extension via
$$
y'\le 1+y^2\implies y(t)\le \tan(t-a+\arctan(y(a)))\le \tan(t-\tfrac12+\tfrac\pi4)
$$
This upper bound has a pole at $t_*-\tfrac12+\tfrac\pi4=\frac\pi2$ or $t_*=\frac12+\frac\pi4=1.285398..$
As long as this upper bound exists, the exact solution has to take finite values and can thus be extended. This means the maximal domain of the solution contains the interval $[0,t_*)$.

Inspired by the details of this bound one can consider the parametrization of the solution $y(t)=\tan(u(t))$, $|u(t)|\le\frac\pi2$. Then
$$
u'(t)=\frac{\cos^2t+\tan^2(u(t))}{1+\tan^2(u(t))},~~u(0)=0
$$
so that $M=1$ independent of the domain sizes $a,b$. To not leave the admissible range of $u$ one needs $Ma<\frac\pi2\implies |t|< a=\frac\pi2$. The maximal domain of $y$ thus contains $(-\frac\pi2,\frac\pi2)$.

The Riccati substitution is $y=-\frac{g'}{g}$, so that then $$g''(t)+\cos^2(t)g(t)=0, ~~g(0)=1,~ g'(0)=0.$$ 
This is a linear DE with continuous coefficients, that is, the solution exists unrestricted. Solutions for the original equation exist between roots of the denominator $g$.
The zeroth approximation of the solution of this Matthieu equation is $g(t)=\cos(\sqrt{\frac12}t)$ which has a root at $t=\frac\pi{\sqrt2}=2.221...$ which gives another impression (but not a valid estimate in any way) of the location of the pole in $y$.
Using the Sturm-Picone comparison theorem on this equation tells us that between any two roots of $g$ there has to be a root of $\cos(t)$. Using the symmetry of the equation and IVP gives that $g$ is even symmetric, this means that the smallest positive root of $g$ has to be larger than $\frac\pi2$.
