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 $b$


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*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 $y(0)=0$


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*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 $a'\in [1.88,1.89]$



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


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*some bounds on $a'$?

*or even better its value? 


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 $1.888380846$.
