Largest interval of definition furnished by Picard's theorem Consider the IVP $$y'=\frac{1}{(3-x^2)(9-y^2)},\;y(0)=0.$$
I need to find the largest interval of definition furnished by Picard's theorem. I know the positive endpoint of the interval is given by $\min (a,\frac bM)$ where $|x|<a,|y|<b$ and $M=\sup_B |f(x,y)|$ with $B$ being the aforementioned rectangle.
For any $a,b$, which necessarily satisfy $a<\sqrt 3,b<3$ we have $\frac bM=b(3-a^2)(9-b^2)$. So I think I need to maximize $$\min(a,b(3-a^2)(9-b^2)).$$ How to do this? Alternatively, what am I doing wrong?
 A: Picard's theorem by itself doesn't furnish a particular interval of definition, it just says there is one.  There are many different ways to find an explicit interval.  In this case the d.e. is separable, leading to the implicit solution
$$ - \frac{\text{arctanh}(x/\sqrt{3})}{\sqrt{3}}
 - \frac{y^3}{3} + 9 y = 0 $$
This has $y=\pm 3$ when $x = \pm \sqrt{3} \tanh(18 \sqrt{3})$. So the actual interval is  $(-\sqrt{3} \tanh(18 \sqrt{3}), \sqrt{3} \tanh(18 \sqrt{3}))$.
EDIT: I would say it's not Picard's theorem that furnishes the interval, but it may be a particular proof of Picard's theorem.  Anyway, you want to maximize
$t$ such that $t \le a$, $t \le b (3-a^2)(9-b^2)$, $a \le \sqrt{3}$, $b \le 3$.
Since $a$ is an increasing function of $a$ while $b(3-a^2)(9-b^2)$ is a decreasing function of $a$ in the given region, the maximum should occur at a point where
$a = b(3-a^2)(9-b^2)$.  Thus we're looking for a point on the curve $a = b(3-a^2)(9-b^2)$ with $a$ as large as possible.  That curve looks like this:

On this curve we want a point where the tangent is vertical, so 
with
$$ g(a,b) = a - b(3-a^2)(9-b^2)$$
we take
$$ \dfrac{\partial g}{\partial b} = -3 (a^2 - 3)(b^2-3) = 0$$
Since we can't have $g=0$ with $a^2=3$, we must have $b^2 = 3$, and then
$t = a = \sqrt{3}(\sqrt{1297}-1)/36 \approx 1.684606387$.
