Show that the largest interval of existence of the solution predicted by Picard's Theorem is $[0,\frac{1}{2}]$ Let the $\operatorname{IVP}$ be given by:
$\dfrac{\operatorname{dy}}{\operatorname{dx}}=y^2+\cos^2 x;x>0;y(0)=0$
Show that the largest interval of existence of the solution predicted by Picard's Theorem is $[0,\frac{1}{2}]$
By Picard's Existence and Uniqueness Theorem;If $f$ is continuous on a domain $D$ and $f$ satisfies Lipschitz condition on $D$ .If $R=\{|x-x_0|\leq a;|y-y_0|\leq b\}$ lies in $D$ and $M=\sup |f(x,y)|,h=\min\{a,\frac{b}{M}\}$. Then the $\operatorname{IVP}$ has a unique solution on the interval $|x-x_0|\leq h$.
Obviously the Lipscitz Condition is satisfied here.But I can't find the rectangle which is needed to apply the theorem.How should I do it?
 A: In your case $x_0 = y_0 = 0$. For any $a > 0, b > 0$, the function
$f(x, y) = y^2 + \cos^2 x$ is defined and Lipschitz continous on
the rectangle $R=\{|x|\leq a, |y|\leq b\}$.
On this rectangle, $$|f(x, y)| \le |y|^2 + |\cos x|^2 \le b^2 + 1$$
with equality for $x = 0$ and $y = b$, so
the supremum is $M = b^2 + 1$ and therefore 
$$h= \min\{a,\frac{b}{b^2 + 1}\}$$
The Picard existence theorem states that the IVP has a (unique)
solution on the interval $[-h, h]$, or – if you restrict the problem
to $x \ge 0$ – on $[0, h]$.
The task is now to choose $a$ and $b$ such that $h$ becomes as
large as possible. From the AM-GM inequality is follows that
$$
 b = \sqrt{1 \cdot b^2} \le \frac{b^2 + 1}2 \Longrightarrow \frac{b}{b^2 + 1} \le \frac 12
$$
with equality for $b=1$.
It follows that $h \le \frac 12$ for any choice of $a, b$, and 
$h = \frac 12$ for $a = \frac 12, b = 1$. So
$h = \frac 12$  is the largest value that can be obtained by this
method.
A: Hint:
You could choose $M = b^2 + 1$. Then, try to find $a,b$ which maximizes $h = \min\{a, b/(b^2+1)\}$.
