ODE $\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$ Consider the ODE $$\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$$ I have to find the interval of unique solution by using Picard method. As in my local book it is solved as $$|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M. $$ $$h=\min\{a,\frac{b}{M}\}$$ where $M=a^{2}+b^{2}$. So $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$. Then my main problem is the author do like $$a=\frac{b}{a^{2}+b^{2}}$$ which gives a quadratic in $b$ as $ab^{2}-b+a^{3}=0$ with discriminant as  $1-4a^{4}=0$ gives $a=\frac{1}{\sqrt{2}}$ so $|h|\leq\frac{1}{\sqrt{2}}.$ Please suggest me how he  find $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$? Why he put $a=\frac{b}{a^{2}+b^{2}}$? Thanks in advance.
 A: We want to find the interval for a unique solution using the Picard–Lindelöf theorem given 
$$\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$$ 
In this case, $x_0 = y_0 = 0$. For some $a, b $, the function
$f(x, y) = x^2 + y^2 $ is defined and Lipschitz continuous on the rectangle $R=\{|x|\leq a, |y|\leq b\}$.
The Picard existence theorem states that the IVP has a unique solution on the interval $[-h, h]$.
By Theorem 1 or the Wiki, the supremum is given by $|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M$, and therefore 
$$h= \min\left\{a,\dfrac{b}{M}\right\} = \min\left\{a,\frac{b}{a^2 + b^2}\right\}$$
We now have to choose $a$ and $b$ such that $h$ becomes as large as possible, that is, we want
$$ h = \max_{a, b}~ \min\left\{a,\frac{b}{a^2 + b^2}\right\}$$
Among all rectangles of a given perimeter the square has the largest area.
Using the 
AM-GM inequality, we want to find when $\sqrt{l~ w} \le \dfrac{l + w}{2}$.  This occurs if and only if $l = w$, that is, the rectangle is a square and the sides are equal (that is why the author equates them). As an aside, this is a consequence of Is there a "simple" proof of the isoperimetric theorem for squares?.
Thus in our case
$$a = \dfrac{b}{a^2 + b^2} \implies a(a^2 + b^2) - b = 0 \implies a \ne 0, b = \dfrac{1-\sqrt{1-4a^4}}{2 a}$$
The expression under the radical must be positive, hence
$$1 - 4a^4 \ge 0 \implies | a | \le \dfrac{1}{\sqrt{2}}$$
We could have stopped here, but lets use this result for $a$ and do an additional step to validate that we get the same result for $b$. Using $a$, we find 
$$ h = \max_b~\min\left\{\pm~ \dfrac{1}{\sqrt{2}},\frac{b}{\frac{1}{2} + b^2}\right\} \implies |b| \le \dfrac{1}{\sqrt{2}}$$ 
Conclude that for any choice of $a, b$, such that $|a| \le \frac 12, a \ne 0, |b| \le \frac 12$, the maximum $h$ is given by
$$| h | \le \dfrac{1}{\sqrt{2}}$$ 
