Interval for unique solution for first order I.V.P. Find largest interval in which the I.V.P $$\frac{dy}{dx}=y^{2}+\cos^{2}x,x>0,y(0)=0.$$ has unique solution.
My attempt is as follows 
Firstly I apply Picard Theorem for the rectangle $\{(x,y):|x|\leq a,|y|\leq b\}$
$M=b^{2}+1, h=\min\{\frac{b}{b^{2}+1},a\}$,  I found $h=1/2$ and hence our required interval is $[0,1/2]$
But my problem is that the function $f(x,y)=y^{2}+\cos^{2}x$ is globally Lipschiz on the region of the type $[0,\infty)\times [-k,k]$ so can i say that the given I.V.P. has unique solution on $[0,\infty)?$
Please suggest where I am wrong? Am I using the wrong result in last one? What is the way to find largest interval of uniqueness of solution on the above I.V.P.? Thank you.
 A: The function $f(x,y)=y^{2}+\cos^{2}x$ is of class $C^{1}$ so it is locally
Lipschitz in a neighborhood of every point $(x_{0},y_{0})$. Hence, as long as
the solution exists, the solution is unique. Since $f$ is continuous, the
solution exists at least for a short time. Let $[0,T)$ be the maximal time of
existence on the right (the case $x<0$ is similar). We want to prove that
$T<\infty$. Assume by contradiction that $y(x)$ exists for all times. Since
$f\geq0$, we have that $y(x)$ is increasing. Actually, it is strictly
increasing, since $y^{\prime}(0)=1>0$ and so for $x>0$ sufficiently small, say
$0<x\leq a$ (for some small $a>0$), we have that $y(x)>y(0)=0$. But then
$y^{\prime}(x)\geq y^{2}(x)\geq y^{2}(a)>0$ for all $x\geq a$ and so $y(x)$ is
actually strictly increasing for all $x>0$. Then for $x>0$, $y^{\prime}(x)\geq
y^{2}(x)>0$ and so
$$
\frac{y^{\prime}}{y^{2}}\geq1
$$
for all $x>0$. Integrating both sides over the interval $[1,x]$ we get
$$
\frac{1}{y(1)}\geq-\frac{1}{y(x)}+\frac{1}{y(1)}\geq x-1.
$$
Letting $x\rightarrow\infty$ we get a contradiction because the left-hand side
is bounded and the right-hand side goes to infinity. This means that
$T<\infty$. Thus the solution exists in $[0,T)$ and since it is strictly
increasing
$$
\lim_{x\rightarrow T^{-}}y(x)=\infty.
$$
You can do the same for $x<0$. 
