TITLE: Solution of the IVP $\dot{y}=x^2+y^2,y(0)=0$
QUESTION [6 upvotes]: The solution of the IVP $$\dot{y}=x^2+y^2,y(0)=0$$ 
I know this IVP has a unique solution but I can't find the interval in which it has a unique solution can someone help me in finding the interval.

REPLY [2 votes]: Since $-y(-x)$ solves the same equation, the solution is an odd function of $x$. Therefore, the interval of existence is symmetric about $0$. One could give upper and lower estimates on size of the interval, but if the goal is to find the exact interval, I'm afraid you need the exact solution. Which is findable with Maple: 
$$
y = -x\frac{J_{-3/4}(x^2/2)-Y_{-3/4}(x^2/2)}{J_{1/4}(x^2/2)-Y_{1/4}(x^2/2)} \tag1
$$ 
where $J$ and $K$ are Bessel functions. With sufficient motivation, (1) can be checked by hand using the recurrence formulas for the derivatives of $J_\alpha$ and $Y_\alpha$.
Unfortunately, the smallest positive root of the function $J_{1/4}-Y_{1/4}$ does not appear to  be anything nice.  Numerically, it's  $2.00629967..$, which yields the interval of existence $(-2.003147...,2.003147..)$.
It may be interesting to prove without a computer that the solution exists on $[-2,2]$, but this is probably hard.