Riccati D.E., vertical asymptotes 
For the D.E. 
  $$y'=x^2+y^2$$
  show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:

$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^2$$
$$y'\geq a^2+y^2$$
$$\frac{y}{a^2+y^2}\geq 1$$
$$\int \frac{dy}{a^2+y^2}\geq \int dx=x+c$$
$$\frac{1}{a}\arctan \frac{y}{a}\geq x+c$$
$$\arctan \frac{y}{a}\geq a(x+c)$$
$$\frac{y}{a}\geq\tan a(x+c)$$
$$y\geq a\tan a(x+c)$$
$$a(x+c)\simeq \frac{\pi}{2}$$
But where to from here?
 A: 1. $x_0$ exists
First note that $y'''(x)$ is increasing$^{[1]}$.  It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{[2]}$, so by Taylor's theorem$^{[3]}$,
$$
y(x)=\frac{x^3}{6}y'''(c)\ge \frac{x^3}{3},\qquad (*)
$$
for all $x>0$ such that $y$ is defined.  Choose one such $x=\epsilon>0$.  Then if $x>\epsilon$, we get
$$
y'(x)\ge \epsilon^2+y(x)^2,
$$
which, since $y(\epsilon)>0$, implies $y(x)\to\infty$ as $x\to x_0<\infty$ for some $x_0>\epsilon$.
Edits:
$[1]$: Since $y'(x)=x^2+y(x)^2\ge 0$, $y$ is increasing.  Since $y\ge 0$ and $x\ge 0$, we have $y''(x)=2x+2y(x)y'(x)\ge 0$, so $y'$ is also increasing.  In a similar way, we deduce that $y'''(x)\ge 0$ and $y^{(4)}(x)\ge 0$.
$[2]$: Since $y(0)=0$, we have $y'(0)=0$.  Therefore, $y''(0)=2x+2y(x)y'(x)|_{x=0}=0$.  On the other hand, $y'''(0)=2+2y'(x)^2+2y(x)y''(x)|_{x=0}=2$.
$[3]$: First note that $y$ is smooth.  Indeed, since $y$ is continuous and $y'(x)=x^2+y(x)^2$, we see that $y'(x)$ is continuous.  Since $y''(x)=2x+2y(x)y'(x)$ and the right hand side is continuous, so is $y''$.  In a similar way, all derivatives of $y$ are continuous.  Since $y$ is smooth, Taylor's theorem can be applied:
$$
y(x)=y(0)+xy'(0)+\frac{1}{2}x^2y''(0)+\frac{1}{6}x^3 y'''(c),\qquad x>0,
$$
where $c\in(0,x)$.  But the first three terms are zero by [2], so (*) holds.
2. Lower bound:
Since a finite $x_0>0$ exists, we get
$$
y'(x)\le x_0^2+y(x)^2,
$$
which, since $y(0)=0$, implies
$$
y(x)\le x_0 \tan (x_0\,x).
$$
If it were true that $x_0^2<\pi/2$, then $y(x_0)<\infty$, so $x_0\ge\sqrt{\pi/2}=:z$.
3. Upper bound
For $x>z$, where $z$ is the lower bound, we have
$$
y'(x)\ge z^2+y(x)^2,
$$
which implies
$$
y(x)\ge z\,\tan z(x+c),
$$
where
$$
c=-z+\frac{1}{z}\arctan\frac{y(z)}{z}\ge-z+\frac{1}{z}\arctan\frac{z^2}{3}
$$
by inequality (*).  Let 
$$
\zeta=\frac{\pi}{2z}-c\le \frac{\pi}{2z}+z-\frac{1}{z}\arctan\frac{z^2}{3}\approx 2.12.
$$
Then $y(\zeta)$ does not exist, so $x_0<\zeta$.  Note that $z\approx 1.25$.
A: The usual trick to get a better manageable equation out of this Riccati equation is to substitute $y=-\frac{u'}{u}$ which results in the linear ODE of second order

$$
u''+x^2u=0,\quad u(0)=1,\, u'(0)=0
$$

While this still does not lead to a symbolic solution without involving (very) special functions (Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$), one can easily find a power series solution 
$$
u(x)=1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}-\frac{x^{12}}{3·4·7·8·11·12}\pm…
$$
This is an alternating series with eventually monotonically falling absolute values of the terms. For $x<\sqrt7$ one gets the bounds by partial sums
$$
1-\frac{x^4}{3·4}\le u(x)\le1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}.
$$
The first positive root of $u(x_0)=0$ is the location of the first pole of $y$. From the bounds one gets the root bounds

$$
\sqrt[4\,]{12}\le x_0\le \sqrt[4\,]{16+4(3-\sqrt7)}
$$

which numerically gives the interval
$$
[1.8612097182041991,\; 2.042882110200651]
$$
while the numerator $-u'(x)=\frac{x^3}{3}(1-\frac{x^4}{4·7}\pm…)$ has its first positive root above $\sqrt[4\,]{28}$.
A: From numerical solution comes out that $x=2$ and $x=-2$ are vertical asymptotes. Trying to solve as a Bernoulli equation gives a mess and the substitution $w=\frac{1}{y}$ gives problems as
$$w=\frac{1}{y};\;w'=-\frac{y'}{y^2}$$
Divide the original equation by $y^2$
$$\frac{y'}{y^2}=\frac{x^2}{y^2}+1\rightarrow -w'=x^2w^2+1$$
$$w'+x^2w^2=-1\rightarrow w(x)=c\;e^{-\frac{x^3}{3}}+\frac{e^{-\frac{x^3}{3}} x \Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right)}{3^{2/3} \sqrt[3]{-x^3}}$$
The problem is now with the initial value, as $w\to\infty$ as $x\to 0$
Anyway the general solution is
$$y=\frac{1}{w(x)}=\frac{3 e^{\frac{x^3}{3}}}{3 c+x E_{\frac{2}{3}}\left(-\frac{x^3}{3}\right)}$$
where $E_k(x)$ is the integral exponential function defined by
$$E_k(x)=\int_1^{\infty }\frac{e^{-tx}}{t^k}\,dt$$
and has a vertical asymptote for any $c\in\mathbb{R}$
Hope this helps
