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?

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 $\sqrt[4\,]{28}$.
