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?

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
