Qualitative study of a differential equation. I am asked to study this differential equation:
$$y'=y^2-\frac{1}{1+x^2} \qquad y(0)=1$$
Let $y$ be a maximal solution and let $[0,b[$ be the interval where it's defined.


*

*Compute the Taylor series centered in $0$ and stopped at the second order:
Clearly $y(0)=1$ and $y'(0)=0$ by simple sostitution. By computing we have:
$$y''(0)=2y(0)y'(0)+\frac{2x}{(1+x^2)^2}=0$$
and so $y(x)=1+o(x^2)$.

*Prove that $y$ is increasing:
$$y'>0 \iff y^2 > \frac{1}{1+x^2}$$
and I guess it's correct, but I can't prove it right. By using the Taylor formula we see that $y'>0$ in a neighbourhood of $0$ but I can't extend it to all the interval. How can I do it?

*Prove that $b<+\infty$ by giving upper and lower bounds:
I don't actually know how to start doing it. I understand that it might have a blow-up but don't know how to write it.


Can you please help me?
 A: By assuming
$$ y(x) = -\frac{f'(x)}{f(x)}=-\frac{d}{dx}\log f(x) $$
the differential equation simply becomes
$$ f(x) = (1+x^2) f''(x) $$
whose solution, by the power series method, is a combination of hypergeometric functions
$$ f(x) = \phantom{}_2 F_1\left(-\tfrac{1+\sqrt{5}}{4},\tfrac{-1+\sqrt{5}}{4};\tfrac{1}{2};-x^2\right)-x\cdot \phantom{}_2 F_1\left(\tfrac{1-\sqrt{5}}{4},\tfrac{1+\sqrt{5}}{4};\tfrac{3}{2};-x^2\right).$$
$y(x)$ is increasing iff $f(x)$ is log-concave, and we may employ the Cauchy-Schwarz inequality applied to the integral representation of $f(x)$ to deduce that $f(x)$ is indeed log-concave (as the opposite of a moment, essentially).
The life time of $y(x)$ is the unique positive root of $f(x)$, which by Newton's method lies between $\frac{3}{2}$ and $2$ (see the comment by Professor Vector about cubic convergence).

An elementary approach. In the region $y>\frac{1}{\sqrt{1+x^2}}$ we have $y'>0$. The solution starts on the boundary of such region and by Taylor series it moves into that region after an infinitesimal time, hence it remains there until it is defined, and it remains increasing and greater than one. 
Assuming that the solution is global we have that the asymptotic behaviour of the solution is the same as the asymptotic behaviour of the solution of
$$ v'(x) = v(x)^2 $$
since the term $\frac{1}{x^2+1}$ becomes negligible. However the solutions of the last DE are of the form $\frac{1}{\alpha-x}$, so necessarily negative from some point on. We know that $y(x)$ is positive, hence it cannot be global.
We may derive an approximation of $y(x)$ from its Maclaurin series. It is not difficult to derive that during the lifespan we have $y(x)\approx \frac{3}{3-x^2}$, hence the lifetime of $y$ is around $\sqrt{3}$.

By the Gauss continued fraction we also have
$$ \frac{1+\frac{x^2}{2}}{\sqrt{1+\frac{x^2}{3}}}\geq \frac{\phantom{}_2 F_1\left(-\tfrac{1+\sqrt{5}}{4},\tfrac{-1+\sqrt{5}}{4};\tfrac{1}{2};-x^2\right)}{\phantom{}_2 F_1\left(\tfrac{1-\sqrt{5}}{4},\tfrac{1+\sqrt{5}}{4};\tfrac{3}{2};-x^2\right)}\geq\frac{1+\frac{x^2}{2}}{1+\frac{x^2}{6}} $$
hence the lifetime of $y$, i.e. the least positive root of the sort-of parabolic cylinder function fulfilling $f(x)=(1+x^2)f''(x)$ and $f(0)+f'(0)=0$, is between $1.59$ and $1.87$.
It is worth mentioning that the substitution $x=\sinh\theta$ converts the DE $f(x)=(1+x^2)f''(x)$ into the DE $g(\theta)=g''(\theta)-\tanh(\theta)g'(\theta)$, relating the original problem to Legendre functions (evaluated at $\frac{x}{\sqrt{1+x^2}}$), too. By approximating the solutions of the $g(\theta)=g''(\theta)-\tanh(\theta)g'(\theta)$ with the solutions of $g(\theta)=g''(\theta)-\theta g'(\theta)$ (which are given by the error function) the lower bound for the lifetime is improved into
$$\text{lifetime}(y) \geq \sinh\left(\sqrt{2}\operatorname{InverseErf}\left(\sqrt{\frac{2}{\pi}}\right)\right)=1.65069\ldots $$
Using Pfaff transformations the lifetime is given by the positive solution of 
$$ \frac{\phantom{}_2 F_1\left(-\tfrac{1+\sqrt{5}}{4},\tfrac{3-\sqrt{5}}{4};\tfrac{1}{2};\frac{x^2}{x^2+1}\right)}{\phantom{}_2 F_1\left(\tfrac{1-\sqrt{5}}{4},\tfrac{5-\sqrt{5}}{4};\tfrac{3}{2};\frac{x^2}{x^2+1}\right)} = \frac{x}{\sqrt{1+x^2}} $$
and by exploiting the Maclaurin series of $\frac{\phantom{}_2 F_1(\ldots;z)}{\phantom{}_2 F_1(\ldots;z)}$ it is enclosed by the solutions of
$$ 1-\tfrac{1}{6}\left(\tfrac{x^2}{1+x^2}\right)=\tfrac{x}{\sqrt{x^2+1}}\quad\text{and}\quad 1-\tfrac{1}{6}\left(\tfrac{x^2}{1+x^2}\right)-\tfrac{1}{72}\left(\tfrac{x^2}{1+x^2}\right)^2=\tfrac{x}{\sqrt{x^2+1}}. $$
This leads to
$$\boxed{1.739 \leq \text{lifetime}(y) \leq \sqrt{\frac{6}{11}(2+\sqrt{15})}\approx 1.79}$$
which can be sharpened through suitable computational resources.
The numerical value of the lifetime actually is $1.7424143972\ldots$
