Are solutions to the differential equation $y' = 1 + x^2y^2$ growing? The task is to determine whether the statement is true or false.
Statement:
All solutions to the differential equation $y' = 1 + x^2y^2$ are growing.
My answer:
I think that all solution to the differential equation is growing due to that the derivative is $x^2$ and $y^2$ which always gives a positive number, which leads to a positive derivative --> growing function.
Am I thinking right or not?
 A: Seems perfectly fine to me. I just want to restate this slightly more formally for clarity's sake.
Since $x^2,y^2 \geq 0$ for all $x,y$, then 
$$y' = 1 +x^2y^2 \geq 1 + 0 = 1 > 0$$
ergo, the solutions should be growing, as $y' > 0$ always.
A: Moreover, for $x\ge1$ you get
$$
y'\ge1+y^2\implies y(x)\ge\tan(x+c)\text{ where }y(1)=\tan(1+c),
$$
so that not only you get growth, but explosive growth to a blow-up (pole) in finite time.

Other inequalities: Still for $x\ge1$ one gets a separable right side in
$$
y'\le x^2(1+y^2)\implies y(x)\le\tan(x^3/3+c) \text{ where }y(1)=\tan(1/3+c)
$$
For $y(a)$ large, the second term will dominate the constant $1$, so that the pole structure is
$$
y'\gtrapprox x^2y^2\implies y(x)\gtrapprox\frac{y(a)}{1-y(a)(x^3-a^3)/3}
$$
so that given $(a,y(a))$ close to the pole an improved estimate for the pole position is $\sqrt[3\,]{3/y(a)+a^3}$.

Moreover, from the parametrization $y=\frac{u}{u'}$ one gets $u''(x)+x^2u(x)=0$, $u(0)=y(0)$, $u'(0)=1$. The solutions of this are oscillating, thus have maxima and minima where then $y$ has a pole. Using the Sturm-Picone comparison theorem one can make this quantitative getting bounds similar to the ones above.
