TITLE: How can I prove that $f'(x)=1+\left[f(x)\right]^2$ has no solution over $\mathbb{R}$ without solving the equation explicitly?
QUESTION [6 upvotes]: It is known that the only solutions to the ODE
$$f'(x)=1+\left[f(x)\right]^2$$
are of the form $f(x)=\tan(c+x)$ (this is also easy to verify by hand). This shows that the differential equation can't have a solution over $\mathbb{R}$ because $\tan (c+x)$ is undefined whenever $c+x=\frac{\pi}{2}+\pi n$ for integer $n$.
But what if I didn't know this? What if I didn't know that
$$f'(x)=1+\left[f(x)\right]^2\iff f(x)=\tan(c+x)$$
Heck, what if I've never even heard of the tangent function nor any of the other trigonometric functions? Presumably, I could prove that there can't be a solution over $\mathbb{R}$ from the ODE alone, but how would I go about doing it?
For the record, I have no idea how to approach this. From the assumption that $f$ is differentiable everywhere, nothing from the equation seems to "break": you get two everywhere-continuous functions, $f'$ and $1+f^2$, and they are equal to each other.

REPLY [4 votes]: Suppose $y(x)$ is an everywhere differentiable function which satisfies the differential equation $y'=1+y^2$.

Our goal is to derive a contradiction (without explicitly solving the ODE).

From $y'=1+y^2$, it follows that $y'\ge 1$ for all $x\in\mathbb{R}$, hence $y$ is increasing.

If $y$ is bounded above, then since $y$ is increasing, we would have
$$
\lim_{x\to\infty}y'(x)=0
$$
contrary to $y'\ge 1$ for all $x\in\mathbb{R}$.

Similarly, if $y$ is bounded below, then since $y$ is increasing, we would have
$$
\lim_{x\to -\infty}y'(x)=0
$$
contrary to $y'\ge 1$ for all $x\in\mathbb{R}$.

Hence the range of $y$ is equal to $\mathbb{R}$.

Let $a\in\mathbb{R}$ be such that $y(a)=1$ and let $b > a$.
\begin{align*}
\text{Then}\;\;&
y'=1+y^2
\\[4pt]
\implies\;&
\frac{1}{1+y^2}\,dy=dx
\\[4pt]
\implies\;&
\int_{y(a)}^{y(b)}\frac{1}{1+y^2}\,dy=\int_a^b 1\,dx
\\[4pt]
\implies\;&
\int_1^{y(b)}\frac{1}{1+y^2}\,dy=b-a
\\[4pt]
\implies\;&
\lim_{b\to\infty}\left(\int_1^{y(b)}\frac{1}{1+y^2}\,dy\right)=\infty
\\[4pt]
\implies\;&
\int_1^\infty \frac{1}{1+y^2}\,dy=\infty
\\[4pt]
\end{align*}
contradiction, since
$$
\int_1^\infty \frac{1}{1+y^2}\,dy
 <
\int_1^\infty \frac{1}{y^2}\,dy
=
1
$$