TITLE: How to this solve this differential equation?
QUESTION [6 upvotes]: $y'=\dfrac{1}{x^2+y^2}$
where $y=f(x)$
and $x$ lies in $[1,\infty)$
and $f(1)=1$ and it is is differentiable in that interval
I don't know how to even proceed in this problem.
Even the range of $y$ is sufficient

REPLY [3 votes]: What is wanted isn't clear. May be this :
$y'=\frac{1}{x^2+y^2}>0 $ implies that $y(x)$ is increasing.
$y(1)=1$ hence $y(x)>1$ . As a consequence :
$$y'=\frac{1}{x^2+y^2}<\frac{1}{x^2+1}$$
$$y(x)<y(1)+\int_1^x \frac{dt}{t^2+1}=1+\tan^{-1}(x)-\tan^{-1}(1)$$
$$y(x)<1+\frac{\pi}{2}-\frac{\pi}{4}=1+\frac{\pi}{4}$$