TITLE: Upper bound of nonlinear first-order ordinary differential equation
QUESTION [5 upvotes]: Given: $y(x)$ is defined for $x \geq 1$ and satisfies
$$y'=\frac{1}{x^2+y^2},  y(1)=1$$
Show that $$ y(x) < \frac{5\pi}{4} $$ for all $x \geq 1$
I don't see an easy way to solve for $y(x)$, and I don't know how to demonstrate that the function is always less than $5\pi/4$ without solving for $y(x)$. Any suggestions on how to approach this?

REPLY [2 votes]: Since $y'(x)\geq 0$ then $y(x)\geq 1$ if $x\geq 1$. This implies that 
$$
y'\leq \frac{1}{x^2+1},
$$
or, by integrating,
$$
y(x)\leq \tan^{-1}x-\frac \pi 4+1\leq 1+\frac \pi 4,
$$
which is somewhat sharper than you are asking.