We have the function $f : \mathbb{R} \times \mathbb{R} \to \mathbb{R}, (x,y) \mapsto \frac{xy}{\sqrt{y^2+1} }$ and the following IVP
\begin{align*} y'=f(x,y), \qquad y(0)=1. \end{align*}
How does one prove that there is a solution on $\mathbb{R}$ without computing it and then show that, for all $x \geq 0$, the following holds?
\begin{align*} y(x) \in \left [1, 1 + \frac{x^2}{2} \right ] \end{align*}
I know how to prove that there is a unique maximal solution on an open interval $I$. I'm just struggling to prove that $I=\mathbb{R}$ and I have no idea how to approach the second part. Help would be appreciated.
You have $$ |y'(x)|\le|x|. $$ The upper bound follows directly, the lower bound from the sign of $f$.