<s>[INST] Showing that this IVP has a solution for all of $\mathbb{R}$ I'm preparing for an exam and came across this problem:
Given the IVP


$f'(x) = \frac{f(x)}{1+x^2+(f(x))^2}$
$f(x_0) = y_0$


-show this has an unique solution $f \in C(\mathbb{R})$ defined on all of $\mathbb{R}$.
-What is the solution if $y_0=0$?
We're using the Picard-Lindelof Theorem, so we need to show Lipschitz of $F(x,y)$ where $f'(x)=F(x,y)$ in y on the region on which the IVP is defined.  Clearly $F(x,y)$ is continuous on $\mathbb{R}^2$, so I assume we'll just have to show Lipschitz and we're done. This is where I run into a snag, however - how do I show $|F(x,y)-F(x,y_1)| \leq C|y-y_1|$?
 [/INST] You have
$$
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|
\leq
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y^2}\right|
+\left|\frac{y_1}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|\\
=\frac{|y-y_1|}{1+x^2+y^2}+\left|\frac{|y_1|(1+x^2+y_1^2)-|y_1|(1+x^2+y^2)}{(1+x^2+y^2)(1+x^2+y_1^2)}\right|\\
=\frac{|y-y_1|}{1+x^2+y^2}+\frac{|y_1||y_1^2-y^2|}{(1+x^2+y^2)(1+x^2+y_1^2)}\\
=\frac{|y-y_1|}{1+x^2+y^2}+\frac{|y_1||y_1+y|\,|y_1-y|}{(1+x^2+y^2)(1+x^2+y_1^2)}\\
\leq|y-y_1|+|y-y_1|\,\frac{(y_1^2+|y|)}{(1+y_1^2)(1+y^2)}\\
\leq2|y_1-y|
$$
</s>