Question on an interval that makes the solution unique http://www.math.uiuc.edu/~tyson/existence.pdf (example 5)
In the example, the solution for the interval is $\delta_2 = |x_0|$. I'm confused on how they got that. What was the algebraic step to solve for $\delta_2$? I mean I can reason it out, and it makes sense. However, how do I actually solve for it algebraically? Thank you! 
 A: To satisfy conditions for uniqueness, solutions in this example have to stay away from $0$. Since the distance from a point $x_0\ne0$ to $0$ is $|x_0|$, an interval on which a unique solution exists cannot go by more than $\delta_2=|x_0|$ away from $x_0$.
A: We know that an ODE of the form $$y' = F(x,y),\quad y(x_{0}) = y_{0}$$ is guaranteed a unique solution when both $F(x,y)$ and $F_{y}(x,y)$ are continuous.  For Example $5$, $$y' = 2y/x,\quad y(x_{0}) = y_{0},$$ we see that both $$F(x,y) = 2y/x\quad \text{and}\quad F_{y}(x,y) = 2/x$$ are continuous so long as $x\neq 0$.  That means that first of all we get an interval of uniqueness for some $\delta_{2} > 0$: $$ x_{0} - \delta_{2} < x < x_{0} + \delta_{2}.$$  Because we need to keep $x$ away from $0$, and we don't know if $x_{0}$ is greater than or less than $0$ we have two cases:
First, if $x_{0} > 0$ we can take $$x_{0} - \delta_{2} = 0 \implies \delta_{2} = x_{0}.$$
Second, if $x_{0} < 0$ we can take $$x_{0} + \delta_{2} = 0 \implies \delta_{2} = |x_{0}|.$$
In either case, it suffices to take $\delta_{2} = |x_{0}|$ to guarantee uniqueness. 
