Example: $f(x,y)$ is not Lipschitz in y but still has a unique solution to initial value problem Consider the differential equation $y^\prime =\sqrt{y}+1$ with $y(0)=0$. My question is how to prove the uniqueness of the solution. 
I found a hint for it as follows:
Consider $z(x)=(\sqrt{y_1(x)}-\sqrt{y_2(x}))^2$ where $y_1,y_2$ are solutions.
I'm trying to use the hint to prove the uniqueness of solution, but I do not know what the hint means. 
My attempt:
$z'(x)=2(\sqrt{y_1(x)}-\sqrt{y_2(x)}[(\sqrt{y_1(x)})'-(\sqrt{y_2(x)})']=(\sqrt{y_1(x)}-\sqrt{y_2(x)})\Big(\frac{y_1'(x)}{\sqrt{y_1(x)}}-\frac{y_2'(x)}{\sqrt{y_2(x)}}\Big)=(y_1'(x)-y_2'(x))\Big(\frac{y_1'(x)}{\sqrt{y_1(x)}}-\frac{y_2'(x)}{\sqrt{y_2(x)}}\Big)$.
I couldn't proceed it further. Please give me slightly more comment or hint for the problem.
Thanks in advance!
 A: Here is an extended hint, followed by a more complete answer: After fully simplifying $z'$ using the differential equations that $y_1$ and $y_2$ satisfy, conclude that $z'(x) \le 0$ for all $x \ne 0$. 
So $z$ is a function satisfying


*

*$z(0) = 0$

*$z(x) \ge 0$

*$z'(x) \le 0$ almost everywhere.


What conclusions can you draw?
Spoilers below.

Notice that

 $$z' = (\sqrt{y_1} - \sqrt{y_2})\left(\frac{y_1'}{\sqrt{y_1}} - \frac{y_2'}{\sqrt{y_2}}\right)$$

After a simplification, the numerator of the term in parentheses is

 $$y_1' \sqrt{y_2} - y_2' \sqrt{y_1} = (\sqrt{y_1} + 1)\sqrt{y_2} - (\sqrt{y_2} + 1)\sqrt{y_1} = \sqrt{y_2} - \sqrt{y_1}$$

and we can conclude that

 $$z' = -\frac{(\sqrt{y_1} - \sqrt{y_2})^2}{\sqrt{y_1 y_2}} \le 0.$$

Therefore, $z \equiv 0$ because

 $z$ is a decreasing nonnegative function which is already zero at zero.

A: Your equation is a separable ODE, with nonzero (and continuous) right-hand side.  For such equations one has local uniqueness (see Existence and uniqueness for separable ODEs where $y'$ diverges, in particular Julián Aguirre's comment, for a proof).
One (very slight) difficulty is that in your example the domain of the rhs is $(-\infty, \infty) \times [0, \infty)$, whereas in (the standard formulation of) the above theorem the domain is the Cartesian product of open intervals.  But you can extend the equation to the whole $\mathbb{R}^2$ by putting $y' = \sqrt{\lvert y \rvert} + 1$.
