Contradiction of the Uniqueness Theorem So I have a differential equation $$\frac{\mathrm{d}y}{\mathrm{d}t}= 2 \sqrt{|y|}$$ with $y(0)=0$. So, I found that the solution to this ODE is $y=(t+c)^2$ and $y=-(t+c)^2$. Doesn't this Differential equation contradict the Uniqueness theorem?
 A: Let $c \ge 0$. Here is a family of solutions all satisfying $y_c(0) = 0$: $y_c (t) = \begin{cases} 0 , & t < c \\
(t-c)^2, & t \ge c \end{cases}$.
The usual proof of uniqueness depends on the right hand side being Lipschitz,
which fails here.
A: Firstly, we have the stipulation that $y(0) = 0$, and so we obtain $c=0$ in both cases. So your question is really "why are $y=\pm t^2$ both solutions?".
Well, the simple answer is that they're not both solutions. Let $y=t^2$. Then $$\mathrm{LHS} = \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2t$$ while $$\mathrm{RHS} = 2 \sqrt{|y|} = 2 |t|$$

As an aside, for completeness: the uniqueness theorem I know for first-order ODEs in general is:

Let $\phi: \mathbb{R}^2 \to \mathbb{R}$ be continuous, and Lipschitz in the second variable. Let $y_0, t_0 \in \mathbb{R}$. Then there is a ball around $t_0$ such that there is a unique solution $y$ to the DE $$y'(t) = \phi(t, y(t)); y(t_0) = y_0$$ in that ball.

This theorem doesn't apply to your ODE because $\phi: (t, y) \mapsto 2 \sqrt{|y|}$ is not Lipschitz in the variable $y$.
