Uniqueness of OdE Solution, with 0 as one of the solutions

Question

I am totally confused, take the ODE $$u-tu'-(u')^2=0$$ Clearly $-t^2/4$ and $ct+c^2$ are both solutions, and if we take $c = 0$, then they both satisfy the initial condition $u(0) = 0$.

Why does this not contradict uniqueness?

Thank you!

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Addendum: The Existence and Uniqueness Theorem.

Consider the initial value problem $X' = F(X)$, $X(0) = X_0$ where $X_0 ∈ \Bbb R^n$. Suppose that $F : \Bbb R^n → \Bbb R^n$ is $C^1$. Then there exists a unique solution of this initial value problem. More precisely, there exists $a > 0$ and a unique solution $X : (−a, a) → \Bbb R^n$ of this differential equation satisfying the initial condition $X(0) = X_0$.

Answer 1

From your cited version of the existence and uniqueness theorem you see that it only applies to explicit differential equations $$X'(t)=F(t,X(t)).$$

The equation you consider is not explicit. You can solve it in two different ways to be explicit by the sign choices of the square root below. Even then, $$ u'=f(t,u)=-\frac t2\pm\sqrt{u+\frac{t^2}4} $$ is neither differentiable nor Lipschitz at $u=-\frac{t^2}4$, so you can not expect uniqueness there.