Nonlinear differential equation existence and uniqueness theorem

Question

What is the nonlinear differential equation existence and uniqueness theorem? I have two solutions to a nonlinear first-order differential equation that both satisfy the same initial value and satisfy the differential equation. They are also both valid for all real numbers. I am told, however, that despite this, the two solutions do not contradict the nonlinear existence and uniqueness theorem. How could this be?

For reference, the differential equation is $dy/dx = \frac{1}{2}(-x+(x^2+4y)^{\frac12})$, and the initial value given is $y(2) = -1$. The two solutions I have are $y = 1-x$ and $y=-x^2/4$.

Answer 1

EDIT

Observe that $y^{\prime}$ is the solution of the following equation \begin{align*} (y^{\prime})^{2} + xy^{\prime} - y = 0 \Longleftrightarrow y^{\prime} = \frac{-x\pm\sqrt{x^{2}+4y}}{2} \end{align*}

Answer 2

At the points where the square root takes the value zero, the assumption for uniqueness is no longer satisfied, as the $y$ derivative goes to infinity and thus there can not be a finite local Lipschitz constant.

Note that the the equation resulting from eliminating the square root, $$ (x+2y')^2=x^2+4y\implies y=xy'+(y')^2 $$ is a Clairaut differential equation that has a solution family of lines $$y=cx+c^2$$ and a singular solution that is the envelope of the lines given by $x+2y'=0$ or $$y=-\frac{x^2}4.$$ Apart from these solutions one can combine pieces of the lines and the singular solution.

The line that is tangent to the singular solution at $x=2$, $y=-1$ is indeed the one with parameter $c=-1$, $y=-x+1$.