Consider the following Cauchy problem:
$$ \frac{\partial y}{\partial t} = \frac{-t + \sqrt{t^2+4y}}{2} \\ y(2) = -1 $$
I propose two valid solutions:
$$ y_1(t) = 1 - t \\ y_2(t) = \frac{-t^2}{4} $$
Why does this not contradict the Existence and Uniqueness Theorem related to this kind of questions?
On
The equation is one of the solution of the equation $$y'(t)^2+t y'(t)-y(t)=0\quad\quad(1)$$ which gives $$y_1'(t)=\frac{-t+\sqrt{t^2+4y}}{2}\quad\quad(2)$$ and $$y_2'(t)=\frac{-t-\sqrt{t^2+4y}}{2}\quad\quad(3)$$ differentiate $(1)$ wrt $t$ $$t y''(t)+2 y'(t) y''(t)=0$$ Collect $y''(t)$ $$y''(t)(t+2y'(t))=0$$ we get two solutions $$y_1(t)=1-t$$ and $$y_2(t)=-\frac{t^2}{4}$$ but solution $y_1(t)$ doesn't satisfy the given equation $(2)$ while $y_2(t)$ does.
Thus the unique solution of the given equation is $y_2(t)=-\frac{t^2}{4}$.
The existence an uniqueness theorems (like the Picard-Lindelof Theorem) have conditions... If the conditions are not met there is no contradiction.
edit: It was noted in comments to the OP that, strictly speaking, $y_1$ is in not a solution of the Cauchy problem, even though it is a solution of the forward problem ($t>2$).