Existence and uniqueness theorem - ODE solutions So we have an ODE namely $$\frac{dy}{dx} =\frac{-x+\sqrt{x^2+4y}}{2}, \ y(2)=-1 $$. 
Ok so we have two solutions $y_1=1-x$ which is valid for $x\geq 2 $ and $y_2=-x^2/4$ which seems to be valid for $x\in \mathbb{R} $. 
Now a question asks how the existence of two solutions to the initial value problem does not contradict the existence and uniqueness theorem. 
Anyone have a simple explanation?
 A: Check the Lipschitz continuity at the initial point.
You could also employ the substitution $z=y+\frac14x^2$,
$$
z'=y'+\frac12x=\sqrt{z}
$$
to reduce the problem to a well-known example.
A: The theorem as it's name implies guarantees there exist a unique general solution for the IVP. This can either be applied to the  maximal interval of existence or to local existence and uniqueness, which doesn't contradict each other because we're essentially considering a different problem by restricting the domain. 
I hope this helps!
A: The $\pm$ seems to way lay the path, better to unify it, so we can trace it backwards for uniqueness and existence of either of them in a compound ODE.
$$ y^{\prime2} + xy^{\prime} - y= 0 \tag{1}$$
Solve for $ y^{\prime}$
$$\frac{dy}{dx} =\frac{-x \pm \sqrt{x^2+4y}}{2} \tag{2}  $$
Consider both the cases. Differentiate (1) and simplify
$$ y^{\prime \prime} \, (x + 2 y^{\prime}) = 0 \tag{3}  $$
with solutions 
$$ y= C_1 x + C_2,\quad y= - x^2/4 + C_3 \tag{4} $$
With given boundary condition $ (x,y)=( 2,-1) $ we have
$$  C_2 = -1-2 C_1, \quad C_3 =0, $$
$$ y_1 = C_1 x -1- 2\, C_1 = C_1 (x-2) -1, \quad y_2= x^2/4. \tag{5} $$
$C_1=-1 $ in your first case. So there is no contradiction of E&U between the two different solutions as above.
