$y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem 
$y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$
are:
$y_1=1-x$
$y_2={{-x^2}\over 4}$
And explain why the two solutions not contradiction with the theory of existence and uniqueness.
( sorry I don't speak English well )
My answer:
For the initial condition :
$y_1=1-x$ Satisfy the initial condition
$y(2)=-1$
$y_1=1-2=-1$ 
And Satisfy the differential equation
${y_1}^{\prime}=-1$
${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$
$={{-x+\sqrt{(x-2)^2}}\over 2}$
$={{-x+x-2}\over 2}=-1$
L.h.s.=R.h.s.
$y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$
Also $y_2$ Satisfy the initial condition
$y_2={{-x^2}\over4}={-4\over4}=-1$


*

*And satisfy the differential equation.


$y_2^{\prime}={-x\over 2}$
${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$
And for second part of exercise:
${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$
and this not defined when $x=2, \ y=-1$
So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness.
True ?
If this also wrong I well be delet it , I answered by the same way of my teacher :(

Let $u=\sqrt{x^2+4y}$ ,
Then $y=\dfrac{u^2-x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{u}{2}$ with $u(2)=0$
$\dfrac{du}{dx}=1$ or $u=0$ with $u(2)=0$
$u=x-2$ or $u=0$
$\sqrt{x^2+4y}=x-2$ or $\sqrt{x^2+4y}=0$
$x^2+4y=x^2-4x+4$ or $x^2+4y=0$
$4y=-4x+4$ or $4y=-x^2$
$y=1-x$ or $y=-\dfrac{x^2}{4}$
