\begin {align*} y^{\prime } & =2\sqrt {y}\\ y\left ( 0\right ) & =0 \end {align*}
First we find the region where solution exists and is unique. Domain of \(f\left ( x,y\right ) =2\sqrt {y}\) is \(y\geq 0\) (since we do not want complex numbers). Since \(y_{0}=0\) is inside this domain, then we know solution exists. The domain of \(f_{y}=\frac {1}{\sqrt {y}}\) is \(y>0\). We see that the region is all \(x\) and \(y>0\). i.e. the top half of the plane not including \(x\)-axis.
Since the point given is \(\left ( 0,0\right ) \) then the theory do not apply. The point \(x_{0},y_{0}\) have to be inside the region and not on the edge.
There is no guarantee that solution will be unique. Solving this ode gives\begin {align*} 2\sqrt {y} & =2x+c\\ \sqrt {y} & =x+c_{1} \end {align*}
At IC\[ 0=c_{1}\] Hence solution is\begin {align*} \sqrt {y} & =x\\ y & =x^{2} \end {align*}
But \(y=0\) is another solution. Notice that \(y=0\) can not be obtained from \(\sqrt {y}=x+c_{1}\) for any choice of \(c_{1}\). So it is a singular solution and not trivial solution. This shows that solution exists but is not unique. In this example, theory predicted that solution exists but did not say anything about uniqueness. Only by solving it, we found the solution is not unique.