\begin {align*} y^{\prime } & =\sqrt {1-y^{2}}\\ y\left ( 0\right ) & =2 \end {align*}
\(f\left ( x,y\right ) =\sqrt {1-y^{2}}\) is continuous in \(x\) everywhere. For \(y\) we want \(1-y^{2}\geq 0\) or \(y^{2}\leq 1\). The point \(y_{0}=2\) does not satisfy. Hence theorem does not apply. We just need any solution that satisfies the ode. Since the ode has form \(y^{\prime }=f\left ( y\right ) \) and not \(y^{\prime }=f\left ( x,y\right ) \) then we always try \(y\left ( x\right ) =y_{0}\) to see if it satisfies the ode. Substituting \(y=2\) into the ode gives
\begin {align*} 0 & =\sqrt {1-y^{2}}\\ & =\sqrt {1-4} \end {align*}
Therefore this solution did not work. In this case we have to solve the ode by integration which gives\begin {align*} \frac {dy}{\sqrt {1-y^{2}}} & =dx\\ \arcsin \left ( y\right ) & =x+c\\ y & =\sin \left ( x+c\right ) \end {align*}
At initial conditions the above gives \(2=\sin c\). Or \(c=\arcsin \left ( 2\right ) \). Hence the solution is\[ y\left ( x\right ) =\sin \left ( x+\arcsin \left ( 2\right ) \right ) \]