3.1.5 Example 5

$$\begin{array}{rl}{y}^{\prime }& =\sqrt{1-y^2}\\ y\left(0\right)& =1\end{array}$$

$f(x,y)=\sqrt{1-y^2}$ is continuous in $x$ everywhere. For $y$ we want $1-y^2\ge 0$ or $y^2\le 1$. The point $y_0=1$ satisfies this. Now $f_y=\frac{-2y}{2\sqrt{1-y^2}}$. We want $1-y^2>1$ or $y^2<1$. The point $y_0$ does not satisfy this. Hence theory says nothing about uniqueness. Solution can be unique or not. When the ode has form $y^{\prime }=f\left(y\right)$ we always check if IC satisfies the ode. In this case $y\left(x\right)=1$ does satisfy the ode. So this means $y\left(x\right)=1$ is solution. We do not need to solve by integration. But if we did, we will obtain the following

$$\begin{array}{rl}\frac{dy}{\sqrt{1-y^2}}& =dx\\ \arcsin \left(y\right)& =x+c\\ y& =\sin (x+c)\end{array}$$

At initial conditions the above gives $1=\sin c$. Hence $c=\frac{\pi }{2}$. Therefore solution is $y=\sin (x+\frac{\pi }{2})=\cos x$. So this is another solution that satisfies the ode. Solution is not unique.