3.1.1.6 Example 6

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

$f(x,y)=\sqrt{1-y^2}+x$ 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 does not apply.

In this case the ode has form $y^{\prime }=f(x,y)$ and not $y^{\prime }=f\left(y\right)$. So we can not just check if initial conditions satisfies the ode and use that as solution. If we did, we see that $y\left(x\right)=1$ does satisfy the ode at $x=0$ but this will be wrong solution. In this case we have to go ahead and solve the ode. In this case we will find that no general solution exists.