3.1.1.8 Example 8

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

By Existence and uniqueness, we see $f(x,y)$ is not defined at $y_0=0$. Hence theorem does not apply. Since ode has form $y^{\prime }=f\left(y\right)$ we now check if IC satisfies the ode itself. Plugging in $y=0$ into the ode is not satisfied due to $\frac{1}{0}$. So we have to solve the ode in this case. integrating gives

$$\begin{array}{rl}\int ydy& =\int dx\\ \frac{1}{2}{y}^2& =x+c\end{array}$$

At IC this gives

$$\begin{array}{rl}0& =1+c\\ c& =-1\end{array}$$

Hence solution is

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

We see solution is not unique.