3.1.1.8 Example 8

\begin{align*} y^{\prime } & =\frac {1}{y}\\ y\left ( 1\right ) & =0 \end{align*}

By Existence and uniqueness, we see \(f\left ( x,y\right ) \) 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{align*} \int ydy & =\int dx\\ \frac {1}{2}y^{2} & =x+c \end{align*}

At IC this gives

\begin{align*} 0 & =1+c\\ c & =-1 \end{align*}

Hence solution is

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

We see solution is not unique.