Example 4

Solve

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

We see that \(f=\frac {y}{x}\) is not continuous at \(x=0\). Hence by uniqueness and existence theorem, there is no guarantee that solution exist. (Notice we do not say that no solution exist, as there might be one, but there is no guarantee that one exists using the theorem). Integrating gives

\begin{align*} \int \frac {dy}{y} & =\int \frac {1}{x}dx\qquad y\neq 0\\ \ln y & =\ln x+c\\ y & =cx \end{align*}

Applying IC gives \(1=0\), hence no solution exist. When no solution exist, we do not need to consider singular solutions.