Example 3

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

In standard form \(y^{\prime }-p\left ( x\right ) y=q\left ( x\right ) \). So \(p=\frac {-1}{x},q=0\). The domain of \(p\) is all \(x\) except \(x=0\). Domain of \(q\) is all \(x\). Since IC does not include \(x=0\) then solution is guaranteed to exist and be unique in some region near \(x=1\). Solving gives

\[ y=cx \]

As solution. Applying I.C. gives

\[ 0=c \]

Hence the unique solution is

\[ y=0\qquad x>0 \]

Solution exists and is unique. Solution can only be in the right hand plan which includes \(x=1\) and it can not cross \(x=0\). i.e. solution is \(y=0\) for all \(x>0\).  If IC was \(y\left ( -1\right ) =0\) then the solution would have been \(y=0\) for all \(x<0\).