6.1 Example 1. \(xy^{\prime }+y=0\)

\begin{equation} xy^{\prime }+y=0\tag {1}\end{equation}

With expansion around \(x=0\). Since \(x\) is regular singular point, then let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\end{align*}

Then (1) becomes

\begin{align} x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\tag {2}\\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\tag {3A}\end{align}

All \(x\) in sums start from same index, so there is no adjustments needed. For \(n=0\), EQ (3) gives

\begin{align*} ra_{0}x^{r}+a_{0}x^{r} & =0\\ \left ( r+1\right ) a_{0}x^{r} & =0 \end{align*}

But \(a_{0}\neq 0\), hence \(r+1=0\) or

\[ r=-1 \]

Therefore (3) becomes

\begin{equation} \sum _{n=0}^{\infty }\left ( n-1\right ) a_{n}x^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n-1}=0\tag {3B}\end{equation}

For \(n>0\) (because \(n=0\) was used to find \(r\)) EQ (4) gives the recursive relation

\begin{align} \left ( n-1\right ) a_{n}+a_{n} & =0\nonumber \\ na_{n} & =0\tag {3C}\end{align}

We see that for all \(n>0\) then \(a_{n}=0\). Hence solution is for \(n=0\) only, which is

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ & =a_{0}x^{-1}\\ & =\frac {a_{0}}{x}\end{align*}