Example 5 \[ y^{\prime }+\frac {y}{x}=0 \] Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) =\frac {1}{x}\) is not analytic at \(x=0\) but \(\lim _{x\rightarrow 0}xp\left ( x\right ) =0\) is analytic. Therefore we must use Frobenius series in this case.  Let \begin {align} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\tag {A}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\nonumber \end {align}

The ode becomes\begin {align} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\frac {1}{x}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) a_{n}+a_{n}\right ) x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r+1\right ) a_{n}x^{n+r-1} & =0 \tag {1} \end {align}

For \(n=0\)\[ \left ( r+1\right ) a_{0}=0 \] Hence \(r=-1\) since \(a_{0}\neq 0\). Eq (1) becomes, where \(r=-1\) now\begin {align} \sum _{n=0}^{\infty }na_{n}x^{n} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2} \end {align}

\(n=0\) is not used since that was used to find \(r\). Therefore we start from \(n=1\). For \(n=1\) the above gives \(a_{1}=0\) and same for all \(n\geq 1\). Hence from Eq (A), since \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\) then (note: When there is only one \(\sum \) term left in (1) as in this case, then this means there is no recurrence relation and all \(a_{n}=0\) for \(n>0\)).

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

Letting \(a_{0}=1\) the above becomes\[ y=c_{1}\left ( x^{-1}+O\left ( x\right ) \right ) \]