Example 3 \[ y^{\prime }=\frac {1}{x}\] Expansion is around \(x=0\). The (homogenous) ode has the form \(y^{\prime }+p\left ( x\right ) y=0\). We see that \(p\left ( x\right ) =0\) is analytic at \(x=0\). However the RHS has no series expansion at \(x=0\) (not analytic there). Therefore we must use Frobenius series in this case.  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*}

The (homogenous) ode becomes\begin {equation} \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}=0 \tag {1} \end {equation} For \(n=0\)\[ ra_{0}x^{r-1}=0 \] Hence \(r=0\) since \(a_{0}\neq 0\). Therefore the ode satisfies\[ y^{\prime }=ra_{0}x^{r-1}\] Eq (1) becomes\begin {align} \sum _{n=0}^{\infty }na_{n}x^{n-1} & =0\nonumber \\ na_{n}x^{n-1} & =0 \tag {2} \end {align}

Therefore for all \(n\geq 1\) we have \(a_{n}=0\). Hence\[ y_{h}=a_{0}\] Now we need to find \(y_{p}\) using the balance equation. From above we found that\[ ra_{0}x^{r-1}=\frac {1}{x}\] Changing \(r\) to \(m\) and \(a_{0}\) to \(c_{0}\) so not to confuse notation gives\[ mc_{0}x^{m-1}=x^{-1}\] Hence \(m-1=-1\) or \(m=0\). Therefore there is no solution for \(c_{0}\). Unable to find \(y_{p}\) therefore no series solution exists. Asymptotic methods are needed to solve this. Mathematica AsymptoticDSolveValue gives the solution as \(y\left ( x\right ) =c+\ln x\).