Example 4 \[ y^{\prime }=\frac {1}{x^{2}}\] 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 balance equation is \[ ra_{0}x^{r-1}=\frac {1}{x^{2}}\] Or by changing \(r\) to \(m\) and \(a_{0}\) to \(c_{0}\) so not to confuse notation with \(y_{h}\) gives\begin {equation} mc_{0}x^{m-1}=x^{-2} \tag {2} \end {equation} Eq (1) becomes, where \(r=0\) now\begin {align} \sum _{n=0}^{\infty }na_{n}x^{n-1} & =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 all \(n\geq 1\) we see from (2) that \(a_{n}=0\). Hence \[ y_{h}=c_{1}\left ( a_{0}+O\left ( x\right ) \right ) \] Letting \(a_{0}=1\) the above becomes\[ y_{h}=c_{1}\left ( 1+O\left ( x\right ) \right ) \] Now we need to find \(y_{p}\) using the balance equation. From (2) above we found that\[ mc_{0}x^{m-1}=x^{-2}\] To balance, we need \(m-1=-2\) or \(m=-1\) and \(mc_{0}=1\) or \(c_{0}=-1\). Therefore \[ y_{p}=x^{m}\sum _{n=0}^{\infty }c_{0}x^{n}\] Where \(c_{0}=-1\) and all \(c_{n}\) for \(n\geq 1\) are found using the recurrence relation from finding \(y_{h}\). But from above we found that all \(a_{n}=0\) for \(n\geq 1\). Hence \(c_{n}=0\) also for \(n\geq 1\). Therefore\begin {align*} y_{p} & =x^{m}c_{0}\\ & =\frac {-1}{x}+O\left ( x^{2}\right ) \end {align*}

Hence the solution is \begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1+O\left ( x^{2}\right ) \right ) +\left ( \frac {-1}{x}+O\left ( x^{2}\right ) \right ) \end {align*}

If we to ignore the big \(O\), the above becomes\[ y=c_{1}-\frac {1}{x}\] To verify, we see that \(y^{\prime }=\frac {1}{x^{2}}\).