1.1.3.7 Example 7. \(xy^{\prime }+y=\frac {1}{x}\)

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

This is same as example 1, but with nonzero on RHS. The solution is \(y=y_{h}+y_{p}\). Where \(y_{h}\) was found above as

\[ y_{h}=\frac {a_{0}}{x}\]

To find \(y_{p}\) we will use the balance equation, EQ (*) found in the first example when finding \(y_{h}\). We just need to rename \(a_{0}\) to \(c_{0}\) and add the \(x\) on the right side of the balance equation.

\[ \left ( r+1\right ) c_{0}x^{r}=\frac {1}{x}\]

Hence \(r=-1\). Therefore \(\left ( r+1\right ) c_{0}=1\) or \(0c_{0}=1\). No solution exist. Can not find \(y_{p}\). This is an example where there is no series solution. This ode of course can be easily solved directly which gives solution \(y=\frac {c_{1}}{x}+\frac {1}{x}\ln x\), but not using series method. The next problems shows that when changing \(\frac {1}{x}\) to \(\frac {1}{x^{2}}\) then the balance equation is able to find \(c_{0}\).