6.7 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

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

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 assume it has power series

\[ y_{p}=\sum _{n=0}^{\infty }c_{n}x^{n+r}\]

Substituting the above into the ode and simplifying as we did in first example gives EQ. 3 in the first example, but now with \(\sin x\) on RHS

\[ \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r}+\sum _{n=0}^{\infty }c_{n}x^{n+r}=x^{-1}\]

For \(n=0\) the above becomes

\begin{align*} rc_{0}x^{r}+c_{0}x^{r} & =x^{-1}\\ \left ( r+1\right ) c_{0}x^{r} & =x^{-1}\end{align*}

Hence \(r=-1\) and \(\left ( r+1\right ) c_{0}=1\) which gives \(0c_{0}=1\). So not possible to solve for \(c_{0}\). Since we can not find \(c_{0}\), 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.