1.2.2.2.6 Example 6 \(x^{2}y^{\prime \prime }+xy^{\prime }+xy=\frac {1}{x}\)

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=\frac {1}{x}\]

This is the same ode as above but with different RHS. So we will go directly to finding \(y_{p}\). From above we found that the balance equation is

\[ c_{0}r^{2}x^{r}=x^{-1}\]

Which implies \(r=-1\) and therefore \(r^{2}c_{0}=1\) or \(c_{0}=1\). Using the recurrence equation (1) in the above problem using using \(c_{n}\) in place of \(a_{n}\) gives

\[ c_{n}=-\frac {c_{n-1}}{\left ( n+r\right ) ^{2}}\]

For \(m=-1\)

\[ c_{n}=-\frac {c_{n-1}}{\left ( n-1\right ) ^{2}}\]

Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =\frac {1}{x}\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

Now to find few \(c_{n}\) terms. For \(n=1\)

\begin{align*} c_{1} & =-\frac {c_{0}}{\left ( 1-1\right ) ^{2}}\\ & =\infty \end{align*}

Hence series does not converge. No \(y_{p}\) exist. There is no solution in terms of series solution.