1.2.7.2.4 Example 4. \(y^{\prime \prime }=\frac {1}{x}\)

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

With expansion around \(x=0\). This is an ordinary point for the ode itself. Let

\[ y\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n}\]

Therefore

\begin{align*} y^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }na_{n}x^{n-1}\\ & =\sum _{n=1}^{\infty }na_{n}x^{n-1}\\ y^{\prime \prime }\left ( x\right ) & =\sum _{n=1}^{\infty }n\left ( n-1\right ) a_{n}x^{n-2}\\ & =\sum _{n=2}^{\infty }n\left ( n-1\right ) a_{n}x^{n-2}\end{align*}

Hence the ode becomes

\begin{equation} \sum _{n=2}^{\infty }n\left ( n-1\right ) a_{n}x^{n-2}=\frac {1}{x} \tag {2}\end{equation}

The solution is given by \(y=y_{h}+y_{p}\), where \(y_{h}\) is solution to

\[ \sum _{n=2}^{\infty }n\left ( n-1\right ) a_{n}x^{n-2}=0 \]

Recursive equation is

\[ n\left ( n-1\right ) a_{n}=0\hspace {0.5in}n\geq 2 \]

Hence all \(a_{n}=0\) for \(n\geq 2\), therefore

\begin{align*} y_{h} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x \end{align*}

Now we need to find \(y_{p}\). From (2), and now we replace \(a_{n}\) by \(c_{n}\)

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

\(n=2\) gives

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

Hence for balance \(0=-1\) which is not possible. Hence no \(y_{p}\) exist using series method. Solution exist by direct integration.