2.4.1.2 Example 2 \(x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=1\)

\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=1 \]

This is same as last example but with non zero on right side. We can find \(y_{h}\) as

\[ y_{h}=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) +c_{2}\left ( -\frac {1}{12}\left ( x^{2}-\frac {1}{12}x^{4}+\frac {1}{384}x^{6}+\cdots \right ) \ln \left ( x\right ) +x^{-2}\left ( 1+\frac {1}{4}x^{2}-\frac {1}{288}x^{6}+\cdots \right ) \right ) \]

Let solution be \(y=y_{h}+y_{p}\). We just need to find \(y_{p}\). Assuming

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

Substituting this into the ode gives EQ. (2) in the above example but with \(c_{n}\) replacing \(a_{n}\) and with \(1\) now on the right side

\[ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) c_{n}x^{n+r}+\sum _{n=2}^{\infty }c_{n-2}x^{n+r}=1 \]

For \(n=0\)

\[ \left ( r\left ( r-1\right ) +r-4\right ) c_{0}x^{r}=1 \]

For balance we need \(r=0\). Hence the coefficient is

\begin{align*} \left ( r\left ( r-1\right ) +r-4\right ) c_{0} & =1\\ -4c_{0} & =1\\ c_{0} & =-\frac {1}{4}\end{align*}

Hence

\begin{align*} y_{p} & ={\displaystyle \sum \limits _{n=0}^{\infty }} c_{n}x^{n}\\ y_{p}^{\prime } & ={\displaystyle \sum \limits _{n=0}^{\infty }} nc_{n}x^{n-1}\\ & ={\displaystyle \sum \limits _{n=1}^{\infty }} nc_{n}x^{n-1}\\ y_{p}^{\prime \prime } & ={\displaystyle \sum \limits _{n=1}^{\infty }} n\left ( n-1\right ) c_{n}x^{n-2}\\ & ={\displaystyle \sum \limits _{n=2}^{\infty }} n\left ( n-1\right ) c_{n}x^{n-2}\end{align*}

Substituting this into the ode gives

\begin{align*} x^{2}{\displaystyle \sum \limits _{n=2}^{\infty }} n\left ( n-1\right ) c_{n}x^{n-2}+x{\displaystyle \sum \limits _{n=1}^{\infty }} nc_{n}x^{n-1}+\left ( x^{2}-4\right ) {\displaystyle \sum \limits _{n=0}^{\infty }} c_{n}x^{n} & =1\\{\displaystyle \sum \limits _{n=2}^{\infty }} n\left ( n-1\right ) c_{n}x^{n}+{\displaystyle \sum \limits _{n=1}^{\infty }} nc_{n}x^{n}+{\displaystyle \sum \limits _{n=0}^{\infty }} c_{n}x^{n+2}-4{\displaystyle \sum \limits _{n=0}^{\infty }} c_{n}x^{n} & =1 \end{align*}

Adjust all powers to lowest gives

\[{\displaystyle \sum \limits _{n=2}^{\infty }} n\left ( n-1\right ) c_{n}x^{n}+{\displaystyle \sum \limits _{n=1}^{\infty }} nc_{n}x^{n}+{\displaystyle \sum \limits _{n=2}^{\infty }} c_{n-2}x^{n}-{\displaystyle \sum \limits _{n=0}^{\infty }} 4c_{n}x^{n}=1 \]

For \(n>0\) the right side is zero since no balance (no \(x^{n}\) terms with \(n>0\)). The above can then be written as

\[{\displaystyle \sum \limits _{n=2}^{\infty }} n\left ( n-1\right ) c_{n}x^{n}+{\displaystyle \sum \limits _{n=1}^{\infty }} nc_{n}x^{n}+{\displaystyle \sum \limits _{n=2}^{\infty }} c_{n-2}x^{n}-{\displaystyle \sum \limits _{n=0}^{\infty }} 4c_{n}x^{n}=0 \]

For \(n=1\)

\begin{align*} c_{1}-4c_{1} & =0\\ c_{1} & =0 \end{align*}

The recursive relation is for \(n\geq 2\) gives

\begin{align} n\left ( n-1\right ) c_{n}+nc_{n}+c_{n-2}-4c_{n} & =0\nonumber \\ c_{n} & =\frac {-c_{n-2}}{n\left ( n-1\right ) +n-4} \tag {1}\end{align}

For \(n=2\)

\begin{align*} c_{2} & =\frac {-c_{0}}{n\left ( n-1\right ) +n-4}\\ & =\frac {\frac {1}{4}}{0}\\ & =\infty \end{align*}

The series does not converge. Hence no solution in series exist.