2.5.2.9 Example 9. \(2x^{2}y^{\prime \prime }\left ( x\right ) -xy^{\prime }\left ( x\right ) +\left ( 1-x^{2}\right ) y\left ( x\right ) =1+x\)

With expansion around \(x=0\). This is a regular singular ODE. This is same ode solved in example 1, but now with non-zero on the right side. Hence solution is given by

\[ y=y_{h}+y_{p}\]

Where we found \(y_{h}\) to be

\[ y_{h}=c_{1}\left ( x+\frac {1}{10}x^{3}+\frac {1}{360}x^{5}+\cdots \right ) +c_{2}\left ( x^{\frac {1}{2}}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \right ) \]

To find \(y_{p}\), we see there are two terms on the right side. we find \(y_{p_{1}}\) that corresponds to \(1\) on the right side and then \(y_{p_{2}}\) that corresponds to \(x\) on the right side. We will find that \(y_{p_{2}}\) does not exist. Hence no solution exist using series. Let us now find \(y_{p_{1}}\). The ode now is

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

Let

\[ y_{p_{1}}=\sum _{n=0}^{\infty }c_{n}x^{n+r}\]

Using same expansion as in example 1, but now using \(c_{n}\) instead of \(a_{n}\) gives EQ. (1) from example 1 as

\begin{equation} \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r}-\sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r}+\sum _{n=0}^{\infty }c_{n}x^{n+r}-\sum _{n=2}^{\infty }c_{n-2}x^{n+r}=1 \tag {1A}\end{equation}

For \(n=0\)

\begin{align*} 2r\left ( r-1\right ) c_{0}x^{r}-rc_{0}x^{r}+c_{0}x^{r} & =x^{0}\\ \left ( 2r\left ( r-1\right ) c_{0}-rc_{0}+c_{0}\right ) x^{r} & =x^{0}\\ \left ( 2r\left ( r-1\right ) -r+1\right ) c_{0}x^{r} & =x^{0}\\ \left ( 2r^{2}-3r+1\right ) c_{0}x^{r} & =x^{0}\end{align*}

We see that for balance, then \(r=0\). Which implies \(\left ( 2r^{2}-3r+1\right ) c_{0}=1\) or \(c_{0}=1\,.\) Hence

\[ y_{p_{1}}=\sum _{n=0}^{\infty }c_{n}x^{n}\]

Now that we used \(n=0\) to find \(r\) by matching against the RHS which is \(1\), from now on, for all \(n>0\) we will use (1A) to solve for all other \(c_{n}\). From now on, we just need to solve with RHS zero, since there can be no more matches for any \(x^{n}\) on the right.

Then (1A) becomes (using \(r=0\))

\begin{equation} \sum _{n=0}^{\infty }2n\left ( n-1\right ) c_{n}x^{n}-\sum _{n=0}^{\infty }nc_{n}x^{n}+\sum _{n=0}^{\infty }c_{n}x^{n}-\sum _{n=2}^{\infty }c_{n-2}x^{n}=0 \tag {2}\end{equation}

For \(n=1\) the above gives

\[ -c_{1}x+c_{1}x=0 \]

\(c_{1}\) is arbitrary Let \(c_{1}=0\). For \(n=2\) EQ. (2) gives

\begin{align*} 4c_{2}x^{2}-2c_{2}x^{2}+c_{2}x^{2}-c_{0}x^{2} & =0\\ \left ( 4c_{2}-2c_{2}+c_{2}-c_{0}\right ) x^{2} & =0\\ \left ( 3c_{2}-1\right ) x^{2} & =0 \end{align*}

Hence \(3c_{2}-1=0\) since there is no \(x^{2}\) term on the right side. Hence \(c_{2}=\frac {1}{3}\). For \(n=3\) EQ. (2) gives

\begin{align*} \left ( 12c_{3}-3c_{3}+c_{3}-c_{1}\right ) x^{3} & =0\\ \left ( 10c_{3}\right ) x^{3} & =0 \end{align*}

Hence \(c_{3}=0\). For \(n=4\) EQ. (2) gives

\begin{align*} 24c_{4}x^{4}-4c_{4}x^{4}+c_{4}x^{4}-c_{2}x^{4} & =0\\ \left ( 21c_{4}-\frac {1}{3}\right ) x^{4} & =0 \end{align*}

Hence \(21c_{4}-\frac {1}{3}=0\) or \(c_{4}=\frac {1}{63}\) and so on. We find that

\begin{align*} y_{p_{1}} & =\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+\cdots \\ & =1+\frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\cdots \end{align*}

Now we find \(y_{p_{2}}\). EQ. (1A) now is

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

For \(n=0\)

\begin{align*} 2\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r}-\left ( n+r\right ) c_{n}x^{n+r}+c_{n}x^{n+r} & =x\\ 2r\left ( r-1\right ) c_{0}x^{r}-rc_{0}x^{r}+c_{0}x^{r} & =x\\ \left ( 2r\left ( r-1\right ) -r+1\right ) c_{0}x^{r} & =x\\ \left ( 2r^{2}-3r+1\right ) c_{0}x^{r} & =x \end{align*}

For balance we need \(r=1\). This results in

\begin{align*} \left ( 2r^{2}-3r+1\right ) c_{0} & =1\\ \left ( 2-3+1\right ) c_{0} & =1\\ 0c_{0} & =1 \end{align*}

Not possible. We see why there is no series solution. It is not possible to solve for \(y_{p2}\,\).