2.5.2.8 Example 8. \(2x^{2}y^{\prime \prime }\left ( x\right ) -xy^{\prime }\left ( x\right ) +\left ( 1-x^{2}\right ) y\left ( x\right ) =0\)

With expansion around \(x=0\). This is a regular singular ODE. Let

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

Therefore

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

Substituting the above into the ode and simplifying gives

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

The next step is to make all powers of \(x\) to be \(n+r\). This results in

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

The indicial equation is obtained from \(n=0\)

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

The roots are

\begin{align*} r_{1} & =1\\ r_{2} & =\frac {1}{2}\end{align*}

Since the roots differ by non integer, then the solutions are given by

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}=\sum _{n=0}^{\infty }a_{n}x^{n+1}\\ y_{2} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{2}}=\sum _{n=0}^{\infty }a_{n}x^{n+\frac {1}{2}}\end{align*}

We start by finding \(y_{1}\).  EQ. (1) gives for \(r=1\)

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

For \(n=1\) (we skip \(n=0\) as that was used to find \(r\)) and we always let \(a_{0}=1\) as it is arbitrary. Hence

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

For \(n\geq 2\), we have recursion relation. From it we can find that \(a_{2}=\frac {1}{10},a_{3}=0,a_{4}=\frac {1}{360}\) and so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+1}\\ & =a_{0}x+a_{1}x^{2}+a_{2}x^{3}+a_{3}x^{4}+a_{4}x^{5}+\cdots \\ & =x+\frac {1}{10}x^{3}+\frac {1}{360}x^{5}+\cdots \end{align*}

Now we do the same for \(y_{2}\). EQ. (1) now becomes for \(r_{2}=\frac {1}{2}\)

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

For \(n=1\) (we skip \(n=0\) as that was used to find \(r\)) and we always let \(a_{0}=1\) as it is arbitrary. Hence

\begin{align*} 2\left ( 1+\frac {1}{2}\right ) \left ( 1+\frac {1}{2}-1\right ) a_{1}-\left ( 1+\frac {1}{2}\right ) a_{1}+a_{1} & =0\\ a_{1} & =0 \end{align*}

For \(n\geq 2\), we have recursion relation. From it we can find that \(a_{2}=\frac {1}{6},a_{3}=0,a_{4}=\frac {1}{168}\) and so on. Hence

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }a_{n}x^{n+\frac {1}{2}}\\ & =x^{\frac {1}{2}}\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =x^{\frac {1}{2}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =x^{\frac {1}{2}}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end{align*}

Hence the final solution is

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =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 ) \end{align*}