2.2.2.0 Example 9 \(\sin \left ( x\right ) y^{\prime \prime }+y^{\prime }+y=0\)

\[ \sin \left ( x\right ) y^{\prime \prime }+y^{\prime }+y=0 \]

Comparing the ode to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Hence \(p\left ( x\right ) =\frac {1}{\sin \left ( x\right ) },q\left ( x\right ) =\frac {1}{\sin x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x}{x-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=\frac {1}{1-\frac {x}{3!}+\frac {x^{4}}{5!}-}=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x^{2}}{x-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=\frac {x}{1-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=0\). Hence the indicial equation is

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

Therefore \(r_{1}=0,r_{2}=0\).  Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

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

The ode becomes

\begin{align*} \sin \left ( x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \left ( x-\frac {x^{3}}{3!}+\frac {x^{5}}{5!}-\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

Using \(O\left ( x^{7}\right ) \) terms as the Order of the series (if more terms are needed we will use more terms from the \(\sin x\) series). This means we have to now only expand up to \(n=7\) as that is the order used for the series of \(\sin x\). The above becomes

\begin{multline*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-\frac {x^{3}}{3!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\\ +\frac {x^{5}}{5!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

Which becomes

\begin{multline*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+1}\\ +\sum _{n=0}^{\infty }\frac {1}{120}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+3}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

Re indexing to lowest powers on \(x\) gives

\begin{multline*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}x^{n+r-1}\\ +\sum _{n=4}^{\infty }\frac {1}{120}\left ( n+r-4\right ) \left ( n+r-5\right ) a_{n-4}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \end{multline*}

Simplifying gives

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{6}a_{n-2}x^{n+r-1}+\sum _{n=4}^{\infty }\frac {\left ( n+r-4\right ) \left ( n+r-5\right ) }{120}a_{n-4}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \tag {1}\end{equation}

The indicial equation is obtained from \(n=0\). The above reduces to

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

Since \(a_{0}\neq 0\) then

\[ r^{2}=0 \]

Hence \(r_{1}=0,r_{2}=0\) as found earlier. Since the roots are repeated then two linearly independent solutions can be constructed using

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

\(n=1\) gives from (1) and by taking \(a_{0}=1\)

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

For \(n=2\) gives from (1)

\begin{align*} \left ( 2+r\right ) ^{2}a_{2}-\frac {\left ( r\right ) \left ( r-1\right ) }{6}a_{0}+a_{1} & =0\\ \left ( 2+r\right ) ^{2}a_{2} & =-a_{1}+\frac {\left ( r\right ) \left ( r-1\right ) }{6}a_{0}\\ a_{2} & =\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}\end{align*}

For \(n=3\)

\begin{align*} \left ( 3+r\right ) ^{2}a_{3}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6}a_{1}+a_{2} & =0\\ a_{3} & =-\frac {a_{2}}{\left ( 3+r\right ) ^{2}}+\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}}{\left ( 3+r\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}}\frac {1}{\left ( 1+r\right ) ^{2}}\\ & =-\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}\end{align*}

For \(n\geq 4\) the recurrence relation is

\[ \left ( n+r\right ) ^{2}a_{n}-\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{6}a_{n-2}+\frac {\left ( n+r-4\right ) \left ( n+r-5\right ) }{120}a_{n-4}+a_{n-1}=0 \]

Or

\begin{equation} a_{n}=-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}}+\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{6\left ( n+r\right ) ^{2}}a_{n-2}-\frac {\left ( n+r-4\right ) \left ( n+r-5\right ) }{120\left ( n+r\right ) ^{2}}a_{n-4} \tag {2}\end{equation}

Since we need to differentiate \(y_{1}\) to obtain \(y_{2}\) and the differentiation is w.r.t \(r\), we will carry the calculations with \(r\) in place and at the end replace \(r\) by its value (which happened to be \(zero\) in this example). We do this only in the case of repeated roots.

For \(n=4\) then (2) gives

\begin{align*} a_{4} & =-\frac {a_{3}}{\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}a_{2}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}a_{0}\\ & =-\frac {-\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}}}{\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}a_{2}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}a_{0}\\ & =\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}\end{align*}

And so on. Now we replace \(r=0\) to find \(y_{1}\). Just remember not to use anything over \(n=5\) since we cut off the series for \(\sin \left ( x\right ) \) at \(x^{5}\).

Using \(r=0\), then the above values for \(a_{i}\) found become

\begin{align*} a_{1} & =-\frac {1}{\left ( 1+r\right ) ^{2}}=-1\\ a_{2} & =\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}=\frac {1}{4}\\ a_{3} & =-\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}=-\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}}=-\frac {1}{36}\\ a_{4} & =\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}\\ & =\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}\left ( 4\right ) ^{2}}+\frac {\left ( 2\right ) }{6\left ( 4\right ) ^{2}}\frac {1}{\left ( 2\right ) ^{2}}\\ & =\frac {1}{144}\end{align*}

Let find one more term. For \(n=5\) then (2) gives

\begin{align*} a_{5} & =-\frac {a_{4}}{\left ( 5+r\right ) ^{2}}+\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{6\left ( 5+r\right ) ^{2}}a_{3}-\frac {\left ( 1+r\right ) \left ( r\right ) }{120\left ( 5+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{144}}{5^{2}}+\frac {\left ( 3\right ) \left ( 2\right ) }{6\left ( 5\right ) ^{2}}\left ( -\frac {1}{36}\right ) \\ & =-\frac {1}{720}\end{align*}

For \(n=6\) the above recurrence relation gives

\begin{align*} a_{6} & =-\frac {a_{5}}{\left ( 6+r\right ) ^{2}}+\frac {\left ( 4+r\right ) \left ( 3+r\right ) }{6\left ( 6+r\right ) ^{2}}a_{4}-\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{120\left ( 6+r\right ) ^{2}}a_{2}\\ & =-\frac {-\frac {1}{720}}{6^{2}}+\frac {\left ( 4\right ) \left ( 3\right ) }{6\left ( 6\right ) ^{2}}\frac {1}{144}-\frac {\left ( 2\right ) }{120\left ( 6\right ) ^{2}}\frac {1}{4}\\ & =\frac {1}{3240}\end{align*}

For \(n=7\)

\begin{align*} a_{7} & =-\frac {a_{6}}{\left ( 7+r\right ) ^{2}}+\frac {\left ( 5+r\right ) \left ( 4+r\right ) }{6\left ( 7+r\right ) ^{2}}a_{5}-\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{120\left ( 7+r\right ) ^{2}}a_{3}\\ & =-\frac {\frac {1}{3240}}{\left ( 7\right ) ^{2}}+\frac {\left ( 5\right ) \left ( 4\right ) }{6\left ( 7\right ) ^{2}}\left ( -\frac {1}{720}\right ) -\frac {\left ( 3\right ) \left ( 2\right ) }{120\left ( 7\right ) ^{2}}\left ( -\frac {1}{36}\right ) \\ & =-\frac {23}{317\,520}\end{align*}

For \(n=8\)

\begin{align*} a_{8} & =-\frac {a_{7}}{\left ( 8+r\right ) ^{2}}+\frac {\left ( 6+r\right ) \left ( 5+r\right ) }{6\left ( 8+r\right ) ^{2}}a_{6}-\frac {\left ( 4+r\right ) \left ( 3+r\right ) }{120\left ( 8+r\right ) ^{2}}a_{4}\\ & =-\frac {\left ( -\frac {23}{317\,520}\right ) }{\left ( 8\right ) ^{2}}+\frac {\left ( 6\right ) \left ( 5\right ) }{6\left ( 8\right ) ^{2}}\left ( \frac {1}{3240}\right ) -\frac {\left ( 4\right ) \left ( 3\right ) }{120\left ( 8\right ) ^{2}}\left ( \frac {1}{144}\right ) \\ & =\frac {13}{903\,168}\end{align*}

Which is now the wrong value. It should be \(\frac {1}{62720}\). So using 3 terms from \(\sin x\) we obtain up to \(a_{7}\) correct terms. Hence

\begin{align*} y_{1} & =\sum a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}-\frac {1}{720}x^{5}+\frac {1}{3240}x^{6}-\frac {23}{317\,520}x^{7}+\cdots \end{align*}

What would have happened if we expanded \(\sin \left ( x\right ) \) only for two terms? Lets find out. The ode becomes

\begin{align*} \sin \left ( x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \left ( x-\frac {x^{3}}{3!}+\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

The above becomes

\begin{align*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-\frac {x^{3}}{3!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

Reindex

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

For \(n=0\) we obtain the indicial equation as we did above. For \(n=1\)

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

For \(r=0\) this gives

\[ a_{1}=-1 \]

\(n\geq 2\) gives

\begin{align} \left ( n+r\right ) ^{2}a_{n}-\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}+a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{\left ( n+r\right ) ^{2}}a_{n-2} \tag {2A}\end{align}

Hence for \(n=2\)

\begin{align*} a_{2} & =-\frac {a_{1}}{\left ( 2+r\right ) ^{2}}+\frac {1}{6}\frac {r\left ( -1+r\right ) }{\left ( 2+r\right ) ^{2}}a_{0}\\ & =-\frac {-\frac {1}{\left ( 1+r^{2}\right ) }}{\left ( 2+r\right ) ^{2}}+\frac {1}{6}\frac {r\left ( -1+r\right ) }{\left ( 2+r\right ) ^{2}}\end{align*}

For \(r=0\) the above gives

\[ a_{2}=-\frac {-\frac {1}{\left ( 1\right ) }}{\left ( 2\right ) ^{2}}=\frac {1}{4}\]

\(n=3\) gives

\begin{align*} a_{3} & =-\frac {a_{2}}{\left ( 3+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 1+r\right ) \left ( r\right ) }{\left ( 3+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{4}}{\left ( 3+r\right ) ^{2}}-\frac {1}{6}\frac {\left ( 1+r\right ) \left ( r\right ) }{\left ( 3+r\right ) ^{2}}\end{align*}

For \(r=0\)

\[ a_{3}=-\frac {\frac {1}{4}}{\left ( 3\right ) ^{2}}=-\frac {1}{36}\]

For \(n=4\)

\begin{align*} a_{4} & =-\frac {a_{3}}{\left ( 4+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{\left ( 4+r\right ) ^{2}}a_{2}\\ & =-\frac {a_{3}}{\left ( 4\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2\right ) \left ( 1\right ) }{\left ( 4\right ) ^{2}}a_{2}\\ & =-\frac {\left ( -\frac {1}{36}\right ) }{\left ( 4\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2\right ) \left ( 1\right ) }{\left ( 4\right ) ^{2}}\left ( \frac {1}{4}\right ) \\ & =\frac {1}{144}\end{align*}

For \(n=5\)

\begin{align*} a_{5} & =-\frac {a_{4}}{\left ( 5+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{\left ( 5+r\right ) ^{2}}a_{3}\\ & =-\frac {\frac {1}{144}}{\left ( 5\right ) ^{2}}+\frac {1}{6}\frac {\left ( 3\right ) \left ( 2\right ) }{\left ( 5\right ) ^{2}}\left ( -\frac {1}{36}\right ) =-\frac {1}{720}\end{align*}

For \(n=6\)

\begin{align*} a_{6} & =-\frac {a_{5}}{\left ( 6+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 6+r-2\right ) \left ( 6+r-3\right ) }{\left ( 6+r\right ) ^{2}}a_{4}\\ & =-\frac {\left ( -\frac {1}{720}\right ) }{\left ( 6\right ) ^{2}}+\frac {1}{6}\frac {\left ( 4\right ) \left ( 3\right ) }{6^{2}}\frac {1}{144}\\ & =\frac {11}{25\,920}\end{align*}

Which is the wrong value. We see that using two terms only from the \(\sin \left ( x\right ) \) gave up correct \(a_{n}\) values up to \(a_{5}\). What if we used only one term? Lets find out.

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

\(n=0\) gives the indicial equation. For \(n\geq 1\) the recurrence relation is

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

For \(n=1\)

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

For \(r=0\)

\[ a_{1}=-1 \]

For \(n=2\)

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

For \(r=0\)

\[ a_{2}=\frac {1}{4}\]

For \(n=3\)

\[ a_{3}=-\frac {a_{2}}{\left ( 3+r\right ) ^{2}}=-\frac {\frac {1}{4}}{\left ( 3+r\right ) ^{2}}\]

For \(r=0\)

\[ a_{3}=-\frac {\frac {1}{4}}{\left ( 3\right ) ^{2}}=-\frac {1}{36}\]

For \(n=4\)

\[ a_{4}=-\frac {a_{3}}{\left ( 4+r\right ) ^{2}}=-\frac {-\frac {1}{36}}{\left ( 4+r\right ) ^{2}}\]

For \(r=0\)

\[ a_{4}=-\frac {-\frac {1}{36}}{\left ( 4\right ) ^{2}}=\frac {1}{576}\]

We see that this is the wrong value. So when using one term only we obtain correct \(a_{n}\) up to \(a_{3}\). What do we learn from all the above? It is that if we expand \(f\left ( x\right ) \) up to \(O\left ( x^{n}\right ) \) order, then we can only determine correct terms up to \(a_{n}\) and no more. In the above when we used \(\sin \left ( x\right ) =x-\frac {x^{3}}{6}+\frac {x^{5}}{120}+O\left ( x^{7}\right ) \) then we obtained correct terms up to \(a_{7}\). And when we used \(\sin \left ( x\right ) =x-\frac {x^{3}}{6}+O\left ( x^{5}\right ) \) then we obtained correct terms up to \(a_{5}\) and when we used \(\sin \left ( x\right ) =x+O\left ( x^{3}\right ) \) then we obtained correct terms up to \(a_{3}\). So we should keep this in mind from now on,.

To find \(y_{2}\) we use \(b_{n}=\frac {d}{dr}a_{n}\) and evaluate this at \(r=r_{2}\) which in this case is zero. Hence

\begin{align*} b_{1} & =\frac {d}{dr}a_{1}=\frac {d}{dr}\left ( -\frac {1}{\left ( 1+r\right ) ^{2}}\right ) _{r=0}=\frac {2}{\left ( r+1\right ) ^{3}}=2\\ b_{2} & =\frac {d}{dr}a_{2}=\frac {d}{dr}\left ( \frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}\right ) =\left ( \frac {5r^{4}+13r^{3}+9r^{2}-25r-38}{6\left ( r^{2}+3r+2\right ) ^{3}}\right ) _{r=0}=\frac {-38}{6\left ( 2\right ) ^{3}}=-\frac {19}{24}\\ b_{3} & =\frac {d}{dr}a_{3}\\ & =\frac {d}{dr}\left ( -\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}\right ) \\ & =\left ( \frac {\left ( 4r^{6}+18r^{5}+20r^{4}-15r^{3}-18r^{2}+93r+114\right ) }{6\left ( r^{3}+6r^{2}+11r+6\right ) ^{3}}\right ) _{r=0}\\ & =\frac {114}{6\left ( 6\right ) ^{3}}\\ & =\frac {19}{216}\end{align*}

And so on. Hence

\begin{align*} y_{1} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}}\\ & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {19}{24}x^{2}+\frac {19}{216}x^{3}+\cdots \right ) \end{align*}

Therefore the complete solution is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}+\cdots \right ) \\ & +c_{2}\left ( \left ( 1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}+\cdots \right ) \ln \left ( x\right ) +\left ( 2x-\frac {19}{24}x^{2}+\frac {19}{216}x^{3}+\cdots \right ) \right ) \end{align*}