Example 9 sin (x) y′′ + y′ + y = 0

1.2.2.2.9 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

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

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

\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

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*}

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