4.3.2.0 Regular singular point using Frobenius series method.

[next] [prev] [prev-tail] [tail] [up]

Regular singular point using Frobenius series method. expansion point is regular singular point. Four sub methods depending on type of roots of the indicial equations.

Roots of indicial equation are complex ode internal name "second_order_series_method_regular_singular_point_complex_roots"

In this case the solution is\[ y=c_{1}y_{1}+c_{2}y_{2}\] Where \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}} \end {align*}

Where \(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary.

Example 1\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=1 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {1}{x^{2}}\). There is one singular point at \(x_{0}=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}1=1\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r+1 & =0\\ r^{2}+1 & =0\\ r & =\pm i \end {align*}

Hence \(r_{1}=i,r_{2}=-i\). Expansion around \(x=0\). This is regular singular point. 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*}

Solving ﬁrst for the homogenous ode. \begin {align*} x^{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}+\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}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end {align*}

For \(n=0\)\begin {equation} \left ( r\left ( r-1\right ) +r+1\right ) a_{0}x^{r}=0 \tag {1} \end {equation} Since \(a_{0}\neq 0\), then \(\left ( r\left ( r-1\right ) +r+1\right ) =0\) or \(\allowbreak r^{2}+1=0\). Therefore \(r=\pm i\) as was found above. The homogenous ode therefore satisﬁes\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=\left ( r^{2}+1\right ) a_{0}x^{r}\] Since when \(r=\pm i\), the RHS is zero. For \(n\geq 1\) the recurrence relation is\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n} & =0\nonumber \\ \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) +1\right ) a_{n} & =0\nonumber \\ \left ( n^{2}+2nr+r^{2}+1\right ) a_{n} & =0 \tag {2} \end {align}

Let \(a_{0}=1\). For \(r=i\). For \(n=1\)\[ \left ( 1+2i-1+1\right ) a_{1}=0 \] Hence \(a_{1}=0\). Similarly all \(a_{n}=0\) for \(n\geq 1\). Hence \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+i}\\ & =x^{i}\left ( a_{0}+a_{1}x+\cdots \right ) \\ & =a_{0}x^{i}\\ & =x^{i} \end {align*}

For \(r=-i\). For \(n=1\) and using \(b\) instead of \(a\), we obtain (also using \(b_{0}=1\))\[ \left ( 1-2i+1+1\right ) b_{n}=0 \] Hence \(b_{1}=0\). Similarly all \(b_{n}=0\) for \(n\geq 1\). Hence \begin {align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-i}\\ & =x^{-i}\left ( b_{0}+b_{1}x+\cdots \right ) \\ & =b_{0}x^{-i}\\ & =x^{-i} \end {align*}

Therefore \begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x^{i}+c_{2}x^{-i} \end {align*}

To ﬁnd \(y_{p}\) since the ode satisﬁes\[ x^{2}y^{\prime \prime }+xy^{\prime }+y=\left ( r^{2}+1\right ) a_{0}x^{r}\] Relabel \(r=m,a_{0}=c_{0}\) to avoid confusion with terms used above, then we balance RHS, hence\[ \left ( m^{2}+1\right ) c_{0}x^{m}=1 \] This implies \(m=0\) and \(c_{0}=1\). Therefore \begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

Using the recurrence relation (2) found above, but now using the values found \(m=0\) and \(c_{0}=1\), then (2) becomes\begin {align*} \left ( n^{2}+2nm+m^{2}+1\right ) c_{n} & =0\\ \left ( n^{2}+1\right ) c_{n} & =0 \end {align*}

Hence all \(c_{n}=0\) except for \(c_{0}\). Therefore\begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}\\ & =1 \end {align*}

Hence the solution is \begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{i}+c_{2}x^{-i}+1 \end {align*}

Roots of indicial equation diﬀer by non integer ode internal name "second_order_series_method_regular_singular_point_diﬀerence_not_integer"

If one of the roots is an integer, and the ode is inhomogeneous. ode, then we do not need to split the solution into \(y_{h},y_{p}\) and can use the integer root to ﬁnd \(y_{p}\) directly. If both roots are non-integer, we have to split the problem into \(y_{h},y_{p}\). This is because it will not be possible to match powers on \(x\) from the left side to the right side. Because the RHS will be polynomial in \(x\), but the LHS will not be polynomial in \(x\) because of the non integer powers on \(x.\)In this case the solution is\[ y=c_{1}y_{1}+c_{2}y_{2}\] Where\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}} \end {align*}

And \(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary.

Example 1\[ 2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=x^{2}+2x \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {3}{2x},q\left ( x\right ) =\frac {-1}{2x}\). There is one singular point at \(x=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{2}=\frac {3}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}-\frac {x}{2}=0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {3}{2}r+0 & =0\\ r\left ( 2r+1\right ) & =0\\ r & =0,-\frac {1}{2} \end {align*}

Therefore \(r_{1}=0,r_{2}=-\frac {1}{2}\).

Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence\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*} 2x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-x\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 }3\left ( n+r\right ) a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}-\sum _{n=1}^{\infty }a_{n-1}x^{n+r}=0 \] When \(n=0\)\begin {align*} 2\left ( r\right ) \left ( r-1\right ) a_{0}x^{r}+3\left ( r\right ) a_{0}x^{r} & =0\\ \left ( r\left ( 2r+1\right ) \right ) a_{0}x^{r} & =0 \end {align*}

Since \(a_{0}\neq 0\) then \(r\left ( 2r+1\right ) =0\) and \(r=0,r=-\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes\[ 2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=\left ( r\left ( 2r+1\right ) \right ) a_{0}x^{r}\] Where the RHS will be zero when \(r=0\) or \(r=-\frac {1}{2}\). For \(n\geq 1\) the recurrence relation is \begin {align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+3\left ( n+r\right ) a_{n}-a_{n-1} & =0\nonumber \\ a_{n} & =\frac {a_{n-1}}{2\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) }\nonumber \\ & =\frac {a_{n-1}}{2n^{2}+4nr+n+2r^{2}+r} \tag {1} \end {align}

For \(r=0\) the above becomes\[ a_{n}=\frac {a_{n-1}}{2n^{2}+n}\] For \(n=1\) and letting \(a_{0}=1\)\[ a_{1}=\frac {1}{3}\] For \(n=2\)\[ a_{2}=\frac {a_{1}}{8+2}=\frac {a_{1}}{10}=\frac {1}{30}\] For \(n=3\)\[ a_{3}=\frac {a_{2}}{18+3}=\frac {a_{2}}{21}=\frac {1}{21\left ( 30\right ) }=\frac {1}{630}\] And so on. Hence \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \end {align*}

And for \(r=-\frac {1}{2}\) the recurrence relation (2) becomes, and using \(b\) instead of \(a\)\[ b_{n}=\frac {b_{n-1}}{2n^{2}+4n\left ( -\frac {1}{2}\right ) +n+\frac {1}{2}-\frac {1}{2}}=-\frac {b_{n-1}}{n-2n^{2}}\] For \(n=1\) and using \(b_{0}=1\)\[ b_{1}=-\frac {b_{0}}{1-2}=1 \] For \(n=2\)\[ b_{2}=-\frac {b_{1}}{2-8}=-\frac {1}{2-8}=\frac {1}{6}\] For \(n=3\)\[ b_{3}=-\frac {b_{2}}{3-18}=-\frac {\frac {1}{6}}{3-18}=\frac {1}{90}\] And so on. Hence\begin {align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =\frac {1}{\sqrt {x}}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end {align*}

Hence \begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end {align*}

Now we ﬁnd \(y_{p}\). Since ode satisﬁes\[ 2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=\left ( r\left ( 2r+1\right ) \right ) a_{0}x^{r}\] To ﬁnd \(y_{p}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used for \(y_{h}\). Then the above becomes\[ 2x^{2}y^{\prime \prime }+3xy^{\prime }-xy=\left ( m\left ( 2m+1\right ) \right ) c_{0}x^{m}\] The RHS is \(x^{2}+2x\). We balance each term at a time, this ﬁnds a particular solution for each term on the RHS, then these particular solutions are added at the end. For the input \(2x\) the balance equation is\[ \left ( m\left ( 2m+1\right ) \right ) c_{0}x^{m}=2x \] This implies that\[ m=1 \] Therefore \(\left ( m\left ( 2m+1\right ) \right ) c_{0}=2\), or \(c_{0}\left ( 1\left ( 2+1\right ) \right ) =2\) or \(3c_{0}=2\) or\[ c_{0}=\frac {2}{3}\] The recurrence relation now becomes (using \(m\) for \(r\) and \(c_{0}\) for \(a_{0}\))\[ c_{n}=\frac {c_{n-1}}{2n^{2}+4nm+n+2m^{2}+m}\] For \(m=1\) the above becomes\[ c_{n}=\frac {c_{n-1}}{2n^{2}+5n+3}\] For \(n=1\) and using \(c_{0}=\frac {2}{3}\)\[ c_{1}=\frac {\frac {2}{3}}{2+5+3}=\frac {1}{15}\] For \(n=2\)\[ c_{2}=\frac {c_{1}}{8+10+3}=\frac {\frac {1}{15}}{8+10+3}=\frac {1}{315}\] For \(n=3\)\[ c_{3}=\frac {c_{2}}{18+15+3}=\frac {\frac {1}{315}}{18+15+3}=\frac {1}{11\,340}\] And so on. Hence \begin {align*} y_{p_{1}} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x\left ( \frac {2}{3}+\frac {1}{15}x+\frac {1}{315}x^{2}+\frac {1}{11\,340}x^{3}+\cdots \right ) \\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) \end {align*}

The second term \(x^{2}\) is now balanced \(x^{2}\). The balance equation is \[ \left ( m\left ( 2m+1\right ) \right ) c_{0}x^{m}=x^{2}\] Therefore \(m=2\) and \(\left ( m\left ( 2m+1\right ) \right ) c_{0}=1\). Hence\begin {align*} \left ( 2\left ( 4+1\right ) \right ) c_{0} & =1\\ c_{0} & =\frac {1}{10} \end {align*}

The recurrence relation becomes for \(m=2\)\[ c_{n}=\frac {c_{n-1}}{2n^{2}+4nm+n+2m^{2}+m}\] For \(m=2\) the above becomes\[ c_{n}=\frac {c_{n-1}}{2n^{2}+9n+10}\] For \(n=1\) and using \(c_{0}=\frac {1}{10}\)\[ c_{1}=\frac {\frac {1}{10}}{2+9+10}=\frac {1}{210}\] For \(n=2\)\[ c_{2}=\frac {c_{1}}{8+18+10}=\frac {\frac {1}{210}}{8+18+10}=\frac {1}{7560}\] For \(n=3\)\[ c_{3}=\frac {c_{2}}{18+27+10}=\frac {\frac {1}{7560}}{18+27+10}=\frac {1}{415\,800}\] And so on. Hence \begin {align*} y_{p_{2}} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{10}+\frac {1}{210}x+\frac {1}{7560}x^{2}+\frac {1}{415\,800}x^{3}+\cdots \right ) \\ & =\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \end {align*}

The particular solution is the sum of all the particular solutions found above, which is\begin {align*} y_{p} & =y_{p_{1}}+y_{p_{2}}\\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) +\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \\ & =\frac {2}{3}x+\left ( \frac {1}{15}+\frac {1}{10}\right ) x^{2}+\left ( \frac {1}{315}+\frac {1}{210}\right ) x^{3}+\left ( \frac {1}{11\,340}+\frac {1}{7560}\right ) x^{4}+\cdots \\ & =\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end {align*}

Hence the complete solution is \begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) +\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end {align*}

Example 2\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=5 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {\left ( x+1\right ) }{2x},q\left ( x\right ) =\frac {3}{2x}\). There is one singular point at \(x=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {\left ( x+1\right ) }{2}=\frac {1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3x}{2}=0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {1}{2}r+0 & =0\\ r\left ( 2r-1\right ) & =0\\ r & =0,\frac {1}{2} \end {align*}

Therefore \(r_{1}=0,r_{2}=\frac {1}{2}\).

Expansion around \(x=x_{0}=0\). 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 homogenous ode becomes\begin {align*} 2x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\left ( x+1\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+3\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-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3a_{n}x^{n+r} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }3a_{n-1}x^{n+r-1}=0 \] For \(n=0\)\begin {align*} \left ( 2\left ( r\right ) \left ( r-1\right ) a_{0}+ra_{0}\right ) x^{r-1} & =0\\ \left ( 2r\left ( r-1\right ) +r\right ) a_{0} & =0 \end {align*}

Since \(a_{0}\neq 0\) then the ﬁrst term above will vanish only when \(2r\left ( r-1\right ) +r=0\) or \(r\left ( 2r-1\right ) =0\). Hence \(r=0,r=\frac {1}{2}\) as was found above. For \(n\geq 1\)\begin {align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}+\left ( n+r\right ) a_{n}+3a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {n+r+2}{\left ( n+r\right ) \left ( 2r+2n-1\right ) }a_{n-1} \tag {1} \end {align}

Therefore the diﬀerential equation satisﬁes\begin {equation} 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=r\left ( 2r-1\right ) a_{0}x^{r-1} \tag {2} \end {equation} The RHS above will be zero when \(r=0\) or \(r=\frac {1}{2}\). When \(r=0\) the recurrence relation (1) becomes\[ a_{n}=-\frac {n+2}{\left ( n\right ) \left ( 2n-1\right ) }a_{n-1}\] Which gives (for \(a_{0}=1\)) (working out few terms using the above)\[ y_{1}=1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \] And when \(r=\frac {1}{2}\) the recurrence relation is (using \(b\) in place of \(a\) and letting \(b_{0}=1\) also)\[ b_{n}=-\frac {n+\frac {5}{2}}{\left ( n+\frac {1}{2}\right ) \left ( 1+2n-1\right ) }b_{n-1}\] Which gives (working out few terms)\[ y_{2}=\sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \] Hence the solution is \begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \right ) \end {align*}

Now we ﬁnd \(y_{p}\). From (2), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=m\left ( 2m-1\right ) c_{0}x^{m-1}\] Therefore we need to balance \(m\left ( 2m-1\right ) c_{0}x^{m-1}=5\) since the RHS is \(5\). This implies \(m-1=0\) or \(m=1\). Therefore \(m\left ( 2m-1\right ) c_{0}=5\) or \(\left ( 2-1\right ) c_{0}=5\) which gives \(c_{0}=5\). Hence \begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

To ﬁnd \(c_{n}\), the same recurrence relation (1) is used by with \(r\) replaced by \(m\) and \(a\) replaced by \(c\). This gives\[ c_{n}=-\frac {n+m+2}{\left ( n+m\right ) \left ( 2m+2n-1\right ) }c_{n-1}\] For \(m=1\) the above becomes\[ c_{n}=-\frac {n+3}{\left ( n+1\right ) \left ( 1+2n\right ) }c_{n-1}\] For \(n=1\)\[ c_{1}=-\frac {1+3}{\left ( 1+1\right ) \left ( 1+2\right ) }c_{0}=-\frac {2}{3}c_{0}=-\frac {2}{3}\left ( 5\right ) =-\frac {10}{3}\] For \(n=2\)\[ c_{2}=-\frac {2+3}{\left ( 2+1\right ) \left ( 1+4\right ) }c_{1}=-\frac {1}{3}c_{1}=-\frac {1}{3}\left ( -\frac {10}{3}\right ) =\frac {10}{9}\] For \(n=3\)\[ c_{1}=-\frac {3+3}{\left ( 3+1\right ) \left ( 1+6\right ) }c_{2}=-\frac {3}{14}\left ( \frac {10}{9}\right ) =-\frac {2}{3}\left ( 5\right ) =-\frac {5}{21}\] And so on. Hence

\begin {align*} y_{p} & =x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) \\ & =x\left ( 5-\frac {10}{3}x+\frac {10}{9}x^{2}-\frac {5}{21}x^{3}+\cdots \right ) \\ & =\left ( 5x-\frac {10}{3}x^{2}+\frac {10}{9}x^{3}-\frac {5}{21}x^{4}+\cdots \right ) \end {align*}

Hence the ﬁnal solution\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +\sqrt {x}c_{2}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) +\left ( 5x-\frac {10}{3}x^{2}+\frac {10}{9}x^{3}-\frac {5}{21}x^{4}+\cdots \right ) \end {align*}

Example 3\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=x \] This is the same problem as above but diﬀerent RHS. As shown above, we obtained that the diﬀerential equation satisﬁes\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=r\left ( 2r-1\right ) a_{0}x^{r-1}\] To ﬁnd \(y_{p}\), and using \(m\) in place of \(r\) and \(c\) in place of \(a\) so not to confuse terms with the \(y_{h}\) terms, then the above becomes\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=m\left ( 2m-1\right ) c_{0}x^{m-1}\] The RHS above will be zero when \(m=0\) or \(m=\frac {1}{2}\). We now need to balance the RHS against given RHS which is \(x\). Hence \[ m\left ( 2m-1\right ) c_{0}x^{m-1}=x \] To balance this we need \(m-1=1\) or \(m=2\). Hence \(2\left ( 4-1\right ) c_{0}=1\) or \(c_{0}=\frac {1}{6}\). Using the recurrence relation we found above, which is for \(n\geq 1\) (again, calling \(r\) as \(m\) so not to confuse \(y_{h}\) terms with \(y_{p}\) terms), we obtain\[ c_{n}=-\frac {n+m+2}{\left ( n+r\right ) \left ( 2m+2n-1\right ) }c_{n-1}\] But now using \(m=2\)\[ c_{n}=-\frac {n+4}{\left ( n+2\right ) \left ( 4+2n-1\right ) }c_{n-1}\] Hence for \(n=1\)\begin {align*} c_{1} & =-\frac {1+4}{\left ( 1+2\right ) \left ( 4+2-1\right ) }c_{0}\\ & =-\frac {1}{3}c_{0}\\ & =-\frac {1}{3}\left ( \frac {1}{6}\right ) =-\frac {1}{18} \end {align*}

for \(n=2\)\begin {align*} c_{2} & =-\frac {6}{\left ( 2+2\right ) \left ( 4+4-1\right ) }c_{1}\\ & =-\frac {3}{14}c_{1}=-\frac {3}{14}\left ( -\frac {1}{18}\right ) =\frac {1}{84} \end {align*}

For \(n=3\)\begin {align*} c_{3} & =-\frac {3+4}{\left ( 3+2\right ) \left ( 4+6-1\right ) }c_{2}\\ & =-\frac {7}{45}c_{2}=-\frac {7}{45}\left ( \frac {1}{84}\right ) =-\frac {1}{540} \end {align*}

For \(n=4\)\begin {align*} c_{4} & =-\frac {4+4}{\left ( 4+2\right ) \left ( 4+8-1\right ) }c_{3}\\ & =-\frac {4}{33}c_{3}=-\frac {4}{33}\left ( -\frac {1}{540}\right ) =\frac {1}{4455} \end {align*}

And so on. Hence\begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{6}-\frac {1}{18}x+\frac {1}{84}x^{2}-\frac {1}{540}x^{3}+\frac {1}{4455}x^{4}+\cdots \right ) \end {align*}

Hence the solution is (\(y_{h}\) was found in the earlier problem)\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \right ) +x^{2}\left ( \frac {1}{6}-\frac {1}{18}x+\frac {1}{84}x^{2}-\frac {1}{540}x^{3}+\frac {1}{4455}x^{4}+\cdots \cdots \right ) \end {align*}

Example 4\[ x^{2}y^{\prime \prime }+\left ( x+1\right ) y^{\prime }+y=5 \] Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {x+1}{x^{2}},q\left ( x\right ) =\frac {1}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x+1}{x}\) which is not deﬁned. Hence not possible to solve this using series solution.

Example 5\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=x^{2}\] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {-x}{2x^{2}}=-\frac {1}{2x},q\left ( x\right ) =\frac {\left ( 1-x^{2}\right ) }{2x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {-1}{2}=\frac {-1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {\left ( 1-x^{2}\right ) }{2}=\frac {1}{2}\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -\frac {1}{2}r+\frac {1}{2} & =0\\ r^{2}-\frac {3}{2}r+\frac {1}{2} & =0\\ r & =1,\frac {1}{2} \end {align*}

Therefore \(r_{1}=0,r_{2}=-\frac {1}{2}\). Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence\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 homogenous ode becomes\begin {align*} 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y & =0\\ 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}-\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}-\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r}=0 \] When \(n=0\)\begin {align*} \left ( 2\left ( n+r\right ) \left ( n+r-1\right ) a_{0}-\left ( n+r\right ) a_{0}+a_{0}\right ) x^{r} & =0\\ \left ( 2r\left ( r-1\right ) -r+1\right ) a_{0}x^{r} & =0\\ \left ( 2r^{2}-3r+1\right ) a_{0}x^{r} & =0 \end {align*}

Since \(a_{0}\neq 0\) then \(2r^{2}-3r+1=0\), hence \(r=1,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2r^{2}-3r+1\right ) a_{0}x^{r}\] Where the RHS will be zero when \(r=1,r=\frac {1}{2}\). When \(n=1\) \begin {align*} 2\left ( 1+r\right ) \left ( 1+r-1\right ) a_{1}-\left ( 1+r\right ) a_{1}+a_{1} & =0\\ \left ( 2\left ( 1+r\right ) \left ( 1+r-1\right ) -\left ( 1+r\right ) +1\right ) a_{1} & =0\\ r\left ( 2r+1\right ) a_{1} & =0 \end {align*}

Hence \(a_{1}=0\). For \(n\geq 2\) the recurrence relation is \begin {align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}-\left ( n+r\right ) a_{n}+a_{n}-a_{n-2} & =0\nonumber \\ a_{n} & =\frac {a_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1}\nonumber \\ & =\frac {a_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1} \tag {1} \end {align}

For \(r=1\) the above becomes\[ a_{n}=\frac {a_{n-2}}{n\left ( 2n+1\right ) }\] For \(n=2\) and letting \(a_{0}=1\)\[ a_{2}=\frac {a_{0}}{2\left ( 4+1\right ) }=\frac {1}{10}\] For \(n=3\)\[ a_{3}=\frac {a_{1}}{n\left ( 2n+1\right ) }=0 \] For \(n=4\)\[ a_{4}=\frac {a_{2}}{4\left ( 8+1\right ) }=\frac {\frac {1}{10}}{4\left ( 8+1\right ) }=\frac {1}{360}\] And so on. Hence \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=x\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \end {align*}

And for \(r=\frac {1}{2}\) the recurrence relation (1) becomes, and using \(b\) instead of \(a\)\begin {align*} b_{n} & ==\frac {b_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1}\\ & =\frac {b_{n-2}}{2\left ( n+\frac {1}{2}\right ) \left ( n+\frac {1}{2}-1\right ) -\left ( n+\frac {1}{2}\right ) +1}\\ & =\frac {b_{n-2}}{n\left ( 2n-1\right ) } \end {align*}

Notice also that \(b_{1}=0\) just like \(a_{1}=0\) from above. Now, for \(n=2\) and using \(b_{0}=1\)\[ b_{2}=\frac {b_{0}}{2\left ( 4-1\right ) }=\frac {1}{6}\] For \(n=3\)\[ b_{2}=-\frac {b_{1}}{2-8}=-\frac {1}{2-8}=\frac {1}{6}\] For \(n=3\)\[ b_{3}=\frac {b_{1}}{n\left ( 2n-1\right ) }=0 \] For \(n=4\)\[ b_{n}=\frac {b_{2}}{4\left ( 8-1\right ) }=\frac {\frac {1}{6}}{4\left ( 8-1\right ) }=\frac {1}{168}\] And so on. Hence\begin {align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =\sqrt {x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\sqrt {x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end {align*}

Hence \begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end {align*}

Now we ﬁnd \(y_{p}\). Since ode satisﬁes\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2r^{2}-3r+1\right ) a_{0}x^{r}\] To ﬁnd \(y_{p}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used for \(y_{h}\). Then the above becomes\[ 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y=\left ( 2m^{2}-3m+1\right ) c_{0}x^{m}\] The RHS is \(x^{2}\). Hence the balance equation is\[ \left ( 2m^{2}-3m+1\right ) c_{0}x^{m}=x^{2}\] This implies that\[ m=2 \] Therefore \(\left ( 2m^{2}-3m+1\right ) c_{0}=1\), or \(\left ( 8-6+1\right ) c_{0}=1\) or \[ c_{0}=\frac {1}{3}\] The recurrence relation (1) from above now becomes (using \(m\) for \(r\) and \(c_{0}\) for \(a_{0}\))\[ c_{n}=\frac {c_{n-2}}{2\left ( n+m\right ) \left ( n+m-1\right ) -\left ( n+m\right ) +1}\] For \(m=2\) the above becomes\begin {align*} c_{n} & =\frac {c_{n-2}}{2\left ( n+2\right ) \left ( n+1\right ) -\left ( n+2\right ) +1}\\ & =\frac {c_{n-2}}{2n^{2}+5n+3} \end {align*}

For \(n=1\) we use \(c_{1}=0\) the same as was found for \(a_{1},b_{1}\). For \(n\geq 2\) the above is used. Hence for \(n=2\)\[ c_{2}=\frac {c_{0}}{8+10+3}=\frac {\frac {1}{3}}{8+10+3}=\frac {1}{63}\] For \(n=3\)\[ c_{3}=\frac {c_{1}}{18+15+3}=0 \] For \(n=4\)\[ c_{4}=\frac {c_{2}}{32+20+3}=\frac {\frac {1}{63}}{32+20+3}=\frac {1}{3465}\] And so on. Hence \begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{3}+\frac {1}{63}x^{2}+\frac {1}{3465}x^{4}+\cdots \right ) \\ & =\frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \end {align*}

Hence the complete solution is \begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) +\left ( \frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \right ) \end {align*}

Alternative way to ﬁnd \(y_{p}\) is the the following. Let \(y_{p}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \) then \(y_{p}^{\prime }=c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \) and \(y_{p}^{\prime \prime }=2c_{2}+6c_{3}x+\cdots \). Hence the ode becomes\begin {align*} 2x^{2}\left ( 2c_{2}+6c_{3}x+\cdots \right ) -x\left ( c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \right ) +\left ( 1-x^{2}\right ) \left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) & =x^{2}\\ c_{0}+x\left ( -c_{1}+c_{1}\right ) +x^{2}\left ( 4c_{2}-2c_{2}+c_{2}-c_{0}\right ) +x^{3}\left ( \cdots \right ) & =x^{2} \end {align*}

Hence \(c_{0}=0,4c_{2}-2c_{2}+c_{2}-c_{0}=1\) or \(3c_{2}-c_{0}=1\) or \(c_{2}=\frac {1}{3}\). We need to keep adding more equations and solving them simultaneously. This method is not as easy to use as the method used above, which uses the balance equation to ﬁnd to \(y_{p}\). Also this method could fail, since in practice we should not use undetermined coeﬃcients method (which is what this does) on an ode with variable coeﬃcients. So I will not use this any more.

Example 6\[ 2xy^{\prime \prime }+y^{\prime }+y=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{2x},q\left ( x\right ) =\frac {1}{2x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {1}{2}=\frac {1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x}{2}=0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {1}{2}r & =0\\ r\left ( 2r-1\right ) & =0\\ r & =0,\frac {1}{2} \end {align*}

Therefore \(r_{1}=0,r_{2}=\frac {1}{2}\). Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence\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*} xy^{\prime \prime }+y^{\prime }+y & =0\\ 2x\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 }2\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*}

Re indexing to lowest powers on \(x\) gives\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \] When \(n=0\)\begin {align*} 2\left ( r\right ) \left ( r-1\right ) a_{0}x^{r-1}+ra_{0}x^{r-1} & =0\\ \left ( 2r\left ( r-1\right ) +r\right ) a_{0}x^{r-1} & =0\\ \left ( r\left ( 2r-1\right ) \right ) a_{0}x^{r-1} & =0 \end {align*}

Since \(a_{0}\neq 0\) then \(r\left ( 2r-1\right ) =0\), hence \(r=0,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes\[ 2xy^{\prime \prime }+y^{\prime }+y=\left ( r\left ( 2r-1\right ) \right ) a_{0}x^{r-1}\] Where the RHS will be zero when \(r=1,r=\frac {1}{2}\). For \(n\geq 1\) the recurrence relation is\begin {align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n} & =-a_{n-1}\nonumber \\ a_{n} & =\frac {-a_{n-1}}{2\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) }\nonumber \\ & =\frac {-a_{n-1}}{2n^{2}+4nr-n+2r^{2}-r} \tag {1} \end {align}

For \(r=0\) the above becomes\[ a_{n}=\frac {-a_{n-1}}{n\left ( 2n-1\right ) }\] For \(n=1\) and using \(a_{0}=1\)\[ a_{1}=\frac {-a_{0}}{n\left ( 2n-1\right ) }=-1 \] For \(n=2\)\[ a_{2}=\frac {-a_{1}}{2\left ( 3\right ) }=\frac {1}{6}\] For \(n=3\)\[ a_{3}=\frac {-a_{2}}{3\left ( 5\right ) }=\frac {-\frac {1}{6}}{15}=-\frac {1}{90}\] And so on. Hence \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \end {align*}

To ﬁnd \(y_{2}\), using (1) but replacing \(a\) by \(b\) and using \(r=\frac {1}{2}\) and letting \(b_{0}=1\) and following the above process gives\[ b_{n}=\frac {-b_{n-1}}{2n^{2}+4n\left ( \frac {1}{2}\right ) -n+2\left ( \frac {1}{2}\right ) ^{2}-\frac {1}{2}}=-\frac {b_{n-1}}{2n^{2}+n}\] For \(n=1\)\[ b_{1}=-\frac {b_{0}}{3}=-\frac {1}{3}\] For \(n=2\)\[ b_{2}=-\frac {b_{1}}{8+2}=-\frac {b_{1}}{10}=-\frac {-\frac {1}{3}}{10}=\frac {1}{30}\] And so on. Hence we obtain\begin {align*} y_{2} & =\sqrt {x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\sqrt {x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \end {align*}

Therefore the solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{1}\\ & =c_{1}\left ( 1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \right ) \end {align*}

Example 7\[ 4xy^{\prime \prime }+3y^{\prime }+3y=\sqrt {x}\] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {3}{4x},q\left ( x\right ) =\frac {3}{4x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{4}=\frac {3}{4}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3x}{4}=0\). Hence \(x=0\) is regular singular point. The indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {3}{4}r+0 & =0\\ r\left ( r-1\right ) +\frac {3}{4}r & =0\\ r & =\frac {1}{4},0 \end {align*}

Frobenius is now used. Roots diﬀer by non integer. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\).\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 homogenous ode becomes\begin {align*} 4x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+3\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3a_{n}x^{n+r} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\[ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }3a_{n-1}x^{n+r-1}=0 \] When \(n=0\)\begin {align*} 4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+3\left ( n+r\right ) a_{n}x^{n+r-1} & =0\\ 4r\left ( r-1\right ) a_{0}+3ra_{0} & =0\\ \left ( 4r\left ( r-1\right ) +3r\right ) a_{0} & =0 \end {align*}

Since \(a_{0}\neq 0\) then \(4r\left ( r-1\right ) +3r=0\), hence \(r=0,r=\frac {1}{4}\) as was found above. Therefore the homogenous ode satisﬁes\[ 4xy^{\prime \prime }+3y^{\prime }+3y=\left ( 4r\left ( r-1\right ) +3r\right ) a_{0}x^{r-1}\] Hence the balance equation is that we will use to ﬁnd the particular solution is \[ \left ( 4m\left ( m-1\right ) +3m\right ) c_{0}x^{m-1}=\sqrt {x}\] We will get back to the above after ﬁnding \(y_{h}\). Going over the same steps as before, we ﬁnd the recurrence relation \[ a_{n}=-\frac {3a_{n-1}}{4n^{2}+8nr+4r^{2}-n-r}\] For \(r=\frac {1}{4},n>0\) and similarly\[ b_{n}=-\frac {3a_{n-1}}{4n^{2}+8nr+4r^{2}-n-r}\] For \(r=0,n>0\). Finding few terms using the above gives the solution as\begin {align*} y_{h} & =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) \end {align*}

Now we need to ﬁnd \(y_{p}\). From the balance equation \[ \left ( 4m\left ( m-1\right ) +3m\right ) c_{0}x^{m-1}=\sqrt {x}\] Hence \(m-1=\frac {1}{2}\) or \(m=\frac {3}{2}\). And \(\left ( 4m\left ( m-1\right ) +3m\right ) c_{0}=1\), hence \(\left ( 4\left ( \frac {3}{2}\right ) \left ( \frac {3}{2}-1\right ) +3\left ( \frac {3}{2}\right ) \right ) c_{0}=1\), which gives \(c_{0}=\frac {2}{15}\). Therefore\begin {align*} y_{p} & =x^{m}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {3}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \end {align*}

We now just need to determined \(c_{n}\) for \(n>0\). For this we use the same recurrence relation as found above. We can use \(a_{n}\) or \(b_{n}\) as they are the same, but change \(a_{n}\) to \(c_{n}\) and \(r\) to \(c\) (so not to confuse notations). This gives\[ c_{n}=-\frac {3c_{n-1}}{4n^{2}+8nm+4m^{2}-n-m}\] For \(n>0\) and \(m=\frac {3}{2}\). Hence for \(n=1\) the above gives\begin {align*} c_{1} & =-\frac {3c_{0}}{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {3\left ( \frac {2}{15}\right ) }{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {4}{225} \end {align*}

For \(n=2\)\begin {align*} c_{1} & =-\frac {3c_{1}}{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =-\frac {3\left ( -\frac {4}{225}\right ) }{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =\frac {8}{6825} \end {align*}

And so on. Hence \begin {align*} y_{p} & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end {align*}

Hence the complete solution is \begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) +x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end {align*}

Roots of indicial equation diﬀer by integer. Good case ode internal name "second_order_series_method_regular_singular_point_diﬀerence_is_integer_good_case".

In this case the solution is\[ y=c_{1}y_{1}+c_{2}y_{2}\] There are two sub cases that show up when roots diﬀer by integer. First sub case is when the second solution \(y_{2}\) is obtained similar to how \(y_{1}\) is obtained. i.e. using standard Frobenius series but with the second root. The second sub case is the harder one, this is when \(y_{2}\) fails to be obtained using the standard method due to \(b_{N}\) being undeﬁned where \(N\) is the diﬀerence between the roots. In this sub case we need to use a modiﬁed Frobenius series method where, which is explained more using examples below. Therefore for sub case one (called the good case) we have

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

For the second subcase (called the bad case) ﬁrst we will ﬁnd the bad root \(r\) of the indicial equation which causes the recurrence relation to become undeﬁned at some \(n\). Call it \(r_{bad}\), then we ﬁrst ﬁnd \(\overline {y}\) deﬁned as \[ \overline {y}=x^{r}\sum _{n=0}^{\infty }\left ( r-r_{bad}\right ) a_{n}x^{n}\] Where \(a_{n}\) is found using the recurrence relation (but \(r\) is kept symbolic). \(y_{1}\) is then found from by evaluating it \(r=r_{bad}\)\[ y_{1}=\overline {y}_{r=r_{bad}}\] And also setting \(a_{0}=1\). Note that some terms will vanish above but not all, since there will be cancellation of \(\left ( r-r_{bad}\right ) \) during the process. \(y_{2}\) is next found using\begin {align*} y_{2} & =\left ( \frac {d}{dr}\overline {y}\right ) _{r=r_{bad}}\\ & =y_{1}\ln \left ( x\right ) +x^{r_{bad}}\sum _{n=0}^{\infty }\left ( \frac {d}{dr}\left ( \left ( r-r_{bad}\right ) a_{n}x^{n}\right ) \right ) _{r=r_{bad}} \end {align*}

Example 1\[ \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y=3x^{2}\] Comparing the above to \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0\) shows that \(p\left ( x\right ) =\frac {3}{x\left ( 1-x\right ) },q\left ( x\right ) =\frac {2}{x\left ( x-1\right ) }\). Hence there are two singular points, one at \(x=0\) and one at \(x=1\). Let the expansion be around \(x=0\). This means the solution will deﬁne up to \(x=1\), which is the next nearest singular point.

\[ p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x\frac {3}{x\left ( 1-x\right ) }=3 \] And \[ q_{0}=\lim _{x\rightarrow 0}x^{2}\frac {2}{x\left ( 1-x\right ) }=0 \] Hence \(x_{0}=0\) is a regular singular point. The indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +3r & =0\\ r^{2}-r+3r & =0\\ r^{2}+2r & =0\\ r\left ( r+2\right ) & =0 \end {align*}

Therefore \(r=0,r=-2\). They diﬀer by an integer \(N=2\). Therefore two linearly independent solutions can be constructed using\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}} \end {align*}

Where \(C\) above can be zero depending on a condition given below. Now we will work out the solution for a general \(r\). 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 homogeneous ode becomes\begin {align*} \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y & =0\\ \left ( x-x^{2}\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+2\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 ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }2a_{n}x^{n+r} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=1}^{\infty }\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }2a_{n-1}x^{n+r-1}=0 \tag {1A} \end {equation} For \(n=0\)\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+3\left ( n+r\right ) a_{n}x^{n+r-1} & =0\nonumber \\ \left ( r\left ( r-1\right ) +3r\right ) a_{0}x^{r-1} & =\nonumber \\ \left ( r^{2}+2r\right ) a_{0}x^{r-1} & =0 \tag {1B} \end {align}

Since \(a_{0}\neq 0\), then \(r=0,r=-2\) as was found above. Hence \(N=2\) which is the diﬀerence between the two roots. The homogenous ode therefore satisﬁes\[ \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y=\left ( r^{2}+2r\right ) a_{0}x^{r-1}\] Since when \(r=0,r=-2\) the RHS is zero. The term on the right of the above is important as it will be used to determine the particular solution. The recurrence relation is when \(n\geq 1\) from (1A) and is given by\[ \left ( n+r\right ) \left ( n+r-1\right ) a_{n}-\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}+3\left ( n+r\right ) a_{n}+2a_{n-1}=0 \] Keeping larger \(a_{n}\) on the left and all lower \(a_{n}\) on the right gives\begin {align} a_{n} & =\frac {-2+\left ( n+r-1\right ) \left ( n+r-2\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) }a_{n-1}\nonumber \\ a_{n} & =\frac {n+r-3}{n+r+2}a_{n-1} \tag {1} \end {align}

Now we ﬁnd \(y_{h}=c_{1}y_{1}+c_{2}y_{2}\). For \(r=0\) then (1) becomes\begin {equation} a_{n}=\frac {n-3}{n+2}a_{n-1} \tag {2} \end {equation} For \(n=1\) and letting \(a_{0}=1\) then (2) gives \[ a_{1}=\frac {1-3}{1+2}a_{0}=\frac {-2}{3}\] For \(n=2\) Eq. (2) gives \[ a_{2}=\frac {2-3}{2+2}a_{1}=\frac {2-3}{2+2}\left ( \frac {-2}{3}\right ) =\frac {1}{6}\] For \(n=3\) Eq. (2) gives \[ a_{3}=\frac {3-3}{3+2}a_{2}=0 \] And all other higher \(a_{n}=0\). Hence\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}\\ & =1-\frac {2}{3}x+\frac {1}{6}x^{2} \end {align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we calculate \(b_{N}=b_{2}\) using same recurrence relation (1) to see if it is deﬁned or not. If it is deﬁned, then we continue, else we have to use the modiﬁed Frobenius method. From (1) and using \(b\) instead of \(a\) and using \(r=r_{2}=-2\) gives\begin {align*} b_{n} & =\frac {n+r-3}{n+r+2}b_{n-1}\\ & =\frac {n-2-3}{n-2+2}b_{n-1}\\ & =\frac {n-5}{n}b_{n-1} \end {align*}

Hence for \(n=1\) and using \(b_{0}=1\) as we did for \(a_{0}\) gives\[ b_{1}=-4b_{0}=-4 \] For \(n=N=2\)\[ b_{n}=\frac {-3}{2}b_{1}=6 \] Since \(b_{N}\) is deﬁned, we can continue and \(y_{2}\) is found using same recurrence relation. Hence this is subcase one. For \(n=3\)\[ b_{3}=\frac {-2}{3}b_{2}=-4 \] For \(n=4\)\[ b_{4}=\frac {-1}{4}b_{3}=1 \] And so on. Hence\begin {align*} y_{2} & =\frac {1}{x^{2}}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{x^{2}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}\right ) \\ & =\frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \end {align*}

Therefore\begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {2}{3}x+\frac {1}{6}x^{2}\right ) +c_{2}\left ( \frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \right ) \end {align*}

Now we ﬁnd \(y_{p}\). From earlier we found in (1B) the balance equation which gives\[ \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y=\left ( r^{2}+2r\right ) a_{0}x^{r-1}\] Relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used in ﬁnding \(y_{h}\) the above becomes\[ \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y=\left ( m^{2}+2m\right ) c_{0}x^{m-1}\] Therefore we need to balance \(\left ( m^{2}+2m\right ) c_{0}x^{m-1}=3x^{2}\). This implies \(m-1=2\) or \(m=3\). Therefore \(\left ( m^{2}+2m\right ) c_{0}=3\) or \(\left ( 9+6\right ) c_{0}=3\) which gives \(c_{0}=\frac {3}{15}=\frac {1}{5}\). Hence \begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x^{3}\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

To ﬁnd \(c_{n}\), the same recurrence relation given in (1) is used again but now \(r\) is replaced by \(m\) and \(a\) replaced by \(c\). This gives the recurrence relation to ﬁnd coeﬃcients of the particular solution as\[ c_{n}=\frac {n+m-3}{n+m+2}c_{n-1}\] For \(m=3\) the above becomes\begin {align*} c_{n} & =\frac {n+3-3}{n+3+2}c_{n-1}\\ & =\frac {n}{n+5}c_{n-1} \end {align*}

For \(n=1\)\[ c_{1}=\frac {1}{6}c_{0}=\frac {1}{6}\left ( \frac {1}{5}\right ) =\frac {1}{30}\] For \(n=2\)\[ c_{2}=\frac {2}{2+5}c_{1}=\frac {2}{7}\left ( \frac {1}{30}\right ) =\frac {1}{105}\] And so on. Hence\begin {align*} y_{p} & =x^{3}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{3}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{3}\left ( \frac {1}{5}+\frac {1}{30}x+\frac {1}{105}x^{2}+\cdots \right ) \end {align*}

Hence the ﬁnal solution\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-\frac {2}{3}x+\frac {1}{6}x^{2}\right ) +c_{2}\left ( \frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \right ) +\left ( \frac {1}{5}x^{3}+\frac {1}{30}x^{4}+\frac {1}{105}x^{5}+\cdots \right ) \end {align*}

If we try to ﬁnd \(y_{p}\) by assuming \(y_{p}=\sum _{n=0}^{\infty }c_{n}x^{n}\) and substituting into the ode and try to match coeﬃcients, we can not always be successful. The above method using the balance equation always works and that is what I am using in my solver.

Example 2\[ 4x^{2}y^{\prime \prime }+4xy^{\prime }+\left ( 4x^{2}-1\right ) y=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {4x^{2}-1}{4x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {4x^{2}-1}{4}=-\frac {1}{4}\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-\frac {1}{4} & =0\\ r^{2}-\frac {1}{4} & =0\\ r & =-\frac {1}{2},\frac {1}{2} \end {align*}

Therefore \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\).

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} 4x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+4x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+4x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) \left ( n+r-1\right ) +4\left ( n+r\right ) -1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4n^{2}+8nr+4r^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0 \tag {1} \end {align}

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }4a_{n-2}x^{n+r}=0 \tag {2} \end {equation} \(n=0\) gives\[ \left ( 4r^{2}-1\right ) a_{0}x^{r}=0 \] Since \(a_{0}\neq 0\), then \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\) as was found above. The ode therefore satisﬁes\[ 4x^{2}y^{\prime \prime }+4xy^{\prime }+\left ( 4x^{2}-1\right ) y=\left ( 4r^{2}-1\right ) a_{0}x^{r}\] Since when \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\) the RHS is zero. When \(n=1\) then (2) gives\begin {equation} \left ( 4\left ( 1+r\right ) ^{2}-1\right ) a_{1}=0 \tag {3} \end {equation} The recurrence relation is when \(n\geq 2\) from (2) is given by\begin {align} \left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}+4a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-4}{4\left ( n+r\right ) ^{2}-1}a_{n-2} \tag {4} \end {align}

Since roots diﬀer by an integer \(N=1\) then there two linearly independent solutions can be constructed using\begin {align*} y_{1} & =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +x^{r_{2}}\sum _{n=0}^{\infty }b_{n}x^{n} \end {align*}

\(C\) above can come out to be zero. We start by ﬁnding \(y_{1}\) (the one with the larger \(r\)).

Now, using \(r=\frac {1}{2}\). For \(n=1\) and from (3) \begin {align*} \left ( 4\left ( 1+\frac {1}{2}\right ) ^{2}-1\right ) a_{1} & =0\\ 8a_{1} & =0\\ a_{1} & =0 \end {align*}

From \(n=2\) from (4) and using \(r=\frac {1}{2}\) it becomes\begin {align} a_{n} & =\frac {-4}{4\left ( n+\frac {1}{2}\right ) ^{2}-1}a_{n-2}\nonumber \\ & =-\frac {1}{n^{2}+n}a_{n-2} \tag {5} \end {align}

For \(n=2\) then (5) gives (and using \(a_{0}=1\))\begin {align*} a_{2} & =-\frac {1}{6}a_{0}\\ & =-\frac {1}{6} \end {align*}

For \(n=3\) Eq (5) gives\begin {align*} a_{3} & =-\frac {1}{12}a_{1}\\ & =0 \end {align*}

For \(n=4\) Eq (5) gives\begin {align*} a_{4} & =-\frac {1}{20}a_{2}\\ & =-\frac {1}{20}\left ( -\frac {1}{6}\right ) =\frac {1}{120} \end {align*}

And so on. Hence\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+\frac {1}{2}}\\ & =x^{\frac {1}{2}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \end {align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we calculate \(b_{N}=b_{1}\) using same recurrence relation (1) to see if it is deﬁned or not. If it is deﬁned, then we continue, else we have to use the modiﬁed Frobenius method. From (1) and using \(b\) instead of \(a\) and using \(r=r_{2}=-\frac {1}{2}\) gives\begin {align*} \left ( 4\left ( 1-\frac {1}{2}\right ) ^{2}-1\right ) b_{1} & =0\\ 0b_{1} & =0 \end {align*}

Hence \(b_{1}\) is arbitrary. Let \(b_{1}=0\). Since \(b_{N}=b_{1}\) is deﬁned, we can continue and \(y_{2}\) is found using same recurrence relation. Hence this is subcase one. From (4) and using \(r=-\frac {1}{2}\) it becomes\begin {align} b_{n} & =\frac {-4}{4\left ( n-\frac {1}{2}\right ) ^{2}-1}b_{n-2}\nonumber \\ & =-\frac {1}{n\left ( n-1\right ) }b_{n-2} \tag {6} \end {align}

For \(n=2\) Eq (6) gives (and using \(b_{0}=1\))\begin {align*} b_{2} & =-\frac {1}{2\left ( 2-1\right ) }b_{0}\\ & =-\frac {1}{2} \end {align*}

For \(n=3\) Eq (6) gives\begin {align*} b_{3} & =-\frac {1}{3\left ( 3-1\right ) }b_{1}\\ & =0 \end {align*}

For \(n=4\) Eq (6) gives\begin {align*} b_{4} & =-\frac {1}{4\left ( 4-1\right ) }b_{2}\\ & =--\frac {1}{12}\left ( -\frac {1}{2}\right ) =\frac {1}{24} \end {align*}

And so on. Hence\begin {align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-\frac {1}{2}}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end {align*}

Therefore the ﬁnal solution is \begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end {align*}

Example 3\[ y^{\prime \prime }+y^{\prime }+y=\sqrt {x}\] This ode is here because the RHS has no series expansion at \(x=0\). Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =1,q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x=0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) & =0\\ r & =0,1 \end {align*}

Therefore \(r_{1}=1,r_{2}=0\).

Expansion around \(x=0\). This is regular singular point (due to the RHS not having series expansion). 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 {equation} \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 \tag {1} \end {equation} Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-2}=0 \tag {2} \end {equation} \(n=0\) gives\[ r\left ( r-1\right ) a_{0}x^{r-2}=0 \] Since \(a_{0}\neq 0\), then \(r_{1}=1,r_{2}=0\) as was found above. The ode therefore satisﬁes\[ y^{\prime \prime }+y^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r-2}\] When \(n=1\) then (2) gives\begin {align} \left ( 1+r\right ) \left ( r\right ) a_{1}+ra_{0} & =0\nonumber \\ a_{1} & =\frac {-a_{0}}{1+r} \tag {3} \end {align}

The recurrence relation is when \(n\geq 2\) from (2) is given by\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-\left ( n+r-1\right ) a_{n-1}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {4} \end {align}

\(C\) above can come out to be zero. We start by ﬁnding \(y_{1}\) (the one with the larger \(r\)).

Now, using \(r=1\). For \(n=1\) and from (3) and using \(a_{0}=1\) gives \begin {align*} a_{1} & =\frac {-a_{0}}{2}\\ a_{1} & =\frac {-1}{2} \end {align*}

From \(n=2\) from (4) and using \(r=1\) it becomes\begin {equation} a_{2}=\frac {-2a_{1}-a_{0}}{\left ( 2+1\right ) \left ( 2\right ) }=\frac {-2a_{1}-a_{0}}{6}=\frac {-2\left ( \frac {-1}{2}\right ) -1}{6}=0\nonumber \end {equation} For \(n=3\) then (5) gives\[ a_{3}=\frac {-\left ( 3\right ) a_{2}-a_{1}}{\left ( 3+1\right ) \left ( 3\right ) }=\frac {-a_{1}}{12}=\frac {-\left ( \frac {-1}{2}\right ) }{12}=\frac {1}{24}\] And so on. Hence\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+1}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) \end {align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we look at \(a_{1}=\frac {-a_{0}}{1+r}\) and see this is deﬁned for \(r=0\). Next we look at the recurrence relation \(a_{n}=\frac {-\left ( n+r-1\right ) a_{n-1}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }\) and see this is also deﬁned for \(r=1\). Hence \(C=0\) and we can ﬁnd \(y_{2}\) using same series expansion and using \(b_{0}=1\).\[ b_{1}=\frac {-b_{0}}{1+r}=\frac {-1}{1}=-1 \] For \(n\geq 2\) we have\[ b_{n}=\frac {-\left ( n+r-1\right ) b_{n-1}-b_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }\] Which for \(r=0\) becomes\begin {equation} b_{n}=\frac {-\left ( n-1\right ) b_{n-1}-b_{n-2}}{n\left ( n-1\right ) } \tag {5} \end {equation} For \(n=2\)\[ b_{2}=\frac {-\left ( 2-1\right ) b_{1}-b_{0}}{2}=\frac {-\left ( 2-1\right ) \left ( -1\right ) -1}{2}=0 \] For \(n=3\)\[ b_{3}=\frac {-\left ( 3-1\right ) b_{2}-b_{1}}{3\left ( 3-1\right ) }=\frac {1}{6}\] For \(n=4\)\[ b_{4}=\frac {-\left ( 3\right ) b_{3}-b_{2}}{4\left ( 3\right ) }=\frac {-\left ( 3\right ) \left ( \frac {1}{6}\right ) }{4\left ( 3\right ) }=-\frac {1}{24}\] And so on. Hence\begin {align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+0}\\ & =\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \end {align*}

Therefore \(y_{h}\)\begin {align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \right ) \end {align*}

Now we ﬁnd \(y_{p}\). From above \(y^{\prime \prime }+y^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r-2}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used\[ y^{\prime \prime }+y^{\prime }+y=m\left ( m-1\right ) c_{0}x^{m-2}\] Therefore we need to balance \(m\left ( m-1\right ) c_{0}x^{m-2}=x^{\frac {1}{2}}\) since the RHS is \(\sqrt {x}\). This implies \(m-2=\frac {1}{2}\) or \(m=\frac {5}{2}\). Therefore \(m\left ( m-1\right ) c_{0}=1\) or \(\frac {5}{2}\left ( \frac {5}{2}-1\right ) c_{0}=1\), \(c_{0}=\frac {4}{15}\). Hence \begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x^{\frac {5}{2}}\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

To ﬁnd \(c_{n}\), the same recurrence relation (4) is used by with \(r\) replaced by \(m\) and \(a\) replaced by \(c\). This gives\begin {align} c_{n} & =\frac {-\left ( n+m-1\right ) c_{n-1}-c_{n-2}}{\left ( n+m\right ) \left ( n+m-1\right ) }\nonumber \\ & =\frac {-\left ( n+\frac {5}{2}-1\right ) c_{n-1}-c_{n-2}}{\left ( n+\frac {5}{2}\right ) \left ( n+\frac {5}{2}-1\right ) }\nonumber \\ & =-4\frac {\frac {3}{2}c_{n-1}+c_{n-2}+nc_{n-1}}{\left ( 2n+3\right ) \left ( 2n+5\right ) } \tag {6} \end {align}

The above is only for \(n\geq 2\). For \(n=1\), using \(a_{1}=\frac {-a_{0}}{1+r}\) and replacing \(a\) by \(c\) and \(r\) by \(m\) gives\[ c_{1}=\frac {-c_{0}}{1+m}=\frac {-\frac {4}{15}}{1+\left ( \frac {5}{2}\right ) }=-\frac {8}{105}\] For \(n=2\) from (6)\[ c_{2}=-4\frac {\frac {3}{2}c_{1}+c_{0}+2c_{1}}{\left ( 4+3\right ) \left ( 4+5\right ) }=-4\left ( \frac {\frac {3}{2}\left ( -\frac {8}{105}\right ) +\frac {4}{15}+2\left ( -\frac {8}{105}\right ) }{\left ( 4+3\right ) \left ( 4+5\right ) }\right ) =0 \] For \(n=3\)\[ c_{3}=-4\frac {\frac {3}{2}c_{2}+c_{1}+3c_{2}}{\left ( 6+3\right ) \left ( 6+5\right ) }=-4\left ( \frac {-\frac {8}{105}}{\left ( 6+3\right ) \left ( 6+5\right ) }\right ) =\frac {32}{10\,395}\] And so on. Hence

\begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+\frac {5}{2}}\\ & =x^{\frac {5}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {5}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end {align*}

Hence the ﬁnal solution\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \right ) +x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end {align*}

Roots of indicial equation diﬀer by integer. Bad case ode internal name "second_order_series_method_regular_singular_point_diﬀerence_is_integer_bad_case".

The description is given above. Only examples are given below.

Example 1\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {x^{2}-4}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}-4=-4\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-4 & =0\\ r^{2}-4 & =0\\ r & =2,-2 \end {align*}

Therefore \(r_{1}=2,r_{2}=-2\). 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} x^{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 ( x^{2}-4\right ) \sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\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}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-4\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\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+2}-\sum _{n=0}^{\infty }4a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0\nonumber \end {align}

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r}=0 \tag {2} \end {equation} \(n=0\) gives\begin {align*} \left ( r\left ( r-1\right ) +r-4\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-4\right ) a_{0}x^{r} & =0 \end {align*}

Since \(a_{0}\neq 0\), then \(r^{2}=4\) or \(r_{1}=2,r_{2}=-2\) as was found above. The ode therefore satisﬁes\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-4\right ) y=\left ( r^{2}-4\right ) a_{0}x^{r}\] Since when \(r_{1}=2\) or \(r_{2}=-2\) then the RHS is zero. When \(n=1\) then (2) gives\begin {align*} \left ( \left ( 1+r\right ) r+\left ( 1+r\right ) -4\right ) a_{1} & =0\\ \left ( r^{2}+2r-3\right ) a_{1} & =0 \end {align*}

Hence\[ a_{1}=0 \] The recurrence relation is when \(n\geq 2\) from (2) is given by\begin {align} \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) } \tag {4} \end {align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=2\) gives\[ a_{n}=\frac {-a_{n-2}}{\left ( \left ( n+2\right ) \left ( n+2-1\right ) +\left ( n+2\right ) -4\right ) }=-\frac {1}{n}\frac {a_{n-2}}{n+4}\] We see that it is deﬁned for all \(n\geq 2\). Now we check the other root \(r_{2}=-2\). (4) now becomes\[ a_{n}=\frac {-a_{n-2}}{\left ( \left ( n-2\right ) \left ( n-3\right ) +\left ( n-2\right ) -4\right ) }=-\frac {1}{n}\frac {a_{n-2}}{n-4}\] We see that this is the diﬃcult root as at \(n=4\) it is not deﬁned as it gives \(1/0\) error. Hence \[ r_{bad}=-2 \] Therefore this is subcase two. For this case we do the following. We ﬁrst ﬁnd the solution using symbolic \(r\) using (4), and at the end replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}=\left ( r+2\right ) b_{0}\). From (4) and for \(n=2\) \[ a_{2}=\frac {-a_{0}}{\left ( \left ( 2+r\right ) \left ( 1+r\right ) +\left ( 2+r\right ) -4\right ) }=-\frac {1}{r}\frac {a_{0}}{r+4}\] Since \(a_{1}=0\) then all odd \(a_{n}=0\). For \(n=4\)\[ a_{4}=\frac {-a_{2}}{\left ( \left ( 4+r\right ) \left ( 3+r\right ) +\left ( 4+r\right ) -4\right ) }=-\frac {a_{2}}{\left ( r+6\right ) \left ( r+2\right ) }=-\frac {-\frac {1}{r}\frac {a_{0}}{r+4}}{\left ( r+6\right ) \left ( r+2\right ) }=\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }\] For \(n=6\)\begin {align*} a_{6} & =\frac {-a_{4}}{\left ( \left ( 6+r\right ) \left ( 5+r\right ) +\left ( 6+r\right ) -4\right ) }=-\frac {a_{4}}{\left ( r+8\right ) \left ( r+4\right ) }\\ & =-\frac {\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }}{\left ( r+8\right ) \left ( r+4\right ) }\\ & =-\frac {1}{r}\frac {a_{0}}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) } \end {align*}

And so on. Hence\begin {align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r}\frac {1}{r+4}x^{2}+\frac {1}{r}\frac {1}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {1}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \end {align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r+2\right ) \) the above becomes\begin {equation} \overline {y}=x^{r}b_{0}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \tag {5} \end {equation} Now\begin {align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=-2}\\ & =x^{-2}b_{0}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) _{r=-2}\\ & =x^{-2}b_{0}\left ( \frac {1}{r}\frac {1}{\left ( r+4\right ) \left ( r+6\right ) }x^{4}-\frac {1}{r}\frac {1}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) }x^{6}+\cdots \right ) _{r=-2}\\ & =x^{-2}b_{0}\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}-\cdots \right ) \end {align*}

But \(b_{0}=1\). Hence\begin {align*} y_{1} & =\left ( -\frac {1}{16}x^{2}+\frac {1}{192}x^{4}-\cdots \right ) \\ & =-\frac {1}{16}\left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) \end {align*}

We can removing the leading \(-\frac {1}{16}\) since it will be absorbed by the \(c_{1}\) constant. Hence \[ y_{1}=c_{1}\left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) \] Now we ﬁnd \(y_{2}\) using\[ y_{2}=\left ( \frac {d\overline {y}}{dr}\right ) _{r=r_{bad}}\] Notice the derivative is evaluated also at the bad root \(r=r_{bad}=-2\) same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives\begin {align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \right ) _{r=-2}\\ & =\overline {y}_{r=-2}\ln x+x^{r}\frac {d}{dr}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) _{r=-2} \end {align*}

But \[ y_{1}=\overline {y}_{r=-2}\] Therefore, evaluating all the derivatives gives\begin {align*} y_{2} & =y_{1}\ln x+x^{r}\left ( 1+\frac {\left ( r^{2}+4r+8\right ) }{r^{2}\left ( r+4\right ) ^{2}}x^{2}-\frac {1}{r^{2}}\frac {3r^{2}+20r+24}{\left ( r^{2}+10r+24\right ) ^{2}}x^{4}+\frac {\left ( 5r^{3}+68r^{2}+256r+192\right ) }{r^{2}\left ( r+4\right ) ^{3}\left ( r^{2}+14r+48\right ) ^{2}}x^{6}+\cdots \right ) _{r=-2}\\ & =y_{1}\ln x+x^{-2}\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \end {align*}

Hence

\[ y_{2}=y_{1}\ln x+\left ( \frac {1}{4}+\frac {1}{x^{2}}+\frac {1}{64}x^{2}-\frac {11}{2304}x^{4}+\cdots \cdots \right ) \] Therefore the ﬁnal solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) +\left ( \frac {1}{4}+\frac {1}{x^{2}}+\frac {1}{64}x^{2}-\frac {11}{2304}x^{4}+\cdots \cdots \right ) \right ) \end {align*}

Example 2\[ xy^{\prime \prime }-3y^{\prime }+xy=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {-3}{x},q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\left ( -3\right ) =-3\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -3r & =0\\ r^{2}-4r & =0\\ r\left ( r-4\right ) & =0\\ r & =0,4 \end {align*}

Therefore \(r_{1}=4,r_{2}=0\). 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*} xy^{\prime \prime }-3y^{\prime }+xy & =0\\ x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x\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 }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1}=0 \tag {2} \end {equation} \(n=0\) gives\begin {align*} r\left ( r-1\right ) a_{0}x^{r-1}-3ra_{0}x^{r-1} & =0\\ \left ( r\left ( r-4\right ) \right ) a_{0}x^{r-1} & =0 \end {align*}

Since \(a_{0}\neq 0\), then \(r\left ( r-4\right ) =0\) or \(r_{1}=0,r_{2}=4\) as was found above. The ode therefore satisﬁes\[ xy^{\prime \prime }-3y^{\prime }+xy=\left ( r\left ( r-4\right ) \right ) a_{0}x^{r-1}\] Since when \(r_{1}=4\) or \(r_{2}=0\) then the RHS is zero. When \(n=1\) then (2) gives\begin {align*} \left ( 1+r\right ) \left ( r\right ) a_{1}-3\left ( 1+r\right ) a_{1} & =0\\ \left ( r^{2}-2r-3\right ) a_{1} & =0 \end {align*}

Hence \[ a_{1}=0 \] The recurrence relation is when \(n\geq 2\) from (2) is given by\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}-3\left ( n+r\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) -3\left ( n+r\right ) } \tag {4} \end {align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=4\) gives\[ a_{n}=\frac {-a_{n-2}}{\left ( n+4\right ) \left ( n+3\right ) -3\left ( n+4\right ) }=-\frac {1}{n}\frac {a_{n-2}}{n+4}\] Which is deﬁned for all \(n\geq 2\). Checking the second root \(r=0\) gives \[ a_{n}=\frac {-a_{n-2}}{\left ( n+0\right ) \left ( n+0-1\right ) -3\left ( n+0\right ) }=-\frac {1}{n}\frac {a_{n-2}}{n-4}\] Which is not deﬁned for \(n=4\). Hence this is subcase two, where \(y_{2}\) does not exist using standard method. Hence\[ r_{bad}=0 \] For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}\). From (4) and for \(n=2\) \[ a_{2}=\frac {-a_{0}}{\left ( 2+r\right ) \left ( 1+r\right ) -3\left ( 2+r\right ) }=-\frac {a_{0}}{r^{2}-4}\] Since \(a_{1}=0\) then all odd \(a_{n}=0\). For \(n=4\)\[ a_{4}=\frac {-a_{2}}{\left ( 4+r\right ) \left ( 4+r-1\right ) -3\left ( 4+r\right ) }=\frac {\frac {a_{0}}{r^{2}-4}}{r\left ( r+4\right ) }=\frac {a_{0}}{r\left ( r+4\right ) \left ( r^{2}-4\right ) }\] For \(n=6\)\begin {align*} a_{6} & =\frac {-a_{4}}{\left ( 6+r\right ) \left ( 5+r\right ) -3\left ( 6+r\right ) }=\frac {-\frac {a_{0}}{r\left ( r+4\right ) \left ( r^{2}-4\right ) }}{r^{2}+8r+12}\\ & =\frac {-a_{0}}{\left ( r^{2}+8r+12\right ) r\left ( r+4\right ) \left ( r^{2}-4\right ) } \end {align*}

And so on. Hence\begin {align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r^{2}-4}x^{2}+\frac {1}{r\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) r\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \end {align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{2}\right ) \) or \(b_{0}r\) since \(r_{2}=0\), the above becomes\begin {equation} \overline {y}=x^{r}b_{0}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \tag {5} \end {equation} Now\begin {align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=0}\\ & =b_{0}\left ( \frac {1}{\left ( 4\right ) \left ( -4\right ) }x^{4}-\frac {1}{\left ( 12\right ) \left ( 4\right ) \left ( -4\right ) }x^{6}+\cdots \right ) \\ & =b_{0}\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}+\cdots \right ) \end {align*}

But \(b_{0}=1\). Hence\begin {align*} y_{1} & =\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}+\cdots \right ) \\ & =-\frac {1}{16}\left ( x^{4}-\frac {1}{12}x^{6}+\cdots \right ) \end {align*}

We can removing the leading \(-\frac {1}{16}\) since it will be absorbed by the \(c_{1}\) constant. Hence \begin {align*} y_{1} & =c_{1}\left ( x^{4}-\frac {1}{12}x^{6}+\cdots \right ) \\ & =x^{4}c_{1}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \end {align*}

Now we ﬁnd \(y_{2}\) using\[ y_{2}=\left ( \frac {d\overline {y}}{dr}\right ) _{r=r_{bad}}\] Notice the derivative is evaluated also at root \(r=r_{bad}=0\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives\begin {align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{r}\frac {d}{dr}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{0}\left ( 1+\frac {\left ( r^{2}+4\right ) }{\left ( r^{2}-4\right ) ^{2}}x^{2}-\frac {3r^{2}+8r-4}{\left ( r^{3}+4r^{2}-4r-16\right ) ^{2}}x^{4}+\frac {1}{\left ( r+2\right ) ^{3}}\frac {5r^{3}+38r^{2}+44r-88}{\left ( r^{3}+8r^{2}+4r-48\right ) ^{2}}x^{6}-\cdots \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \end {align*}

But \[ \overline {y}_{r=0}=y_{1}\] Therefore\[ y_{2}=y_{1}\ln x+\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \] The complete solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =x^{4}c_{1}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( x^{4}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \right ) +\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \right ) \end {align*}

Example 3\[ x^{2}y^{\prime \prime }+\left ( x^{2}-2x\right ) y^{\prime }+2y=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Show that \(p\left ( x\right ) =\frac {\left ( x^{2}-2x\right ) }{x^{2}}=\frac {\left ( x-2\right ) }{x},q\left ( x\right ) =\frac {2}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\left ( x-2\right ) =-2\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}2=2\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -2r+2 & =0\\ r^{2}-3r+2 & =0\\ r & =2,1 \end {align*}

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

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r}-\sum _{n=0}^{\infty }2\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }2a_{n}x^{n+r}=0 \tag {2} \end {equation} \(n=0\) gives\begin {align*} r\left ( r-1\right ) a_{0}x^{r}-2ra_{0}x^{r}+2a_{0}x^{r} & =0\\ \left ( r\left ( r-1\right ) -2r+2\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-3r+2\right ) a_{0}x^{r} & =0 \end {align*}

Since \(a_{0}\neq 0\), then \(r^{2}-3r+2=0,\)or \(r_{1}=2,r_{2}=1\) as was found above. The ode therefore satisﬁes\[ x^{2}y^{\prime \prime }+\left ( x^{2}-2x\right ) y^{\prime }+2y=\left ( r^{2}-3r+2\right ) a_{0}x^{r}\] Recurrence relation is when \(n\geq 1\). From (2)\begin {equation} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}-2\left ( n+r\right ) a_{n}+2a_{n}=0\nonumber \end {equation} Therefore\begin {align} a_{n} & =-\frac {\left ( n+r-1\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) -2\left ( n+r\right ) +2}a_{n-1}\nonumber \\ & =-\frac {1}{n+r-2}a_{n-1} \tag {3} \end {align}

We check ﬁrst if this is subcase one or two. To do this, we check if the above recurrence relation is deﬁned for both roots for all \(n\geq 1\). The above for \(r=r_{1}=2\) gives\[ a_{n}=-\frac {1}{n}a_{n-1}\] Which is deﬁned for all \(n\geq 1\). Checking the second root \(r=1\) gives \[ a_{n}=-\frac {1}{n-1}a_{n-1}\] Which is not deﬁned for \(n=1\). Hence this is subcase two, where \(y_{2}\) does not exist using standard method. Hence\[ r_{bad}=1 \] For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}\). From (3) and for \(n=1\) \[ a_{1}=-\frac {1}{r-1}a_{0}\] For \(n=2\)\[ a_{2}=-\frac {1}{r}a_{1}=\frac {1}{\left ( r\right ) \left ( r-1\right ) }a_{0}\] For \(n=3\)\[ a_{3}=-\frac {1}{r+1}a_{2}=-\frac {a_{0}}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) }\] For \(n=4\)\[ a_{4}=-\frac {1}{2+r}a_{3}=\frac {a_{0}}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) \left ( r+2\right ) }\] And so on. Hence\begin {align*} \overline {y} & =x^{r}\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r-1}x+\frac {1}{\left ( r\right ) \left ( r-1\right ) }x^{2}-\frac {1}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) }x^{3}+\frac {1}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \end {align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r-1\right ) \) since \(r_{bad}=1\), the above becomes\begin {align} \overline {y} & =x^{r}b_{0}\left ( \left ( r-1\right ) -\frac {\left ( r-1\right ) }{r-1}x+\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) }x^{2}-\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) }x^{3}+\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \tag {5} \end {align}

Now\begin {align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=1}\\ & =xb_{0}\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{\left ( 1\right ) \left ( 2\right ) \left ( 3\right ) }x^{4}-\cdots \right ) \\ & =xb_{0}\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{6}x^{4}-\cdots \right ) \end {align*}

But \(b_{0}=1\). Hence\begin {align*} y_{1} & =x\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{6}x^{4}-\cdots \right ) \\ & =-x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \end {align*}

Now we ﬁnd \(y_{2}\) using\[ y_{2}=\left ( \frac {d\overline {y}}{dr}\right ) _{r=r_{bad}}\] Notice the derivative is evaluated also at root \(r=r_{bad}=1,\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives\begin {align*} y_{2} & =\frac {d}{dr}\left ( x^{r}b_{0}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \right ) _{r=1}\\ & =\overline {y}_{r=1}\ln x+x^{r=1}\frac {d}{dr}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\frac {d}{dr}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\left ( 1-\frac {1}{r^{2}}x^{2}+\frac {1}{r^{2}}\frac {2r+1}{\left ( r+1\right ) ^{2}}x^{3}-\frac {1}{r^{2}}\frac {3r^{2}+6r+2}{\left ( r^{2}+3r+2\right ) ^{2}}x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\left ( 1-x^{2}+\frac {3}{4}x^{3}-\frac {11}{36}x^{4}-\cdots \right ) \end {align*}

Therefore\[ y_{2}=y_{1}\ln x+\left ( x-x^{3}+\frac {3}{4}x^{4}-\frac {11}{36}x^{5}-\cdots \right ) \] The complete solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( -x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( -x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \right ) +\left ( x-x^{3}+\frac {3}{4}x^{4}-\frac {11}{36}x^{5}-\cdots \right ) \right ) \end {align*}

Example 4\[ \left ( x-1\right ) y^{\prime \prime }+xy^{\prime }+\frac {y}{x}=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {x}{x-1},q\left ( x\right ) =\frac {1}{x\left ( x-1\right ) }\). There is a singular point at \(x=0\) and at \(x=1\). For \(x=0\), \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =0\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) & =0\\ r & =0,1 \end {align*}

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

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=1}^{\infty }\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}x^{n+r-2}-\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=2}^{\infty }\left ( n+r-2\right ) a_{n-2}x^{n+r-2}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-2}=0 \tag {2} \end {equation} \(n=0\) gives\[ \left ( r\left ( r-1\right ) \right ) a_{0}=0 \] Since \(a_{0}\neq 0\), then \(r_{1}=0,r_{2}=1\) as was found above. For \(n=1\)\begin {align*} \left ( r\right ) \left ( r-1\right ) a_{0}-\left ( 1+r\right ) \left ( r\right ) a_{1}+a_{0} & =0\\ a_{1} & =\frac {a_{0}+\left ( r\right ) \left ( r-1\right ) a_{0}}{\left ( 1+r\right ) \left ( r\right ) }=\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0} \end {align*}

For \(r=0\) the above is not deﬁned. Therefore this falls into case two (diﬃcult case). Hence \(r_{bad}=0\). For \(r=1\) we see \(a_{1}\) is deﬁned.

For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}=rb_{0}\). For \(n=1\) \[ a_{1}=\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0}\] For \(n\geq 2\), the recurrence relation is\[ \left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}-\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-2\right ) a_{n-2}+a_{n-1}=0 \] Or\begin {equation} a_{n}=\frac {\left ( n+r-1\right ) \left ( n+r-2\right ) +1}{\left ( n+r\right ) \left ( n+r-1\right ) }a_{n-1}+\frac {\left ( n+r-2\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) }a_{n-2} \tag {3} \end {equation} For \(n=2\)\begin {align*} a_{2} & =\frac {\left ( 1+r\right ) \left ( r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{1}+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{0}\\ & =\frac {r\left ( 1+r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }\left ( \frac {1+r\left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0}\right ) +\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{0}\\ & =\left ( \frac {r\left ( 1+r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }\frac {1+r\left ( r-1\right ) }{r\left ( 1+r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) a_{0}\\ & =\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) a_{0} \end {align*}

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

And so on. Hence\begin {align*} \overline {y} & =x^{r}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \end {align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}r\) since \(r_{bad}=0\), the above becomes\begin {align} \overline {y} & =x^{r}b_{0}\left ( r+r\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }x+r\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \tag {5} \end {align}

Now\begin {align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=0}\\ & =x^{0}b_{0}\left ( x+\left ( \frac {1}{\left ( 2\right ) \left ( 1\right ) \left ( 1\right ) }\right ) x^{2}+\left [ \frac {\left ( 2\right ) \left ( 1\right ) +1}{\left ( 3\right ) \left ( 2\right ) }\left ( \frac {\left ( 1\right ) \left ( 1\right ) }{\left ( 2\right ) \left ( 1\right ) \left ( 1\right ) }\right ) +\frac {\left ( 1\right ) }{\left ( 3\right ) \left ( 2\right ) }\right ] x^{3}\cdots \right ) \\ & =b_{0}\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) \end {align*}

But \(b_{0}=1\). Hence\[ y_{1}=x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \] \(y_{2}\) is found using\[ y_{2}=\left ( \frac {d\overline {y}}{dr}\right ) _{r=r_{bad}}\] Notice the derivative is evaluated also at root \(r=r_{bad}=0,\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives\begin {align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{r=0}\frac {d}{dr}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) _{r=0} \end {align*}

But \(\overline {y}_{r=0}=y_{1}\). The above becomes\[ y_{2}=y_{1}\ln x+\frac {d}{dr}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) _{r=0}\] Carrying out the derivatives gives\[ y_{2}=y_{1}\ln x+\left ( 1+\frac {1}{\left ( r+1\right ) ^{2}}\left ( r^{2}+2r-2\right ) x+\left ( \frac {\left ( r^{5}+7r^{4}+10r^{3}+8r^{2}+5r-5\right ) }{\left ( r+1\right ) ^{3}\left ( r+2\right ) ^{2}}\right ) x^{2}+\cdots \right ) _{r=0}\] Evaluating at \(r=0\)\[ y_{2}=y_{1}\ln x+\left ( 1-2x-\frac {5}{4}x^{2}+\cdots \right ) \] Therefore the complete solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) +\left ( 1-2x-\frac {5}{4}x^{2}+\cdots \right ) \right ) \end {align*}

Example 5\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {x^{2}-4}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\left ( x^{2}-1\right ) =-1\). Hence the indicial equation is \begin {align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-1 & =0\\ r^{2}-1 & =0\\ r & =1,-1 \end {align*}

Therefore \(r_{1}=1,r_{2}=-1\). 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} x^{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 ( x^{2}-1\right ) \sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\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}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\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+2}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \end {align}

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\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=2}^{\infty }a_{n-2}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \tag {2} \end {equation} \(n=0\) gives\begin {align*} r\left ( r-1\right ) a_{0}x^{r}+ra_{0}x^{r}-a_{0} & =0\\ \left ( r\left ( r-1\right ) +r-1\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-1\right ) a_{0}x^{r} & =0 \end {align*}

Since \(a_{0}\neq 0\), then \(r^{2}=1\) or \(r_{1}=,r_{2}=-1\) as was found above. The ode therefore satisﬁes\begin {equation} x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=\left ( r^{2}-1\right ) a_{0}x^{r} \tag {2A} \end {equation}

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

Hence\[ a_{1}=0 \] The recurrence relation is when \(n\geq 2\) from (2) is given by\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n-2}-a_{n} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -1} \tag {4} \end {align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=1\) gives\[ a_{n}=\frac {-a_{n-2}}{\left ( n+1\right ) n+n}\] We see that it is deﬁned for all \(n\geq 2\). Now we check the other root \(r_{2}=-1\). (4) now becomes\[ a_{n}=\frac {-a_{n-2}}{\left ( n-1\right ) \left ( n-2\right ) +\left ( n-2\right ) }\] We see that this is the diﬃcult root as at \(n=2\) it is not deﬁned as it gives \(1/0\) error. Hence \[ r_{bad}=-1 \] Therefore this is subcase two. For this case we do the following. We ﬁrst ﬁnd the solution using symbolic \(r\) using (4), and at the end replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}=\left ( r+1\right ) b_{0}\). From (4) and for \(n=2\) \[ a_{2}=\frac {-a_{0}}{\left ( \left ( 2+r\right ) \left ( 1+r\right ) +\left ( 2+r\right ) -1\right ) }=\frac {-a_{0}}{\left ( r+1\right ) \left ( r+3\right ) }\] Since \(a_{1}=0\) then all odd \(a_{n}=0\). For \(n=4\)\[ a_{4}=\frac {-a_{2}}{\left ( \left ( 4+r\right ) \left ( 3+r\right ) +\left ( 4+r\right ) -1\right ) }=-\frac {a_{2}}{\left ( r+5\right ) \left ( r+3\right ) }=-\frac {\frac {-a_{0}}{\left ( r+1\right ) \left ( r+3\right ) }}{\left ( r+5\right ) \left ( r+3\right ) }=\frac {a_{0}}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }\] For \(n=6\)\begin {align*} a_{6} & =\frac {-a_{4}}{\left ( \left ( 6+r\right ) \left ( 5+r\right ) +\left ( 6+r\right ) -1\right ) }=-\frac {a_{4}}{\left ( r+7\right ) \left ( r+5\right ) }=-\frac {\frac {a_{0}}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }}{\left ( r+7\right ) \left ( r+5\right ) }\\ & =-\frac {a_{0}}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) } \end {align*}

And so on. Hence\begin {align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{\left ( r+1\right ) \left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \end {align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r+1\right ) \) the above becomes\begin {align} \overline {y} & =x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {\left ( r+1\right ) }{\left ( r+1\right ) \left ( r+3\right ) }x^{2}+\frac {\left ( r+1\right ) }{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{4}-\frac {\left ( r+1\right ) }{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \tag {5} \end {align}

Now\begin {align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=-1}\\ & =x^{-1}b_{0}\left ( -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1}\\ & =x^{-1}b_{0}\left ( -\frac {1}{\left ( -1+3\right ) }x^{2}+\frac {1}{\left ( -1+5\right ) \left ( -1+3\right ) \left ( -1+3\right ) }x^{4}-\frac {1}{\left ( -1+7\right ) \left ( -1+5\right ) \left ( -1+5\right ) \left ( -1+3\right ) \left ( -1+3\right ) }x^{6}+\cdots \right ) \\ & =x^{-1}b_{0}\left ( -\frac {1}{2}x^{2}+\frac {1}{16}x^{4}-\frac {1}{384}x^{6}+\cdots \right ) \end {align*}

\(b_{0}=1\). Hence\begin {align*} y_{1} & =\frac {1}{x}\left ( -\frac {1}{2}x^{2}+\frac {1}{16}x^{4}-\frac {1}{384}x^{6}+\cdots \right ) \\ & =\left ( -\frac {1}{2}x+\frac {1}{16}x^{3}-\frac {1}{384}x^{5}+\cdots \right ) \\ & =-\frac {1}{2}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \end {align*}

We can remove the leading \(-\frac {1}{2}\) since it will be absorbed by the \(c_{1}\) constant. Hence \[ y_{1}=\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \] Now we ﬁnd \(y_{2}\) using\[ y_{2}=\left ( \frac {d\overline {y}}{dr}\right ) _{r=r_{bad}}\] Notice the derivative is evaluated also at the bad root \(r=r_{bad}=-2\) same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives\begin {align*} y_{2} & =\frac {d}{dr}\left ( x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \right ) _{r=-1}\\ & =\overline {y}_{r=-1}\ln x+x^{r}\frac {d}{dr}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1} \end {align*}

But \[ y_{1}=\overline {y}_{r=-2}\] Therefore, evaluating all the derivatives gives\begin {align*} y_{2} & =y_{1}\ln x+x^{-1}\frac {d}{dr}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1}\\ & =y_{1}\ln x+x^{-1}\left ( 1+\frac {1}{\left ( r+3\right ) ^{2}}x^{2}-\frac {3r+13}{\left ( r+3\right ) ^{3}\left ( r+5\right ) ^{2}}x^{4}+\frac {1}{\left ( r+7\right ) ^{2}}\frac {5r^{2}+52r+127}{\left ( r^{2}+8r+15\right ) ^{3}}x^{6}+\cdots \right ) _{r=-1}\\ & =y_{1}\ln x+x^{-1}\left ( 1+\frac {1}{4}x^{2}-\frac {5}{64}x^{4}+\frac {5}{1152}x^{6}+\cdots \right ) \end {align*}

Hence\[ y_{2}=y_{1}\ln x+\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \] Therefore the ﬁnal solution is\begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \end {align*}

Example 6\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=1 \] This is same example as above but with non zero in the RHS. So we can use the solution for \(y_{h}\) obtained above, but need to ﬁnd \(y_{p}\) here and add these to obtain the general solution. From above we found that\begin {align*} y_{h} & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \end {align*}

And from (2A) in the above example we also found the balance equation, which is always the starting point to ﬁnding \(y_{p}\), which is \[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=\left ( r^{2}-1\right ) a_{0}x^{r}\] Therefore, and as we did all the time, relabel \(r\) as \(m\) and \(a\) as \(c\) so not to confuse notations. Therefore we have\[ \left ( m^{2}-1\right ) c_{0}x^{m}=1 \] Hence \[ m=0 \] This implies \(\left ( m^{2}-1\right ) c_{0}=1\) or \[ c_{0}=-1 \] Now we ﬁnd \(y_{p}\) using the same recursive relation found when ﬁnding \(y_{h}\) terms but using \(r=m=0\) now and using \(a_{0}=c_{0}=-1\) (instead of \(a_{0}=1\,\) as is always done when ﬁnding \(y_{h}\)). Also let \(c_{1}=0\) as that is the same as \(a_{1}\). Now we get to the recurrence relation (4) in last example which is \[ a_{n}=\frac {-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -1}\] Using \(c\) in place of \(a\) and using \(m\) in place \(r\) it becomes for \(n\geq 2\)\[ c_{n}=\frac {-c_{n-2}}{\left ( n+m\right ) \left ( n+m-1\right ) +\left ( n+m\right ) -1}\] But \(m=0\)\[ c_{n}=\frac {-c_{n-2}}{n\left ( n-1\right ) +\left ( n-1\right ) }\] For \(n=2\)\[ c_{2}=\frac {-c_{0}}{2+1}=-\frac {c_{0}}{3}\] But \(c_{0}=-1\). The above becomes\[ c_{2}=\frac {-c_{0}}{2+1}=\frac {1}{3}\] For \(n=4\) (since all odd \(c_{n}=0\))\[ c_{4}=\frac {-c_{2}}{4\left ( 3\right ) +\left ( 3\right ) }=\frac {-\frac {1}{3}}{4\left ( 3\right ) +\left ( 3\right ) }=-\frac {1}{45}\] For \(n=6\)\[ c_{6}=\frac {-c_{4}}{6\left ( 5\right ) +\left ( 5\right ) }=\frac {\frac {1}{45}}{6\left ( 5\right ) +\left ( 5\right ) }=\frac {1}{1575}\] And so on. Hence\begin {align*} y_{p} & =x^{m}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}+c_{2}x^{2}+c_{4}x^{4}+\cdots \\ & =-1+\frac {1}{3}x^{2}-\frac {1}{45}x^{4}+\frac {1}{1575}x^{6}+\cdots \end {align*}

Hence the general solution is\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\\ & c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \\ & +\left ( -1+\frac {1}{3}x^{2}-\frac {1}{45}x^{4}+\frac {1}{1575}x^{6}+\cdots \right ) \end {align*}

Example 7\[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=\frac {1}{x}\] This is same example as above but with \(\frac {1}{x}\) instead of \(1\) in the RHS to show that there will not be a series solution in this. From (2A) in the above example we found the balance equation, which is always the starting point to ﬁnding \(y_{p}\), which is \[ x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=\left ( r^{2}-1\right ) a_{0}x^{r}\] Therefore, and as we did all the time, relabel \(r\) as \(m\) and \(a\) as \(c\) so not to confuse notations. Therefore we have\[ \left ( m^{2}-1\right ) c_{0}x^{m}=x^{-1}\] Hence \[ m=-1 \] This implies \(\left ( m^{2}-1\right ) c_{0}=1\) or \begin {align*} \left ( \left ( -1\right ) ^{2}-1\right ) c_{0} & =1\\ 0c_{0} & =1 \end {align*}

Therefore no solution exists. This is why there is no series solution for this ode. If we try to solve this using Maple, will will get no answer and the above explains why.

Roots of indicial equation are repeated ode internal name "second_order_series_method_regular_singular_point_repeated_root".

In this case the solution is \[ y=c_{1}y_{1}+c_{2}y_{2}\] Where \begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}} \end {align*}

\(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary. The coeﬃcients \(b_{n}\) are not found from the recurrence relation but found using using \(b_{n}=\frac {d}{dr}a_{n}\left ( r\right ) \) after ﬁnding \(a_{n}\) ﬁrst, and the result evaluated at root \(r_{2}\). (notice that \(r=r_{1}=r_{2}\) in this case). Notice there is no \(C\) term in from of the \(\ln \) in this case as when root diﬀer by an integer and the sum on \(b_{n}\) starts at \(1\) since \(b_{0}\) is always zero due to \(\frac {d}{dr}a_{0}\left ( r\right ) =0\) always as \(a_{0}=1\) by default.

Example 1

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {1}{x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x=0\). 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 \(r_{1}=0,r_{2}=0\).

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

Re indexing to lowest powers on \(x\) gives\begin {equation} \sum _{n=0}^{\infty }\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=1}^{\infty }a_{n-1}x^{n+r}=0 \tag {1} \end {equation} The indicial equation is obtained from \(n=0\). The above reduces to\begin {align*} \sum _{n=0}^{\infty }\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} & =0\\ \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n} & =0\\ \left ( r\right ) \left ( r-1\right ) a_{0}+ra_{0} & =0\\ a_{0}\left ( \left ( r^{2}-r\right ) +r\right ) & =0\\ a_{0}r^{2} & =0 \end {align*}

Since \(a_{0}\neq 0\) then\[ r^{2}=0 \] Hence \(r_{1}=0,r_{2}=0\). 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*}

For \(n\geq 1\) the recurrence relation is\begin {align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {a_{n-1}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) }\nonumber \\ & =-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}} \tag {1} \end {align}

Starting with \(y_{1}\). From (1) with \(r=0\) gives\[ a_{n}=-\frac {a_{n-1}}{n^{2}}\] For \(n=1\) and using \(a_{0}=1\)\[ a_{1}=-1 \] For \(n=2\)\[ a_{2}=-\frac {a_{1}}{4}=\frac {1}{4}\] And so on. Hence\begin {align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \end {align*}

In the case of duplicate roots, \(b_{n}\) is found using \(b_{n}=\frac {d}{dr}a_{n}\left ( r\right ) \). And this is evaluated at \(r=r_{0}=0\) in this case since \(r_{0}=0\) here. So we need to ﬁnd \(a_{n}\left ( r\right ) \). This is done from (1). For \(n=1\)\begin {align*} b_{1} & =\frac {d}{dr}\left ( a_{1}\left ( r\right ) \right ) \\ b_{1} & =\frac {d}{dr}\left ( -\frac {a_{0}}{\left ( 1+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( -\frac {1}{\left ( 1+r\right ) ^{2}}\right ) =\frac {2}{\left ( r+1\right ) ^{3}} \end {align*}

Evaluated at \(r=0\) gives\[ b_{1}=2 \] For \(n=2\,\ \) then (2) becomes\begin {align*} b_{2} & =\frac {d}{dr}\left ( a_{2}\left ( r\right ) \right ) \\ b_{2} & =\frac {d}{dr}\left ( -\frac {a_{1}}{\left ( 2+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( -\frac {-\frac {1}{\left ( 1+r\right ) ^{2}}}{\left ( 2+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( \frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\right ) =-2\frac {2r+3}{\left ( r^{2}+3r+2\right ) ^{3}} \end {align*}

At \(r=0\) the above becomes\[ b_{2}=-2\frac {3}{\left ( 2\right ) ^{3}}=-\frac {3}{4}\] And so on. Just remember when replacing the \(a_{n}\) in the above, is to use the original \(a_{n}\left ( r\right ) \) as function of \(r\) and not the actual \(a_{n}\) values from above. It has to be function of \(r\) ﬁrst before taking derivatives, Hence\begin {align*} y_{2} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +b_{1}x+b_{2}x^{2}+b_{3}x^{3}+\cdots \\ & =y_{1}\ln \left ( x\right ) +2x-\frac {3}{4}x^{2}+\cdots \\ & =y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {3}{4}x^{2}+\cdots \right ) \end {align*}

Therefore the general solution is \begin {align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \right ) +c_{2}\left ( y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {3}{4}x^{2}+\cdots \right ) \right ) \end {align*}

Example 2

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=1 \] The homogenous ode was solved up, so we just need to ﬁnd \(y_{p}\). To ﬁnd \(y_{p}\), and using \(m\) in place of \(r\) and \(c\) in place of \(a\) so not to confuse terms with the \(y_{h}\) terms, then from the above problem, we found the indicial equation. Hence the balance equation is\[ c_{0}m^{2}x^{m}=1 \] To balance this we need \(m=0\). Hence \(0c_{0}=1\) which is not possible. Hence no particular solution exists. No solution in series exists.

Example 3

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=\frac {1}{x}\] This is the same ode as above but with diﬀerent RHS. So we will go directly to ﬁnding \(y_{p}\). From above we found that the balance equation is \[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=m^{2}c_{0}x^{m}\] Hence\[ m^{2}c_{0}x^{m}=x^{-1}\] Which implies \(m=-1\) and therefore \(m^{2}c_{0}=1\) or \(c_{0}=1\). Using the recurrence equation (1) in the above problem using using \(c_{n}\) in place of \(a_{n}\) and \(m\) in place or \(r\) gives\[ c_{n}=-\frac {c_{n-1}}{\left ( n+m\right ) ^{2}}\] For \(m=-1\)\[ c_{n}=-\frac {c_{n-1}}{\left ( n-1\right ) ^{2}}\] Hence\begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\frac {1}{x}\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

Now to ﬁnd few \(c_{n}\) terms. For \(n=1\)\[ c_{1}=-\frac {c_{0}}{\left ( 1-1\right ) ^{2}}\] Which is not deﬁned. Hence no \(y_{p}\) exist. There is no solution in terms of series solution.

Example 4

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=x \] This is the same ode as above, where we found \(y_{h}\) but with diﬀerent RHS. So we will go directly to ﬁnding \(y_{p}\). From above we found that the balance equation is \[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=m^{2}c_{0}x^{m}\] Hence\[ m^{2}c_{0}x^{m}=x \] Which implies \(m=1\) and therefore \(m^{2}c_{0}=1\) or \(c_{0}=1\). Using the recurrence equation (1) in the above problem and using \(c_{n}\) in place of \(a_{n}\) and \(m\) in place or \(r\) gives\[ c_{n}=-\frac {c_{n-1}}{\left ( n+m\right ) ^{2}}\] For \(m=1\)\[ c_{n}=-\frac {c_{n-1}}{\left ( n+1\right ) ^{2}}\] Hence\begin {align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}

Now to ﬁnd few \(c_{n}\) terms. For \(n=1\)\[ c_{1}=-\frac {c_{0}}{\left ( 2\right ) ^{2}}=-\frac {1}{4}\] For \(n=2\)\[ c_{2}=-\frac {c_{1}}{\left ( 2+1\right ) ^{2}}=\frac {\frac {1}{4}}{9}=\frac {1}{36}\] For \(n=3\)\[ c_{3}=-\frac {c_{2}}{\left ( 3+1\right ) ^{2}}=-\frac {\frac {1}{36}}{16}=-\frac {1}{576}\] And so on. Hence \begin {align*} y_{p} & =x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{4}x+\frac {1}{36}x^{2}-\frac {1}{576}x^{3}+\cdots \right ) \\ & =\left ( x-\frac {1}{4}x^{2}+\frac {1}{36}x^{3}-\frac {1}{576}x^{4}+\cdots \right ) \end {align*}

Using \(y_{h}\) found in the above problem since that does not change, then the general solution is\begin {align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\frac {1}{576}x^{4}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\frac {1}{576}x^{4}+\cdots \right ) +\left ( 2x-\frac {3}{4}x^{2}+\frac {14}{108}x^{3}+\cdots \right ) \right ) \\ & +\left ( x-\frac {1}{4}x^{2}+\frac {1}{36}x^{3}-\frac {1}{576}x^{4}+\cdots \right ) \end {align*}

Example 5\[ xy^{\prime \prime }+y^{\prime }-xy=0 \] Comparing the ode to \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =-1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). 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*} x\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}-x\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+1} & =0 \end {align*}

Re indexing to lowest powers on \(x\) gives\begin {align} \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=2}^{\infty }a_{n-2}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) \right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1} & =0 \tag {1} \end {align}

The indicial equation is obtained from \(n=0\). The above reduces to\[ r^{2}a_{0}x^{n+r-1}=0 \] 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\begin {align*} \left ( 1+r\right ) \left ( r\right ) a_{1}+\left ( 1+r\right ) a_{1} & =0\\ \left ( r+1\right ) ^{2}a_{1} & =0 \end {align*}

Hence \(a_{1}=0\). The recurrence relation is obtained for \(n\geq 2\). From (1)\begin {align} n+r\left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}-a_{n-2} & =0\nonumber \\ a_{n} & =\frac {a_{n-2}}{\left ( n+r\right ) ^{2}} \tag {1} \end {align}

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 \(n=2\)\[ a_{2}=\frac {a_{0}}{\left ( 2+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}}\] For \(n=3\)\[ a_{3}=\frac {a_{1}}{\left ( 3+r\right ) ^{2}}=0 \] For \(n=4\)\[ a_{4}=\frac {a_{2}}{\left ( 4+r\right ) ^{2}}=\frac {\frac {1}{\left ( 2+r\right ) ^{2}}}{\left ( 4+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}}\] For \(n=5\), we will ﬁnd \(a_{5}=0\) (for all odd \(n\) this is the case). For \(n=6\)\[ a_{6}=\frac {a_{4}}{\left ( 6+r\right ) ^{2}}=\frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}\left ( 6+r\right ) ^{2}}\] And so on. We see that \(n^{th}\) term is \(a_{n}=\Pi _{j=1}^{k}\frac {1}{\left ( 2j+r\right ) ^{2}}\). Now we can substitute the \(r=0\) value into the above to obtain\begin {align*} a_{2} & =\frac {1}{4}\\ a_{4} & =\frac {1}{64}\\ a_{6} & =\frac {1}{2304} \end {align*}

Hence\begin {align*} y_{1} & =\sum a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \end {align*}

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_{2} & =\frac {d}{dr}a_{2}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}}\right ) =\left ( -\frac {2}{\left ( r+2\right ) ^{3}}\right ) _{r=0}=-\frac {2}{8}=-\frac {1}{4}\\ b_{4} & =\frac {d}{dr}a_{4}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}}\right ) =\left ( -4\frac {r+3}{\left ( r^{2}+6r+8\right ) ^{3}}\right ) _{r=0}=\left ( -4\frac {3}{\left ( 8\right ) ^{3}}\right ) =-\frac {3}{128}\\ b_{6} & =\frac {d}{dr}a_{6}\\ & =\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}\left ( 6+r\right ) ^{2}}\right ) \\ & =\left ( -2\frac {3r^{2}+24r+44}{\left ( r^{3}+12r^{2}+44r+48\right ) ^{3}}\right ) _{r=0}\\ & =-2\frac {44}{\left ( 48\right ) ^{3}}\\ & =-\frac {11}{13\,824} \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 ( b_{2}x^{2}+b_{4}x^{4}+b_{6}x^{6}+\cdots \right ) \\ & =y_{1}\ln \left ( x\right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\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}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\cdots \right ) \right ) \end {align*}

Example 6\[ \sin \left ( x\right ) y^{\prime \prime }+y^{\prime }+y=0 \] 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*}

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*}

\(\allowbreak \)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} \(\allowbreak \)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*}

[next] [prev] [prev-tail] [front] [up]