Internal problem ID [6498]
Internal file name [OUTPUT/5746_Sunday_June_05_2022_03_52_34_PM_45330625/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Problems for review and
discovert. (B) Challenge Problems . Page 194
Problem number: 1(c).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[_Gegenbauer]
\[ \boxed {\left (-x^{2}+1\right ) y^{\prime \prime }-2 y^{\prime } x +p \left (p +1\right ) y=0} \] With the expansion point for the power series method at \(x = \infty \).
Since expansion is around \(\infty \), then the independent variable \(x\) is replaced by \(\frac {1}{t}\) and the expansion is made around \(t = 0\) and after solving, the solution is changed back to \(x\) using \(x = \frac {1}{t}\). Changing variables results in the new ode \begin {align*} -\left (-\frac {1}{t^{2}}+1\right ) \left (-\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-2 \left (\frac {d}{d t}y \left (t \right )\right ) t \right ) t^{2}+2 \left (\frac {d}{d t}y \left (t \right )\right ) t +\left (p^{2}+p \right ) y \left (t \right ) = 0 \end {align*}
The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (t^{4}-t^{2}\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+2 \left (\frac {d}{d t}y \left (t \right )\right ) t^{3}+\left (p^{2}+p \right ) y \left (t \right ) = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d t^{2}}y \left (t \right )+p(t) \frac {d}{d t}y \left (t \right ) + q(t) y \left (t \right ) &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {2 t}{t^{2}-1}\\ q(t) &= \frac {p \left (p +1\right )}{\left (t^{2}-1\right ) t^{2}}\\ \end {align*}
| \(p(t)=\frac {2 t}{t^{2}-1}\) | |
| singularity | type |
| \(t = -1\) | \(\text {``regular''}\) |
| \(t = 1\) | \(\text {``regular''}\) |
| \(q(t)=\frac {p \left (p +1\right )}{\left (t^{2}-1\right ) t^{2}}\) | |
| singularity | type |
| \(t = -1\) | \(\text {``regular''}\) |
| \(t = 0\) | \(\text {``regular''}\) |
| \(t = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (t^{2}-1\right ) t^{2}+2 \left (\frac {d}{d t}y \left (t \right )\right ) t^{3}+\left (p^{2}+p \right ) y \left (t \right ) = 0 \] Let the solution be represented as Frobenius power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) \left (t^{2}-1\right ) t^{2}+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right ) t^{3}+\left (p^{2}+p \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 t^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (p^{2}+p \right ) a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 t^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) t^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) t^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (p^{2}+p \right ) a_{n} t^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (p^{2}+p \right ) a_{n} t^{n +r} = 0 \] When \(n = 0\) the above becomes \[ -t^{r} a_{0} r \left (-1+r \right )+\left (p^{2}+p \right ) a_{0} t^{r} = 0 \] Or \[ \left (-t^{r} r \left (-1+r \right )+\left (p^{2}+p \right ) t^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (p^{2}-r^{2}+p +r \right ) t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ p^{2}-r^{2}+p +r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -p\\ r_2 &= p +1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (p^{2}-r^{2}+p +r \right ) t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-p, p +1]\).
Assuming the roots differ by non-integer Since \(r_1 - r_2 = -2 p -1\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n -p}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +p +1} \end {align*}
We start by finding \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -2} \left (n +r -2\right )+a_{n} p \left (p +1\right ) = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -2} \left (n^{2}+2 n r +r^{2}-3 n -3 r +2\right )}{n^{2}+2 n r -p^{2}+r^{2}-n -p -r}\tag {4} \] Which for the root \(r = -p\) becomes \[ a_{n} = \frac {a_{n -2} \left (n -1-p \right ) \left (n -p -2\right )}{n \left (n -2 p -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -p\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2} \] Which for the root \(r = -p\) becomes \[ a_{2}=-\frac {\left (-1+p \right ) p}{4 p -2} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )} \] Which for the root \(r = -p\) becomes \[ a_{4}=\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{32 p^{2}-64 p +24} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{32 p^{2}-64 p +24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{32 p^{2}-64 p +24}\) |
| \(a_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )} \] Which for the root \(r = -p\) becomes \[ a_{6}=-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{384 p^{3}-1728 p^{2}+2208 p -720} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{32 p^{2}-64 p +24}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )}\) | \(-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{384 p^{3}-1728 p^{2}+2208 p -720}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(-\frac {\left (-1+p \right ) p}{4 p -2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{32 p^{2}-64 p +24}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )}\) | \(-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{384 p^{3}-1728 p^{2}+2208 p -720}\) |
| \(a_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= t^{-p} \left (a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}+a_{7} t^{7}+a_{8} t^{8}\dots \right ) \\ &= t^{-p} \left (1-\frac {\left (-1+p \right ) p \,t^{2}}{4 p -2}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) t^{4}}{32 p^{2}-64 p +24}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right ) t^{6}}{384 p^{3}-1728 p^{2}+2208 p -720}+O\left (t^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -2} \left (n +r -2\right )+p \left (p +1\right ) b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -2} \left (n^{2}+2 n r +r^{2}-3 n -3 r +2\right )}{n^{2}+2 n r -p^{2}+r^{2}-n -p -r}\tag {4} \] Which for the root \(r = p +1\) becomes \[ b_{n} = \frac {b_{n -2} \left (n +p \right ) \left (n +p -1\right )}{n \left (n +2 p +1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = p +1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2} \] Which for the root \(r = p +1\) becomes \[ b_{2}=\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )} \] Which for the root \(r = p +1\) becomes \[ b_{4}=\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{32 p^{2}+128 p +120} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{32 p^{2}+128 p +120}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{32 p^{2}+128 p +120}\) |
| \(b_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )} \] Which for the root \(r = p +1\) becomes \[ b_{6}=\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{384 p^{3}+2880 p^{2}+6816 p +5040} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{32 p^{2}+128 p +120}\) |
| \(b_{5}\) | \(0\) | \(0\) |
| \(b_{6}\) | \(-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{384 p^{3}+2880 p^{2}+6816 p +5040}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {\left (1+r \right ) r}{p^{2}-r^{2}+p -3 r -2}\) | \(\frac {\left (2+p \right ) \left (p +1\right )}{4 p +6}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{32 p^{2}+128 p +120}\) |
| \(b_{5}\) | \(0\) | \(0\) |
| \(b_{6}\) | \(-\frac {\left (1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (4+r \right )}{\left (p +r +2\right ) \left (p -1-r \right ) \left (p +4+r \right ) \left (p -3-r \right ) \left (p +6+r \right ) \left (p -5-r \right )}\) | \(\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{384 p^{3}+2880 p^{2}+6816 p +5040}\) |
| \(b_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= t^{-p} \left (b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}+b_{7} t^{7}+b_{8} t^{8}\dots \right ) \\ &= t^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right ) t^{2}}{4 p +6}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) t^{4}}{32 p^{2}+128 p +120}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right ) t^{6}}{384 p^{3}+2880 p^{2}+6816 p +5040}+O\left (t^{8}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} t^{-p} \left (1-\frac {\left (-1+p \right ) p \,t^{2}}{4 p -2}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) t^{4}}{32 p^{2}-64 p +24}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right ) t^{6}}{384 p^{3}-1728 p^{2}+2208 p -720}+O\left (t^{8}\right )\right ) + c_{2} t^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right ) t^{2}}{4 p +6}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) t^{4}}{32 p^{2}+128 p +120}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right ) t^{6}}{384 p^{3}+2880 p^{2}+6816 p +5040}+O\left (t^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} t^{-p} \left (1-\frac {\left (-1+p \right ) p \,t^{2}}{4 p -2}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) t^{4}}{32 p^{2}-64 p +24}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right ) t^{6}}{384 p^{3}-1728 p^{2}+2208 p -720}+O\left (t^{8}\right )\right )+c_{2} t^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right ) t^{2}}{4 p +6}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) t^{4}}{32 p^{2}+128 p +120}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right ) t^{6}}{384 p^{3}+2880 p^{2}+6816 p +5040}+O\left (t^{8}\right )\right ) \\ \end{align*} Replacing \(t\) by \(\frac {1}{x}\) gives \begin {align*} y = c_{1} \left (\frac {1}{x}\right )^{-p} \left (1-\frac {\left (-1+p \right ) p}{\left (4 p -2\right ) x^{2}}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{\left (32 p^{2}-64 p +24\right ) x^{4}}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{\left (384 p^{3}-1728 p^{2}+2208 p -720\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\frac {1}{x}\right )^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right )}{\left (4 p +6\right ) x^{2}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{\left (32 p^{2}+128 p +120\right ) x^{4}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{\left (384 p^{3}+2880 p^{2}+6816 p +5040\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (\frac {1}{x}\right )^{-p} \left (1-\frac {\left (-1+p \right ) p}{\left (4 p -2\right ) x^{2}}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{\left (32 p^{2}-64 p +24\right ) x^{4}}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{\left (384 p^{3}-1728 p^{2}+2208 p -720\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\frac {1}{x}\right )^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right )}{\left (4 p +6\right ) x^{2}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{\left (32 p^{2}+128 p +120\right ) x^{4}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{\left (384 p^{3}+2880 p^{2}+6816 p +5040\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (\frac {1}{x}\right )^{-p} \left (1-\frac {\left (-1+p \right ) p}{\left (4 p -2\right ) x^{2}}+\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right )}{\left (32 p^{2}-64 p +24\right ) x^{4}}-\frac {\left (-1+p \right ) p \left (p -3\right ) \left (p -2\right ) \left (p -5\right ) \left (-4+p \right )}{\left (384 p^{3}-1728 p^{2}+2208 p -720\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\frac {1}{x}\right )^{p +1} \left (1+\frac {\left (2+p \right ) \left (p +1\right )}{\left (4 p +6\right ) x^{2}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right )}{\left (32 p^{2}+128 p +120\right ) x^{4}}+\frac {\left (2+p \right ) \left (p +1\right ) \left (p +4\right ) \left (p +3\right ) \left (p +6\right ) \left (5+p \right )}{\left (384 p^{3}+2880 p^{2}+6816 p +5040\right ) x^{6}}+O\left (\frac {1}{x^{8}}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )-2 y^{\prime } x +\left (p^{2}+p \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {p \left (p +1\right ) y}{x^{2}-1}-\frac {2 x y^{\prime }}{x^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {2 x y^{\prime }}{x^{2}-1}-\frac {p \left (p +1\right ) y}{x^{2}-1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {2 x}{x^{2}-1}, P_{3}\left (x \right )=-\frac {p \left (p +1\right )}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=1 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{2}-1\right )+2 y^{\prime } x -p \left (p +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-p^{2}-p \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r^{2} u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (r +1+p +k \right ) \left (r -p +k \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1\right )^{2}+a_{k} \left (1+p +k \right ) \left (-p +k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (1+p +k \right ) \left (-p +k \right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (1+p +k \right ) \left (-p +k \right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {a_{k} \left (1+p +k \right ) \left (-p +k \right )}{2 \left (k +1\right )^{2}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=\frac {a_{k} \left (1+p +k \right ) \left (-p +k \right )}{2 \left (k +1\right )^{2}}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Legendre successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 1124
Order:=8; dsolve((1-x^2)*diff(y(x),x$2)-2*x*diff(y(x),x)+p*(p+1)*y(x)=0,y(x),type='series',x=infinity);
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 2707
AsymptoticDSolveValue[(1-x^2)*y''[x]-2*x*y'[x]+p*(p+1)*y[x]==0,y[x],{x,infinity,7}]
\[ y(x)\to \left (\frac {p^2 x^{-p-7}}{-p^2-p+(p+6) (p+7)}+\frac {3 p x^{-p-7}}{-p^2-p+(p+6) (p+7)}+\frac {p^4 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {6 p^3 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {17 p^2 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {24 p x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {12 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {p^4 x^{-p-7}}{\left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {6 p^3 x^{-p-7}}{\left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {21 p^2 x^{-p-7}}{\left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {36 p x^{-p-7}}{\left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {p^6 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {9 p^5 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {45 p^4 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {135 p^3 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {254 p^2 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {276 p x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {120 x^{-p-7}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {20 x^{-p-7}}{\left (-p^2-p+(p+4) (p+5)\right ) \left (-p^2-p+(p+6) (p+7)\right )}+\frac {2 x^{-p-7}}{-p^2-p+(p+6) (p+7)}+\frac {p^2 x^{-p-5}}{-p^2-p+(p+4) (p+5)}+\frac {3 p x^{-p-5}}{-p^2-p+(p+4) (p+5)}+\frac {p^4 x^{-p-5}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right )}+\frac {6 p^3 x^{-p-5}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right )}+\frac {17 p^2 x^{-p-5}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right )}+\frac {24 p x^{-p-5}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right )}+\frac {12 x^{-p-5}}{\left (-p^2-p+(p+2) (p+3)\right ) \left (-p^2-p+(p+4) (p+5)\right )}+\frac {2 x^{-p-5}}{-p^2-p+(p+4) (p+5)}+\frac {p^2 x^{-p-3}}{-p^2-p+(p+2) (p+3)}+\frac {3 p x^{-p-3}}{-p^2-p+(p+2) (p+3)}+\frac {2 x^{-p-3}}{-p^2-p+(p+2) (p+3)}+x^{-p-1}\right ) c_1+\left (\frac {p^2 x^{p-6}}{-p^2-p+(5-p) (6-p)}-\frac {p x^{p-6}}{-p^2-p+(5-p) (6-p)}+\frac {p^4 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {2 p^3 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {5 p^2 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {4 p x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {p^4 x^{p-6}}{\left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {2 p^3 x^{p-6}}{\left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {9 p^2 x^{p-6}}{\left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {8 p x^{p-6}}{\left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {p^6 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {3 p^5 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {15 p^4 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {25 p^3 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {44 p^2 x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}-\frac {32 p x^{p-6}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right ) \left (-p^2-p+(5-p) (6-p)\right )}+\frac {p^2 x^{p-4}}{-p^2-p+(3-p) (4-p)}-\frac {p x^{p-4}}{-p^2-p+(3-p) (4-p)}+\frac {p^4 x^{p-4}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right )}-\frac {2 p^3 x^{p-4}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right )}+\frac {5 p^2 x^{p-4}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right )}-\frac {4 p x^{p-4}}{\left (-p^2-p+(1-p) (2-p)\right ) \left (-p^2-p+(3-p) (4-p)\right )}+\frac {p^2 x^{p-2}}{-p^2-p+(1-p) (2-p)}-\frac {p x^{p-2}}{-p^2-p+(1-p) (2-p)}+x^p\right ) c_2 \]