Internal problem ID [4707]
Internal file name [OUTPUT/4200_Sunday_June_05_2022_12_40_23_PM_76393033/index.tex]
Book: A treatise on ordinary and partial differential equations by William Woolsey Johnson.
1913
Section: Chapter VII, Solutions in series. Examples XIV. page 177
Problem number: 14.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (1-4 x \right ) y^{\prime \prime }+\left (\left (1-n \right ) x -\left (6-4 n \right ) x^{2}\right ) y^{\prime }+n \left (1-n \right ) x y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (-4 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (\left (4 n -6\right ) x^{2}+\left (1-n \right ) x \right ) y^{\prime }+\left (-n^{2}+n \right ) x y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {4 x n -n -6 x +1}{x \left (4 x -1\right )}\\ q(x) &= \frac {n \left (n -1\right )}{x \left (4 x -1\right )}\\ \end {align*}
| \(p(x)=-\frac {4 x n -n -6 x +1}{x \left (4 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{4}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {n \left (n -1\right )}{x \left (4 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{4}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, {\frac {1}{4}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -x^{2} \left (4 x -1\right ) y^{\prime \prime }+\left (\left (4 n -6\right ) x^{2}+\left (1-n \right ) x \right ) y^{\prime }+\left (-n^{2}+n \right ) x y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} -x^{2} \left (4 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (\left (4 n -6\right ) x^{2}+\left (1-n \right ) x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-n^{2}+n \right ) x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) x^{1+n +r} a_{n} \left (n -\frac {3}{2}\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n -1\right ) \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} x^{r} a_{n} n \left (n -1\right ) x \right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) x^{1+n +r} a_{n} \left (n -\frac {3}{2}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n -\frac {3}{2}\right ) \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} x^{r} a_{n} n \left (n -1\right ) x \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n -1} \left (n -1\right ) x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n -\frac {3}{2}\right ) \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n -1\right ) \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n -1} \left (n -1\right ) x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n -1\right ) \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} \left (n -1\right ) r = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} \left (n -1\right ) r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{r} \left (r -n \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (r -n \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= n \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{r} \left (r -n \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, n]\).
Assuming the roots differ by non-integer Since \(r_1 - r_2 = -n\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} -4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n -\frac {3}{2}\right ) \left (n +r -1\right )-a_{n} \left (n -1\right ) \left (n +r \right )-n a_{n -1} \left (n -1\right ) = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-4 n n -4 n r +4 n^{2}+8 n r +4 r^{2}+3 n -6 n -6 r +2\right )}{n n +n r -n^{2}-2 n r -r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n -2 n +2\right ) \left (n -2 n +1\right )}{n \left (n -n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )} \] Which for the root \(r = 0\) becomes \[ a_{1}=-n \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(-n\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )} \] Which for the root \(r = 0\) becomes \[ a_{2}=\frac {\left (n -3\right ) n}{2} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(-n\) |
| \(a_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n -3\right ) n}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-\frac {\left (n -4\right ) \left (n -5\right ) n}{6} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(-n\) |
| \(a_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n -3\right ) n}{2}\) |
| \(a_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(-\frac {\left (n -4\right ) \left (n -5\right ) n}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n}{24} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(-n\) |
| \(a_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n -3\right ) n}{2}\) |
| \(a_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(-\frac {\left (n -4\right ) \left (n -5\right ) n}{6}\) |
| \(a_{4}\) | \(\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )}\) | \(\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (n -2 r -8\right ) \left (n -2 r -9\right ) \left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right ) \left (r +5\right ) \left (-r +n -5\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n}{120} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(-n\) |
| \(a_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n -3\right ) n}{2}\) |
| \(a_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(-\frac {\left (n -4\right ) \left (n -5\right ) n}{6}\) |
| \(a_{4}\) | \(\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )}\) | \(\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n}{24}\) |
| \(a_{5}\) | \(-\frac {\left (n -2 r -8\right ) \left (n -2 r -9\right ) \left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right ) \left (r +5\right ) \left (-r +n -5\right )}\) | \(-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n}{120}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \\ &= 1-n x +\frac {\left (n -3\right ) n \,x^{2}}{2}-\frac {\left (n -4\right ) \left (n -5\right ) n \,x^{3}}{6}+\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n \,x^{4}}{24}-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n \,x^{5}}{120}+O\left (x^{6}\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) 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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} -4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -1} \left (n -\frac {3}{2}\right ) \left (n +r -1\right )-b_{n} \left (n -1\right ) \left (n +r \right )-n b_{n -1} \left (n -1\right ) = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}-4 n n -4 n r +4 n^{2}+8 n r +4 r^{2}+3 n -6 n -6 r +2\right )}{n n +n r -n^{2}-2 n r -r^{2}}\tag {4} \] Which for the root \(r = n\) becomes \[ b_{n} = \frac {b_{n -1} \left (n +2 n -1\right ) \left (n +2 n -2\right )}{n \left (n +n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = n\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )} \] Which for the root \(r = n\) becomes \[ b_{1}=n \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(n\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )} \] Which for the root \(r = n\) becomes \[ b_{2}=\frac {\left (n +3\right ) n}{2} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(n\) |
| \(b_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n +3\right ) n}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )} \] Which for the root \(r = n\) becomes \[ b_{3}=\frac {n \left (n +5\right ) \left (n +4\right )}{6} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(n\) |
| \(b_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n +3\right ) n}{2}\) |
| \(b_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(\frac {n \left (n +5\right ) \left (n +4\right )}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )} \] Which for the root \(r = n\) becomes \[ b_{4}=\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n}{24} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(n\) |
| \(b_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n +3\right ) n}{2}\) |
| \(b_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(\frac {n \left (n +5\right ) \left (n +4\right )}{6}\) |
| \(b_{4}\) | \(\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )}\) | \(\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (n -2 r -8\right ) \left (n -2 r -9\right ) \left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right ) \left (r +5\right ) \left (-r +n -5\right )} \] Which for the root \(r = n\) becomes \[ b_{5}=\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right )}{120} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right )}\) | \(n\) |
| \(b_{2}\) | \(\frac {\left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right )}\) | \(\frac {\left (n +3\right ) n}{2}\) |
| \(b_{3}\) | \(-\frac {\left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right )}\) | \(\frac {n \left (n +5\right ) \left (n +4\right )}{6}\) |
| \(b_{4}\) | \(\frac {\left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right )}\) | \(\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n}{24}\) |
| \(b_{5}\) | \(-\frac {\left (n -2 r -8\right ) \left (n -2 r -9\right ) \left (n -2 r -6\right ) \left (n -2 r -7\right ) \left (n -2 r -4\right ) \left (n -2 r -5\right ) \left (n -2 r -2\right ) \left (n -2 r -3\right ) \left (n -2 r \right ) \left (n -2 r -1\right )}{\left (r +1\right ) \left (-r +n -1\right ) \left (r +2\right ) \left (-r +n -2\right ) \left (r +3\right ) \left (-r +n -3\right ) \left (r +4\right ) \left (-r +n -4\right ) \left (r +5\right ) \left (-r +n -5\right )}\) | \(\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right )}{120}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= x^{n} \left (1+n x +\frac {\left (n +3\right ) n \,x^{2}}{2}+\frac {n \left (n +5\right ) \left (n +4\right ) x^{3}}{6}+\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n \,x^{4}}{24}+\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right ) x^{5}}{120}+O\left (x^{6}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \left (1-n x +\frac {\left (n -3\right ) n \,x^{2}}{2}-\frac {\left (n -4\right ) \left (n -5\right ) n \,x^{3}}{6}+\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n \,x^{4}}{24}-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n \,x^{5}}{120}+O\left (x^{6}\right )\right ) + c_{2} x^{n} \left (1+n x +\frac {\left (n +3\right ) n \,x^{2}}{2}+\frac {n \left (n +5\right ) \left (n +4\right ) x^{3}}{6}+\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n \,x^{4}}{24}+\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right ) x^{5}}{120}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-n x +\frac {\left (n -3\right ) n \,x^{2}}{2}-\frac {\left (n -4\right ) \left (n -5\right ) n \,x^{3}}{6}+\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n \,x^{4}}{24}-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n \,x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} x^{n} \left (1+n x +\frac {\left (n +3\right ) n \,x^{2}}{2}+\frac {n \left (n +5\right ) \left (n +4\right ) x^{3}}{6}+\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n \,x^{4}}{24}+\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right ) x^{5}}{120}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-n x +\frac {\left (n -3\right ) n \,x^{2}}{2}-\frac {\left (n -4\right ) \left (n -5\right ) n \,x^{3}}{6}+\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n \,x^{4}}{24}-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n \,x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} x^{n} \left (1+n x +\frac {\left (n +3\right ) n \,x^{2}}{2}+\frac {n \left (n +5\right ) \left (n +4\right ) x^{3}}{6}+\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n \,x^{4}}{24}+\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right ) x^{5}}{120}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-n x +\frac {\left (n -3\right ) n \,x^{2}}{2}-\frac {\left (n -4\right ) \left (n -5\right ) n \,x^{3}}{6}+\frac {\left (n -6\right ) \left (n -7\right ) \left (n -5\right ) n \,x^{4}}{24}-\frac {\left (n -8\right ) \left (n -9\right ) \left (n -6\right ) \left (n -7\right ) n \,x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} x^{n} \left (1+n x +\frac {\left (n +3\right ) n \,x^{2}}{2}+\frac {n \left (n +5\right ) \left (n +4\right ) x^{3}}{6}+\frac {\left (n +6\right ) \left (n +7\right ) \left (n +5\right ) n \,x^{4}}{24}+\frac {n \left (n +7\right ) \left (n +6\right ) \left (n +9\right ) \left (n +8\right ) x^{5}}{120}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -x^{2} \left (4 x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\left (4 n -6\right ) x^{2}+\left (1-n \right ) x \right ) y^{\prime }+\left (-n^{2}+n \right ) x 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 {n \left (n -1\right ) y}{x \left (4 x -1\right )}+\frac {\left (4 n x -n -6 x +1\right ) y^{\prime }}{x \left (4 x -1\right )} \\ \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 {\left (4 n x -n -6 x +1\right ) y^{\prime }}{x \left (4 x -1\right )}+\frac {n \left (n -1\right ) y}{x \left (4 x -1\right )}=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 {4 n x -n -6 x +1}{x \left (4 x -1\right )}, P_{3}\left (x \right )=\frac {n \left (n -1\right )}{x \left (4 x -1\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=1-n \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (4 x -1\right )+\left (-4 n x +n +6 x -1\right ) y^{\prime }+n \left (n -1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (r -n \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k +1+r -n \right )+a_{k} \left (2 r +1-n +2 k \right ) \left (2 r -n +2 k \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (r -n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, n\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1+r \right ) \left (k +1+r -n \right )+4 \left (r +\frac {1}{2}-\frac {n}{2}+k \right ) a_{k} \left (r -\frac {n}{2}+k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (2 r +1-n +2 k \right ) a_{k} \left (2 r -n +2 k \right )}{\left (k +1+r \right ) \left (k +1+r -n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {\left (1-n +2 k \right ) a_{k} \left (-n +2 k \right )}{\left (k +1\right ) \left (k +1-n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {\left (1-n +2 k \right ) a_{k} \left (-n +2 k \right )}{\left (k +1\right ) \left (k +1-n \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n \\ {} & {} & a_{k +1}=\frac {\left (n +1+2 k \right ) a_{k} \left (n +2 k \right )}{\left (k +1+n \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n}, a_{k +1}=\frac {\left (n +1+2 k \right ) a_{k} \left (n +2 k \right )}{\left (k +1+n \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +n}\right ), a_{k +1}=\frac {\left (1-n +2 k \right ) a_{k} \left (-n +2 k \right )}{\left (k +1\right ) \left (k +1-n \right )}, b_{k +1}=\frac {\left (n +1+2 k \right ) b_{k} \left (n +2 k \right )}{\left (k +1+n \right ) \left (k +1\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 471
Order:=6; dsolve(x^2*(1-4*x)*diff(y(x),x$2)+((1-n)*x-(6-4*n)*x^2)*diff(y(x),x)+n*(1-n)*x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{n} \left (1+n x +\frac {1}{2} n \left (n +3\right ) x^{2}+\frac {1}{6} \left (n +5\right ) \left (n +4\right ) n x^{3}+\frac {1}{24} n \left (n +5\right ) \left (n +7\right ) \left (n +6\right ) x^{4}+\frac {1}{120} \left (n +9\right ) \left (n +8\right ) \left (n +7\right ) \left (n +6\right ) n x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-n x +\frac {1}{2} n \left (-3+n \right ) x^{2}-\frac {1}{6} \left (-4+n \right ) \left (-5+n \right ) n x^{3}+\frac {1}{24} n \left (-5+n \right ) \left (-6+n \right ) \left (n -7\right ) x^{4}-\frac {1}{120} \left (-6+n \right ) \left (n -7\right ) \left (n -8\right ) \left (n -9\right ) n x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 2114
AsymptoticDSolveValue[x^2*(1-4*x)*y''[x]+((1-n)*x-(6-4*n)*x^2)*y'[x]+n*(1-n)*x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (\frac {\left (512 n-256 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (64 \left (n-n^2\right )-128 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (16 \left (n-n^2\right )-32 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}-\frac {\left (4 \left (n-n^2\right )-8 (n+3)\right ) \left (32 n-16 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (4 \left (n-n^2\right )-8 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (-n^2+n-2 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}\right )}{(1-n) (n+3)+(n+2) (n+3)}-\frac {\left (-n^2+n-2 (n+4)\right ) \left (128 n-64 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (16 \left (n-n^2\right )-32 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (4 \left (n-n^2\right )-8 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}-\frac {\left (-n^2+n-2 (n+3)\right ) \left (32 n-16 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (4 \left (n-n^2\right )-8 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (-n^2+n-2 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}\right )}{(1-n) (n+3)+(n+2) (n+3)}\right )}{(1-n) (n+4)+(n+3) (n+4)}\right ) x^5}{(1-n) (n+5)+(n+4) (n+5)}+\frac {\left (128 n-64 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (16 \left (n-n^2\right )-32 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (4 \left (n-n^2\right )-8 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}-\frac {\left (-n^2+n-2 (n+3)\right ) \left (32 n-16 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (4 \left (n-n^2\right )-8 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (-n^2+n-2 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}\right )}{(1-n) (n+3)+(n+2) (n+3)}\right ) x^4}{(1-n) (n+4)+(n+3) (n+4)}+\frac {\left (32 n-16 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (4 \left (n-n^2\right )-8 (n+1)\right )}{(1-n) (n+1)+n (n+1)}-\frac {\left (-n^2+n-2 (n+2)\right ) \left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right )}{(1-n) (n+2)+(n+1) (n+2)}\right ) x^3}{(1-n) (n+3)+(n+2) (n+3)}+\frac {\left (8 n-4 \left (n-n^2\right )-\frac {\left (n^2+n\right ) \left (-n^2+n-2 (n+1)\right )}{(1-n) (n+1)+n (n+1)}\right ) x^2}{(1-n) (n+2)+(n+1) (n+2)}+\frac {\left (n^2+n\right ) x}{(1-n) (n+1)+n (n+1)}+1\right ) c_1 x^n+\left (\frac {\left (-256 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (64 \left (n-n^2\right )-128\right )}{1-n}-\frac {\left (16 \left (n-n^2\right )-64\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}-\frac {\left (4 \left (n-n^2\right )-24\right ) \left (-16 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (4 \left (n-n^2\right )-8\right )}{1-n}-\frac {\left (-n^2+n-4\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}\right )}{3 (1-n)+6}-\frac {\left (-n^2+n-8\right ) \left (-64 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (16 \left (n-n^2\right )-32\right )}{1-n}-\frac {\left (4 \left (n-n^2\right )-16\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}-\frac {\left (-n^2+n-6\right ) \left (-16 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (4 \left (n-n^2\right )-8\right )}{1-n}-\frac {\left (-n^2+n-4\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}\right )}{3 (1-n)+6}\right )}{4 (1-n)+12}\right ) x^5}{5 (1-n)+20}+\frac {\left (-64 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (16 \left (n-n^2\right )-32\right )}{1-n}-\frac {\left (4 \left (n-n^2\right )-16\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}-\frac {\left (-n^2+n-6\right ) \left (-16 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (4 \left (n-n^2\right )-8\right )}{1-n}-\frac {\left (-n^2+n-4\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}\right )}{3 (1-n)+6}\right ) x^4}{4 (1-n)+12}+\frac {\left (-16 \left (n-n^2\right )-\frac {\left (n^2-n\right ) \left (4 \left (n-n^2\right )-8\right )}{1-n}-\frac {\left (-n^2+n-4\right ) \left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right )}{2 (1-n)+2}\right ) x^3}{3 (1-n)+6}+\frac {\left (-4 \left (n-n^2\right )-\frac {\left (-n^2+n-2\right ) \left (n^2-n\right )}{1-n}\right ) x^2}{2 (1-n)+2}+\frac {\left (n^2-n\right ) x}{1-n}+1\right ) c_2 \]