problem 7.3.101 (d)

Internal problem ID [5528]
Internal ﬁle name [OUTPUT/4776_Sunday_June_05_2022_03_05_34_PM_16023406/index.tex]

Book: Notes on Diﬀy Qs. Diﬀerential Equations for Engineers. By by Jiri Lebl, 2013.
Section: Chapter 7. POWER SERIES METHODS. 7.3.2 The method of Frobenius. Exercises. page 300
Problem number: 7.3.101 (d).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second order series method. Regular singular point. Diﬀerence is integer"

\[ \boxed {\sin \left (x \right ) y^{\prime \prime }-y=0} \] 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. \[ \sin \left (x \right ) y^{\prime \prime }-y = 0 \] 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) &= 0\\ q(x) &= -\frac {1}{\sin \left (x \right )}\\ \end {align*}

Combining everything together gives the following summary of singularities for the ode as

Since \(x = 0\) is regular singular point, then Frobenius power series is used. 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} \sin \left (x \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 (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} \end {align*}

Which simpliﬁes to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) x^{n +r -1}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) x^{n +r -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\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 \(n +r -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) x^{n +r -1}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) x^{n +r -1}}{120}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simpliﬁes to \[ x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).

Since \(r_1 - r_2 = 1\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\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^{1+n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}

Where \(C\) above can be zero. We start by ﬁnding \(y_{1}\). 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 \[ a_{1} = \frac {1}{r \left (1+r \right )} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{4}-r^{2}+6}{6 r \left (1+r \right )^{2} \left (2+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {r^{4}+2 r^{3}+2 r^{2}+r +3}{3 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {7 r^{8}+56 r^{7}+154 r^{6}+140 r^{5}+103 r^{4}+524 r^{3}+1536 r^{2}+1800 r +1080}{360 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {r^{8}+12 r^{7}+59 r^{6}+153 r^{5}+239 r^{4}+273 r^{3}+331 r^{2}+372 r +225}{15 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right )}{5040}+\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right )}{120}-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{6}+a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -6}-42 n^{2} a_{n -4}+840 n^{2} a_{n -2}+2 n r a_{n -6}-84 n r a_{n -4}+1680 n r a_{n -2}+r^{2} a_{n -6}-42 r^{2} a_{n -4}+840 r^{2} a_{n -2}-13 n a_{n -6}+378 n a_{n -4}-4200 n a_{n -2}-13 r a_{n -6}+378 r a_{n -4}-4200 r a_{n -2}+42 a_{n -6}-840 a_{n -4}+5040 a_{n -2}+5040 a_{n -1}}{5040 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (a_{n -6}-42 a_{n -4}+840 a_{n -2}\right ) n^{2}+\left (-11 a_{n -6}+294 a_{n -4}-2520 a_{n -2}\right ) n +30 a_{n -6}-504 a_{n -4}+1680 a_{n -2}+5040 a_{n -1}}{5040 n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.


\(n\)	\(a_{n ,r}\)	\(a_{n}\)

\(a_{0}\)	\(1\)	\(1\)

\(a_{1}\)	\(\frac {1}{r \left (1+r \right )}\)	\(\frac {1}{2}\)

\(a_{2}\)	\(\frac {r^{4}-r^{2}+6}{6 r \left (1+r \right )^{2} \left (2+r \right )}\)	\(\frac {1}{12}\)

\(a_{3}\)	\(\frac {r^{4}+2 r^{3}+2 r^{2}+r +3}{3 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\)	\(\frac {1}{48}\)

\(a_{4}\)	\(\frac {7 r^{8}+56 r^{7}+154 r^{6}+140 r^{5}+103 r^{4}+524 r^{3}+1536 r^{2}+1800 r +1080}{360 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\)	\(\frac {1}{192}\)

\(a_{5}\)	\(\frac {r^{8}+12 r^{7}+59 r^{6}+153 r^{5}+239 r^{4}+273 r^{3}+331 r^{2}+372 r +225}{15 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\)	\(\frac {37}{28800}\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x \left (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 \right ) \\ &= x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right ) \end {align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the diﬀerence between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by ﬁnding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= \frac {1}{r \left (1+r \right )} \end {align*}

Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1}{r \left (1+r \right )}&= \lim _{r\rightarrow 0}\frac {1}{r \left (1+r \right )}\\ &= \textit {undefined} \end {align*}

Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(\sin \left (x \right ) y^{\prime \prime }-y = 0\) gives \[ \sin \left (x \right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-C y_{1}\left (x \right ) \ln \left (x \right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (\sin \left (x \right ) y_{1}^{\prime \prime }\left (x \right )-y_{1}\left (x \right )\right ) \ln \left (x \right )+\sin \left (x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ \sin \left (x \right ) y_{1}^{\prime \prime }\left (x \right )-y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \sin \left (x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \frac {\sin \left (x \right ) \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x^{2}}+\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\sin \left (x \right ) \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (1+n \right )\right ) x -\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\right )\right ) C}{x^{2}}+\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (n -1\right )\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0 \end{equation} Expanding \(\frac {2 \sin \left (x \right ) C}{x}\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} \frac {2 \sin \left (x \right ) C}{x} &= 2 C -\frac {1}{3} C \,x^{2}+\frac {1}{60} C \,x^{4}-\frac {1}{2520} C \,x^{6} + \dots \\ &= 2 C -\frac {1}{3} C \,x^{2}+\frac {1}{60} C \,x^{4}-\frac {1}{2520} C \,x^{6} \end {align*}

Expanding \(-\frac {\sin \left (x \right ) C}{x}\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} -\frac {\sin \left (x \right ) C}{x} &= -C +\frac {1}{6} C \,x^{2}-\frac {1}{120} C \,x^{4}+\frac {1}{5040} C \,x^{6} + \dots \\ &= -C +\frac {1}{6} C \,x^{2}-\frac {1}{120} C \,x^{4}+\frac {1}{5040} C \,x^{6} \end {align*}

Expanding \(\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} \end {align*}

Which simpliﬁes to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +6} a_{n} \left (1+n \right )}{2520}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +4} a_{n} \left (1+n \right )}{60}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +2} a_{n} \left (1+n \right )}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n} C \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n} C \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +2} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +4} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +6} a_{n}}{5040}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +5} b_{n} \left (n -1\right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +3} b_{n} \left (n -1\right )}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{1+n} b_{n} \left (n -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +6} a_{n} \left (1+n \right )}{2520}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{n -7} \left (n -6\right ) x^{n -1}}{2520}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +4} a_{n} \left (1+n \right )}{60} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{-5+n} \left (n -4\right ) x^{n -1}}{60} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +2} a_{n} \left (1+n \right )}{3}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {C a_{n -3} \left (n -2\right ) x^{n -1}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n} C \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n} C \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +2} a_{n}}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {C a_{n -3} x^{n -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {C \,x^{n +4} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{-5+n} x^{n -1}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {C \,x^{n +6} a_{n}}{5040} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {C a_{n -7} x^{n -1}}{5040} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{n +5} b_{n} \left (n -1\right )}{5040}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -6\right ) b_{n -6} \left (n -7\right ) x^{n -1}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {n \,x^{n +3} b_{n} \left (n -1\right )}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -4\right ) b_{n -4} \left (-5+n \right ) x^{n -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {n \,x^{1+n} b_{n} \left (n -1\right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -1}\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 \(n -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {C a_{n -7} \left (n -6\right ) x^{n -1}}{2520}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {C a_{-5+n} \left (n -4\right ) x^{n -1}}{60}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {C a_{n -3} \left (n -2\right ) x^{n -1}}{3}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {C a_{n -3} x^{n -1}}{6}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {C a_{-5+n} x^{n -1}}{120}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {C a_{n -7} x^{n -1}}{5040}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -6\right ) b_{n -6} \left (n -7\right ) x^{n -1}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -4\right ) b_{n -4} \left (-5+n \right ) x^{n -1}}{120}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {\left (n -2\right ) b_{n -2} \left (n -3\right ) x^{n -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -1}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the diﬀerence between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ C -1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=1 \] For \(n=2\), Eq (2B) gives \[ 3 C a_{1}-b_{1}+2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 2 b_{2}+\frac {3}{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {3}{4}} \] For \(n=3\), Eq (2B) gives \[ \frac {\left (-a_{0}+30 a_{2}\right ) C}{6}-b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 1+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-{\frac {1}{6}} \] For \(n=4\), Eq (2B) gives \[ \frac {\left (-3 a_{1}+42 a_{3}\right ) C}{6}-\frac {b_{2}}{3}-b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {5}{16}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {5}{192}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (a_{0}-100 a_{2}+1080 a_{4}\right ) C}{120}-b_{3}-b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {257}{1440}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {257}{28800}} \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C=1\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= 1\eslowast \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {x^{3}}{6}-\frac {5 x^{4}}{192}-\frac {257 x^{5}}{28800}+O\left (x^{6}\right ) \] 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} x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right ) + c_{2} \left (1\eslowast \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {x^{3}}{6}-\frac {5 x^{4}}{192}-\frac {257 x^{5}}{28800}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the ﬁnal solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {x^{3}}{6}-\frac {5 x^{4}}{192}-\frac {257 x^{5}}{28800}+O\left (x^{6}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {x^{3}}{6}-\frac {5 x^{4}}{192}-\frac {257 x^{5}}{28800}+O\left (x^{6}\right )\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}+\frac {x^{4}}{192}+\frac {37 x^{5}}{28800}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}-\frac {x^{3}}{6}-\frac {5 x^{4}}{192}-\frac {257 x^{5}}{28800}+O\left (x^{6}\right )\right ) \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying an equivalence, under non-integer power transformations, 
      to LODEs admitting Liouvillian solutions. 
      -> Trying a Liouvillian solution using Kovacics algorithm 
      <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
   -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
   <- Heun successful: received ODE is equivalent to the  HeunG  ODE, case  a <> 0, e <> 0, g <> 0, c = 0 
   Change of variables used: 
      [x = arccos(t)] 
   Linear ODE actually solved: 
      -(-t^2+1)^(1/2)*u(t)+(t^3-t)*diff(u(t),t)+(t^4-2*t^2+1)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = c_{1} x \left (1+\frac {1}{2} x +\frac {1}{12} x^{2}+\frac {1}{48} x^{3}+\frac {1}{192} x^{4}+\frac {37}{28800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (x +\frac {1}{2} x^{2}+\frac {1}{12} x^{3}+\frac {1}{48} x^{4}+\frac {1}{192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {3}{4} x^{2}-\frac {1}{6} x^{3}-\frac {5}{192} x^{4}-\frac {257}{28800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]

\[ y(x)\to c_1 \left (\frac {1}{48} x \left (x^3+4 x^2+24 x+48\right ) \log (x)+\frac {1}{64} \left (-3 x^4-16 x^3-80 x^2-64 x+64\right )\right )+c_2 \left (\frac {x^5}{192}+\frac {x^4}{48}+\frac {x^3}{12}+\frac {x^2}{2}+x\right ) \]


\(q(x)=-\frac {1}{\sin \left (x \right )}\)

singularity	type

\(x = \pi Z\)	\(\text {``regular''}\)

2.14 problem 7.3.101 (d)