Internal problem ID [6447]
Internal file name [OUTPUT/5695_Sunday_June_05_2022_03_47_26_PM_50063468/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. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 2(c).
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"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+\sin \left (x \right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+\sin \left (x \right ) y = 0 \] 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 {\sin \left (x \right )}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\sin \left (x \right ) y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\sin \left (x \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 \(8\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \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 }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +7} a_{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +9} a_{n}}{362880}\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 }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +7} a_{n}}{5040}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +9} a_{n}}{362880} &= \moverset {\infty }{\munderset {n =9}{\sum }}\frac {a_{n -9} x^{n +r}}{362880} \\ \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} \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 }}a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r}}{120}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =9}{\sum }}\frac {a_{n -9} x^{n +r}}{362880}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ x^{r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r \left (-1+r \right ) = 0 \] 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 \[ x^{r} r \left (-1+r \right ) = 0 \] 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 \[ a_{2} = \frac {1}{r \left (1+r \right )^{2} \left (2+r \right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {r^{4}+4 r^{3}+5 r^{2}+2 r -6}{6 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-r^{4}-6 r^{3}-14 r^{2}-15 r -3}{3 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-r^{8}-16 r^{7}-106 r^{6}-376 r^{5}-709 r^{4}-424 r^{3}+1056 r^{2}+2496 r +1560}{120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {2 r^{8}+40 r^{7}+337 r^{6}+1555 r^{5}+4258 r^{4}+6955 r^{3}+6113 r^{2}+1440 r -1305}{45 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = \frac {r^{12}+36 r^{11}+575 r^{10}+5370 r^{9}+31977 r^{8}+120744 r^{7}+245517 r^{6}-40230 r^{5}-1838278 r^{4}-5602080 r^{3}-8696192 r^{2}-7058640 r -2172240}{5040 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )} \] Substituting \(n = 8\) in Eq. (2B) gives \[ a_{8} = \frac {-r^{12}-42 r^{11}-785 r^{10}-8610 r^{9}-61461 r^{8}-298830 r^{7}-1003386 r^{6}-2292885 r^{5}-3351683 r^{4}-2497173 r^{3}+466901 r^{2}+2574075 r +1562715}{315 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2} \left (8+r \right )} \] For \(9\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1}-\frac {a_{n -3}}{6}+\frac {a_{n -5}}{120}-\frac {a_{n -7}}{5040}+\frac {a_{n -9}}{362880} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {362880 a_{n -1}-60480 a_{n -3}+3024 a_{n -5}-72 a_{n -7}+a_{n -9}}{362880 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-362880 a_{n -1}+60480 a_{n -3}-3024 a_{n -5}+72 a_{n -7}-a_{n -9}}{362880 \left (1+n \right ) n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {1}{r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {r^{4}+4 r^{3}+5 r^{2}+2 r -6}{6 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{144}\) |
| \(a_{4}\) | \(\frac {-r^{4}-6 r^{3}-14 r^{2}-15 r -3}{3 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {13}{2880}}\) |
| \(a_{5}\) | \(\frac {-r^{8}-16 r^{7}-106 r^{6}-376 r^{5}-709 r^{4}-424 r^{3}+1056 r^{2}+2496 r +1560}{120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(\frac {29}{86400}\) |
| \(a_{6}\) | \(\frac {2 r^{8}+40 r^{7}+337 r^{6}+1555 r^{5}+4258 r^{4}+6955 r^{3}+6113 r^{2}+1440 r -1305}{45 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\) | \(\frac {431}{3628800}\) |
| \(a_{7}\) | \(\frac {r^{12}+36 r^{11}+575 r^{10}+5370 r^{9}+31977 r^{8}+120744 r^{7}+245517 r^{6}-40230 r^{5}-1838278 r^{4}-5602080 r^{3}-8696192 r^{2}-7058640 r -2172240}{5040 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )}\) | \(-{\frac {4961}{203212800}}\) |
| \(a_{8}\) | \(\frac {-r^{12}-42 r^{11}-785 r^{10}-8610 r^{9}-61461 r^{8}-298830 r^{7}-1003386 r^{6}-2292885 r^{5}-3351683 r^{4}-2497173 r^{3}+466901 r^{2}+2574075 r +1562715}{315 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2} \left (8+r \right )}\) | \(-{\frac {5197}{4877107200}}\) |
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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\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 \(x^{2} y^{\prime \prime }+\sin \left (x \right ) y = 0\) gives \[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\sin \left (x \right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x^{2}+y_{1}\left (x \right ) \sin \left (x \right )\right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +x^{2} \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 )+\sin \left (x \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 \[ y_{1}^{\prime \prime }\left (x \right ) x^{2}+y_{1}\left (x \right ) \sin \left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +x^{2} \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 )+\sin \left (x \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} \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 +\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 ) x^{2}+\sin \left (x \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} \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 +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right ) x^{2}+\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) = 0 \end{equation} Expanding \(\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} x^{1+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +3} b_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +5} b_{n}}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +7} b_{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +9} b_{n}}{362880}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{1+n} a_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} x^{1+n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +3} b_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +5} b_{n}}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {b_{n -5} x^{n}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +7} b_{n}}{5040}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} x^{n}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +9} b_{n}}{362880} &= \moverset {\infty }{\munderset {n =9}{\sum }}\frac {b_{n -9} x^{n}}{362880} \\ \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 \(n\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {b_{n -5} x^{n}}{120}\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} x^{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =9}{\sum }}\frac {b_{n -9} x^{n}}{362880}\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 \[ 5 C a_{2}+b_{2}+6 b_{3}-\frac {b_{0}}{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {4}{3}+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {2}{9}} \] For \(n=4\), Eq (2B) gives \[ 7 C a_{3}+b_{3}+12 b_{4}-\frac {b_{1}}{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {25}{144}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {25}{1728}} \] For \(n=5\), Eq (2B) gives \[ 9 C a_{4}+b_{4}+20 b_{5}-\frac {b_{2}}{6}+\frac {b_{0}}{120} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {689}{4320}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {689}{86400}} \] For \(n=6\), Eq (2B) gives \[ 11 C a_{5}+b_{5}+30 b_{6}-\frac {b_{3}}{6}+\frac {b_{1}}{120} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {263}{5400}+30 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}={\frac {263}{162000}} \] For \(n=7\), Eq (2B) gives \[ 13 C a_{6}+b_{6}+42 b_{7}-\frac {b_{4}}{6}+\frac {b_{2}}{120}-\frac {b_{0}}{5040} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {71809}{18144000}+42 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}={\frac {71809}{762048000}} \] 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 \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {2 x^{3}}{9}-\frac {25 x^{4}}{1728}-\frac {689 x^{5}}{86400}+\frac {263 x^{6}}{162000}+\frac {71809 x^{7}}{762048000}+O\left (x^{8}\right ) \] 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}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {2 x^{3}}{9}-\frac {25 x^{4}}{1728}-\frac {689 x^{5}}{86400}+\frac {263 x^{6}}{162000}+\frac {71809 x^{7}}{762048000}+O\left (x^{8}\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}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {2 x^{3}}{9}-\frac {25 x^{4}}{1728}-\frac {689 x^{5}}{86400}+\frac {263 x^{6}}{162000}+\frac {71809 x^{7}}{762048000}+O\left (x^{8}\right )\right ) \\ \end{align*}
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}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {2 x^{3}}{9}-\frac {25 x^{4}}{1728}-\frac {689 x^{5}}{86400}+\frac {263 x^{6}}{162000}+\frac {71809 x^{7}}{762048000}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{144}-\frac {13 x^{4}}{2880}+\frac {29 x^{5}}{86400}+\frac {431 x^{6}}{3628800}-\frac {4961 x^{7}}{203212800}-\frac {5197 x^{8}}{4877107200}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {2 x^{3}}{9}-\frac {25 x^{4}}{1728}-\frac {689 x^{5}}{86400}+\frac {263 x^{6}}{162000}+\frac {71809 x^{7}}{762048000}+O\left (x^{8}\right )\right ) \] 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 -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 70
Order:=8; dsolve(x^2*diff(y(x),x$2)+sin(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{2} x +\frac {1}{12} x^{2}+\frac {1}{144} x^{3}-\frac {13}{2880} x^{4}+\frac {29}{86400} x^{5}+\frac {431}{3628800} x^{6}-\frac {4961}{203212800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x +\frac {1}{2} x^{2}-\frac {1}{12} x^{3}-\frac {1}{144} x^{4}+\frac {13}{2880} x^{5}-\frac {29}{86400} x^{6}-\frac {431}{3628800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (1-\frac {3}{4} x^{2}+\frac {2}{9} x^{3}-\frac {25}{1728} x^{4}-\frac {689}{86400} x^{5}+\frac {263}{162000} x^{6}+\frac {71809}{762048000} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 121
AsymptoticDSolveValue[x^2*y''[x]+Sin[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {2539 x^6-16185 x^5-9750 x^4+396000 x^3-1620000 x^2+1296000 x+1296000}{1296000}-\frac {x \left (29 x^5-390 x^4+600 x^3+7200 x^2-43200 x+86400\right ) \log (x)}{86400}\right )+c_2 \left (\frac {431 x^7}{3628800}+\frac {29 x^6}{86400}-\frac {13 x^5}{2880}+\frac {x^4}{144}+\frac {x^3}{12}-\frac {x^2}{2}+x\right ) \]