Internal problem ID [5008]
Internal file name [OUTPUT/4501_Sunday_June_05_2022_02_59_34_PM_7114731/index.tex]
Book: Fundamentals of Differential Equations. By Nagle, Saff and Snider. 9th edition. Boston.
Pearson 2018.
Section: Chapter 8, Series solutions of differential equations. Section 8.3. page 443
Problem number: 7.
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]]
\[ \boxed {\sin \left (x \right ) y^{\prime \prime }+y \cos \left (x \right )=0} \] 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. \[ \sin \left (x \right ) y^{\prime \prime }+y \cos \left (x \right ) = 0 \] 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 {\cos \left (x \right )}{\sin \left (x \right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([\pi Z]\)
Irregular singular points : \([\infty ]\)
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 ) \cos \left (x \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*}
Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \cos \left (x \right ) &= -\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 + \dots \\ &= -\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 \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 (-\frac {x^{n +r +6} a_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 +r -7\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 +r -5\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 +r -3\right ) x^{n +r -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r -1}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r -1}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \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 +r -7\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 +r -5\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 +r -3\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 =7}{\sum }}\left (-\frac {a_{n -7} x^{n +r -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r -1}}{24}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ a_{3} = \frac {r^{4}+8 r^{3}+11 r^{2}+4 r -6}{6 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {7 r^{8}+56 r^{7}+154 r^{6}+140 r^{5}-257 r^{4}-1636 r^{3}-3504 r^{2}-3600 r -1080}{360 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {7 r^{8}+104 r^{7}+578 r^{6}+1556 r^{5}+2303 r^{4}+2756 r^{3}+4312 r^{2}+5664 r +3240}{120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {31 r^{12}+744 r^{11}+7595 r^{10}+42780 r^{9}+142935 r^{8}+280224 r^{7}+311861 r^{6}+305940 r^{5}+536074 r^{4}-24888 r^{3}-2631456 r^{2}-4415040 r -2585520}{15120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )} \] For \(7\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {a_{n -6} \left (n +r -6\right ) \left (n +r -7\right )}{5040}+\frac {a_{n -4} \left (-4+n +r \right ) \left (n +r -5\right )}{120}-\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )}{6}+a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -7}}{720}+\frac {a_{n -5}}{24}-\frac {a_{n -3}}{2}+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -6}-42 n^{2} a_{n -4}+840 n^{2} a_{n -2}+2 n r a_{n -6}-84 n r a_{n -4}+1680 n r a_{n -2}+r^{2} a_{n -6}-42 r^{2} a_{n -4}+840 r^{2} a_{n -2}-13 n a_{n -6}+378 n a_{n -4}-4200 n a_{n -2}-13 r a_{n -6}+378 r a_{n -4}-4200 r a_{n -2}+7 a_{n -7}+42 a_{n -6}-210 a_{n -5}-840 a_{n -4}+2520 a_{n -3}+5040 a_{n -2}-5040 a_{n -1}}{5040 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (a_{n -6}-42 a_{n -4}+840 a_{n -2}\right ) n^{2}+\left (-11 a_{n -6}+294 a_{n -4}-2520 a_{n -2}\right ) n +7 a_{n -7}+30 a_{n -6}-210 a_{n -5}-504 a_{n -4}+2520 a_{n -3}+1680 a_{n -2}-5040 a_{n -1}}{5040 n \left (1+n \right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r \left (1+r \right )}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r^{4}-r^{2}+6}{6 r \left (1+r \right )^{2} \left (2+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {r^{4}+8 r^{3}+11 r^{2}+4 r -6}{6 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )}\) | \(\frac {1}{48}\) |
| \(a_{4}\) | \(\frac {7 r^{8}+56 r^{7}+154 r^{6}+140 r^{5}-257 r^{4}-1636 r^{3}-3504 r^{2}-3600 r -1080}{360 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )}\) | \(-{\frac {3}{320}}\) |
| \(a_{5}\) | \(\frac {7 r^{8}+104 r^{7}+578 r^{6}+1556 r^{5}+2303 r^{4}+2756 r^{3}+4312 r^{2}+5664 r +3240}{120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )}\) | \(\frac {19}{9600}\) |
| \(a_{6}\) | \(\frac {31 r^{12}+744 r^{11}+7595 r^{10}+42780 r^{9}+142935 r^{8}+280224 r^{7}+311861 r^{6}+305940 r^{5}+536074 r^{4}-24888 r^{3}-2631456 r^{2}-4415040 r -2585520}{15120 r \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2} \left (6+r \right )}\) | \(-{\frac {59}{403200}}\) |
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 {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+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 \(\sin \left (x \right ) y^{\prime \prime }+y \cos \left (x \right ) = 0\) gives \[ \sin \left (x \right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) \cos \left (x \right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) \sin \left (x \right )+y_{1}\left (x \right ) \cos \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 ) \cos \left (x \right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (x \right ) \sin \left (x \right )+y_{1}\left (x \right ) \cos \left (x \right ) = 0 \] 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 ) \cos \left (x \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 ) \cos \left (x \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 ) \cos \left (x \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*}
Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \cos \left (x \right ) &= -\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 + \dots \\ &= -\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 \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 (-\frac {x^{n +6} b_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +4} b_{n}}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +2} b_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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_{n -5} \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_{n -5} 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 (n -5\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 (-\frac {x^{n +6} b_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {b_{n -7} x^{n -1}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +4} b_{n}}{24} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {b_{n -5} x^{n -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +2} b_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n -1}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -1} \\ \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_{n -5} \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_{n -5} 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 (n -5\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 =7}{\sum }}\left (-\frac {b_{n -7} x^{n -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {b_{n -5} x^{n -1}}{24}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {b_{n -3} x^{n -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ \frac {\left (-a_{0}+30 a_{2}\right ) C}{6}-\frac {b_{0}}{2}+b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {3}{2}+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {1}{4}} \] For \(n=4\), Eq (2B) gives \[ \frac {\left (-3 a_{1}+42 a_{3}\right ) C}{6}-\frac {b_{1}}{2}-\frac {b_{2}}{3}+b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {5}{48}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {5}{576}} \] For \(n=5\), Eq (2B) gives \[ \frac {\left (a_{0}-100 a_{2}+1080 a_{4}\right ) C}{120}+\frac {b_{0}}{24}-\frac {b_{2}}{2}-b_{3}+b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {437}{1440}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {437}{28800}} \] 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 \[ y_{2}\left (x \right )= \left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {x^{3}}{4}-\frac {5 x^{4}}{576}-\frac {437 x^{5}}{28800}+O\left (x^{6}\right ) \] 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 {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-1\right )\eslowast \left (x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {x^{3}}{4}-\frac {5 x^{4}}{576}-\frac {437 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 {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {x^{3}}{4}-\frac {5 x^{4}}{576}-\frac {437 x^{5}}{28800}+O\left (x^{6}\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}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {x^{3}}{4}-\frac {5 x^{4}}{576}-\frac {437 x^{5}}{28800}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right )+c_{2} \left (-x \left (1-\frac {x}{2}+\frac {x^{2}}{12}+\frac {x^{3}}{48}-\frac {3 x^{4}}{320}+\frac {19 x^{5}}{9600}-\frac {59 x^{6}}{403200}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-\frac {3 x^{2}}{4}+\frac {x^{3}}{4}-\frac {5 x^{4}}{576}-\frac {437 x^{5}}{28800}+O\left (x^{6}\right )\right ) \] 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 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: 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 -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> 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: 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`[0, u]
✓ Solution by Maple
Time used: 0.203 (sec). Leaf size: 58
Order:=6; dsolve(sin(x)*diff(y(x),x$2)+cos(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {1}{2} x +\frac {1}{12} x^{2}+\frac {1}{48} x^{3}-\frac {3}{320} x^{4}+\frac {19}{9600} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (-x +\frac {1}{2} x^{2}-\frac {1}{12} x^{3}-\frac {1}{48} x^{4}+\frac {3}{320} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {3}{4} x^{2}+\frac {1}{4} x^{3}-\frac {5}{576} x^{4}-\frac {437}{28800} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 85
AsymptoticDSolveValue[Sin[x]*y''[x]+Cos[x]*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{576} \left (7 x^4+192 x^3-720 x^2+576 x+576\right )-\frac {1}{48} x \left (x^3+4 x^2-24 x+48\right ) \log (x)\right )+c_2 \left (-\frac {3 x^5}{320}+\frac {x^4}{48}+\frac {x^3}{12}-\frac {x^2}{2}+x\right ) \]