Internal problem ID [6046]
Internal file name [OUTPUT/5294_Sunday_June_05_2022_03_29_54_PM_91142448/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 4. Linear equations with Regular Singular Points. Page 154
Problem number: 1(g).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+y \cos \left (x \right )=0} \] 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. \[ x^{2} y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+y \cos \left (x \right ) = 0 \] 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) &= \frac {\sin \left (x \right )}{x^{2}}\\ q(x) &= \frac {\cos \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 \[ x^{2} y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+y \cos \left (x \right ) = 0 \] 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 }}\left (n +r \right ) a_{n} x^{n +r -1}\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 \(8\) 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*}
Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} \cos \left (x \right ) &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4} + \dots \\ &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4} \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 }}\frac {x^{n +r +8} a_{n} \left (n +r \right )}{362880}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24}\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 }}\frac {x^{n +r +8} a_{n} \left (n +r \right )}{362880} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} \left (n -8+r \right ) x^{n +r}}{362880} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right )}{5040}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n} \left (n +r \right )}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{40320} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n}}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n}}{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} x^{n +r}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{24} \\ \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 =8}{\sum }}\frac {a_{n -8} \left (n -8+r \right ) x^{n +r}}{362880}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{120}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r}}{40320}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r}}{24}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= i\\ r_2 &= -i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([i, -i]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -i} \end {align*}
\(y_{1}\left (x \right )\) is found first. Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {3+r}{6 r^{2}+24 r +30} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-3 r^{3}-17 r^{2}+5 r +75}{360 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = 0 \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {3 r^{5}+15 r^{4}-230 r^{3}-1818 r^{2}-3805 r -2037}{15120 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+12 r +37\right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = 0 \] For \(8\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n -8} \left (n -8+r \right )}{362880}-\frac {a_{n -6} \left (n -6+r \right )}{5040}+\frac {a_{n -4} \left (n -4+r \right )}{120}-\frac {a_{n -2} \left (n +r -2\right )}{6}+a_{n} \left (n +r \right )+\frac {a_{n -8}}{40320}-\frac {a_{n -6}}{720}+a_{n}-\frac {a_{n -2}}{2}+\frac {a_{n -4}}{24} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -8}-72 n a_{n -6}+3024 n a_{n -4}-60480 n a_{n -2}+r a_{n -8}-72 r a_{n -6}+3024 r a_{n -4}-60480 r a_{n -2}+a_{n -8}-72 a_{n -6}+3024 a_{n -4}-60480 a_{n -2}}{362880 \left (n^{2}+2 n r +r^{2}+1\right )}\tag {4} \] Which for the root \(r = i\) becomes \[ a_{n} = -\frac {\left (1+i+n \right ) \left (a_{n -8}-72 a_{n -6}+3024 a_{n -4}-60480 a_{n -2}\right )}{362880 n \left (2 i+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {3+r}{6 r^{2}+24 r +30}\) | \(\frac {1}{12}-\frac {i}{24}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {-3 r^{3}-17 r^{2}+5 r +75}{360 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right )}\) | \(\frac {29}{28800}-\frac {67 i}{28800}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(\frac {3 r^{5}+15 r^{4}-230 r^{3}-1818 r^{2}-3805 r -2037}{15120 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+12 r +37\right )}\) | \(-\frac {893}{14515200}+\frac {17 i}{4838400}\) |
| \(a_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{i} \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^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \] 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^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) + c_{2} x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right )+c_{2} x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+O\left (x^{8}\right )\right ) \] 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 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 a symmetry of the form [xi=0, eta=F(x)] 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 2nd order, integrating factor of the form mu(x,y) 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 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 <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] 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: 0.203 (sec). Leaf size: 53
Order:=8; dsolve(x^2*diff(y(x),x$2)+sin(x)*diff(y(x),x)+cos(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-i} \left (1+\left (\frac {1}{12}+\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^{6}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{i} \left (1+\left (\frac {1}{12}-\frac {i}{24}\right ) x^{2}+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^{4}+\left (-\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^{6}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.048 (sec). Leaf size: 112
AsymptoticDSolveValue[x^2*y''[x]+Sin[x]*y'[x]+Cos[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 x^{-i} \left (\left (-\frac {26459}{59222016000}-\frac {12449 i}{7402752000}\right ) x^8-\left (\frac {893}{14515200}+\frac {17 i}{4838400}\right ) x^6+\left (\frac {29}{28800}+\frac {67 i}{28800}\right ) x^4+\left (\frac {1}{12}+\frac {i}{24}\right ) x^2+1\right )+c_2 x^i \left (\left (-\frac {26459}{59222016000}+\frac {12449 i}{7402752000}\right ) x^8-\left (\frac {893}{14515200}-\frac {17 i}{4838400}\right ) x^6+\left (\frac {29}{28800}-\frac {67 i}{28800}\right ) x^4+\left (\frac {1}{12}-\frac {i}{24}\right ) x^2+1\right ) \]