Internal problem ID [1798]
Internal file name [OUTPUT/1799_Sunday_June_05_2022_02_32_20_AM_20248527/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.2, Regular singular points, the method of Frobenius. Page 214
Problem number: 6.
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]]
\[ \boxed {t^{3} y^{\prime \prime }+\sin \left (t^{3}\right ) y^{\prime }+y t=0} \] With the expansion point for the power series method at \(t = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ t^{3} y^{\prime \prime }+\sin \left (t^{3}\right ) y^{\prime }+y t = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {\sin \left (t^{3}\right )}{t^{3}}\\ q(t) &= \frac {1}{t^{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 \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t^{3} y^{\prime \prime }+\sin \left (t^{3}\right ) y^{\prime }+y t = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) t^{3}+\sin \left (t^{3}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) t = 0 \end{equation} Expanding \(\sin \left (t^{3}\right )\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} \sin \left (t^{3}\right ) &= t^{3}-\frac {1}{6} t^{9} + \dots \\ &= t^{3}-\frac {1}{6} t^{9} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +8} a_{n} \left (n +r \right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(1+n +r\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{1+n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +8} a_{n} \left (n +r \right )}{6}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} \left (-7+n +r \right ) t^{1+n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{1+n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(1+n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} \left (-7+n +r \right ) t^{1+n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{1+n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+t^{1+n +r} a_{n} = 0 \] When \(n = 0\) the above becomes \[ t^{1+r} a_{0} r \left (-1+r \right )+t^{1+r} a_{0} = 0 \] Or \[ \left (t^{1+r} r \left (-1+r \right )+t^{1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r +1\right ) t^{1+r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2}-r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\frac {i \sqrt {3}}{2}\\ r_2 &= \frac {1}{2}-\frac {i \sqrt {3}}{2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r +1\right ) t^{1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\frac {i \sqrt {3}}{2}, \frac {1}{2}-\frac {i \sqrt {3}}{2}\right ]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +\frac {1}{2}+\frac {i \sqrt {3}}{2}}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +\frac {1}{2}-\frac {i \sqrt {3}}{2}} \end {align*}
\(y_{1}\left (t \right )\) 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 \[ a_{1} = -\frac {r}{r^{2}+r +1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = -\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = -\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )} \] For \(7\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -7} \left (-7+n +r \right )}{6}+a_{n -1} \left (n +r -1\right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n a_{n -7}-6 n a_{n -1}+r a_{n -7}-6 r a_{n -1}-7 a_{n -7}+6 a_{n -1}}{6 n^{2}+12 n r +6 r^{2}-6 n -6 r +6}\tag {4} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{n} = \frac {i \left (a_{n -7}-6 a_{n -1}\right ) \sqrt {3}+2 \left (a_{n -7}-6 a_{n -1}\right ) n -13 a_{n -7}+6 a_{n -1}}{12 n \left (i \sqrt {3}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )}\) | \(\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}}\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )}\) | \(\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}\) | \(\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
| \(a_{5}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}\) | \(-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right )}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
| \(a_{6}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right ) \left (r^{2}+11 r +31\right )}\) | \(\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right )}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin{align*} y_{1}\left (t \right )&= t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \right ) \\ &= t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (i \sqrt {3}+3\right ) t^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) t^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) t^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) t^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right ) t^{6}}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (t \right )\) is found by taking the complex conjugate of \(y_{1}\left (t \right )\) which gives \[ y_{2}\left (t \right )= t^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (-i \sqrt {3}+3\right ) t^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) t^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) t^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) t^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) \left (-i \sqrt {3}+11\right ) t^{6}}{46080 \left (6-i \sqrt {3}\right ) \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (i \sqrt {3}+3\right ) t^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) t^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) t^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) t^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right ) t^{6}}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) + c_{2} t^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (-i \sqrt {3}+3\right ) t^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) t^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) t^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) t^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) \left (-i \sqrt {3}+11\right ) t^{6}}{46080 \left (6-i \sqrt {3}\right ) \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (i \sqrt {3}+3\right ) t^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) t^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) t^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) t^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right ) t^{6}}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (t^{6}\right )\right )+c_{2} t^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (-i \sqrt {3}+3\right ) t^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) t^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) t^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) t^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) \left (-i \sqrt {3}+11\right ) t^{6}}{46080 \left (6-i \sqrt {3}\right ) \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (i \sqrt {3}+3\right ) t^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) t^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) t^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) t^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right ) t^{6}}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (t^{6}\right )\right )+c_{2} t^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (-i \sqrt {3}+3\right ) t^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) t^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) t^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) t^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) \left (-i \sqrt {3}+11\right ) t^{6}}{46080 \left (6-i \sqrt {3}\right ) \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (i \sqrt {3}+3\right ) t^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) t^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) t^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) t^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) \left (i \sqrt {3}+11\right ) t^{6}}{46080 \left (6+i \sqrt {3}\right ) \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (t^{6}\right )\right )+c_{2} t^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {t}{2}+\frac {\left (-i \sqrt {3}+3\right ) t^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) t^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) t^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) t^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) \left (-i \sqrt {3}+11\right ) t^{6}}{46080 \left (6-i \sqrt {3}\right ) \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (t^{6}\right )\right ) \] 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 <- 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 <- 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: 907
Order:=6; dsolve(t^3*diff(y(t),t$2)+sin(t^3)*diff(y(t),t)+t*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \sqrt {t}\, \left (c_{2} t^{\frac {i \sqrt {3}}{2}} \left (1-\frac {1}{2} t +\frac {i \sqrt {3}+3}{8 i \sqrt {3}+16} t^{2}+\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96} t^{3}+\frac {1}{384} \frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{\left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} t^{4}-\frac {1}{3840} \frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right )}{\left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+c_{1} t^{-\frac {i \sqrt {3}}{2}} \left (1-\frac {1}{2} t +\frac {\sqrt {3}+3 i}{8 \sqrt {3}+16 i} t^{2}+\frac {-\sqrt {3}-5 i}{48 \sqrt {3}+96 i} t^{3}+\frac {3 i \sqrt {3}-8}{576 i \sqrt {3}-480} t^{4}-\frac {1}{3840} \frac {\left (\sqrt {3}+7 i\right ) \left (\sqrt {3}+9 i\right )}{\left (\sqrt {3}+4 i\right ) \left (\sqrt {3}+2 i\right )} t^{5}+\operatorname {O}\left (t^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 886
AsymptoticDSolveValue[t^3*y''[t]+Sin[t^3]*y'[t]+t*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to \left (\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right ) t^5}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right ) \left (1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )\right ) \left (1+\left (4-(-1)^{2/3}\right ) \left (5-(-1)^{2/3}\right )\right )}-\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right ) t^4}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right ) \left (1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )\right )}+\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) t^3}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right )}-\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) t^2}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}+\frac {(-1)^{2/3} t}{1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )}+1\right ) c_1 t^{-(-1)^{2/3}}+\left (-\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right ) t^5}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right ) \left (1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )\right ) \left (1+\left (4+\sqrt [3]{-1}\right ) \left (5+\sqrt [3]{-1}\right )\right )}+\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right ) t^4}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right ) \left (1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )\right )}-\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) t^3}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right )}+\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) t^2}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}-\frac {\sqrt [3]{-1} t}{1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )}+1\right ) c_2 t^{\sqrt [3]{-1}} \]