Internal problem ID [1818]
Internal file name [OUTPUT/1819_Sunday_June_05_2022_02_33_45_AM_87091496/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: 27.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {2 \sin \left (t \right ) y^{\prime \prime }+\left (1-t \right ) y^{\prime }-2 y=0} \] 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. \[ 2 \sin \left (t \right ) y^{\prime \prime }+\left (1-t \right ) y^{\prime }-2 y = 0 \] 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 {t -1}{2 \sin \left (t \right )}\\ q(t) &= -\frac {1}{\sin \left (t \right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([\pi Z, \pi Z]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] 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} 2 \sin \left (t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (1-t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Expanding \(2 \sin \left (t \right )\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} 2 \sin \left (t \right ) &= 2 t -\frac {1}{3} t^{3}+\frac {1}{60} t^{5}-\frac {1}{2520} t^{7} + \dots \\ &= 2 t -\frac {1}{3} t^{3}+\frac {1}{60} t^{5}-\frac {1}{2520} t^{7} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2520}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{60}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2520}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) t^{n +r -1}}{2520}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{60} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) t^{n +r -1}}{60} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{3}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r -1}}{3}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} t^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} t^{n +r -1}\right ) \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) t^{n +r -1}}{2520}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) t^{n +r -1}}{60}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r -1}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (2 t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,t^{-1+r} \left (2 r -1\right ) = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 2 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,t^{-1+r} \left (2 r -1\right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not an integer, then we can construct two linearly independent solutions \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}}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n} \end {align*}
We start by finding \(y_{1}\left (t \right )\). 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 \[ a_{1} = \frac {2+r}{2 r^{2}+3 r +1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {2 r^{4}+r^{3}+r^{2}+14 r +18}{12 r^{4}+60 r^{3}+105 r^{2}+75 r +18} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {4 r^{5}+22 r^{4}+36 r^{3}+50 r^{2}+86 r +72}{24 r^{6}+252 r^{5}+1050 r^{4}+2205 r^{3}+2436 r^{2}+1323 r +270} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {56 r^{8}+492 r^{7}+1902 r^{6}+5985 r^{5}+17589 r^{4}+37533 r^{3}+59293 r^{2}+62910 r +32400}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {192 r^{9}+3240 r^{8}+22588 r^{7}+88190 r^{6}+227593 r^{5}+451550 r^{4}+769987 r^{3}+1088360 r^{2}+1014900 r +436320}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right )}{2520}+\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right )}{60}-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{3}+2 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -6}-42 n^{2} a_{n -4}+840 n^{2} a_{n -2}+2 n r a_{n -6}-84 n r a_{n -4}+1680 n r a_{n -2}+r^{2} a_{n -6}-42 r^{2} a_{n -4}+840 r^{2} a_{n -2}-13 n a_{n -6}+378 n a_{n -4}-4200 n a_{n -2}+2520 n a_{n -1}-13 r a_{n -6}+378 r a_{n -4}-4200 r a_{n -2}+2520 r a_{n -1}+42 a_{n -6}-840 a_{n -4}+5040 a_{n -2}+2520 a_{n -1}}{5040 n^{2}+10080 n r +5040 r^{2}-2520 n -2520 r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (4 a_{n -6}-168 a_{n -4}+3360 a_{n -2}\right ) n^{2}+\left (-48 a_{n -6}+1344 a_{n -4}-13440 a_{n -2}+10080 a_{n -1}\right ) n +143 a_{n -6}-2646 a_{n -4}+12600 a_{n -2}+15120 a_{n -1}}{20160 n^{2}+10080 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{2 r^{2}+3 r +1}\) | \(\frac {5}{6}\) |
| \(a_{2}\) | \(\frac {2 r^{4}+r^{3}+r^{2}+14 r +18}{12 r^{4}+60 r^{3}+105 r^{2}+75 r +18}\) | \(\frac {17}{60}\) |
| \(a_{3}\) | \(\frac {4 r^{5}+22 r^{4}+36 r^{3}+50 r^{2}+86 r +72}{24 r^{6}+252 r^{5}+1050 r^{4}+2205 r^{3}+2436 r^{2}+1323 r +270}\) | \(\frac {89}{1260}\) |
| \(a_{4}\) | \(\frac {56 r^{8}+492 r^{7}+1902 r^{6}+5985 r^{5}+17589 r^{4}+37533 r^{3}+59293 r^{2}+62910 r +32400}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {941}{45360}\) |
| \(a_{5}\) | \(\frac {192 r^{9}+3240 r^{8}+22588 r^{7}+88190 r^{6}+227593 r^{5}+451550 r^{4}+769987 r^{3}+1088360 r^{2}+1014900 r +436320}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(\frac {14989}{2494800}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= \sqrt {t} \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 ) \\ &= \sqrt {t}\, \left (1+\frac {5 t}{6}+\frac {17 t^{2}}{60}+\frac {89 t^{3}}{1260}+\frac {941 t^{4}}{45360}+\frac {14989 t^{5}}{2494800}+O\left (t^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {2+r}{2 r^{2}+3 r +1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {2 r^{4}+r^{3}+r^{2}+14 r +18}{12 r^{4}+60 r^{3}+105 r^{2}+75 r +18} \] Substituting \(n = 3\) in Eq. (2B) gives \[ b_{3} = \frac {4 r^{5}+22 r^{4}+36 r^{3}+50 r^{2}+86 r +72}{24 r^{6}+252 r^{5}+1050 r^{4}+2205 r^{3}+2436 r^{2}+1323 r +270} \] Substituting \(n = 4\) in Eq. (2B) gives \[ b_{4} = \frac {56 r^{8}+492 r^{7}+1902 r^{6}+5985 r^{5}+17589 r^{4}+37533 r^{3}+59293 r^{2}+62910 r +32400}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ b_{5} = \frac {192 r^{9}+3240 r^{8}+22588 r^{7}+88190 r^{6}+227593 r^{5}+451550 r^{4}+769987 r^{3}+1088360 r^{2}+1014900 r +436320}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {b_{n -6} \left (n +r -6\right ) \left (n -7+r \right )}{2520}+\frac {b_{n -4} \left (-4+n +r \right ) \left (n -5+r \right )}{60}-\frac {b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{3}+2 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )+\left (n +r \right ) b_{n}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -6}-42 n^{2} b_{n -4}+840 n^{2} b_{n -2}+2 n r b_{n -6}-84 n r b_{n -4}+1680 n r b_{n -2}+r^{2} b_{n -6}-42 r^{2} b_{n -4}+840 r^{2} b_{n -2}-13 n b_{n -6}+378 n b_{n -4}-4200 n b_{n -2}+2520 n b_{n -1}-13 r b_{n -6}+378 r b_{n -4}-4200 r b_{n -2}+2520 r b_{n -1}+42 b_{n -6}-840 b_{n -4}+5040 b_{n -2}+2520 b_{n -1}}{5040 n^{2}+10080 n r +5040 r^{2}-2520 n -2520 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (b_{n -6}-42 b_{n -4}+840 b_{n -2}\right ) n^{2}+\left (-13 b_{n -6}+378 b_{n -4}-4200 b_{n -2}+2520 b_{n -1}\right ) n +42 b_{n -6}-840 b_{n -4}+5040 b_{n -2}+2520 b_{n -1}}{5040 n^{2}-2520 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2+r}{2 r^{2}+3 r +1}\) | \(2\) |
| \(b_{2}\) | \(\frac {2 r^{4}+r^{3}+r^{2}+14 r +18}{12 r^{4}+60 r^{3}+105 r^{2}+75 r +18}\) | \(1\) |
| \(b_{3}\) | \(\frac {4 r^{5}+22 r^{4}+36 r^{3}+50 r^{2}+86 r +72}{24 r^{6}+252 r^{5}+1050 r^{4}+2205 r^{3}+2436 r^{2}+1323 r +270}\) | \(\frac {4}{15}\) |
| \(b_{4}\) | \(\frac {56 r^{8}+492 r^{7}+1902 r^{6}+5985 r^{5}+17589 r^{4}+37533 r^{3}+59293 r^{2}+62910 r +32400}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{14}\) |
| \(b_{5}\) | \(\frac {192 r^{9}+3240 r^{8}+22588 r^{7}+88190 r^{6}+227593 r^{5}+451550 r^{4}+769987 r^{3}+1088360 r^{2}+1014900 r +436320}{180 \left (8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(\frac {101}{4725}\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= 1+2 t +t^{2}+\frac {4 t^{3}}{15}+\frac {t^{4}}{14}+\frac {101 t^{5}}{4725}+O\left (t^{6}\right ) \end {align*}
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} \sqrt {t}\, \left (1+\frac {5 t}{6}+\frac {17 t^{2}}{60}+\frac {89 t^{3}}{1260}+\frac {941 t^{4}}{45360}+\frac {14989 t^{5}}{2494800}+O\left (t^{6}\right )\right ) + c_{2} \left (1+2 t +t^{2}+\frac {4 t^{3}}{15}+\frac {t^{4}}{14}+\frac {101 t^{5}}{4725}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {t}\, \left (1+\frac {5 t}{6}+\frac {17 t^{2}}{60}+\frac {89 t^{3}}{1260}+\frac {941 t^{4}}{45360}+\frac {14989 t^{5}}{2494800}+O\left (t^{6}\right )\right )+c_{2} \left (1+2 t +t^{2}+\frac {4 t^{3}}{15}+\frac {t^{4}}{14}+\frac {101 t^{5}}{4725}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {t}\, \left (1+\frac {5 t}{6}+\frac {17 t^{2}}{60}+\frac {89 t^{3}}{1260}+\frac {941 t^{4}}{45360}+\frac {14989 t^{5}}{2494800}+O\left (t^{6}\right )\right )+c_{2} \left (1+2 t +t^{2}+\frac {4 t^{3}}{15}+\frac {t^{4}}{14}+\frac {101 t^{5}}{4725}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {t}\, \left (1+\frac {5 t}{6}+\frac {17 t^{2}}{60}+\frac {89 t^{3}}{1260}+\frac {941 t^{4}}{45360}+\frac {14989 t^{5}}{2494800}+O\left (t^{6}\right )\right )+c_{2} \left (1+2 t +t^{2}+\frac {4 t^{3}}{15}+\frac {t^{4}}{14}+\frac {101 t^{5}}{4725}+O\left (t^{6}\right )\right ) \] 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 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 -> 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: 0.265 (sec). Leaf size: 44
Order:=6; dsolve(2*sin(t)*diff(y(t),t$2)+(1-t)*diff(y(t),t)-2*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} \sqrt {t}\, \left (1+\frac {5}{6} t +\frac {17}{60} t^{2}+\frac {89}{1260} t^{3}+\frac {941}{45360} t^{4}+\frac {14989}{2494800} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} \left (1+2 t +t^{2}+\frac {4}{15} t^{3}+\frac {1}{14} t^{4}+\frac {101}{4725} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 1303
AsymptoticDSolveValue[2*sin(t)*y''[t]+(1-t)*y'[t]-2*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to \left (\frac {\left (\frac {2 \sin -1}{4 \sin ^2}+\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+3}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+4}{2 \sin }-\frac {1}{\sin }\right ) t^5}{\left (\frac {(2 \sin -1) \left (\frac {2 \sin -1}{2 \sin }+1\right )}{2 \sin }+\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+1\right ) \left (\frac {2 \sin -1}{2 \sin }+2\right )+\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+2\right ) \left (\frac {2 \sin -1}{2 \sin }+3\right )+\frac {\frac {2 \sin -1}{2 \sin }+3}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+3\right ) \left (\frac {2 \sin -1}{2 \sin }+4\right )+\frac {\frac {2 \sin -1}{2 \sin }+4}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+4\right ) \left (\frac {2 \sin -1}{2 \sin }+5\right )+\frac {\frac {2 \sin -1}{2 \sin }+5}{2 \sin }\right )}-\frac {\left (\frac {2 \sin -1}{4 \sin ^2}+\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+3}{2 \sin }-\frac {1}{\sin }\right ) t^4}{\left (\frac {(2 \sin -1) \left (\frac {2 \sin -1}{2 \sin }+1\right )}{2 \sin }+\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+1\right ) \left (\frac {2 \sin -1}{2 \sin }+2\right )+\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+2\right ) \left (\frac {2 \sin -1}{2 \sin }+3\right )+\frac {\frac {2 \sin -1}{2 \sin }+3}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+3\right ) \left (\frac {2 \sin -1}{2 \sin }+4\right )+\frac {\frac {2 \sin -1}{2 \sin }+4}{2 \sin }\right )}+\frac {\left (\frac {2 \sin -1}{4 \sin ^2}+\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }-\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }-\frac {1}{\sin }\right ) t^3}{\left (\frac {(2 \sin -1) \left (\frac {2 \sin -1}{2 \sin }+1\right )}{2 \sin }+\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+1\right ) \left (\frac {2 \sin -1}{2 \sin }+2\right )+\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+2\right ) \left (\frac {2 \sin -1}{2 \sin }+3\right )+\frac {\frac {2 \sin -1}{2 \sin }+3}{2 \sin }\right )}-\frac {\left (\frac {2 \sin -1}{4 \sin ^2}+\frac {1}{\sin }\right ) \left (-\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }-\frac {1}{\sin }\right ) t^2}{\left (\frac {(2 \sin -1) \left (\frac {2 \sin -1}{2 \sin }+1\right )}{2 \sin }+\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }\right ) \left (\left (\frac {2 \sin -1}{2 \sin }+1\right ) \left (\frac {2 \sin -1}{2 \sin }+2\right )+\frac {\frac {2 \sin -1}{2 \sin }+2}{2 \sin }\right )}+\frac {\left (\frac {2 \sin -1}{4 \sin ^2}+\frac {1}{\sin }\right ) t}{\frac {(2 \sin -1) \left (\frac {2 \sin -1}{2 \sin }+1\right )}{2 \sin }+\frac {\frac {2 \sin -1}{2 \sin }+1}{2 \sin }}+1\right ) c_1 t^{\frac {2 \sin -1}{2 \sin }}+\left (\frac {45 t^5}{\left (2+\frac {1}{\sin }\right ) \left (6+\frac {3}{2 \sin }\right ) \left (12+\frac {2}{\sin }\right ) \left (20+\frac {5}{2 \sin }\right ) \sin ^4}+\frac {15 t^4}{\left (2+\frac {1}{\sin }\right ) \left (6+\frac {3}{2 \sin }\right ) \left (12+\frac {2}{\sin }\right ) \sin ^3}+\frac {6 t^3}{\left (2+\frac {1}{\sin }\right ) \left (6+\frac {3}{2 \sin }\right ) \sin ^2}+\frac {3 t^2}{\left (2+\frac {1}{\sin }\right ) \sin }+2 t+1\right ) c_2 \]