Internal problem ID [1295]
Internal file name [OUTPUT/1296_Sunday_June_05_2022_02_08_46_AM_84605145/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 1.
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]]
\[ \boxed {2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+x \left (5 x^{2}+3 x +3\right ) y^{\prime }+y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (2 x^{4}+2 x^{3}+2 x^{2}\right ) y^{\prime \prime }+\left (5 x^{3}+3 x^{2}+3 x \right ) y^{\prime }+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {5 x^{2}+3 x +3}{2 x \left (x^{2}+x +1\right )}\\ q(x) &= \frac {1}{2 x^{2} \left (x^{2}+x +1\right )}\\ \end {align*}
| \(p(x)=\frac {5 x^{2}+3 x +3}{2 x \left (x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{2 x^{2} \left (x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -\frac {1}{2}-\frac {i \sqrt {3}}{2}, -\frac {1}{2}+\frac {i \sqrt {3}}{2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+\left (5 x^{3}+3 x^{2}+3 x \right ) y^{\prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 2 x^{2} \left (x^{2}+x +1\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 (5 x^{3}+3 x^{2}+3 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 ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\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 \(n +r\) 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 }}2 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \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 =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )+3 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}+r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}+r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -\frac {1}{4}+\frac {i \sqrt {7}}{4}\\ r_2 &= -\frac {i \sqrt {7}}{4}-\frac {1}{4} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}+r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-\frac {1}{4}+\frac {i \sqrt {7}}{4}, -\frac {i \sqrt {7}}{4}-\frac {1}{4}\right ]\).
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 -\frac {1}{4}+\frac {i \sqrt {7}}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {i \sqrt {7}}{4}-\frac {1}{4}} \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 \[ a_{1} = \frac {-2 r^{2}-r}{2 r^{2}+5 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -2} \left (n +r -2\right )+3 a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 n^{2} a_{n -2}+2 n^{2} a_{n -1}+4 n r a_{n -2}+4 n r a_{n -1}+2 r^{2} a_{n -2}+2 r^{2} a_{n -1}-5 n a_{n -2}-3 n a_{n -1}-5 r a_{n -2}-3 r a_{n -1}+2 a_{n -2}+a_{n -1}}{2 n^{2}+4 n r +2 r^{2}+n +r +1}\tag {4} \] Which for the root \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) becomes \[ a_{n} = \frac {i \left (2 \left (-a_{n -2}-a_{n -1}\right ) n +3 a_{n -2}+2 a_{n -1}\right ) \sqrt {7}+4 \left (-a_{n -2}-a_{n -1}\right ) n^{2}+4 \left (3 a_{n -2}+2 a_{n -1}\right ) n -5 a_{n -2}-2 a_{n -1}}{2 n \left (i \sqrt {7}+2 n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r^{2}-r}{2 r^{2}+5 r +4}\) | \(\frac {1}{i \sqrt {7}+2}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=-\frac {r \left (2 r +3\right )^{2}}{4 r^{4}+28 r^{3}+75 r^{2}+91 r +44} \] Which for the root \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) becomes \[ a_{2}=-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right )}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r^{2}-r}{2 r^{2}+5 r +4}\) | \(\frac {1}{i \sqrt {7}+2}\) |
| \(a_{2}\) | \(-\frac {r \left (2 r +3\right )^{2}}{4 r^{4}+28 r^{3}+75 r^{2}+91 r +44}\) | \(-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right )}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 r^{6}+76 r^{5}+282 r^{4}+505 r^{3}+433 r^{2}+145 r}{8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968} \] Which for the root \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) becomes \[ a_{3}=\frac {49 i \sqrt {7}+89}{432-444 i \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r^{2}-r}{2 r^{2}+5 r +4}\) | \(\frac {1}{i \sqrt {7}+2}\) |
| \(a_{2}\) | \(-\frac {r \left (2 r +3\right )^{2}}{4 r^{4}+28 r^{3}+75 r^{2}+91 r +44}\) | \(-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right )}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}\) |
| \(a_{3}\) | \(\frac {8 r^{6}+76 r^{5}+282 r^{4}+505 r^{3}+433 r^{2}+145 r}{8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968}\) | \(\frac {49 i \sqrt {7}+89}{432-444 i \sqrt {7}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=-\frac {\left (16 r^{7}+240 r^{6}+1480 r^{5}+4800 r^{4}+8645 r^{3}+8269 r^{2}+3466 r +273\right ) r}{\left (2 r^{2}+17 r +37\right ) \left (8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968\right )} \] Which for the root \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) becomes \[ a_{4}=\frac {1553 i+395 \sqrt {7}}{26256 i+12480 \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r^{2}-r}{2 r^{2}+5 r +4}\) | \(\frac {1}{i \sqrt {7}+2}\) |
| \(a_{2}\) | \(-\frac {r \left (2 r +3\right )^{2}}{4 r^{4}+28 r^{3}+75 r^{2}+91 r +44}\) | \(-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right )}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}\) |
| \(a_{3}\) | \(\frac {8 r^{6}+76 r^{5}+282 r^{4}+505 r^{3}+433 r^{2}+145 r}{8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968}\) | \(\frac {49 i \sqrt {7}+89}{432-444 i \sqrt {7}}\) |
| \(a_{4}\) | \(-\frac {\left (16 r^{7}+240 r^{6}+1480 r^{5}+4800 r^{4}+8645 r^{3}+8269 r^{2}+3466 r +273\right ) r}{\left (2 r^{2}+17 r +37\right ) \left (8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968\right )}\) | \(\frac {1553 i+395 \sqrt {7}}{26256 i+12480 \sqrt {7}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {64 r \left (r +\frac {9}{2}\right ) \left (r^{7}+18 r^{6}+136 r^{5}+559 r^{4}+\frac {21571}{16} r^{3}+\frac {7619}{4} r^{2}+\frac {23343}{16} r +\frac {15003}{32}\right )}{\left (2 r^{2}+17 r +37\right ) \left (8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968\right ) \left (2 r^{2}+21 r +56\right )} \] Which for the root \(r = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) becomes \[ a_{5}=\frac {42423 i \sqrt {7}+45275}{492720 i \sqrt {7}-1749600} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r^{2}-r}{2 r^{2}+5 r +4}\) | \(\frac {1}{i \sqrt {7}+2}\) |
| \(a_{2}\) | \(-\frac {r \left (2 r +3\right )^{2}}{4 r^{4}+28 r^{3}+75 r^{2}+91 r +44}\) | \(-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right )}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}\) |
| \(a_{3}\) | \(\frac {8 r^{6}+76 r^{5}+282 r^{4}+505 r^{3}+433 r^{2}+145 r}{8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968}\) | \(\frac {49 i \sqrt {7}+89}{432-444 i \sqrt {7}}\) |
| \(a_{4}\) | \(-\frac {\left (16 r^{7}+240 r^{6}+1480 r^{5}+4800 r^{4}+8645 r^{3}+8269 r^{2}+3466 r +273\right ) r}{\left (2 r^{2}+17 r +37\right ) \left (8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968\right )}\) | \(\frac {1553 i+395 \sqrt {7}}{26256 i+12480 \sqrt {7}}\) |
| \(a_{5}\) | \(-\frac {64 r \left (r +\frac {9}{2}\right ) \left (r^{7}+18 r^{6}+136 r^{5}+559 r^{4}+\frac {21571}{16} r^{3}+\frac {7619}{4} r^{2}+\frac {23343}{16} r +\frac {15003}{32}\right )}{\left (2 r^{2}+17 r +37\right ) \left (8 r^{6}+108 r^{5}+602 r^{4}+1773 r^{3}+2921 r^{2}+2574 r +968\right ) \left (2 r^{2}+21 r +56\right )}\) | \(\frac {42423 i \sqrt {7}+45275}{492720 i \sqrt {7}-1749600}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \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^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {x}{i \sqrt {7}+2}-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}+\frac {\left (49 i \sqrt {7}+89\right ) x^{3}}{432-444 i \sqrt {7}}+\frac {\left (1553 i+395 \sqrt {7}\right ) x^{4}}{26256 i+12480 \sqrt {7}}+\frac {\left (42423 i \sqrt {7}+45275\right ) x^{5}}{492720 i \sqrt {7}-1749600}+O\left (x^{6}\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 \[ y_{2}\left (x \right )= x^{-\frac {i \sqrt {7}}{4}-\frac {1}{4}} \left (1+\frac {x}{-i \sqrt {7}+2}-\frac {5 \left (-i \sqrt {7}-1\right ) \left (-i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (-i \sqrt {7}+2\right ) \left (-i \sqrt {7}+4\right )}+\frac {\left (-49 i \sqrt {7}+89\right ) x^{3}}{432+444 i \sqrt {7}}+\frac {\left (-1553 i+395 \sqrt {7}\right ) x^{4}}{-26256 i+12480 \sqrt {7}}+\frac {\left (-42423 i \sqrt {7}+45275\right ) x^{5}}{-492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) \] 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^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {x}{i \sqrt {7}+2}-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}+\frac {\left (49 i \sqrt {7}+89\right ) x^{3}}{432-444 i \sqrt {7}}+\frac {\left (1553 i+395 \sqrt {7}\right ) x^{4}}{26256 i+12480 \sqrt {7}}+\frac {\left (42423 i \sqrt {7}+45275\right ) x^{5}}{492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) + c_{2} x^{-\frac {i \sqrt {7}}{4}-\frac {1}{4}} \left (1+\frac {x}{-i \sqrt {7}+2}-\frac {5 \left (-i \sqrt {7}-1\right ) \left (-i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (-i \sqrt {7}+2\right ) \left (-i \sqrt {7}+4\right )}+\frac {\left (-49 i \sqrt {7}+89\right ) x^{3}}{432+444 i \sqrt {7}}+\frac {\left (-1553 i+395 \sqrt {7}\right ) x^{4}}{-26256 i+12480 \sqrt {7}}+\frac {\left (-42423 i \sqrt {7}+45275\right ) x^{5}}{-492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {x}{i \sqrt {7}+2}-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}+\frac {\left (49 i \sqrt {7}+89\right ) x^{3}}{432-444 i \sqrt {7}}+\frac {\left (1553 i+395 \sqrt {7}\right ) x^{4}}{26256 i+12480 \sqrt {7}}+\frac {\left (42423 i \sqrt {7}+45275\right ) x^{5}}{492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {i \sqrt {7}}{4}-\frac {1}{4}} \left (1+\frac {x}{-i \sqrt {7}+2}-\frac {5 \left (-i \sqrt {7}-1\right ) \left (-i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (-i \sqrt {7}+2\right ) \left (-i \sqrt {7}+4\right )}+\frac {\left (-49 i \sqrt {7}+89\right ) x^{3}}{432+444 i \sqrt {7}}+\frac {\left (-1553 i+395 \sqrt {7}\right ) x^{4}}{-26256 i+12480 \sqrt {7}}+\frac {\left (-42423 i \sqrt {7}+45275\right ) x^{5}}{-492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {x}{i \sqrt {7}+2}-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}+\frac {\left (49 i \sqrt {7}+89\right ) x^{3}}{432-444 i \sqrt {7}}+\frac {\left (1553 i+395 \sqrt {7}\right ) x^{4}}{26256 i+12480 \sqrt {7}}+\frac {\left (42423 i \sqrt {7}+45275\right ) x^{5}}{492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {i \sqrt {7}}{4}-\frac {1}{4}} \left (1+\frac {x}{-i \sqrt {7}+2}-\frac {5 \left (-i \sqrt {7}-1\right ) \left (-i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (-i \sqrt {7}+2\right ) \left (-i \sqrt {7}+4\right )}+\frac {\left (-49 i \sqrt {7}+89\right ) x^{3}}{432+444 i \sqrt {7}}+\frac {\left (-1553 i+395 \sqrt {7}\right ) x^{4}}{-26256 i+12480 \sqrt {7}}+\frac {\left (-42423 i \sqrt {7}+45275\right ) x^{5}}{-492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{-\frac {1}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {x}{i \sqrt {7}+2}-\frac {5 \left (-1+i \sqrt {7}\right ) \left (i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (i \sqrt {7}+2\right ) \left (i \sqrt {7}+4\right )}+\frac {\left (49 i \sqrt {7}+89\right ) x^{3}}{432-444 i \sqrt {7}}+\frac {\left (1553 i+395 \sqrt {7}\right ) x^{4}}{26256 i+12480 \sqrt {7}}+\frac {\left (42423 i \sqrt {7}+45275\right ) x^{5}}{492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {i \sqrt {7}}{4}-\frac {1}{4}} \left (1+\frac {x}{-i \sqrt {7}+2}-\frac {5 \left (-i \sqrt {7}-1\right ) \left (-i \sqrt {7}+\frac {9}{5}\right ) x^{2}}{16 \left (-i \sqrt {7}+2\right ) \left (-i \sqrt {7}+4\right )}+\frac {\left (-49 i \sqrt {7}+89\right ) x^{3}}{432+444 i \sqrt {7}}+\frac {\left (-1553 i+395 \sqrt {7}\right ) x^{4}}{-26256 i+12480 \sqrt {7}}+\frac {\left (-42423 i \sqrt {7}+45275\right ) x^{5}}{-492720 i \sqrt {7}-1749600}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+\left (5 x^{3}+3 x^{2}+3 x \right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y}{2 x^{2} \left (x^{2}+x +1\right )}-\frac {\left (5 x^{2}+3 x +3\right ) y^{\prime }}{2 x \left (x^{2}+x +1\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (5 x^{2}+3 x +3\right ) y^{\prime }}{2 x \left (x^{2}+x +1\right )}+\frac {y}{2 x^{2} \left (x^{2}+x +1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {5 x^{2}+3 x +3}{2 x \left (x^{2}+x +1\right )}, P_{3}\left (x \right )=\frac {1}{2 x^{2} \left (x^{2}+x +1\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {3}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {1}{2} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+x \left (5 x^{2}+3 x +3\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (2 r^{2}+r +1\right ) x^{r}+\left (\left (2 r^{2}+5 r +4\right ) a_{1}+a_{0} r \left (1+2 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k^{2}+4 k r +2 r^{2}+k +r +1\right )+a_{k -1} \left (k +r -1\right ) \left (2 k -1+2 r \right )+a_{k -2} \left (k -2+r \right ) \left (2 k -1+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r^{2}+r +1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}, -\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (2 r^{2}+5 r +4\right ) a_{1}+a_{0} r \left (1+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} r \left (1+2 r \right )}{2 r^{2}+5 r +4} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (2 a_{k}+2 a_{k -2}+2 a_{k -1}\right ) k^{2}+\left (\left (4 a_{k}+4 a_{k -2}+4 a_{k -1}\right ) r +a_{k}-5 a_{k -2}-3 a_{k -1}\right ) k +\left (2 a_{k}+2 a_{k -2}+2 a_{k -1}\right ) r^{2}+\left (a_{k}-5 a_{k -2}-3 a_{k -1}\right ) r +a_{k}+2 a_{k -2}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (2 a_{k +2}+2 a_{k}+2 a_{k +1}\right ) \left (k +2\right )^{2}+\left (\left (4 a_{k +2}+4 a_{k}+4 a_{k +1}\right ) r +a_{k +2}-5 a_{k}-3 a_{k +1}\right ) \left (k +2\right )+\left (2 a_{k +2}+2 a_{k}+2 a_{k +1}\right ) r^{2}+\left (a_{k +2}-5 a_{k}-3 a_{k +1}\right ) r +a_{k +2}+2 a_{k}+a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{k +1}+4 k r a_{k}+4 k r a_{k +1}+2 r^{2} a_{k}+2 r^{2} a_{k +1}+3 k a_{k}+5 k a_{k +1}+3 r a_{k}+5 r a_{k +1}+3 a_{k +1}}{2 k^{2}+4 k r +2 r^{2}+9 k +9 r +11} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{k +1}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k +1}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{k}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{k +1}+3 k a_{k}+5 k a_{k +1}+3 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+5 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k +1}+3 a_{k +1}}{2 k^{2}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2}+9 k +\frac {35}{4}+\frac {9 \,\mathrm {I} \sqrt {7}}{4}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}}, a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{k +1}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k +1}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{k}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{k +1}+3 k a_{k}+5 k a_{k +1}+3 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+5 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k +1}+3 a_{k +1}}{2 k^{2}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2}+9 k +\frac {35}{4}+\frac {9 \,\mathrm {I} \sqrt {7}}{4}}, a_{1}=-\frac {a_{0} \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) \left (\frac {\mathrm {I} \sqrt {7}}{2}+\frac {1}{2}\right )}{2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2}+\frac {11}{4}+\frac {5 \,\mathrm {I} \sqrt {7}}{4}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{k +1}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k +1}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} a_{k}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} a_{k +1}+3 k a_{k}+5 k a_{k +1}+3 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k}+5 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k +1}+3 a_{k +1}}{2 k^{2}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2}+9 k -\frac {9 \,\mathrm {I} \sqrt {7}}{4}+\frac {35}{4}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}}, a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{k +1}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k +1}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} a_{k}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} a_{k +1}+3 k a_{k}+5 k a_{k +1}+3 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k}+5 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) a_{k +1}+3 a_{k +1}}{2 k^{2}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2}+9 k -\frac {9 \,\mathrm {I} \sqrt {7}}{4}+\frac {35}{4}}, a_{1}=-\frac {a_{0} \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{2}\right )}{2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2}-\frac {5 \,\mathrm {I} \sqrt {7}}{4}+\frac {11}{4}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}}\right ), a_{k +2}=-\frac {2 k^{2} a_{k}+2 k^{2} a_{1+k}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{1+k}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{k}+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2} a_{1+k}+3 k a_{k}+5 k a_{1+k}+3 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{k}+5 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) a_{1+k}+3 a_{1+k}}{2 k^{2}+4 k \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )+2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2}+9 k +\frac {35}{4}+\frac {9 \,\mathrm {I} \sqrt {7}}{4}}, a_{1}=-\frac {a_{0} \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right ) \left (\frac {\mathrm {I} \sqrt {7}}{2}+\frac {1}{2}\right )}{2 \left (-\frac {1}{4}+\frac {\mathrm {I} \sqrt {7}}{4}\right )^{2}+\frac {11}{4}+\frac {5 \,\mathrm {I} \sqrt {7}}{4}}, b_{k +2}=-\frac {2 k^{2} b_{k}+2 k^{2} b_{1+k}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) b_{k}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) b_{1+k}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} b_{k}+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2} b_{1+k}+3 k b_{k}+5 k b_{1+k}+3 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) b_{k}+5 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) b_{1+k}+3 b_{1+k}}{2 k^{2}+4 k \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )+2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2}+9 k -\frac {9 \,\mathrm {I} \sqrt {7}}{4}+\frac {35}{4}}, b_{1}=-\frac {b_{0} \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right ) \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {7}}{2}\right )}{2 \left (-\frac {\mathrm {I} \sqrt {7}}{4}-\frac {1}{4}\right )^{2}-\frac {5 \,\mathrm {I} \sqrt {7}}{4}+\frac {11}{4}}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> 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 trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 1243
Order:=6; dsolve(2*x^2*(1+x+x^2)*diff(y(x),x$2)+x*(3+3*x+5*x^2)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {i \sqrt {7}}{4}} \left (1+\frac {1}{2+i \sqrt {7}} x +\frac {1}{4} \frac {11-i \sqrt {7}}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right )} x^{2}-\frac {1}{12} \frac {49 i \sqrt {7}+89}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right )} x^{3}+\frac {1}{48} \frac {395 i \sqrt {7}-1553}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right ) \left (i \sqrt {7}+8\right )} x^{4}+\frac {1}{240} \frac {42423 i \sqrt {7}+45275}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right ) \left (i \sqrt {7}+8\right ) \left (i \sqrt {7}+10\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} x^{-\frac {i \sqrt {7}}{4}} \left (1+\frac {1}{2-i \sqrt {7}} x +\frac {-11-i \sqrt {7}}{-4+24 i \sqrt {7}} x^{2}+\frac {49 \sqrt {7}+89 i}{432 i-444 \sqrt {7}} x^{3}-\frac {1}{48} \frac {395 i \sqrt {7}+1553}{\left (2 i+\sqrt {7}\right ) \left (\sqrt {7}+4 i\right ) \left (\sqrt {7}+6 i\right ) \left (\sqrt {7}+8 i\right )} x^{4}+\frac {-42423 \sqrt {7}-45275 i}{1749600 i-492720 \sqrt {7}} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{4}}} \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 4838
AsymptoticDSolveValue[2*x^2*(1+x+x^2)*y''[x]+x*(3+3*x+5*x^2)*y'[x]+y[x]==0,y[x],{x,0,5}]
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