Internal problem ID [13978]
Internal file name [OUTPUT/13150_Friday_February_23_2024_06_57_51_AM_19704798/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 35. Modified Power series solutions and basic method of Frobenius. Additional
Exercises. page 715
Problem number: 35.3 (a).
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]]
\[ \boxed {x^{2} y^{\prime \prime }+\frac {x y^{\prime }}{x -2}+\frac {2 y}{x +2}=0} \] 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. \[ x^{2} y^{\prime \prime }+\frac {x y^{\prime }}{x -2}+\frac {2 y}{x +2} = 0 \] 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 {1}{x \left (x -2\right )}\\ q(x) &= \frac {2}{\left (x +2\right ) x^{2}}\\ \end {align*}
| \(p(x)=\frac {1}{x \left (x -2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 2\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2}{\left (x +2\right ) x^{2}}\) | |
| singularity | type |
| \(x = -2\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 2, -2, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x^{2} \left (x -2\right ) \left (x +2\right )+y^{\prime } x \left (x +2\right )+\left (2 x -4\right ) y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (x -2\right ) \left (x +2\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x \left (x +2\right )+\left (2 x -4\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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} 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 }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right )-4 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ -4 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r -4 a_{0} x^{r} = 0 \] Or \[ \left (-4 x^{r} r \left (-1+r \right )+2 x^{r} r -4 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-4 r^{2}+6 r -4\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -4 r^{2}+6 r -4 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {3}{4}-\frac {i \sqrt {7}}{4}\\ r_2 &= \frac {3}{4}+\frac {i \sqrt {7}}{4} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-4 r^{2}+6 r -4\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {3}{4}-\frac {i \sqrt {7}}{4}, \frac {3}{4}+\frac {i \sqrt {7}}{4}\right ]\).
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 {3}{4}-\frac {i \sqrt {7}}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {3}{4}+\frac {i \sqrt {7}}{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 \[ a_{1} = \frac {r +2}{4 r^{2}+2 r +2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-4 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )+2 a_{n -1}-4 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}+n a_{n -1}-5 r a_{n -2}+r a_{n -1}+6 a_{n -2}+a_{n -1}}{4 n^{2}+8 n r +4 r^{2}-6 n -6 r +4}\tag {4} \] Which for the root \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) becomes \[ a_{n} = \frac {-i \left (4 n a_{n -2}-7 a_{n -2}+2 a_{n -1}\right ) \sqrt {7}+8 n^{2} a_{n -2}-4 \left (7 a_{n -2}-2 a_{n -1}\right ) n +19 a_{n -2}+14 a_{n -1}}{16 \left (-i \sqrt {7}+2 n \right ) n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r +2}{4 r^{2}+2 r +2}\) | \(\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{4}-2 r^{3}+r^{2}+3 r +6}{16 r^{4}+48 r^{3}+60 r^{2}+36 r +16} \] Which for the root \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) becomes \[ a_{2}=\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r +2}{4 r^{2}+2 r +2}\) | \(\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i}\) |
| \(a_{2}\) | \(\frac {4 r^{4}-2 r^{3}+r^{2}+3 r +6}{16 r^{4}+48 r^{3}+60 r^{2}+36 r +16}\) | \(\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 r^{5}+36 r^{4}+39 r^{3}+51 r^{2}+34 r +24}{8 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right )} \] Which for the root \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) becomes \[ a_{3}=\frac {223 i \sqrt {7}+43}{9216+9472 i \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r +2}{4 r^{2}+2 r +2}\) | \(\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i}\) |
| \(a_{2}\) | \(\frac {4 r^{4}-2 r^{3}+r^{2}+3 r +6}{16 r^{4}+48 r^{3}+60 r^{2}+36 r +16}\) | \(\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+36 r^{4}+39 r^{3}+51 r^{2}+34 r +24}{8 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right )}\) | \(\frac {223 i \sqrt {7}+43}{9216+9472 i \sqrt {7}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{8}+112 r^{7}+288 r^{6}+358 r^{5}+389 r^{4}+692 r^{3}+1143 r^{2}+938 r +384}{16 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right ) \left (2 r^{2}+13 r +22\right )} \] Which for the root \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) becomes \[ a_{4}=\frac {-7577 \sqrt {7}-979 i}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r +2}{4 r^{2}+2 r +2}\) | \(\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i}\) |
| \(a_{2}\) | \(\frac {4 r^{4}-2 r^{3}+r^{2}+3 r +6}{16 r^{4}+48 r^{3}+60 r^{2}+36 r +16}\) | \(\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+36 r^{4}+39 r^{3}+51 r^{2}+34 r +24}{8 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right )}\) | \(\frac {223 i \sqrt {7}+43}{9216+9472 i \sqrt {7}}\) |
| \(a_{4}\) | \(\frac {16 r^{8}+112 r^{7}+288 r^{6}+358 r^{5}+389 r^{4}+692 r^{3}+1143 r^{2}+938 r +384}{16 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right ) \left (2 r^{2}+13 r +22\right )}\) | \(\frac {-7577 \sqrt {7}-979 i}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\frac {3}{2} r^{9}+\frac {45}{2} r^{8}+\frac {1089}{8} r^{7}+\frac {3555}{8} r^{6}+\frac {28389}{32} r^{5}+\frac {4869}{4} r^{4}+\frac {42603}{32} r^{3}+\frac {9699}{8} r^{2}+\frac {6003}{8} r +270}{\left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right ) \left (2 r^{2}+13 r +22\right ) \left (2 r^{2}+17 r +37\right )} \] Which for the root \(r = \frac {3}{4}-\frac {i \sqrt {7}}{4}\) becomes \[ a_{5}=\frac {-553875 i \sqrt {7}-1249007}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r +2}{4 r^{2}+2 r +2}\) | \(\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i}\) |
| \(a_{2}\) | \(\frac {4 r^{4}-2 r^{3}+r^{2}+3 r +6}{16 r^{4}+48 r^{3}+60 r^{2}+36 r +16}\) | \(\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {8 r^{5}+36 r^{4}+39 r^{3}+51 r^{2}+34 r +24}{8 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right )}\) | \(\frac {223 i \sqrt {7}+43}{9216+9472 i \sqrt {7}}\) |
| \(a_{4}\) | \(\frac {16 r^{8}+112 r^{7}+288 r^{6}+358 r^{5}+389 r^{4}+692 r^{3}+1143 r^{2}+938 r +384}{16 \left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right ) \left (2 r^{2}+13 r +22\right )}\) | \(\frac {-7577 \sqrt {7}-979 i}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}\) |
| \(a_{5}\) | \(\frac {\frac {3}{2} r^{9}+\frac {45}{2} r^{8}+\frac {1089}{8} r^{7}+\frac {3555}{8} r^{6}+\frac {28389}{32} r^{5}+\frac {4869}{4} r^{4}+\frac {42603}{32} r^{3}+\frac {9699}{8} r^{2}+\frac {6003}{8} r +270}{\left (2 r^{2}+9 r +11\right ) \left (4 r^{4}+12 r^{3}+15 r^{2}+9 r +4\right ) \left (2 r^{2}+13 r +22\right ) \left (2 r^{2}+17 r +37\right )}\) | \(\frac {-553875 i \sqrt {7}-1249007}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {3}{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 {3}{4}-\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}+11 i\right ) x}{8 \sqrt {7}+16 i}+\frac {\left (7 i \sqrt {7}-45\right ) x^{2}}{-64+384 i \sqrt {7}}+\frac {\left (223 i \sqrt {7}+43\right ) x^{3}}{9216+9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}-979 i\right ) x^{4}}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}+\frac {\left (-553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}+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 \[ y_{2}\left (x \right )= x^{\frac {3}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}-11 i\right ) x}{8 \sqrt {7}-16 i}+\frac {\left (-7 i \sqrt {7}-45\right ) x^{2}}{-64-384 i \sqrt {7}}+\frac {\left (-223 i \sqrt {7}+43\right ) x^{3}}{9216-9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}+979 i\right ) x^{4}}{4096 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right )}+\frac {\left (553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right ) \left (\sqrt {7}-10 i\right )}+O\left (x^{6}\right )\right ) \] 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 {3}{4}-\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}+11 i\right ) x}{8 \sqrt {7}+16 i}+\frac {\left (7 i \sqrt {7}-45\right ) x^{2}}{-64+384 i \sqrt {7}}+\frac {\left (223 i \sqrt {7}+43\right ) x^{3}}{9216+9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}-979 i\right ) x^{4}}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}+\frac {\left (-553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {3}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}-11 i\right ) x}{8 \sqrt {7}-16 i}+\frac {\left (-7 i \sqrt {7}-45\right ) x^{2}}{-64-384 i \sqrt {7}}+\frac {\left (-223 i \sqrt {7}+43\right ) x^{3}}{9216-9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}+979 i\right ) x^{4}}{4096 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right )}+\frac {\left (553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right ) \left (\sqrt {7}-10 i\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{4}-\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}+11 i\right ) x}{8 \sqrt {7}+16 i}+\frac {\left (7 i \sqrt {7}-45\right ) x^{2}}{-64+384 i \sqrt {7}}+\frac {\left (223 i \sqrt {7}+43\right ) x^{3}}{9216+9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}-979 i\right ) x^{4}}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}+\frac {\left (-553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {3}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}-11 i\right ) x}{8 \sqrt {7}-16 i}+\frac {\left (-7 i \sqrt {7}-45\right ) x^{2}}{-64-384 i \sqrt {7}}+\frac {\left (-223 i \sqrt {7}+43\right ) x^{3}}{9216-9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}+979 i\right ) x^{4}}{4096 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right )}+\frac {\left (553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right ) \left (\sqrt {7}-10 i\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{4}-\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}+11 i\right ) x}{8 \sqrt {7}+16 i}+\frac {\left (7 i \sqrt {7}-45\right ) x^{2}}{-64+384 i \sqrt {7}}+\frac {\left (223 i \sqrt {7}+43\right ) x^{3}}{9216+9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}-979 i\right ) x^{4}}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}+\frac {\left (-553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {3}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}-11 i\right ) x}{8 \sqrt {7}-16 i}+\frac {\left (-7 i \sqrt {7}-45\right ) x^{2}}{-64-384 i \sqrt {7}}+\frac {\left (-223 i \sqrt {7}+43\right ) x^{3}}{9216-9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}+979 i\right ) x^{4}}{4096 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right )}+\frac {\left (553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right ) \left (\sqrt {7}-10 i\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{4}-\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}+11 i\right ) x}{8 \sqrt {7}+16 i}+\frac {\left (7 i \sqrt {7}-45\right ) x^{2}}{-64+384 i \sqrt {7}}+\frac {\left (223 i \sqrt {7}+43\right ) x^{3}}{9216+9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}-979 i\right ) x^{4}}{4096 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right )}+\frac {\left (-553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}+6 i\right ) \left (6 \sqrt {7}+i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {3}{4}+\frac {i \sqrt {7}}{4}} \left (1+\frac {\left (\sqrt {7}-11 i\right ) x}{8 \sqrt {7}-16 i}+\frac {\left (-7 i \sqrt {7}-45\right ) x^{2}}{-64-384 i \sqrt {7}}+\frac {\left (-223 i \sqrt {7}+43\right ) x^{3}}{9216-9472 i \sqrt {7}}+\frac {\left (-7577 \sqrt {7}+979 i\right ) x^{4}}{4096 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right )}+\frac {\left (553875 i \sqrt {7}-1249007\right ) x^{5}}{81920 \left (\sqrt {7}-6 i\right ) \left (6 \sqrt {7}-i\right ) \left (\sqrt {7}-8 i\right ) \left (\sqrt {7}-10 i\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
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: 0.031 (sec). Leaf size: 1283
Order:=6; dsolve(x^2*diff(y(x),x$2)+x/(x-2)*diff(y(x),x)+2/(x+2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {3}{4}} \left (c_{2} x^{\frac {i \sqrt {7}}{4}} \left (1+\frac {i \sqrt {7}+11}{8 i \sqrt {7}+16} x +\frac {1}{64} \frac {7 i \sqrt {7}+45}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right )} x^{2}+\frac {1}{256} \frac {223 i \sqrt {7}-43}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right )} x^{3}+\frac {1}{4096} \frac {7577 i \sqrt {7}+979}{\left (2+i \sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right ) \left (i \sqrt {7}+8\right )} x^{4}+\frac {1}{81920} \frac {553875 \sqrt {7}+1249007 i}{\left (-2 i+\sqrt {7}\right ) \left (i \sqrt {7}+4\right ) \left (i \sqrt {7}+6\right ) \left (i \sqrt {7}+8\right ) \left (i \sqrt {7}+10\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} x^{-\frac {i \sqrt {7}}{4}} \left (1+\frac {\sqrt {7}+11 i}{8 \sqrt {7}+16 i} x +\frac {7 i \sqrt {7}-45}{-64+384 i \sqrt {7}} x^{2}+\frac {-223 \sqrt {7}+43 i}{9216 i-9472 \sqrt {7}} x^{3}+\frac {7577 \sqrt {7}+979 i}{-2240512 i+1064960 \sqrt {7}} x^{4}+\frac {1}{81920} \frac {553875 \sqrt {7}-1249007 i}{\left (4 i+\sqrt {7}\right ) \left (\sqrt {7}+2 i\right ) \left (\sqrt {7}+6 i\right ) \left (\sqrt {7}+8 i\right ) \left (\sqrt {7}+10 i\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 6290
AsymptoticDSolveValue[x^2*y''[x]+x/(x-2)*y'[x]+2/(x+2)*y[x]==0,y[x],{x,0,5}]
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