Internal problem ID [2942]
Internal file name [OUTPUT/2434_Sunday_June_05_2022_03_09_54_AM_68650022/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.5. page
771
Problem number: (c).
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]]
\[ \boxed {x^{2} y^{\prime \prime }+\left (-2 x^{5}+9 x \right ) y^{\prime }+\left (10 x^{4}+5 x^{2}+25\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (-2 x^{5}+9 x \right ) y^{\prime }+\left (10 x^{4}+5 x^{2}+25\right ) y = 0 \] 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 {2 x^{4}-9}{x}\\ q(x) &= \frac {10 x^{4}+5 x^{2}+25}{x^{2}}\\ \end {align*}
| \(p(x)=-\frac {2 x^{4}-9}{x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
| \(q(x)=\frac {10 x^{4}+5 x^{2}+25}{x^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (-2 x^{5}+9 x \right ) y^{\prime }+\left (10 x^{4}+5 x^{2}+25\right ) y = 0 \] 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} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-2 x^{5}+9 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (10 x^{4}+5 x^{2}+25\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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +4} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +4} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 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 }}\left (-2 x^{n +r +4} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-2 a_{n -4} \left (n -4+r \right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +4} a_{n} &= \moverset {\infty }{\munderset {n =4}{\sum }}10 a_{n -4} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} 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 =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-2 a_{n -4} \left (n -4+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}10 a_{n -4} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}25 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+9 x^{n +r} a_{n} \left (n +r \right )+25 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+9 x^{r} a_{0} r +25 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+9 x^{r} r +25 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+8 r +25\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+8 r +25 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -4+3 i\\ r_2 &= -4-3 i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+8 r +25\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-4+3 i, -4-3 i]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -4+3 i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -4-3 i} \end {align*}
\(y_{1}\left (x \right )\) is found first. Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = -\frac {5}{r^{2}+12 r +45} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] For \(4\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -4} \left (n -4+r \right )+9 a_{n} \left (n +r \right )+10 a_{n -4}+5 a_{n -2}+25 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 n a_{n -4}+2 r a_{n -4}-18 a_{n -4}-5 a_{n -2}}{n^{2}+2 n r +r^{2}+8 n +8 r +25}\tag {4} \] Which for the root \(r = -4+3 i\) becomes \[ a_{n} = \frac {\left (-26+6 i+2 n \right ) a_{n -4}-5 a_{n -2}}{n \left (n +6 i\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -4+3 i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {5}{r^{2}+12 r +45}\) | \(-\frac {1}{8}+\frac {3 i}{8}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {2 r^{3}+14 r^{2}-30 r -425}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right )} \] Which for the root \(r = -4+3 i\) becomes \[ a_{4}=-\frac {179}{832}+\frac {483 i}{832} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {5}{r^{2}+12 r +45}\) | \(-\frac {1}{8}+\frac {3 i}{8}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {2 r^{3}+14 r^{2}-30 r -425}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right )}\) | \(-\frac {179}{832}+\frac {483 i}{832}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {5}{r^{2}+12 r +45}\) | \(-\frac {1}{8}+\frac {3 i}{8}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {2 r^{3}+14 r^{2}-30 r -425}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right )}\) | \(-\frac {179}{832}+\frac {483 i}{832}\) |
| \(a_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-20 r^{3}-200 r^{2}-100 r +4315}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right ) \left (r^{2}+20 r +109\right )} \] Which for the root \(r = -4+3 i\) becomes \[ a_{6}=-\frac {433}{3744}-\frac {3943 i}{29952} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {5}{r^{2}+12 r +45}\) | \(-\frac {1}{8}+\frac {3 i}{8}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {2 r^{3}+14 r^{2}-30 r -425}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right )}\) | \(-\frac {179}{832}+\frac {483 i}{832}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(\frac {-20 r^{3}-200 r^{2}-100 r +4315}{\left (r^{2}+12 r +45\right ) \left (r^{2}+16 r +73\right ) \left (r^{2}+20 r +109\right )}\) | \(-\frac {433}{3744}-\frac {3943 i}{29952}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{-4+3 i} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}\dots \right ) \\ &= x^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\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 \[ y_{2}\left (x \right )= x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) \] 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^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) + c_{2} x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right )+c_{2} x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right )+c_{2} x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right )+c_{2} x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+O\left (x^{7}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (-2 x^{5}+9 x \right ) y^{\prime }+\left (10 x^{4}+5 x^{2}+25\right ) 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 {5 \left (2 x^{4}+x^{2}+5\right ) y}{x^{2}}+\frac {\left (2 x^{4}-9\right ) y^{\prime }}{x} \\ \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 (2 x^{4}-9\right ) y^{\prime }}{x}+\frac {5 \left (2 x^{4}+x^{2}+5\right ) y}{x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x^{4}-9}{x}, P_{3}\left (x \right )=\frac {5 \left (2 x^{4}+x^{2}+5\right )}{x^{2}}\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}}}=9 \\ {} & \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}}}=25 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }-x \left (2 x^{4}-9\right ) y^{\prime }+\left (10 x^{4}+5 x^{2}+25\right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..4 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..5 \\ {} & {} & 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^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}+8 r +25\right ) x^{r}+a_{1} \left (r^{2}+10 r +34\right ) x^{1+r}+\left (\left (r^{2}+12 r +45\right ) a_{2}+5 a_{0}\right ) x^{2+r}+\left (a_{3} \left (r^{2}+14 r +58\right )+5 a_{1}\right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+8 k +8 r +25\right )+5 a_{k -2}-2 a_{k -4} \left (k -9+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+8 r +25=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-4-3 \,\mathrm {I}, -4+3 \,\mathrm {I}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (r^{2}+10 r +34\right )=0, \left (r^{2}+12 r +45\right ) a_{2}+5 a_{0}=0, a_{3} \left (r^{2}+14 r +58\right )+5 a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=-\frac {5 a_{0}}{r^{2}+12 r +45}, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r +8\right ) k +r^{2}+8 r +25\right ) a_{k}+5 a_{k -2}-2 a_{k -4} \left (k -9+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & \left (\left (k +4\right )^{2}+\left (2 r +8\right ) \left (k +4\right )+r^{2}+8 r +25\right ) a_{k +4}+5 a_{k +2}-2 a_{k} \left (k +r -5\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {2 k a_{k}+2 r a_{k}-10 a_{k}-5 a_{k +2}}{k^{2}+2 k r +r^{2}+16 k +16 r +73} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-4-3 \,\mathrm {I} \\ {} & {} & a_{k +4}=\frac {2 k a_{k}-\left (8+6 \,\mathrm {I}\right ) a_{k}-10 a_{k}-5 a_{k +2}}{k^{2}-\left (8+6 \,\mathrm {I}\right ) k +16-24 \,\mathrm {I}+16 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-4-3 \,\mathrm {I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -4-3 \,\mathrm {I}}, a_{k +4}=\frac {2 k a_{k}-\left (8+6 \,\mathrm {I}\right ) a_{k}-10 a_{k}-5 a_{k +2}}{k^{2}-\left (8+6 \,\mathrm {I}\right ) k +16-24 \,\mathrm {I}+16 k}, a_{1}=0, a_{2}=\left (-\frac {1}{8}-\frac {3 \,\mathrm {I}}{8}\right ) a_{0}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-4+3 \,\mathrm {I} \\ {} & {} & a_{k +4}=\frac {2 k a_{k}+\left (-8+6 \,\mathrm {I}\right ) a_{k}-10 a_{k}-5 a_{k +2}}{k^{2}+\left (-8+6 \,\mathrm {I}\right ) k +16+24 \,\mathrm {I}+16 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-4+3 \,\mathrm {I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -4+3 \,\mathrm {I}}, a_{k +4}=\frac {2 k a_{k}+\left (-8+6 \,\mathrm {I}\right ) a_{k}-10 a_{k}-5 a_{k +2}}{k^{2}+\left (-8+6 \,\mathrm {I}\right ) k +16+24 \,\mathrm {I}+16 k}, a_{1}=0, a_{2}=\left (-\frac {1}{8}+\frac {3 \,\mathrm {I}}{8}\right ) a_{0}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -4-3 \,\mathrm {I}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -4+3 \,\mathrm {I}}\right ), a_{k +4}=\frac {2 k a_{k}-\left (8+6 \,\mathrm {I}\right ) a_{k}-10 a_{k}-5 a_{2+k}}{k^{2}-\left (8+6 \,\mathrm {I}\right ) k +16-24 \,\mathrm {I}+16 k}, a_{1}=0, a_{2}=\left (-\frac {1}{8}-\frac {3 \,\mathrm {I}}{8}\right ) a_{0}, a_{3}=0, b_{k +4}=\frac {2 k b_{k}+\left (-8+6 \,\mathrm {I}\right ) b_{k}-10 b_{k}-5 b_{2+k}}{k^{2}+\left (-8+6 \,\mathrm {I}\right ) k +16+24 \,\mathrm {I}+16 k}, b_{1}=0, b_{2}=\left (-\frac {1}{8}+\frac {3 \,\mathrm {I}}{8}\right ) b_{0}, b_{3}=0\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 -> Whittaker -> 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 HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 55
Order:=7; dsolve(x^2*diff(y(x),x$2)+(9*x-2*x^5)*diff(y(x),x)+(25+5*x^2+10*x^4)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-4-3 i} \left (1+\left (-\frac {1}{8}-\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}-\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}+\frac {3943 i}{29952}\right ) x^{6}+\operatorname {O}\left (x^{7}\right )\right )+c_{2} x^{-4+3 i} \left (1+\left (-\frac {1}{8}+\frac {3 i}{8}\right ) x^{2}+\left (-\frac {179}{832}+\frac {483 i}{832}\right ) x^{4}+\left (-\frac {433}{3744}-\frac {3943 i}{29952}\right ) x^{6}+\operatorname {O}\left (x^{7}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 70
AsymptoticDSolveValue[x^2*y''[x]+(9*x-2*x^5)*y'[x]+(25+5*x^2+10*x^4)*y[x]==0,y[x],{x,0,6}]
\[ y(x)\to \left (\frac {1}{832}+\frac {5 i}{832}\right ) c_1 x^{-4+3 i} \left ((86+53 i) x^4+(56+32 i) x^2+(32-160 i)\right )-\left (\frac {5}{832}+\frac {i}{832}\right ) c_2 x^{-4-3 i} \left ((53+86 i) x^4+(32+56 i) x^2-(160-32 i)\right ) \]