Internal problem ID [9576]
Internal file name [OUTPUT/8517_Monday_June_06_2022_03_36_09_AM_67346208/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1248.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (x^{2}-1\right ) y^{\prime \prime }+a x y^{\prime }+\left (b \,x^{2}+c x +d \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+a x y^{\prime }+\left (b \,x^{2}+c x +d \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (b \,x^{2}+c x +d \right ) y}{x^{2}-1}-\frac {a x y^{\prime }}{x^{2}-1} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {a x y^{\prime }}{x^{2}-1}+\frac {\left (b \,x^{2}+c x +d \right ) y}{x^{2}-1}=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 {a x}{x^{2}-1}, P_{3}\left (x \right )=\frac {b \,x^{2}+c x +d}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {a}{2} \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) \left (\frac {d}{d x}y^{\prime }\right )+a x y^{\prime }+\left (b \,x^{2}+c x +d \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a u -a \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (b \,u^{2}-2 b u +c u +b -c +d \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-2+2 r +a \right ) u^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (2 r +a \right )+a_{0} \left (a r +r^{2}+b -c +d -r \right )\right ) u^{r}+\left (-a_{2} \left (2+r \right ) \left (2+2 r +a \right )+a_{1} \left (a r +r^{2}+a +b -c +d +r \right )-a_{0} \left (2 b -c \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +a \right )+a_{k} \left (a k +a r +k^{2}+2 k r +r^{2}+b -c +d -k -r \right )-a_{k -1} \left (2 b -c \right )+a_{k -2} b \right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-2+2 r +a \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1-\frac {a}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) \left (2 r +a \right )+a_{0} \left (a r +r^{2}+b -c +d -r \right )=0, -a_{2} \left (2+r \right ) \left (2+2 r +a \right )+a_{1} \left (a r +r^{2}+a +b -c +d +r \right )-a_{0} \left (2 b -c \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (a r +r^{2}+b -c +d -r \right )}{a r +2 r^{2}+a +2 r}, a_{2}=\frac {a_{0} \left (a^{2} r^{2}+2 a \,r^{3}+r^{4}+a^{2} r -c a r +2 a d r +a \,r^{2}-2 b \,r^{2}+2 d \,r^{2}-a b +a d -a r +b^{2}-2 b c +2 b d -4 b r +c^{2}-2 d c +2 c r +d^{2}-r^{2}\right )}{a^{2} r^{2}+4 a \,r^{3}+4 r^{4}+3 a^{2} r +14 a \,r^{2}+16 r^{3}+2 a^{2}+14 a r +20 r^{2}+4 a +8 r}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (a +2 r -1\right ) k +r^{2}+\left (a -1\right ) r +b -c +d \right ) a_{k}-a_{k +1} \left (k +1+r \right ) \left (2 k +2 r +a \right )+\left (a_{k -2}-2 a_{k -1}\right ) b +a_{k -1} c =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+\left (a +2 r -1\right ) \left (k +2\right )+r^{2}+\left (a -1\right ) r +b -c +d \right ) a_{k +2}-a_{k +3} \left (k +3+r \right ) \left (2 k +4+2 r +a \right )+\left (a_{k}-2 a_{k +1}\right ) b +a_{k +1} c =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {a k a_{k +2}+a r a_{k +2}+k^{2} a_{k +2}+2 k r a_{k +2}+r^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+3 r a_{k +2}+2 a_{k +2}}{\left (k +3+r \right ) \left (2 k +4+2 r +a \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {a k a_{k +2}+k^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +4+a \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +3}=\frac {a k a_{k +2}+k^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +4+a \right )}, a_{1}=\frac {a_{0} \left (b -c +d \right )}{a}, a_{2}=\frac {a_{0} \left (-a b +a d +b^{2}-2 b c +2 b d +c^{2}-2 d c +d^{2}\right )}{2 a^{2}+4 a}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +3}=\frac {a k a_{k +2}+k^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +4+a \right )}, a_{1}=\frac {a_{0} \left (b -c +d \right )}{a}, a_{2}=\frac {a_{0} \left (-a b +a d +b^{2}-2 b c +2 b d +c^{2}-2 d c +d^{2}\right )}{2 a^{2}+4 a}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-\frac {a}{2} \\ {} & {} & a_{k +3}=\frac {a k a_{k +2}+a \left (1-\frac {a}{2}\right ) a_{k +2}+k^{2} a_{k +2}+2 k \left (1-\frac {a}{2}\right ) a_{k +2}+\left (1-\frac {a}{2}\right )^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+3 \left (1-\frac {a}{2}\right ) a_{k +2}+2 a_{k +2}}{\left (k +4-\frac {a}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-\frac {a}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1-\frac {a}{2}}, a_{k +3}=\frac {a k a_{k +2}+a \left (1-\frac {a}{2}\right ) a_{k +2}+k^{2} a_{k +2}+2 k \left (1-\frac {a}{2}\right ) a_{k +2}+\left (1-\frac {a}{2}\right )^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+3 \left (1-\frac {a}{2}\right ) a_{k +2}+2 a_{k +2}}{\left (k +4-\frac {a}{2}\right ) \left (2 k +6\right )}, a_{1}=\frac {a_{0} \left (a \left (1-\frac {a}{2}\right )+\left (1-\frac {a}{2}\right )^{2}+b -c +d -1+\frac {a}{2}\right )}{a \left (1-\frac {a}{2}\right )+2 \left (1-\frac {a}{2}\right )^{2}+2}, a_{2}=\frac {a_{0} \left (a^{2} \left (1-\frac {a}{2}\right )^{2}+2 a \left (1-\frac {a}{2}\right )^{3}+\left (1-\frac {a}{2}\right )^{4}+a^{2} \left (1-\frac {a}{2}\right )-c a \left (1-\frac {a}{2}\right )+2 a d \left (1-\frac {a}{2}\right )+a \left (1-\frac {a}{2}\right )^{2}-2 b \left (1-\frac {a}{2}\right )^{2}+2 d \left (1-\frac {a}{2}\right )^{2}-a b +a d -a \left (1-\frac {a}{2}\right )+b^{2}-2 b c +2 b d -4 b \left (1-\frac {a}{2}\right )+c^{2}-2 d c +2 c \left (1-\frac {a}{2}\right )+d^{2}-\left (1-\frac {a}{2}\right )^{2}\right )}{a^{2} \left (1-\frac {a}{2}\right )^{2}+4 a \left (1-\frac {a}{2}\right )^{3}+4 \left (1-\frac {a}{2}\right )^{4}+3 a^{2} \left (1-\frac {a}{2}\right )+14 a \left (1-\frac {a}{2}\right )^{2}+16 \left (1-\frac {a}{2}\right )^{3}+2 a^{2}+14 a \left (1-\frac {a}{2}\right )+20 \left (1-\frac {a}{2}\right )^{2}+8}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +1-\frac {a}{2}}, a_{k +3}=\frac {a k a_{k +2}+a \left (1-\frac {a}{2}\right ) a_{k +2}+k^{2} a_{k +2}+2 k \left (1-\frac {a}{2}\right ) a_{k +2}+\left (1-\frac {a}{2}\right )^{2} a_{k +2}+2 a a_{k +2}+a_{k} b -2 b a_{k +1}+b a_{k +2}+a_{k +1} c -c a_{k +2}+d a_{k +2}+3 k a_{k +2}+3 \left (1-\frac {a}{2}\right ) a_{k +2}+2 a_{k +2}}{\left (k +4-\frac {a}{2}\right ) \left (2 k +6\right )}, a_{1}=\frac {a_{0} \left (a \left (1-\frac {a}{2}\right )+\left (1-\frac {a}{2}\right )^{2}+b -c +d -1+\frac {a}{2}\right )}{a \left (1-\frac {a}{2}\right )+2 \left (1-\frac {a}{2}\right )^{2}+2}, a_{2}=\frac {a_{0} \left (a^{2} \left (1-\frac {a}{2}\right )^{2}+2 a \left (1-\frac {a}{2}\right )^{3}+\left (1-\frac {a}{2}\right )^{4}+a^{2} \left (1-\frac {a}{2}\right )-c a \left (1-\frac {a}{2}\right )+2 a d \left (1-\frac {a}{2}\right )+a \left (1-\frac {a}{2}\right )^{2}-2 b \left (1-\frac {a}{2}\right )^{2}+2 d \left (1-\frac {a}{2}\right )^{2}-a b +a d -a \left (1-\frac {a}{2}\right )+b^{2}-2 b c +2 b d -4 b \left (1-\frac {a}{2}\right )+c^{2}-2 d c +2 c \left (1-\frac {a}{2}\right )+d^{2}-\left (1-\frac {a}{2}\right )^{2}\right )}{a^{2} \left (1-\frac {a}{2}\right )^{2}+4 a \left (1-\frac {a}{2}\right )^{3}+4 \left (1-\frac {a}{2}\right )^{4}+3 a^{2} \left (1-\frac {a}{2}\right )+14 a \left (1-\frac {a}{2}\right )^{2}+16 \left (1-\frac {a}{2}\right )^{3}+2 a^{2}+14 a \left (1-\frac {a}{2}\right )+20 \left (1-\frac {a}{2}\right )^{2}+8}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} \left (x +1\right )^{k +1-\frac {a}{2}}\right ), e_{k +3}=\frac {a k e_{k +2}+k^{2} e_{k +2}+2 a e_{k +2}+b e_{k}-2 b e_{k +1}+b e_{k +2}+c e_{k +1}-c e_{k +2}+d e_{k +2}+3 k e_{k +2}+2 e_{k +2}}{\left (k +3\right ) \left (2 k +4+a \right )}, e_{1}=\frac {e_{0} \left (b -c +d \right )}{a}, e_{2}=\frac {e_{0} \left (-a b +a d +b^{2}-2 b c +2 b d +c^{2}-2 d c +d^{2}\right )}{2 a^{2}+4 a}, f_{k +3}=\frac {a k f_{k +2}+a \left (1-\frac {a}{2}\right ) f_{k +2}+k^{2} f_{k +2}+2 k \left (1-\frac {a}{2}\right ) f_{k +2}+\left (1-\frac {a}{2}\right )^{2} f_{k +2}+2 a f_{k +2}+f_{k} b -2 b f_{k +1}+b f_{k +2}+f_{k +1} c -c f_{k +2}+d f_{k +2}+3 k f_{k +2}+3 \left (1-\frac {a}{2}\right ) f_{k +2}+2 f_{k +2}}{\left (k +4-\frac {a}{2}\right ) \left (2 k +6\right )}, f_{1}=\frac {f_{0} \left (a \left (1-\frac {a}{2}\right )+\left (1-\frac {a}{2}\right )^{2}+b -c +d -1+\frac {a}{2}\right )}{a \left (1-\frac {a}{2}\right )+2 \left (1-\frac {a}{2}\right )^{2}+2}, f_{2}=\frac {f_{0} \left (a^{2} \left (1-\frac {a}{2}\right )^{2}+2 a \left (1-\frac {a}{2}\right )^{3}+\left (1-\frac {a}{2}\right )^{4}+a^{2} \left (1-\frac {a}{2}\right )-c a \left (1-\frac {a}{2}\right )+2 a d \left (1-\frac {a}{2}\right )+a \left (1-\frac {a}{2}\right )^{2}-2 b \left (1-\frac {a}{2}\right )^{2}+2 d \left (1-\frac {a}{2}\right )^{2}-a b +a d -a \left (1-\frac {a}{2}\right )+b^{2}-2 b c +2 b d -4 b \left (1-\frac {a}{2}\right )+c^{2}-2 d c +2 c \left (1-\frac {a}{2}\right )+d^{2}-\left (1-\frac {a}{2}\right )^{2}\right )}{a^{2} \left (1-\frac {a}{2}\right )^{2}+4 a \left (1-\frac {a}{2}\right )^{3}+4 \left (1-\frac {a}{2}\right )^{4}+3 a^{2} \left (1-\frac {a}{2}\right )+14 a \left (1-\frac {a}{2}\right )^{2}+16 \left (1-\frac {a}{2}\right )^{3}+2 a^{2}+14 a \left (1-\frac {a}{2}\right )+20 \left (1-\frac {a}{2}\right )^{2}+8}\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 HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.5 (sec). Leaf size: 126
dsolve((x^2-1)*diff(diff(y(x),x),x)+a*x*diff(y(x),x)+(b*x^2+c*x+d)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{x \sqrt {-b}} \left (2 \operatorname {HeunC}\left (4 \sqrt {-b}, \frac {a}{2}-1, \frac {a}{2}-1, 2 c , d -c -\frac {a^{2}}{8}+b +\frac {1}{2}, \frac {x}{2}+\frac {1}{2}\right ) c_{1} +c_{2} \left (\frac {x}{2}-\frac {1}{2}\right )^{\frac {a}{4}} \left (x +1\right ) \left (\frac {x}{2}+\frac {1}{2}\right )^{-\frac {a}{4}} \left (x^{2}-1\right )^{-\frac {a}{4}} \operatorname {HeunC}\left (4 \sqrt {-b}, 1-\frac {a}{2}, \frac {a}{2}-1, 2 c , d -c -\frac {a^{2}}{8}+b +\frac {1}{2}, \frac {x}{2}+\frac {1}{2}\right )\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.662 (sec). Leaf size: 192
DSolve[(d + c*x + b*x^2)*y[x] + a*x*y'[x] + (-1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} e^{\sqrt {-b} x} \left (c_2 (x-1)^{a/4} \left (x^2-1\right )^{-a/4} (x+1)^{1-\frac {a}{4}} \text {HeunC}\left [\frac {1}{4} a \left (a-4 \sqrt {-b}-2\right )-b+4 \sqrt {-b}+c-d,2 \left (2 \sqrt {-b}+c\right ),2-\frac {a}{2},\frac {a}{2},4 \sqrt {-b},\frac {x+1}{2}\right ]+2 c_1 \text {HeunC}\left [a \sqrt {-b}-b+c-d,2 \left (a \sqrt {-b}+c\right ),\frac {a}{2},\frac {a}{2},4 \sqrt {-b},\frac {x+1}{2}\right ]\right ) \]