Internal problem ID [9690]
Internal file name [OUTPUT/8632_Monday_June_06_2022_04_32_35_AM_3304448/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1363.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (a \,x^{2}+a -2\right ) y^{\prime }}{x \left (x^{2}-1\right )}+\frac {b y}{x^{2}}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{4}-x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (a \,x^{2}+a -2\right ) y^{\prime }+\left (b \,x^{2}-b \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 (a \,x^{2}+a -2\right ) y^{\prime }}{x \left (x^{2}-1\right )}-\frac {b y}{x^{2}} \\ \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 {\left (a \,x^{2}+a -2\right ) y^{\prime }}{x \left (x^{2}-1\right )}+\frac {b y}{x^{2}}=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^{2}+a -2}{x \left (x^{2}-1\right )}, P_{3}\left (x \right )=\frac {b}{x^{2}}\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}}}=a -1 \\ {} & \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 (\frac {d}{d x}y^{\prime }\right ) x^{2} \left (x^{2}-1\right )+x \left (a \,x^{2}+a -2\right ) y^{\prime }+b \left (x^{2}-1\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^{4}-4 u^{3}+5 u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a \,u^{3}-3 a \,u^{2}+4 a u -2 a -2 u +2\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (b \,u^{2}-2 b u \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 =1..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..3 \\ {} & {} & 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..4 \\ {} & {} & 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} \\ {} & {} & -2 a_{0} r \left (-2+r +a \right ) u^{-1+r}+\left (-2 a_{1} \left (1+r \right ) \left (-1+r +a \right )+a_{0} r \left (-7+5 r +4 a \right )\right ) u^{r}+\left (-2 a_{2} \left (2+r \right ) \left (r +a \right )+a_{1} \left (1+r \right ) \left (-2+5 r +4 a \right )-a_{0} \left (3 a r +4 r^{2}+2 b -4 r \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right ) \left (k -1+r +a \right )+a_{k} \left (k +r \right ) \left (5 k +5 r -7+4 a \right )-a_{k -1} \left (3 a \left (k -1\right )+3 a r +4 \left (k -1\right )^{2}+8 \left (k -1\right ) r +4 r^{2}+2 b -4 k +4-4 r \right )+a_{k -2} \left (a \left (k -2\right )+a r +\left (k -2\right )^{2}+2 \left (k -2\right ) r +r^{2}+b -k +2-r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (-2+r +a \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2-a \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-2 a_{1} \left (1+r \right ) \left (-1+r +a \right )+a_{0} r \left (-7+5 r +4 a \right )=0, -2 a_{2} \left (2+r \right ) \left (r +a \right )+a_{1} \left (1+r \right ) \left (-2+5 r +4 a \right )-a_{0} \left (3 a r +4 r^{2}+2 b -4 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} r \left (-7+5 r +4 a \right )}{2 \left (a r +r^{2}+a -1\right )}, a_{2}=\frac {a_{0} \left (10 a^{2} r +26 a \,r^{2}+17 r^{3}-4 b a -22 a r -4 b r -29 r^{2}+4 b +6 r \right )}{4 \left (a^{2} r +2 a \,r^{2}+r^{3}+2 a^{2}+3 a r +r^{2}-2 a -2 r \right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) k^{2}+\left (2 \left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r +\left (4 a_{k}+a_{k -2}-3 a_{k -1}-2 a_{k +1}\right ) a -7 a_{k}-5 a_{k -2}+12 a_{k -1}\right ) k +\left (5 a_{k}+a_{k -2}-4 a_{k -1}-2 a_{k +1}\right ) r^{2}+\left (\left (4 a_{k}+a_{k -2}-3 a_{k -1}-2 a_{k +1}\right ) a -7 a_{k}-5 a_{k -2}+12 a_{k -1}\right ) r +\left (-2 a_{k -2}+3 a_{k -1}-2 a_{k +1}\right ) a +\left (b +6\right ) a_{k -2}+2 \left (-b -4\right ) a_{k -1}+2 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) \left (k +2\right )^{2}+\left (2 \left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r +\left (4 a_{k +2}+a_{k}-3 a_{k +1}-2 a_{k +3}\right ) a -7 a_{k +2}-5 a_{k}+12 a_{k +1}\right ) \left (k +2\right )+\left (5 a_{k +2}+a_{k}-4 a_{k +1}-2 a_{k +3}\right ) r^{2}+\left (\left (4 a_{k +2}+a_{k}-3 a_{k +1}-2 a_{k +3}\right ) a -7 a_{k +2}-5 a_{k}+12 a_{k +1}\right ) r +\left (-2 a_{k}+3 a_{k +1}-2 a_{k +3}\right ) a +\left (b +6\right ) a_{k}+2 \left (-b -4\right ) a_{k +1}+2 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {a k a_{k}-3 a k a_{k +1}+4 a k a_{k +2}+a r a_{k}-3 a r a_{k +1}+4 a r a_{k +2}+k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}+2 k r a_{k}-8 k r a_{k +1}+10 k r a_{k +2}+r^{2} a_{k}-4 r^{2} a_{k +1}+5 r^{2} a_{k +2}-3 a a_{k +1}+8 a a_{k +2}+a_{k} b -2 b a_{k +1}-k a_{k}-4 k a_{k +1}+13 k a_{k +2}-r a_{k}-4 r a_{k +1}+13 r a_{k +2}+6 a_{k +2}}{2 \left (a k +a r +k^{2}+2 k r +r^{2}+3 a +4 k +4 r +3\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {a k a_{k}-3 a k a_{k +1}+4 a k a_{k +2}+k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-3 a a_{k +1}+8 a a_{k +2}+a_{k} b -2 b a_{k +1}-k a_{k}-4 k a_{k +1}+13 k a_{k +2}+6 a_{k +2}}{2 \left (a k +k^{2}+3 a +4 k +3\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}-3 a k a_{k +1}+4 a k a_{k +2}+k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-3 a a_{k +1}+8 a a_{k +2}+a_{k} b -2 b a_{k +1}-k a_{k}-4 k a_{k +1}+13 k a_{k +2}+6 a_{k +2}}{2 \left (a k +k^{2}+3 a +4 k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0} \left (-4 b a +4 b \right )}{4 \left (2 a^{2}-2 a \right )}\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}-3 a k a_{k +1}+4 a k a_{k +2}+k^{2} a_{k}-4 k^{2} a_{k +1}+5 k^{2} a_{k +2}-3 a a_{k +1}+8 a a_{k +2}+a_{k} b -2 b a_{k +1}-k a_{k}-4 k a_{k +1}+13 k a_{k +2}+6 a_{k +2}}{2 \left (a k +k^{2}+3 a +4 k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0} \left (-4 b a +4 b \right )}{4 \left (2 a^{2}-2 a \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2-a \\ {} & {} & a_{k +3}=\frac {6 a_{k +2}+5 k^{2} a_{k +2}+a_{k} b +4 a k a_{k +2}+\left (2-a \right )^{2} a_{k}-4 \left (2-a \right )^{2} a_{k +1}+5 \left (2-a \right )^{2} a_{k +2}-\left (2-a \right ) a_{k}-4 \left (2-a \right ) a_{k +1}+13 \left (2-a \right ) a_{k +2}-4 k^{2} a_{k +1}-3 a a_{k +1}-k a_{k}+a k a_{k}-3 a k a_{k +1}+a \left (2-a \right ) a_{k}-3 a \left (2-a \right ) a_{k +1}+4 a \left (2-a \right ) a_{k +2}+2 k \left (2-a \right ) a_{k}-8 k \left (2-a \right ) a_{k +1}+10 k \left (2-a \right ) a_{k +2}+k^{2} a_{k}+8 a a_{k +2}-2 b a_{k +1}-4 k a_{k +1}+13 k a_{k +2}}{2 \left (a k +a \left (2-a \right )+k^{2}+2 k \left (2-a \right )+\left (2-a \right )^{2}-a +4 k +11\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2-a \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2-a}, a_{k +3}=\frac {6 a_{k +2}+5 k^{2} a_{k +2}+a_{k} b +4 a k a_{k +2}+\left (2-a \right )^{2} a_{k}-4 \left (2-a \right )^{2} a_{k +1}+5 \left (2-a \right )^{2} a_{k +2}-\left (2-a \right ) a_{k}-4 \left (2-a \right ) a_{k +1}+13 \left (2-a \right ) a_{k +2}-4 k^{2} a_{k +1}-3 a a_{k +1}-k a_{k}+a k a_{k}-3 a k a_{k +1}+a \left (2-a \right ) a_{k}-3 a \left (2-a \right ) a_{k +1}+4 a \left (2-a \right ) a_{k +2}+2 k \left (2-a \right ) a_{k}-8 k \left (2-a \right ) a_{k +1}+10 k \left (2-a \right ) a_{k +2}+k^{2} a_{k}+8 a a_{k +2}-2 b a_{k +1}-4 k a_{k +1}+13 k a_{k +2}}{2 \left (a k +a \left (2-a \right )+k^{2}+2 k \left (2-a \right )+\left (2-a \right )^{2}-a +4 k +11\right )}, a_{1}=\frac {a_{0} \left (2-a \right ) \left (-a +3\right )}{2 \left (a \left (2-a \right )+\left (2-a \right )^{2}+a -1\right )}, a_{2}=\frac {a_{0} \left (10 a^{2} \left (2-a \right )+26 a \left (2-a \right )^{2}+17 \left (2-a \right )^{3}-4 b a -22 a \left (2-a \right )-4 b \left (2-a \right )-29 \left (2-a \right )^{2}+4 b +12-6 a \right )}{4 \left (a^{2} \left (2-a \right )+2 a \left (2-a \right )^{2}+\left (2-a \right )^{3}+2 a^{2}+3 a \left (2-a \right )+\left (2-a \right )^{2}-4\right )}\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 +2-a}, a_{k +3}=\frac {6 a_{k +2}+5 k^{2} a_{k +2}+a_{k} b +4 a k a_{k +2}+\left (2-a \right )^{2} a_{k}-4 \left (2-a \right )^{2} a_{k +1}+5 \left (2-a \right )^{2} a_{k +2}-\left (2-a \right ) a_{k}-4 \left (2-a \right ) a_{k +1}+13 \left (2-a \right ) a_{k +2}-4 k^{2} a_{k +1}-3 a a_{k +1}-k a_{k}+a k a_{k}-3 a k a_{k +1}+a \left (2-a \right ) a_{k}-3 a \left (2-a \right ) a_{k +1}+4 a \left (2-a \right ) a_{k +2}+2 k \left (2-a \right ) a_{k}-8 k \left (2-a \right ) a_{k +1}+10 k \left (2-a \right ) a_{k +2}+k^{2} a_{k}+8 a a_{k +2}-2 b a_{k +1}-4 k a_{k +1}+13 k a_{k +2}}{2 \left (a k +a \left (2-a \right )+k^{2}+2 k \left (2-a \right )+\left (2-a \right )^{2}-a +4 k +11\right )}, a_{1}=\frac {a_{0} \left (2-a \right ) \left (-a +3\right )}{2 \left (a \left (2-a \right )+\left (2-a \right )^{2}+a -1\right )}, a_{2}=\frac {a_{0} \left (10 a^{2} \left (2-a \right )+26 a \left (2-a \right )^{2}+17 \left (2-a \right )^{3}-4 b a -22 a \left (2-a \right )-4 b \left (2-a \right )-29 \left (2-a \right )^{2}+4 b +12-6 a \right )}{4 \left (a^{2} \left (2-a \right )+2 a \left (2-a \right )^{2}+\left (2-a \right )^{3}+2 a^{2}+3 a \left (2-a \right )+\left (2-a \right )^{2}-4\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +1\right )^{k +2-a}\right ), c_{k +3}=\frac {a k c_{k}-3 a k c_{k +1}+4 a k c_{k +2}+k^{2} c_{k}-4 k^{2} c_{k +1}+5 k^{2} c_{k +2}-3 a c_{k +1}+8 a c_{k +2}+b c_{k}-2 b c_{k +1}-k c_{k}-4 k c_{k +1}+13 k c_{k +2}+6 c_{k +2}}{2 \left (a k +k^{2}+3 a +4 k +3\right )}, c_{1}=0, c_{2}=\frac {c_{0} \left (-4 b a +4 b \right )}{4 \left (2 a^{2}-2 a \right )}, d_{k +3}=\frac {6 d_{k +2}+5 k^{2} d_{k +2}+d_{k} b +\left (2-a \right )^{2} d_{k}-4 \left (2-a \right )^{2} d_{k +1}+5 \left (2-a \right )^{2} d_{k +2}-\left (2-a \right ) d_{k}-4 \left (2-a \right ) d_{k +1}+13 \left (2-a \right ) d_{k +2}-4 k^{2} d_{k +1}-3 a d_{k +1}-k d_{k}+k^{2} d_{k}+8 a d_{k +2}-2 b d_{k +1}-4 k d_{k +1}+13 k d_{k +2}+4 a k d_{k +2}+a k d_{k}-3 a k d_{k +1}+a \left (2-a \right ) d_{k}-3 a \left (2-a \right ) d_{k +1}+4 a \left (2-a \right ) d_{k +2}+2 k \left (2-a \right ) d_{k}-8 k \left (2-a \right ) d_{k +1}+10 k \left (2-a \right ) d_{k +2}}{2 \left (a k +a \left (2-a \right )+k^{2}+2 k \left (2-a \right )+\left (2-a \right )^{2}-a +4 k +11\right )}, d_{1}=\frac {d_{0} \left (2-a \right ) \left (-a +3\right )}{2 \left (a \left (2-a \right )+\left (2-a \right )^{2}+a -1\right )}, d_{2}=\frac {d_{0} \left (10 a^{2} \left (2-a \right )+26 a \left (2-a \right )^{2}+17 \left (2-a \right )^{3}-4 b a -22 a \left (2-a \right )-4 b \left (2-a \right )-29 \left (2-a \right )^{2}+4 b +12-6 a \right )}{4 \left (a^{2} \left (2-a \right )+2 a \left (2-a \right )^{2}+\left (2-a \right )^{3}+2 a^{2}+3 a \left (2-a \right )+\left (2-a \right )^{2}-4\right )}\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 <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.172 (sec). Leaf size: 253
dsolve(diff(diff(y(x),x),x) = -1/x*(a*x^2+a-2)/(x^2-1)*diff(y(x),x)-b/x^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = x^{\frac {a}{2}-\frac {1}{2}} \left (-x^{2}+1\right )^{\frac {1}{2}+\frac {a}{2}} \left (x^{2}-1\right )^{-a} \left (x^{\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}\right )^{-\frac {\sqrt {a^{2}-2 a -4 b +1}}{4}} \operatorname {LegendreP}\left (\frac {a}{2}-\frac {3}{2}, -\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}, \frac {-x^{2}-1}{x^{2}-1}\right ) c_{1} \Gamma \left (1+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right )+\frac {\csc \left (\frac {\pi \sqrt {a^{2}-2 a -4 b +1}}{2}\right ) x^{-\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}} \left (x^{2}\right )^{\frac {\sqrt {a^{2}-2 a -4 b +1}}{4}} \operatorname {LegendreP}\left (\frac {a}{2}-\frac {3}{2}, \frac {\sqrt {a^{2}-2 a -4 b +1}}{2}, \frac {-x^{2}-1}{x^{2}-1}\right ) \sqrt {a^{2}-2 a -4 b +1}\, \pi c_{2}}{2 \Gamma \left (1+\frac {\sqrt {a^{2}-2 a -4 b +1}}{2}\right )}\right ) \]
✓ Solution by Mathematica
Time used: 0.733 (sec). Leaf size: 212
DSolve[y''[x] == -((b*y[x])/x^2) - ((-2 + a + a*x^2)*y'[x])/(x*(-1 + x^2)),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -(-1)^{\frac {1}{4} \left (-\sqrt {a^2-2 a-4 b+1}+a+3\right )} x^{\frac {1}{2} \left (-\sqrt {a^2-2 a-4 b+1}+a-1\right )} \left (c_1 \operatorname {Hypergeometric2F1}\left (\frac {a-1}{2},\frac {1}{2} \left (a-\sqrt {a^2-2 a-4 b+1}-1\right ),1-\frac {1}{2} \sqrt {a^2-2 a-4 b+1},x^2\right )+c_2 i^{\sqrt {a^2-2 a-4 b+1}} x^{\sqrt {a^2-2 a-4 b+1}} \operatorname {Hypergeometric2F1}\left (\frac {a-1}{2},\frac {1}{2} \left (a+\sqrt {a^2-2 a-4 b+1}-1\right ),\frac {1}{2} \left (\sqrt {a^2-2 a-4 b+1}+2\right ),x^2\right )\right ) \]