32.17 problem 226

Internal problem ID [11050]
Internal file name [OUTPUT/10307_Wednesday_January_24_2024_10_07_00_PM_26256630/index.tex]

Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-7 Equation of form \((a_4 x^4+a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 226.
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.

\[ \boxed {\left (-x^{2}+1\right )^{2} y^{\prime \prime }-2 x \left (-x^{2}+1\right ) y^{\prime }+\left (\nu \left (\nu +1\right ) \left (-x^{2}+1\right )-\mu ^{2}\right ) y=0} \]

32.17.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{4}-2 x^{2}+1\right )+\left (2 x^{3}-2 x \right ) y^{\prime }+\left (-\nu ^{2} x^{2}-\nu \,x^{2}-\mu ^{2}+\nu ^{2}+\nu \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 (\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu \right ) y}{x^{4}-2 x^{2}+1}-\frac {2 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 {2 x y^{\prime }}{x^{2}-1}-\frac {\left (\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu \right ) y}{x^{4}-2 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 {2 x}{x^{2}-1}, P_{3}\left (x \right )=-\frac {\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu }{x^{4}-2 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}}}=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}}}=-\frac {\mu ^{2}}{4} \\ {} & \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 ) \left (x^{2}-1\right ) \left (x^{4}-2 x^{2}+1\right )+2 y^{\prime } x \left (x^{4}-2 x^{2}+1\right )-\left (\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu \right ) \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^{6}-6 u^{5}+12 u^{4}-8 u^{3}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (2 u^{5}-10 u^{4}+16 u^{3}-8 u^{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-\nu ^{2} u^{4}+4 \nu ^{2} u^{3}-\nu \,u^{4}-\mu ^{2} u^{2}-4 \nu ^{2} u^{2}+4 \nu \,u^{3}+2 \mu ^{2} u -4 \nu \,u^{2}\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..4 \\ {} & {} & 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 =2..5 \\ {} & {} & 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 =3..6 \\ {} & {} & 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} \left (\mu +2 r \right ) \left (-\mu +2 r \right ) u^{1+r}+\left (-2 a_{1} \left (2+\mu +2 r \right ) \left (2-\mu +2 r \right )+a_{0} \left (-\mu ^{2}-4 \nu ^{2}+12 r^{2}-4 \nu +4 r \right )\right ) u^{2+r}+\left (-2 a_{2} \left (4+\mu +2 r \right ) \left (4-\mu +2 r \right )+a_{1} \left (-\mu ^{2}-4 \nu ^{2}+12 r^{2}-4 \nu +28 r +16\right )-2 a_{0} \left (-2 \nu ^{2}+3 r^{2}-2 \nu +2 r \right )\right ) u^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (-2 a_{k -1} \left (2 k -2+\mu +2 r \right ) \left (2 k -2-\mu +2 r \right )+a_{k -2} \left (12 \left (k -2\right )^{2}+24 \left (k -2\right ) r -\mu ^{2}-4 \nu ^{2}+12 r^{2}+4 k -8-4 \nu +4 r \right )-2 a_{k -3} \left (3 \left (k -3\right )^{2}+6 \left (k -3\right ) r -2 \nu ^{2}+3 r^{2}+2 k -6-2 \nu +2 r \right )+a_{k -4} \left (r -3+\nu +k \right ) \left (r -\nu +k -4\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 \left (\mu +2 r \right ) \left (-\mu +2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {\mu }{2}, \frac {\mu }{2}\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 (2+\mu +2 r \right ) \left (2-\mu +2 r \right )+a_{0} \left (-\mu ^{2}-4 \nu ^{2}+12 r^{2}-4 \nu +4 r \right )=0, -2 a_{2} \left (4+\mu +2 r \right ) \left (4-\mu +2 r \right )+a_{1} \left (-\mu ^{2}-4 \nu ^{2}+12 r^{2}-4 \nu +28 r +16\right )-2 a_{0} \left (-2 \nu ^{2}+3 r^{2}-2 \nu +2 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (\mu ^{2}+4 \nu ^{2}-12 r^{2}+4 \nu -4 r \right )}{2 \left (\mu ^{2}-4 r^{2}-8 r -4\right )}, a_{2}=\frac {a_{0} \left (\mu ^{4}-12 \mu ^{2} r^{2}+16 \nu ^{4}-64 \nu ^{2} r^{2}+96 r^{4}-24 \mu ^{2} r +32 \nu ^{3}-64 \nu ^{2} r -64 \nu \,r^{2}+256 r^{3}-16 \mu ^{2}-16 \nu ^{2}-64 \nu r +192 r^{2}-32 \nu +32 r \right )}{4 \left (\mu ^{4}-8 \mu ^{2} r^{2}+16 r^{4}-24 \mu ^{2} r +96 r^{3}-20 \mu ^{2}+208 r^{2}+192 r +64\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (a_{k -4}-6 a_{k -3}+12 a_{k -2}-8 a_{k -1}\right ) k^{2}+\left (2 \left (a_{k -4}-6 a_{k -3}+12 a_{k -2}-8 a_{k -1}\right ) r -7 a_{k -4}+32 a_{k -3}-44 a_{k -2}+16 a_{k -1}\right ) k +\left (a_{k -4}-6 a_{k -3}+12 a_{k -2}-8 a_{k -1}\right ) r^{2}+\left (-7 a_{k -4}+32 a_{k -3}-44 a_{k -2}+16 a_{k -1}\right ) r +\left (-\mu ^{2}-4 \nu ^{2}-4 \nu +40\right ) a_{k -2}+\left (-\nu ^{2}-\nu +12\right ) a_{k -4}+2 \left (2 \nu ^{2}+2 \nu -21\right ) a_{k -3}+2 a_{k -1} \left (\mu ^{2}-4\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & \left (a_{k}-6 a_{k +1}+12 a_{k +2}-8 a_{k +3}\right ) \left (k +4\right )^{2}+\left (2 \left (a_{k}-6 a_{k +1}+12 a_{k +2}-8 a_{k +3}\right ) r -7 a_{k}+32 a_{k +1}-44 a_{k +2}+16 a_{k +3}\right ) \left (k +4\right )+\left (a_{k}-6 a_{k +1}+12 a_{k +2}-8 a_{k +3}\right ) r^{2}+\left (-7 a_{k}+32 a_{k +1}-44 a_{k +2}+16 a_{k +3}\right ) r +\left (-\mu ^{2}-4 \nu ^{2}-4 \nu +40\right ) a_{k +2}+\left (-\nu ^{2}-\nu +12\right ) a_{k}+2 \left (2 \nu ^{2}+2 \nu -21\right ) a_{k +1}+2 a_{k +3} \left (\mu ^{2}-4\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}+2 k r a_{k}-12 k r a_{k +1}+24 k r a_{k +2}-\mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+r^{2} a_{k}-6 r^{2} a_{k +1}+12 r^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}+r a_{k}-16 r a_{k +1}+52 r a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}+8 k r -\mu ^{2}+4 r^{2}+24 k +24 r +36\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {\mu }{2} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k \mu a_{k}+6 k \mu a_{k +1}-12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-\frac {1}{2} \mu a_{k}+8 \mu a_{k +1}-26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}-4 k \mu +24 k -12 \mu +36\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {\mu }{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {\mu }{2}}, a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k \mu a_{k}+6 k \mu a_{k +1}-12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-\frac {1}{2} \mu a_{k}+8 \mu a_{k +1}-26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}-4 k \mu +24 k -12 \mu +36\right )}, a_{1}=\frac {a_{0} \left (-2 \mu ^{2}+4 \nu ^{2}+2 \mu +4 \nu \right )}{2 \left (4 \mu -4\right )}, a_{2}=\frac {a_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}-20 \mu ^{3}-16 \mu ^{2} \nu +32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}+32 \mu \nu -16 \nu ^{2}-16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}-96 \mu +64\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 -\frac {\mu }{2}}, a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k \mu a_{k}+6 k \mu a_{k +1}-12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-\frac {1}{2} \mu a_{k}+8 \mu a_{k +1}-26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}-4 k \mu +24 k -12 \mu +36\right )}, a_{1}=\frac {a_{0} \left (-2 \mu ^{2}+4 \nu ^{2}+2 \mu +4 \nu \right )}{2 \left (4 \mu -4\right )}, a_{2}=\frac {a_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}-20 \mu ^{3}-16 \mu ^{2} \nu +32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}+32 \mu \nu -16 \nu ^{2}-16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}-96 \mu +64\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\mu }{2} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}+k \mu a_{k}-6 k \mu a_{k +1}+12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}+\frac {1}{2} \mu a_{k}-8 \mu a_{k +1}+26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}+4 k \mu +24 k +12 \mu +36\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {\mu }{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {\mu }{2}}, a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}+k \mu a_{k}-6 k \mu a_{k +1}+12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}+\frac {1}{2} \mu a_{k}-8 \mu a_{k +1}+26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}+4 k \mu +24 k +12 \mu +36\right )}, a_{1}=\frac {a_{0} \left (-2 \mu ^{2}+4 \nu ^{2}-2 \mu +4 \nu \right )}{2 \left (-4 \mu -4\right )}, a_{2}=\frac {a_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}+20 \mu ^{3}-16 \mu ^{2} \nu -32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}-32 \mu \nu -16 \nu ^{2}+16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}+96 \mu +64\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 +\frac {\mu }{2}}, a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}+k \mu a_{k}-6 k \mu a_{k +1}+12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}+\frac {1}{2} \mu a_{k}-8 \mu a_{k +1}+26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}+4 k \mu +24 k +12 \mu +36\right )}, a_{1}=\frac {a_{0} \left (-2 \mu ^{2}+4 \nu ^{2}-2 \mu +4 \nu \right )}{2 \left (-4 \mu -4\right )}, a_{2}=\frac {a_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}+20 \mu ^{3}-16 \mu ^{2} \nu -32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}-32 \mu \nu -16 \nu ^{2}+16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}+96 \mu +64\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {\mu }{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k +\frac {\mu }{2}}\right ), a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k \mu a_{k}+6 k \mu a_{k +1}-12 k \mu a_{k +2}+\frac {1}{4} a_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} a_{k +1}+2 \mu ^{2} a_{k +2}-a_{k} \nu ^{2}+4 \nu ^{2} a_{k +1}-4 \nu ^{2} a_{k +2}+k a_{k}-16 k a_{k +1}+52 k a_{k +2}-\frac {1}{2} \mu a_{k}+8 \mu a_{k +1}-26 \mu a_{k +2}-a_{k} \nu +4 \nu a_{k +1}-4 \nu a_{k +2}-10 a_{k +1}+56 a_{k +2}}{2 \left (4 k^{2}-4 k \mu +24 k -12 \mu +36\right )}, a_{1}=\frac {a_{0} \left (-2 \mu ^{2}+4 \nu ^{2}+2 \mu +4 \nu \right )}{2 \left (4 \mu -4\right )}, a_{2}=\frac {a_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}-20 \mu ^{3}-16 \mu ^{2} \nu +32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}+32 \mu \nu -16 \nu ^{2}-16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}-96 \mu +64\right )}, b_{k +3}=\frac {k^{2} b_{k}-6 k^{2} b_{k +1}+12 k^{2} b_{k +2}+k \mu b_{k}-6 k \mu b_{k +1}+12 k \mu b_{k +2}+\frac {1}{4} b_{k} \mu ^{2}-\frac {3}{2} \mu ^{2} b_{k +1}+2 \mu ^{2} b_{k +2}-b_{k} \nu ^{2}+4 \nu ^{2} b_{k +1}-4 \nu ^{2} b_{k +2}+k b_{k}-16 k b_{k +1}+52 k b_{k +2}+\frac {1}{2} \mu b_{k}-8 \mu b_{k +1}+26 \mu b_{k +2}-b_{k} \nu +4 \nu b_{k +1}-4 \nu b_{k +2}-10 b_{k +1}+56 b_{k +2}}{2 \left (4 k^{2}+4 k \mu +24 k +12 \mu +36\right )}, b_{1}=\frac {b_{0} \left (-2 \mu ^{2}+4 \nu ^{2}-2 \mu +4 \nu \right )}{2 \left (-4 \mu -4\right )}, b_{2}=\frac {b_{0} \left (4 \mu ^{4}-16 \mu ^{2} \nu ^{2}+16 \nu ^{4}+20 \mu ^{3}-16 \mu ^{2} \nu -32 \mu \,\nu ^{2}+32 \nu ^{3}+32 \mu ^{2}-32 \mu \nu -16 \nu ^{2}+16 \mu -32 \nu \right )}{4 \left (32 \mu ^{2}+96 \mu +64\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 
   <- Legendre successful 
<- special function solution successful`
 

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 17

dsolve((1-x^2)^2*diff(y(x),x$2)-2*x*(1-x^2)*diff(y(x),x)+(nu*(nu+1)*(1-x^2)-mu^2)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \operatorname {LegendreP}\left (\nu , \mu , x\right )+c_{2} \operatorname {LegendreQ}\left (\nu , \mu , x\right ) \]

✓ Solution by Mathematica

Time used: 0.046 (sec). Leaf size: 20

DSolve[(1-x^2)^2*y''[x]-2*x*(1-x^2)*y'[x]+(\[Nu]*(\[Nu]+1)*(1-x^2)-\[Mu]^2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 P_{\nu }^{\mu }(x)+c_2 Q_{\nu }^{\mu }(x) \]