2.33.17 Problem 226

Internal problem ID [13887]
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
Problem number : 226
Date solved : Friday, December 19, 2025 at 07:26:56 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} \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 \\ \end{align*}

2.33.17.1 ✓ Maple. Time used: 0.010 (sec). Leaf size: 17

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

\[ y = c_1 \operatorname {LegendreP}\left (\nu , \mu , x\right )+c_2 \operatorname {LegendreQ}\left (\nu , \mu , x\right ) \]

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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-2 x \left (-x^{2}+1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (\nu \left (\nu +1\right ) \left (-x^{2}+1\right )-\mu ^{2}\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {\left (\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu \right ) y \left (x \right )}{\left (x^{2}-1\right )^{2}}-\frac {2 x \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\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^{2}}{d x^{2}}y \left (x \right )+\frac {2 x \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}-1}-\frac {\left (\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu \right ) y \left (x \right )}{\left (x^{2}-1\right )^{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 {2 x}{x^{2}-1}, P_{3}\left (x \right )=-\frac {\nu ^{2} x^{2}+\nu \,x^{2}+\mu ^{2}-\nu ^{2}-\nu }{\left (x^{2}-1\right )^{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}}}=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 (x^{2}-1\right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+2 \left (x^{2}-1\right ) x \left (\frac {d}{d x}y \left (x \right )\right )+\left (-\nu ^{2} x^{2}-\nu \,x^{2}-\mu ^{2}+\nu ^{2}+\nu \right ) y \left (x \right )=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}+4 u^{2}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (2 u^{3}-6 u^{2}+4 u \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-\nu ^{2} u^{2}+2 \nu ^{2} u -\nu \,u^{2}-\mu ^{2}+2 \nu 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 =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 =1..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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{2}}{d u^{2}}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} \left (\mu +2 r \right ) \left (-\mu +2 r \right ) u^{r}+\left (a_{1} \left (2+\mu +2 r \right ) \left (2-\mu +2 r \right )-2 a_{0} \left (-\nu ^{2}+2 r^{2}-\nu +r \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +\mu +2 r \right ) \left (2 k -\mu +2 r \right )-2 a_{k -1} \left (2 \left (k -1\right )^{2}+4 \left (k -1\right ) r -\nu ^{2}+2 r^{2}+k -1-\nu +r \right )+a_{k -2} \left (r -1+\nu +k \right ) \left (r -\nu +k -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \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 {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2+\mu +2 r \right ) \left (2-\mu +2 r \right )-2 a_{0} \left (-\nu ^{2}+2 r^{2}-\nu +r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {2 a_{0} \left (\nu ^{2}-2 r^{2}+\nu -r \right )}{\mu ^{2}-4 r^{2}-8 r -4} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -2} \left (r -1+\nu +k \right ) \left (r -\nu +k -2\right )+\left (-4 k^{2}+\left (-8 r +6\right ) k -4 r^{2}+2 \nu ^{2}+6 r +2 \nu -2\right ) a_{k -1}+4 \left (k -\frac {\mu }{2}+r \right ) a_{k} \left (k +\frac {\mu }{2}+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k} \left (r +1+\nu +k \right ) \left (r -\nu +k \right )+\left (-4 \left (k +2\right )^{2}+\left (-8 r +6\right ) \left (k +2\right )-4 r^{2}+2 \nu ^{2}+6 r +2 \nu -2\right ) a_{k +1}+4 \left (k +2-\frac {\mu }{2}+r \right ) a_{k +2} \left (k +2+\frac {\mu }{2}+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+2 k r a_{k}-8 k r a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+r^{2} a_{k}-4 r^{2} a_{k +1}+k a_{k}-10 k a_{k +1}-a_{k} \nu +2 \nu a_{k +1}+r a_{k}-10 r a_{k +1}-6 a_{k +1}}{\left (2 k +4-\mu +2 r \right ) \left (2 k +4+\mu +2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {\mu }{2} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k \mu a_{k}+4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}-\frac {1}{2} \mu a_{k}+5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4-2 \mu \right ) \left (2 k +4\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 +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k \mu a_{k}+4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}-\frac {1}{2} \mu a_{k}+5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4-2 \mu \right ) \left (2 k +4\right )}, a_{1}=\frac {2 a_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}+\frac {1}{2} \mu +\nu \right )}{4 \mu -4}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {\mu }{2}}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k \mu a_{k}+4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}-\frac {1}{2} \mu a_{k}+5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4-2 \mu \right ) \left (2 k +4\right )}, a_{1}=\frac {2 a_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}+\frac {1}{2} \mu +\nu \right )}{4 \mu -4}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\mu }{2} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k \mu a_{k}-4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}+\frac {1}{2} \mu a_{k}-5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4\right ) \left (2 k +4+2 \mu \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 +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k \mu a_{k}-4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}+\frac {1}{2} \mu a_{k}-5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4\right ) \left (2 k +4+2 \mu \right )}, a_{1}=\frac {2 a_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}-\frac {1}{2} \mu +\nu \right )}{-4 \mu -4}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +\frac {\mu }{2}}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k \mu a_{k}-4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}+\frac {1}{2} \mu a_{k}-5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4\right ) \left (2 k +4+2 \mu \right )}, a_{1}=\frac {2 a_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}-\frac {1}{2} \mu +\nu \right )}{-4 \mu -4}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\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 +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k \mu a_{k}+4 k \mu a_{k +1}+\frac {1}{4} a_{k} \mu ^{2}-\mu ^{2} a_{k +1}-a_{k} \nu ^{2}+2 \nu ^{2} a_{k +1}+k a_{k}-10 k a_{k +1}-\frac {1}{2} \mu a_{k}+5 \mu a_{k +1}-a_{k} \nu +2 \nu a_{k +1}-6 a_{k +1}}{\left (2 k +4-2 \mu \right ) \left (2 k +4\right )}, a_{1}=\frac {2 a_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}+\frac {1}{2} \mu +\nu \right )}{4 \mu -4}, b_{k +2}=-\frac {k^{2} b_{k}-4 k^{2} b_{k +1}+k \mu b_{k}-4 k \mu b_{k +1}+\frac {1}{4} b_{k} \mu ^{2}-\mu ^{2} b_{k +1}-b_{k} \nu ^{2}+2 \nu ^{2} b_{k +1}+k b_{k}-10 k b_{k +1}+\frac {1}{2} \mu b_{k}-5 \mu b_{k +1}-b_{k} \nu +2 \nu b_{k +1}-6 b_{k +1}}{\left (2 k +4\right ) \left (2 k +4+2 \mu \right )}, b_{1}=\frac {2 b_{0} \left (-\frac {1}{2} \mu ^{2}+\nu ^{2}-\frac {1}{2} \mu +\nu \right )}{-4 \mu -4}\right ] \end {array} \]

2.33.17.2 ✓ Mathematica. Time used: 0.031 (sec). Leaf size: 20

ode=(1-x^2)^2*D[y[x],{x,2}]-2*x*(1-x^2)*D[y[x],x]+(\[Nu]*(\[Nu]+1)*(1-x^2)-\[Mu]^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 P_{\nu }^{\mu }(x)+c_2 Q_{\nu }^{\mu }(x) \end{align*}

2.33.17.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
mu = symbols("mu") 
nu = symbols("nu") 
y = Function("y") 
ode = Eq(-2*x*(1 - x**2)*Derivative(y(x), x) + (1 - x**2)**2*Derivative(y(x), (x, 2)) + (-mu**2 + nu*(1 - x**2)*(nu + 1))*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False