2.31.20 Problem 168

Internal problem ID [13829]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-5
Problem number : 168
Date solved : Friday, December 19, 2025 at 02:18:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} \left (-x^{2}+1\right ) y^{\prime \prime }+\left (x a +b \right ) y^{\prime }+y c&=0 \\ \end{align*}

2.31.20.1 ✓ Maple. Time used: 0.014 (sec). Leaf size: 134

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

\[ y = c_1 \operatorname {hypergeom}\left (\left [-\frac {1}{2}-\frac {a}{2}-\frac {\sqrt {a^{2}+2 a +4 c +1}}{2}, -\frac {1}{2}-\frac {a}{2}+\frac {\sqrt {a^{2}+2 a +4 c +1}}{2}\right ], \left [-\frac {a}{2}+\frac {b}{2}\right ], \frac {x}{2}+\frac {1}{2}\right )+c_2 \left (\frac {x}{2}+\frac {1}{2}\right )^{1+\frac {a}{2}-\frac {b}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {\sqrt {a^{2}+2 a +4 c +1}}{2}-\frac {b}{2}, \frac {1}{2}+\frac {\sqrt {a^{2}+2 a +4 c +1}}{2}-\frac {b}{2}\right ], \left [2+\frac {a}{2}-\frac {b}{2}\right ], \frac {x}{2}+\frac {1}{2}\right ) \]

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 
   -> Kummer 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
   -> hypergeometric 
      -> heuristic approach 
      <- heuristic approach successful 
   <- hypergeometric successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+c 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 {c y \left (x \right )}{x^{2}-1}+\frac {\left (a x +b \right ) \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 {\left (a x +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}-1}-\frac {c y \left (x \right )}{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 +b}{x^{2}-1}, P_{3}\left (x \right )=-\frac {c}{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}+\frac {b}{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^{2}}{d x^{2}}y \left (x \right )\right )+\left (-a x -b \right ) \left (\frac {d}{d x}y \left (x \right )\right )-c 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^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-a u +a -b \right ) \left (\frac {d}{d u}y \left (u \right )\right )-c 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 \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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & 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} r \left (2-2 r +a -b \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (-2 k -2 r +a -b \right )-a_{k} \left (a k +a r -k^{2}-2 k r -r^{2}+c +k +r \right )\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 -b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1+\frac {a}{2}-\frac {b}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (-2 k -2 r +a -b \right )-a_{k} \left (-k^{2}+\left (a -2 r +1\right ) k -r^{2}+\left (a +1\right ) r +c \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (a k +a r -k^{2}-2 k r -r^{2}+c +k +r \right )}{\left (k +1+r \right ) \left (-2 k -2 r +a -b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (a k -k^{2}+c +k \right )}{\left (k +1\right ) \left (-2 k +a -b \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 +1}=\frac {a_{k} \left (a k -k^{2}+c +k \right )}{\left (k +1\right ) \left (-2 k +a -b \right )}\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}, a_{k +1}=\frac {a_{k} \left (a k -k^{2}+c +k \right )}{\left (k +1\right ) \left (-2 k +a -b \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+\frac {a}{2}-\frac {b}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (a k +a \left (1+\frac {a}{2}-\frac {b}{2}\right )-k^{2}-2 k \left (1+\frac {a}{2}-\frac {b}{2}\right )-\left (1+\frac {a}{2}-\frac {b}{2}\right )^{2}+c +k +1+\frac {a}{2}-\frac {b}{2}\right )}{\left (k +2+\frac {a}{2}-\frac {b}{2}\right ) \left (-2 k -2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+\frac {a}{2}-\frac {b}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1+\frac {a}{2}-\frac {b}{2}}, a_{k +1}=\frac {a_{k} \left (a k +a \left (1+\frac {a}{2}-\frac {b}{2}\right )-k^{2}-2 k \left (1+\frac {a}{2}-\frac {b}{2}\right )-\left (1+\frac {a}{2}-\frac {b}{2}\right )^{2}+c +k +1+\frac {a}{2}-\frac {b}{2}\right )}{\left (k +2+\frac {a}{2}-\frac {b}{2}\right ) \left (-2 k -2\right )}\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 +1+\frac {a}{2}-\frac {b}{2}}, a_{k +1}=\frac {a_{k} \left (a k +a \left (1+\frac {a}{2}-\frac {b}{2}\right )-k^{2}-2 k \left (1+\frac {a}{2}-\frac {b}{2}\right )-\left (1+\frac {a}{2}-\frac {b}{2}\right )^{2}+c +k +1+\frac {a}{2}-\frac {b}{2}\right )}{\left (k +2+\frac {a}{2}-\frac {b}{2}\right ) \left (-2 k -2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +1\right )^{k +1+\frac {a}{2}-\frac {b}{2}}\right ), d_{k +1}=\frac {d_{k} \left (a k -k^{2}+c +k \right )}{\left (k +1\right ) \left (-2 k +a -b \right )}, e_{k +1}=\frac {e_{k} \left (a k +a \left (1+\frac {a}{2}-\frac {b}{2}\right )-k^{2}-2 k \left (1+\frac {a}{2}-\frac {b}{2}\right )-\left (1+\frac {a}{2}-\frac {b}{2}\right )^{2}+c +k +1+\frac {a}{2}-\frac {b}{2}\right )}{\left (k +2+\frac {a}{2}-\frac {b}{2}\right ) \left (-2 k -2\right )}\right ] \end {array} \]

2.31.20.2 ✓ Mathematica. Time used: 0.116 (sec). Leaf size: 184

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

\begin{align*} y(x)&\to 2^{\frac {1}{2} (-a-b-2)} \left (c_2 (x-1)^{\frac {1}{2} (a+b+2)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (b-\sqrt {a^2+2 a+4 c+1}+1\right ),\frac {1}{2} \left (b+\sqrt {a^2+2 a+4 c+1}+1\right ),\frac {1}{2} (a+b+4),\frac {1-x}{2}\right )+c_1 2^{\frac {1}{2} (a+b+2)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (-a-\sqrt {a^2+2 a+4 c+1}-1\right ),\frac {1}{2} \left (-a+\sqrt {a^2+2 a+4 c+1}-1\right ),\frac {1}{2} (-a-b),\frac {1-x}{2}\right )\right ) \end{align*}

2.31.20.3 ✗ Sympy

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

False