2.34.16 Problem 254

Internal problem ID [13914]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-8. Other equations.
Problem number : 254
Date solved : Friday, December 19, 2025 at 08:31:29 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} x^{2} \left (x^{2 n} a^{2}-1\right ) y^{\prime \prime }+x \left (a p \,x^{n}+q \right ) y^{\prime }+\left (a r \,x^{n}+s \right ) y&=0 \\ \end{align*}

2.34.16.1 ✓ Maple. Time used: 0.271 (sec). Leaf size: 273

ode:=x^2*(a^2*x^(2*n)-1)*diff(diff(y(x),x),x)+x*(a*p*x^n+q)*diff(y(x),x)+(a*r*x^n+s)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = x^{\frac {q}{2}+\frac {1}{2}} \left (c_1 \,x^{\frac {\sqrt {q^{2}+2 q +4 s +1}}{2}} \operatorname {HeunG}\left (-1, \frac {p q +\sqrt {q^{2}+2 q +4 s +1}\, p +p +2 r}{2 n^{2}}, \frac {\sqrt {q^{2}+2 q +4 s +1}+q +1}{2 n}, \frac {\sqrt {q^{2}+2 q +4 s +1}+q -1}{2 n}, \frac {n +\sqrt {q^{2}+2 q +4 s +1}}{n}, -\frac {p -q}{2 n}, -a \,x^{n}\right )+c_2 \,x^{-\frac {\sqrt {q^{2}+2 q +4 s +1}}{2}} \operatorname {HeunG}\left (-1, \frac {-\sqrt {q^{2}+2 q +4 s +1}\, p +\left (q +1\right ) p +2 r}{2 n^{2}}, -\frac {\sqrt {q^{2}+2 q +4 s +1}-q +1}{2 n}, -\frac {\sqrt {q^{2}+2 q +4 s +1}-q -1}{2 n}, \frac {n -\sqrt {q^{2}+2 q +4 s +1}}{n}, -\frac {p -q}{2 n}, -a \,x^{n}\right )\right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying an equivalence, under non-integer power transformations, 
   to LODEs admitting Liouvillian solutions. 
   -> Trying a Liouvillian solution using Kovacics algorithm 
   <- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Whittaker 
      -> 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 @ Moebi\ 
us 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
<- Heun successful: received ODE is equivalent to the  HeunG  ODE, case  a <> 0\ 
, e <> 0, g <> 0, c = 0
 

2.34.16.2 ✗ Mathematica

ode=x^2*(a^2*x^(2*n)-1)*D[y[x],{x,2}]+x*(a*p*x^n+q)*D[y[x],x]+(a*r*x^n+s)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.34.16.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
n = symbols("n") 
p = symbols("p") 
q = symbols("q") 
r = symbols("r") 
s = symbols("s") 
y = Function("y") 
ode = Eq(x**2*(a**2*x**(2*n) - 1)*Derivative(y(x), (x, 2)) + x*(a*p*x**n + q)*Derivative(y(x), x) + (a*r*x**n + s)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : Symbol object cannot be interpreted as an integer
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', '2nd_power_series_regular')