2.31.21 Problem 169

Internal problem ID [13830]
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 : 169
Date solved : Friday, December 19, 2025 at 02:23:16 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.31.21.1 ✓ Maple. Time used: 0.052 (sec). Leaf size: 899

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

\begin{align*} \text {Solution too large to show}\end{align*}

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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Kummer 
         -> 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 @ Mo\ 
ebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunC  ODE, case  \ 
a <> 0, e <> 0, c = 0 
   <- Kovacics algorithm successful
 

2.31.21.2 ✓ Mathematica. Time used: 1.759 (sec). Leaf size: 74

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

\begin{align*} y(x)&\to e^{\lambda (-x)} \left (c_2 \int _1^x\exp \left (\frac {(b c-a d) \arctan \left (\frac {\sqrt {a} K[1]}{\sqrt {b}}\right )}{a^{3/2} \sqrt {b}}+\left (2 \lambda -\frac {c}{a}\right ) K[1]\right )dK[1]+c_1\right ) \end{align*}

2.31.21.3 ✗ Sympy

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

False
 

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_ordinary')