2.32.27 Problem 209

Internal problem ID [13869]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-6
Problem number : 209
Date solved : Friday, December 19, 2025 at 06:23:39 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} \left (a \,x^{3}+b \,x^{2}+c x +d \right ) y^{\prime \prime }+\left (\lambda ^{3}+x^{3}\right ) y^{\prime }-\left (\lambda ^{2}-\lambda x +x^{2}\right ) y&=0 \\ \end{align*}

2.32.27.1 ✓ Maple. Time used: 0.052 (sec). Leaf size: 76

ode:=(a*x^3+b*x^2+c*x+d)*diff(diff(y(x),x),x)+(lambda^3+x^3)*diff(y(x),x)-(lambda^2-lambda*x+x^2)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \left (\lambda +x \right ) \left (c_2 \int {\mathrm e}^{-\int \frac {x^{4}+\left (2 a +\lambda \right ) x^{3}+2 b \,x^{2}+\left (\lambda ^{3}+2 c \right ) x +\lambda ^{4}+2 d}{\left (a \,x^{3}+b \,x^{2}+c x +d \right ) \left (\lambda +x \right )}d x}d x +c_1 \right ) \]

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 
   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 
   No special function solution was found. 
<- Kovacics algorithm successful
 

2.32.27.2 ✓ Mathematica. Time used: 1.294 (sec). Leaf size: 240

ode=(a*x^3+b*x^2+c*x+d)*D[y[x],{x,2}]+(x^3+\[Lambda]^3)*D[y[x],x]-(x^2-\[Lambda]*x+\[Lambda]^2)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {c_2 (\lambda +x) \int _1^x\exp \left (-\frac {\lambda +K[1]+2 a \log (\lambda +K[1])+\text {RootSum}\left [-a \lambda ^3+b \lambda ^2+3 a \text {$\#$1} \lambda ^2-3 a \text {$\#$1}^2 \lambda -c \lambda -2 b \text {$\#$1} \lambda +a \text {$\#$1}^3+b \text {$\#$1}^2+d+c \text {$\#$1}\&,\frac {a \log (\lambda +K[1]-\text {$\#$1}) \lambda ^3-b \log (\lambda +K[1]-\text {$\#$1}) \lambda ^2+c \log (\lambda +K[1]-\text {$\#$1}) \lambda +2 b \log (\lambda +K[1]-\text {$\#$1}) \text {$\#$1} \lambda -b \log (\lambda +K[1]-\text {$\#$1}) \text {$\#$1}^2-d \log (\lambda +K[1]-\text {$\#$1})-c \log (\lambda +K[1]-\text {$\#$1}) \text {$\#$1}}{3 a \lambda ^2-2 b \lambda -6 a \text {$\#$1} \lambda +3 a \text {$\#$1}^2+c+2 b \text {$\#$1}}\&\right ]}{a}\right )dK[1]}{\lambda }+\frac {c_1 (\lambda +x)}{\lambda } \end{align*}

2.32.27.3 ✗ 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_**3 + x**3)*Derivative(y(x), x) - (lambda_**2 - lambda_*x + x**2)*y(x) + (a*x**3 + b*x**2 + c*x + d)*Derivative(y(x), (x, 2)),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')