2.31.33 Problem 181

Internal problem ID [13842]
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 : 181
Date solved : Friday, December 19, 2025 at 03:12:26 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.31.33.1 ✓ Maple. Time used: 0.043 (sec). Leaf size: 246

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

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

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
 

2.31.33.2 ✓ Mathematica. Time used: 2.578 (sec). Leaf size: 137

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

\begin{align*} y(x)&\to \frac {(k+x) \left (c_2 \int _1^x\frac {\exp \left (\frac {\left (b^3-3 a c b-2 a^3 k^3\right ) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{a^3 \sqrt {4 a c-b^2}}-\frac {K[1] (a K[1]-2 b)}{2 a^2}\right ) (c+K[1] (b+a K[1]))^{-\frac {b^2-a c}{2 a^3}}}{(k+K[1])^2}dK[1]+c_1\right )}{k} \end{align*}

2.31.33.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
k = symbols("k") 
y = Function("y") 
ode = Eq((k**3 + x**3)*Derivative(y(x), x) - (k**2 - k*x + x**2)*y(x) + (a*x**2 + b*x + c)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False