2.32.10 Problem 191

Internal problem ID [13852]
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 : 191
Date solved : Friday, December 19, 2025 at 03:45:57 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.32.10.1 ✓ Maple. Time used: 0.102 (sec). Leaf size: 169

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

\[ y = \frac {\left (a x +b \right )^{\frac {3 a -c}{a}} \left (c_1 \,x^{\frac {-3 a +c}{a}} \operatorname {HeunC}\left (\frac {\lambda a}{b}, 1-\frac {c}{a}, 3-\frac {c}{a}, 0, -\frac {\lambda a}{b}+\frac {c \lambda }{2 b}+\frac {5}{2}-\frac {2 c}{a}+\frac {c^{2}}{2 a^{2}}, -\frac {b}{a x}\right ) x^{2}+c_2 \operatorname {HeunC}\left (\frac {\lambda a}{b}, -1+\frac {c}{a}, 3-\frac {c}{a}, 0, -\frac {\lambda a}{b}+\frac {c \lambda }{2 b}+\frac {5}{2}-\frac {2 c}{a}+\frac {c^{2}}{2 a^{2}}, -\frac {b}{a x}\right )\right )}{x^{2}} \]

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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (c \,x^{2}+\left (a \lambda +2 b \right ) x +b \lambda \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\lambda \left (c -2 a \right ) 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 {\lambda \left (-c +2 a \right ) y \left (x \right )}{x^{2} \left (a x +b \right )}-\frac {\left (a \lambda x +c \,x^{2}+b \lambda +2 b x \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2} \left (a x +b \right )} \\ \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 \lambda x +c \,x^{2}+b \lambda +2 b x \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2} \left (a x +b \right )}-\frac {\lambda \left (-c +2 a \right ) y \left (x \right )}{x^{2} \left (a x +b \right )}=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 \lambda x +c \,x^{2}+b \lambda +2 b x}{x^{2} \left (a x +b \right )}, P_{3}\left (x \right )=-\frac {\lambda \left (-c +2 a \right )}{x^{2} \left (a x +b \right )}\right ] \\ {} & \circ & \left (x +\frac {b}{a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b}{a} \\ {} & {} & \left (\left (x +\frac {b}{a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b}{a}}}}=\frac {\left (\frac {c \,b^{2}}{a^{2}}-\frac {2 b^{2}}{a}\right ) a}{b^{2}} \\ {} & \circ & \left (x +\frac {b}{a}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b}{a} \\ {} & {} & \left (\left (x +\frac {b}{a}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b}{a}}}}=0 \\ {} & \circ & x =-\frac {b}{a}\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}=-\frac {b}{a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a \lambda x +c \,x^{2}+b \lambda +2 b x \right ) \left (\frac {d}{d x}y \left (x \right )\right )-\lambda \left (-c +2 a \right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {b}{a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{3}-2 u^{2} b +\frac {u \,b^{2}}{a}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (a \lambda u +c \,u^{2}-\frac {2 c u b}{a}+\frac {c \,b^{2}}{a^{2}}+2 b u -\frac {2 b^{2}}{a}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 a \lambda +c \lambda \right ) 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..2 \\ {} & {} & 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..3 \\ {} & {} & 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} \\ {} & {} & \frac {a_{0} r \,b^{2} \left (a r -3 a +c \right ) u^{r -1}}{a^{2}}+\left (\frac {a_{1} \left (1+r \right ) b^{2} \left (a r -2 a +c \right )}{a^{2}}+\frac {a_{0} \left (a r -2 a +c \right ) \left (a \lambda -2 b r \right )}{a}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) b^{2} \left (a \left (k +1\right )+a r -3 a +c \right )}{a^{2}}+\frac {a_{k} \left (a k +a r -2 a +c \right ) \left (a \lambda -2 b k -2 b r \right )}{a}+a_{k -1} \left (k +r -1\right ) \left (a \left (k -1\right )+a r -a +c \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \,b^{2} \left (a r -3 a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3 a -c}{a}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \frac {a_{1} \left (1+r \right ) b^{2} \left (a r -2 a +c \right )}{a^{2}}+\frac {a_{0} \left (a r -2 a +c \right ) \left (a \lambda -2 b r \right )}{a}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\left (\left (k a_{k -1}+\lambda a_{k}+r a_{k -1}-a_{k -1}\right ) a^{2}-2 b a_{k} \left (k +r \right ) a +a_{k +1} \left (k +1+r \right ) b^{2}\right ) \left (\left (k +r -2\right ) a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {\left (\left (\left (k +1\right ) a_{k}+\lambda a_{k +1}+r a_{k}-a_{k}\right ) a^{2}-2 b a_{k +1} \left (k +1+r \right ) a +a_{k +2} \left (k +2+r \right ) b^{2}\right ) \left (\left (k +r -1\right ) a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+a r a_{k}-2 b k a_{k +1}-2 b r a_{k +1}-2 b a_{k +1}\right )}{\left (k +2+r \right ) b^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}} \\ \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 +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {a_{1} b^{2} \left (c -2 a \right )}{a^{2}}+a_{0} \left (c -2 a \right ) \lambda =0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b}{a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b}{a}\right )^{k}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {a_{1} b^{2} \left (c -2 a \right )}{a^{2}}+a_{0} \left (c -2 a \right ) \lambda =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3 a -c}{a} \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3 a -c}{a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {3 a -c}{a}}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {a_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+a_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\right )=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b}{a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b}{a}\right )^{k +\frac {3 a -c}{a}}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {a_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+a_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\right )=0\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 +\frac {b}{a}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +\frac {b}{a}\right )^{k +\frac {3 a -c}{a}}\right ), d_{k +2}=-\frac {a \left (a k d_{k}+a \lambda d_{k +1}-2 b k d_{k +1}-2 b d_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {d_{1} b^{2} \left (c -2 a \right )}{a^{2}}+d_{0} \left (c -2 a \right ) \lambda =0, e_{k +2}=-\frac {a \left (a k e_{k}+a \lambda e_{k +1}+\left (3 a -c \right ) e_{k}-2 b k e_{k +1}-\frac {2 b \left (3 a -c \right ) e_{k +1}}{a}-2 b e_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {e_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+e_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\right )=0\right ] \end {array} \]

2.32.10.2 ✓ Mathematica. Time used: 0.797 (sec). Leaf size: 55

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

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

2.32.10.3 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (2*a*lambda_*y(x) - a*x**3*Derivative(y(x)
 

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',)