2.29.25 Problem 85

Internal problem ID [13746]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-3
Problem number : 85
Date solved : Friday, December 19, 2025 at 11:41:52 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.29.25.1 ✓ Maple. Time used: 0.009 (sec). Leaf size: 143

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

\[ y = \frac {{\mathrm e}^{-\frac {x \left (a^{2} x +2 b a -2 c \right )}{2 a}} \left (c_1 \operatorname {hypergeom}\left (\left [\frac {2 a^{3}-d \,a^{2}+a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {\left (a^{2} x +b a -2 c \right )^{2}}{2 a^{3}}\right )+c_2 \operatorname {hypergeom}\left (\left [\frac {3 a^{3}-d \,a^{2}+a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {\left (a^{2} x +b a -2 c \right )^{2}}{2 a^{3}}\right ) \left (a^{2} x +b a -2 c \right )\right )}{x} \]

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 
<- No Liouvillian solutions exists 
-> 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 
      <- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Moebius`` i\ 
s resolved 
   <- hypergeometric successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a \,x^{2}+b x +2\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (c \,x^{2}+d x +b \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 {\left (c \,x^{2}+d x +b \right ) y \left (x \right )}{x}-\frac {\left (a \,x^{2}+b x +2\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x} \\ \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 \,x^{2}+b x +2\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {\left (c \,x^{2}+d x +b \right ) y \left (x \right )}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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 \,x^{2}+b x +2}{x}, P_{3}\left (x \right )=\frac {c \,x^{2}+d x +b}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=2 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a \,x^{2}+b x +2\right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (c \,x^{2}+d x +b \right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+r \right )+a_{0} b \left (1+r \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (3+r \right )+a_{1} b \left (2+r \right )+a_{0} \left (a r +d \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )+a_{k} b \left (k +1+r \right )+a_{k -1} \left (a \left (k -1\right )+a r +d \right )+a_{k -2} c \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (2+r \right )+a_{0} b \left (1+r \right )=0, a_{2} \left (2+r \right ) \left (3+r \right )+a_{1} b \left (2+r \right )+a_{0} \left (a r +d \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} b}{2+r}, a_{2}=-\frac {a_{0} \left (a r -b^{2}+d \right )}{r^{2}+5 r +6}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )+a_{k} b \left (k +1+r \right )+a_{k -1} \left (\left (k +r -1\right ) a +d \right )+a_{k -2} c =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +4+r \right )+a_{k +2} b \left (k +3+r \right )+a_{k +1} \left (\left (k +1+r \right ) a +d \right )+a_{k} c =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+a r a_{k +1}+b k a_{k +2}+b r a_{k +2}+a a_{k +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3+r \right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+2 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+2 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +2\right ) \left (k +3\right )}, a_{1}=-a_{0} b , a_{2}=-\frac {a_{0} \left (-b^{2}-a +d \right )}{2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+a a_{k +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+a a_{k +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3\right ) \left (k +4\right )}, a_{1}=-\frac {a_{0} b}{2}, a_{2}=-\frac {a_{0} \left (-b^{2}+d \right )}{6}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k}\right ), e_{k +3}=-\frac {a k e_{k +1}+b k e_{k +2}+2 b e_{k +2}+c e_{k}+d e_{k +1}}{\left (k +2\right ) \left (k +3\right )}, e_{1}=-e_{0} b , e_{2}=-\frac {e_{0} \left (-b^{2}-a +d \right )}{2}, f_{k +3}=-\frac {a k f_{k +1}+b k f_{k +2}+a f_{k +1}+3 b f_{k +2}+c f_{k}+d f_{k +1}}{\left (k +3\right ) \left (k +4\right )}, f_{1}=-\frac {f_{0} b}{2}, f_{2}=-\frac {f_{0} \left (-b^{2}+d \right )}{6}\right ] \end {array} \]

2.29.25.2 ✓ Mathematica. Time used: 0.054 (sec). Leaf size: 134

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

\begin{align*} y(x)&\to \frac {e^{-\frac {1}{2} x \left (-\frac {2 c}{a}+a x+2 b\right )} \left (c_2 \operatorname {Hypergeometric1F1}\left (-\frac {-2 a^3+d a^2-b c a+c^2}{2 a^3},\frac {1}{2},\frac {\left (x a^2+b a-2 c\right )^2}{2 a^3}\right )+c_1 \operatorname {HermiteH}\left (\frac {-2 a^3+d a^2-b c a+c^2}{a^3},\frac {x a^2+b a-2 c}{\sqrt {2} a^{3/2}}\right )\right )}{x} \end{align*}

2.29.25.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
d = symbols("d") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), (x, 2)) + (b + c*x**2 + d*x)*y(x) + (a*x**2 + b*x + 2)*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_regular')