2.31.24 Problem 172

Internal problem ID [13833]
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 : 172
Date solved : Friday, December 19, 2025 at 02:35:41 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.31.24.1 ✓ Maple. Time used: 0.010 (sec). Leaf size: 230

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

\[ y = c_2 \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )^{-\frac {\operatorname {csgn}\left (a \right ) \left (\left (b -2\right ) \operatorname {csgn}\left (a \right ) a +b a -2 c \right )}{2 a}} \operatorname {hypergeom}\left (\left [\frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +\sqrt {b^{2}-2 b -4 d +1}\, \operatorname {csgn}\left (a \right ) a -b a +2 c \right )}{2 a}, -\frac {\operatorname {csgn}\left (a \right ) \left (\sqrt {b^{2}-2 b -4 d +1}\, \operatorname {csgn}\left (a \right ) a -\operatorname {csgn}\left (a \right ) a +b a -2 c \right )}{2 a}\right ], \left [-\frac {\left (\left (b -4\right ) \operatorname {csgn}\left (a \right ) a +b a -2 c \right ) \operatorname {csgn}\left (a \right )}{2 a}\right ], \frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )}{2 a}\right )+c_1 \operatorname {hypergeom}\left (\left [-\frac {1}{2}+\frac {b}{2}-\frac {\sqrt {b^{2}-2 b -4 d +1}}{2}, -\frac {1}{2}+\frac {b}{2}+\frac {\sqrt {b^{2}-2 b -4 d +1}}{2}\right ], \left [\frac {\left (b \,\operatorname {csgn}\left (a \right ) a +b a -2 c \right ) \operatorname {csgn}\left (a \right )}{2 a}\right ], \frac {\operatorname {csgn}\left (a \right ) \left (\operatorname {csgn}\left (a \right ) a +a +2 x \right )}{2 a}\right ) \]

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 
<- 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 
      <- heuristic approach successful 
   <- hypergeometric successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x +a \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b x +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+d 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 {d y \left (x \right )}{x \left (x +a \right )}-\frac {\left (b x +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +a \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 (b x +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x +a \right )}+\frac {d y \left (x \right )}{x \left (x +a \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 {b x +c}{x \left (x +a \right )}, P_{3}\left (x \right )=\frac {d}{x \left (x +a \right )}\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}}}=\frac {c}{a} \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x +a \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b x +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+d 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 \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \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 ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (a r -a +c \right ) x^{r -1}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (a \left (k +1\right )+a r -a +c \right )+a_{k} \left (b k +b r +k^{2}+2 k r +r^{2}+d -k -r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (a r -a +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {a -c}{a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (a k +a r +c \right ) a_{k +1}+\left (k^{2}+\left (b +2 r -1\right ) k +r^{2}+\left (-1+b \right ) r +d \right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {\left (b k +b r +k^{2}+2 k r +r^{2}+d -k -r \right ) a_{k}}{\left (k +1+r \right ) \left (a k +a r +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {\left (b k +k^{2}+d -k \right ) a_{k}}{\left (k +1\right ) \left (a k +c \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 +1}=-\frac {\left (b k +k^{2}+d -k \right ) a_{k}}{\left (k +1\right ) \left (a k +c \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & a_{k +1}=-\frac {\left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right ) a_{k}}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {a -c}{a}}, a_{k +1}=-\frac {\left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right ) a_{k}}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \right )}\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}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k +\frac {a -c}{a}}\right ), e_{k +1}=-\frac {\left (b k +k^{2}+d -k \right ) e_{k}}{\left (k +1\right ) \left (a k +c \right )}, f_{k +1}=-\frac {\left (b k +\frac {b \left (a -c \right )}{a}+k^{2}+\frac {2 k \left (a -c \right )}{a}+\frac {\left (a -c \right )^{2}}{a^{2}}+d -k -\frac {a -c}{a}\right ) f_{k}}{\left (k +1+\frac {a -c}{a}\right ) \left (a k +a \right )}\right ] \end {array} \]

2.31.24.2 ✓ Mathematica. Time used: 0.218 (sec). Leaf size: 165

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

\begin{align*} y(x)&\to c_2 a^{\frac {c}{a}-1} x^{1-\frac {c}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (b-\frac {2 c}{a}+\sqrt {b^2-2 b-4 d+1}+1\right ),\frac {b a-\sqrt {b^2-2 b-4 d+1} a+a-2 c}{2 a},2-\frac {c}{a},-\frac {x}{a}\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (b-\sqrt {b^2-2 b-4 d+1}-1\right ),\frac {1}{2} \left (b+\sqrt {b^2-2 b-4 d+1}-1\right ),\frac {c}{a},-\frac {x}{a}\right ) \end{align*}

2.31.24.3 ✗ Sympy

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

ValueError : Expected Expr or iterable but got None
 

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_hypergeometric', '2nd_hypergeometric_Integral', '2nd_power_series_regular')