2.32.9 Problem 190

Internal problem ID [13851]
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 : 190
Date solved : Friday, December 19, 2025 at 03:40:32 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.32.9.1 ✓ Maple. Time used: 0.017 (sec). Leaf size: 175

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

\[ y = \left (x^{2}+a \right )^{\frac {\left (-b +2\right ) a +c}{2 a}} \left (x^{\frac {a -c}{a}} \operatorname {hypergeom}\left (\left [-\frac {b}{4}+\frac {5}{4}-\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}, -\frac {b}{4}+\frac {5}{4}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}\right ], \left [\frac {3 a -c}{2 a}\right ], -\frac {x^{2}}{a}\right ) c_1 +\operatorname {hypergeom}\left (\left [-\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}-\frac {b}{4}+\frac {3}{4}+\frac {c}{2 a}, -\frac {b}{4}+\frac {3}{4}+\frac {c}{2 a}+\frac {\sqrt {b^{2}-2 b -4 s +1}}{4}\right ], \left [\frac {1}{2}+\frac {c}{2 a}\right ], -\frac {x^{2}}{a}\right ) c_2 \right ) \]

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^{2}+a \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b \,x^{2}+c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+s x 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 {s y \left (x \right )}{x^{2}+a}-\frac {\left (b \,x^{2}+c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}+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^{2}+c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x \left (x^{2}+a \right )}+\frac {s y \left (x \right )}{x^{2}+a}=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^{2}+c}{x \left (x^{2}+a \right )}, P_{3}\left (x \right )=\frac {s}{x^{2}+a}\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^{2}+a \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (b \,x^{2}+c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+s x 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 \cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y \left (x \right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} 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^{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..3 \\ {} & {} & 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}+a_{1} \left (1+r \right ) \left (a r +c \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (a \left (k +1\right )+a r -a +c \right )+a_{k -1} \left (b \left (k -1\right )+b r +\left (k -1\right )^{2}+2 \left (k -1\right ) r +r^{2}-k +1-r +s \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 must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (a r +c \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (b +2 r -3\right ) k +r^{2}+\left (b -3\right ) r -b +s +2\right ) a_{k -1}+a_{k +1} \left (k +r +1\right ) \left (a k +a r +c \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (b +2 r -3\right ) \left (k +1\right )+r^{2}+\left (b -3\right ) r -b +s +2\right ) a_{k}+a_{k +2} \left (k +2+r \right ) \left (a \left (k +1\right )+a r +c \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {\left (b k +b r +k^{2}+2 k r +r^{2}-k -r +s \right ) a_{k}}{\left (k +2+r \right ) \left (a k +a r +a +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {\left (b k +k^{2}-k +s \right ) a_{k}}{\left (k +2\right ) \left (a k +a +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 +2}=-\frac {\left (b k +k^{2}-k +s \right ) a_{k}}{\left (k +2\right ) \left (a k +a +c \right )}, a_{1} c =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a -c}{a} \\ {} & {} & a_{k +2}=-\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}}-k -\frac {a -c}{a}+s \right ) a_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 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 +2}=-\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}}-k -\frac {a -c}{a}+s \right ) a_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 a \right )}, a_{1} \left (1+\frac {a -c}{a}\right ) a =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} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {a -c}{a}}\right ), d_{k +2}=-\frac {\left (b k +k^{2}-k +s \right ) d_{k}}{\left (k +2\right ) \left (a k +a +c \right )}, d_{1} c =0, e_{k +2}=-\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}}-k -\frac {a -c}{a}+s \right ) e_{k}}{\left (k +2+\frac {a -c}{a}\right ) \left (a k +2 a \right )}, e_{1} \left (1+\frac {a -c}{a}\right ) a =0\right ] \end {array} \]

2.32.9.2 ✓ Mathematica. Time used: 0.501 (sec). Leaf size: 185

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

\begin{align*} y(x)&\to c_2 a^{\frac {1}{2} \left (\frac {c}{a}-1\right )} x^{1-\frac {c}{a}} \operatorname {Hypergeometric2F1}\left (\frac {a \left (b+\sqrt {b^2-2 b-4 s+1}+1\right )-2 c}{4 a},\frac {b a-\sqrt {b^2-2 b-4 s+1} a+a-2 c}{4 a},\frac {3}{2}-\frac {c}{2 a},-\frac {x^2}{a}\right )+c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2-2 b-4 s+1}-1\right ),\frac {1}{4} \left (b+\sqrt {b^2-2 b-4 s+1}-1\right ),\frac {a+c}{2 a},-\frac {x^2}{a}\right ) \end{align*}

2.32.9.3 ✗ Sympy

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

ValueError : Expected Expr or iterable but got None