2.29.24 Problem 84

Internal problem ID [13745]
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 : 84
Date solved : Friday, December 19, 2025 at 11:40:46 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (A \,x^{2}+B x +\operatorname {C0} \right ) y&=0 \\ \end{align*}

2.29.24.1 ✓ Maple. Time used: 0.036 (sec). Leaf size: 186

ode:=x*diff(diff(y(x),x),x)+(a*x^2+b*x+c)*diff(y(x),x)+(A*x^2+B*x+C0)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = {\mathrm e}^{\frac {x \left (-a^{2} x -2 b a +2 A \right )}{2 a}} \left (x^{-c +1} \operatorname {HeunB}\left (-c +1, -\frac {\sqrt {2}\, \left (-b a +2 A \right )}{a^{{3}/{2}}}, \frac {\left (-c -1\right ) a^{3}+2 B \,a^{2}-2 A a b +2 A^{2}}{a^{3}}, \frac {\left (b c -2 \operatorname {C0} \right ) \sqrt {2}}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) c_2 +\operatorname {HeunB}\left (c -1, -\frac {\sqrt {2}\, \left (-b a +2 A \right )}{a^{{3}/{2}}}, \frac {\left (-c -1\right ) a^{3}+2 B \,a^{2}-2 A a b +2 A^{2}}{a^{3}}, \frac {\left (b c -2 \operatorname {C0} \right ) \sqrt {2}}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) c_1 \right ) \]

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 
<- 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 
   -> Mathieu 
      -> Equivalence to the rational form of Mathieu ODE under a power @ Moebi\ 
us 
trying a solution in terms of MeijerG functions 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
<- Heun successful: received ODE is equivalent to the  HeunB  ODE, case  c = 0
 

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 +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (A \,x^{2}+B x +\mathit {C0} \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 (A \,x^{2}+B x +\mathit {C0} \right ) y \left (x \right )}{x}-\frac {\left (a \,x^{2}+b x +c \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 +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+\frac {\left (A \,x^{2}+B x +\mathit {C0} \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 +c}{x}, P_{3}\left (x \right )=\frac {A \,x^{2}+B x +\mathit {C0}}{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}}}=c \\ {} & \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 +c \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (A \,x^{2}+B x +\mathit {C0} \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 +c \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} \left (b r +\mathit {C0} \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (1+r +c \right )+a_{1} \left (b r +\mathit {C0} +b \right )+a_{0} \left (a r +B \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +c \right )+a_{k} \left (b k +b r +\mathit {C0} \right )+a_{k -1} \left (a \left (k -1\right )+a r +B \right )+A a_{k -2}\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 +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -c +1\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 (r +c \right )+a_{0} \left (b r +\mathit {C0} \right )=0, a_{2} \left (2+r \right ) \left (1+r +c \right )+a_{1} \left (b r +\mathit {C0} +b \right )+a_{0} \left (a r +B \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} \left (b r +\mathit {C0} \right )}{r c +r^{2}+c +r}, a_{2}=-\frac {a_{0} \left (a c \,r^{2}+a \,r^{3}-b^{2} r^{2}+B c r +B \,r^{2}-2 \mathit {C0} b r +a c r +a \,r^{2}-b^{2} r +B c +B r -\mathit {C0}^{2}-\mathit {C0} b \right )}{c^{2} r^{2}+2 c \,r^{3}+r^{4}+3 c^{2} r +7 c \,r^{2}+4 r^{3}+2 c^{2}+7 r c +5 r^{2}+2 c +2 r}\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 +r +c \right )+a_{k} \left (\left (k +r \right ) b +\mathit {C0} \right )+\left (\left (k +r -1\right ) a +B \right ) a_{k -1}+A a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r +c \right )+a_{k +2} \left (\left (k +2+r \right ) b +\mathit {C0} \right )+\left (\left (k +1+r \right ) a +B \right ) a_{k +1}+A a_{k}=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}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3+r \right ) \left (k +2+r +c \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}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3\right ) \left (k +2+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 +3}=-\frac {a k a_{k +1}+b k a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3\right ) \left (k +2+c \right )}, a_{1}=-\frac {a_{0} \mathit {C0}}{c}, a_{2}=-\frac {a_{0} \left (B c -\mathit {C0}^{2}-\mathit {C0} b \right )}{2 c^{2}+2 c}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-c +1 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+a \left (-c +1\right ) a_{k +1}+b k a_{k +2}+b \left (-c +1\right ) a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +4-c \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-c +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -c +1}, a_{k +3}=-\frac {a k a_{k +1}+a \left (-c +1\right ) a_{k +1}+b k a_{k +2}+b \left (-c +1\right ) a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +4-c \right ) \left (k +3\right )}, a_{1}=-\frac {a_{0} \left (b \left (-c +1\right )+\mathit {C0} \right )}{\left (-c +1\right ) c +\left (-c +1\right )^{2}+1}, a_{2}=-\frac {a_{0} \left (a \left (-c +1\right )^{2} c +a \left (-c +1\right )^{3}-b^{2} \left (-c +1\right )^{2}+B \left (-c +1\right ) c +B \left (-c +1\right )^{2}-2 b \left (-c +1\right ) \mathit {C0} +a \left (-c +1\right ) c +a \left (-c +1\right )^{2}-b^{2} \left (-c +1\right )+B c +B \left (-c +1\right )-\mathit {C0}^{2}-\mathit {C0} b \right )}{\left (-c +1\right )^{2} c^{2}+2 \left (-c +1\right )^{3} c +\left (-c +1\right )^{4}+3 \left (-c +1\right ) c^{2}+7 \left (-c +1\right )^{2} c +4 \left (-c +1\right )^{3}+2 c^{2}+7 \left (-c +1\right ) c +5 \left (-c +1\right )^{2}+2}\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 -c +1}\right ), d_{k +3}=-\frac {a k d_{k +1}+b k d_{k +2}+A d_{k}+B d_{k +1}+\mathit {C0} d_{k +2}+a d_{k +1}+2 b d_{k +2}}{\left (k +3\right ) \left (k +2+c \right )}, d_{1}=-\frac {d_{0} \mathit {C0}}{c}, d_{2}=-\frac {d_{0} \left (B c -\mathit {C0}^{2}-\mathit {C0} b \right )}{2 c^{2}+2 c}, e_{k +3}=-\frac {a k e_{k +1}+a \left (-c +1\right ) e_{k +1}+b k e_{k +2}+b \left (-c +1\right ) e_{k +2}+A e_{k}+B e_{k +1}+\mathit {C0} e_{k +2}+a e_{k +1}+2 b e_{k +2}}{\left (k +4-c \right ) \left (k +3\right )}, e_{1}=-\frac {e_{0} \left (b \left (-c +1\right )+\mathit {C0} \right )}{\left (-c +1\right ) c +\left (-c +1\right )^{2}+1}, e_{2}=-\frac {e_{0} \left (a \left (-c +1\right )^{2} c +a \left (-c +1\right )^{3}-b^{2} \left (-c +1\right )^{2}+B \left (-c +1\right ) c +B \left (-c +1\right )^{2}-2 b \left (-c +1\right ) \mathit {C0} +a \left (-c +1\right ) c +a \left (-c +1\right )^{2}-b^{2} \left (-c +1\right )+B c +B \left (-c +1\right )-\mathit {C0}^{2}-\mathit {C0} b \right )}{\left (-c +1\right )^{2} c^{2}+2 \left (-c +1\right )^{3} c +\left (-c +1\right )^{4}+3 \left (-c +1\right ) c^{2}+7 \left (-c +1\right )^{2} c +4 \left (-c +1\right )^{3}+2 c^{2}+7 \left (-c +1\right ) c +5 \left (-c +1\right )^{2}+2}\right ] \end {array} \]

2.29.24.2 ✗ Mathematica

ode=x*D[y[x],{x,2}]+(a*x^2+b*x+c)*D[y[x],x]+(A*x^2+B*x+C0)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.29.24.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
A = symbols("A") 
B = symbols("B") 
C0 = symbols("C0") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), (x, 2)) + (A*x**2 + B*x + C0)*y(x) + (a*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