2.32.14 Problem 195

Internal problem ID [13856]
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 : 195
Date solved : Friday, December 19, 2025 at 04:09:21 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} \left (a \,x^{3}+b \,x^{2}+c x \right ) y^{\prime \prime }+\left (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }+\left (\beta -2 b \right ) y&=0 \\ \end{align*}

2.32.14.1 ✓ Maple. Time used: 0.571 (sec). Leaf size: 1493

ode:=(a*x^3+b*x^2+c*x)*diff(diff(y(x),x),x)+(alpha*x^2+beta*x+2*c)*diff(y(x),x)+(beta-2*b)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ \text {Expression too large to display} \]

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 
         <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
      <- hypergeometric successful 
   <- special function solution successful 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form is not straightforward to achieve - returning special \ 
function solution free of uncomputed integrals 
   <- Kovacics algorithm successful
 

Maple step by step

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

2.32.14.2 ✓ Mathematica. Time used: 2.667 (sec). Leaf size: 139

ode=(a*x^3+b*x^2+c*x)*D[y[x],{x,2}]+(\[Alpha]*x^2+\[Beta]*x+2*c)*D[y[x],x]+(\[Beta]-2*b)*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {(2 a x+2 b-\beta -\alpha x) \left (c_2 \int _1^x\frac {\exp \left (\frac {(b \alpha +2 a (b-\beta )) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) (c+K[1] (b+a K[1]))^{1-\frac {\alpha }{2 a}}}{(-2 b+\beta +(\alpha -2 a) K[1])^2}dK[1]+c_1\right )}{x (2 b-\beta )} \end{align*}

2.32.14.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
Alpha = symbols("Alpha") 
BETA = symbols("BETA") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq((BETA - 2*b)*y(x) + (Alpha*x**2 + BETA*x + 2*c)*Derivative(y(x), x) + (a*x**3 + b*x**2 + c*x)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False