2.29.18 Problem 78

Internal problem ID [13739]
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 : 78
Date solved : Friday, December 19, 2025 at 11:32:11 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} x y^{\prime \prime }-\left (2 x a +1\right ) y^{\prime }+y b \,x^{3}&=0 \\ \end{align*}

2.29.18.1 ✓ Maple. Time used: 0.023 (sec). Leaf size: 106

ode:=x*diff(diff(y(x),x),x)-(2*a*x+1)*diff(y(x),x)+b*x^3*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \operatorname {HeunB}\left (2, 0, \frac {a^{2}}{\sqrt {-b}}, -\frac {2 i a}{\left (-b \right )^{{1}/{4}}}, i \left (-b \right )^{{1}/{4}} x \right ) x^{2} {\mathrm e}^{a x +\frac {\sqrt {-b}\, x^{2}}{2}} \left (\int \frac {{\mathrm e}^{-\sqrt {-b}\, x^{2}}}{x^{3} \operatorname {HeunB}\left (2, 0, \frac {a^{2}}{\sqrt {-b}}, -\frac {2 i a}{\left (-b \right )^{{1}/{4}}}, i \left (-b \right )^{{1}/{4}} x \right )^{2}}d x c_2 +c_1 \right ) \]

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 (2 a x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+b \,x^{3} 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 )=-b \,x^{2} y \left (x \right )+\frac {\left (2 a x +1\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 (2 a x +1\right ) \left (\frac {d}{d x}y \left (x \right )\right )}{x}+b \,x^{2} y \left (x \right )=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 {2 a x +1}{x}, P_{3}\left (x \right )=b \,x^{2}\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}}}=-1 \\ {} & \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 (-2 a x -1\right ) \left (\frac {d}{d x}y \left (x \right )\right )+b \,x^{3} 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^{3}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -3 \\ {} & {} & x^{3}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =3}{\sum }}a_{k -3} 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..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 \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 (-2+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-1+r \right )-2 a_{0} a r \right ) x^{r}+\left (a_{2} \left (2+r \right ) r -2 a_{1} \left (1+r \right ) a \right ) x^{1+r}+\left (a_{3} \left (3+r \right ) \left (1+r \right )-2 a_{2} \left (2+r \right ) a \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-2 a_{k} \left (k +r \right ) a +b a_{k -3}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\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 (-1+r \right )-2 a_{0} a r =0, a_{2} \left (2+r \right ) r -2 a_{1} \left (1+r \right ) a =0, a_{3} \left (3+r \right ) \left (1+r \right )-2 a_{2} \left (2+r \right ) a =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {2 a_{0} a r}{r^{2}-1}, a_{2}=\frac {4 a^{2} a_{0}}{r^{2}+r -2}, a_{3}=\frac {8 a^{3} a_{0}}{r^{3}+3 r^{2}-r -3}\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 -1\right )-2 a_{k} \left (k +r \right ) a +b a_{k -3}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +2+r \right )-2 a_{k +3} \left (k +r +3\right ) a +b a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+2 a r a_{k +3}+6 a a_{k +3}-b a_{k}}{\left (k +4+r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+6 a a_{k +3}-b a_{k}}{\left (k +4\right ) \left (k +2\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 +4}=\frac {2 a k a_{k +3}+6 a a_{k +3}-b a_{k}}{\left (k +4\right ) \left (k +2\right )}, a_{1}=0, a_{2}=-2 a^{2} a_{0}, a_{3}=-\frac {8 a^{3} a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+10 a a_{k +3}-b a_{k}}{\left (k +6\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +4}=\frac {2 a k a_{k +3}+10 a a_{k +3}-b a_{k}}{\left (k +6\right ) \left (k +4\right )}, a_{1}=\frac {4 a a_{0}}{3}, a_{2}=a^{2} a_{0}, a_{3}=\frac {8 a^{3} a_{0}}{15}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +2}\right ), c_{k +4}=\frac {2 a k c_{k +3}+6 a c_{k +3}-b c_{k}}{\left (k +4\right ) \left (k +2\right )}, c_{1}=0, c_{2}=-2 a^{2} c_{0}, c_{3}=-\frac {8 a^{3} c_{0}}{3}, d_{k +4}=\frac {2 a k d_{k +3}+10 a d_{k +3}-b d_{k}}{\left (k +6\right ) \left (k +4\right )}, d_{1}=\frac {4 a d_{0}}{3}, d_{2}=a^{2} d_{0}, d_{3}=\frac {8 a^{3} d_{0}}{15}\right ] \end {array} \]

2.29.18.2 ✗ Mathematica

ode=x*D[y[x],{x,2}]-(2*a*x+1)*D[y[x],x]+b*x^3*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.29.18.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(b*x**3*y(x) + x*Derivative(y(x), (x, 2)) - (2*a*x + 1)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

False