2.30.2 Problem 111

Internal problem ID [13772]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-4
Problem number : 111
Date solved : Sunday, January 18, 2026 at 09:14:46 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.30.2.1 second order bessel ode

0.151 (sec)

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

Entering second order bessel ode solverWriting the ode as

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

Bessel ode has the form

\begin{align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} y&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= 2 \sqrt {a}\\ n &= \sqrt {-4 b +1}\\ \gamma &= {\frac {1}{2}} \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} y = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\sqrt {-4 b +1}, 2 \sqrt {a}\, \sqrt {x}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\sqrt {-4 b +1}, 2 \sqrt {a}\, \sqrt {x}\right ) \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\sqrt {-4 b +1}, 2 \sqrt {a}\, \sqrt {x}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\sqrt {-4 b +1}, 2 \sqrt {a}\, \sqrt {x}\right ) \\ \end{align*}

2.30.2.2 ✓ Maple. Time used: 0.001 (sec). Leaf size: 45

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

\[ y = \left (\operatorname {BesselY}\left (\sqrt {1-4 b}, 2 \sqrt {a}\, \sqrt {x}\right ) c_2 +\operatorname {BesselJ}\left (\sqrt {1-4 b}, 2 \sqrt {a}\, \sqrt {x}\right ) c_1 \right ) \sqrt {x} \]

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 
   <- Bessel successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a x +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 (a x +b \right ) y \left (x \right )}{x^{2}} \\ \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 +b \right ) y \left (x \right )}{x^{2}}=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 )=0, P_{3}\left (x \right )=\frac {a x +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}}}=0 \\ {} & \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}}}=b \\ {} & \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^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (a x +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 y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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^{2}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}+b -r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+b -k -r \right )+a_{k -1} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+b -r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}, \frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r -1\right ) k +r^{2}+b -r \right ) a_{k}+a_{k -1} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (2 r -1\right ) \left (k +1\right )+r^{2}+b -r \right ) a_{k +1}+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} a}{k^{2}+2 k r +r^{2}+b +k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {1-4 b}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} a}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {1-4 b}}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}}, a_{k +1}=-\frac {a_{k} a}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\frac {\sqrt {1-4 b}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} a}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\frac {\sqrt {1-4 b}}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}}, a_{k +1}=-\frac {a_{k} a}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}}\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 +\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}}\right ), c_{k +1}=-\frac {c_{k} a}{k^{2}+2 k \left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}-\frac {\sqrt {1-4 b}}{2}}, d_{k +1}=-\frac {d_{k} a}{k^{2}+2 k \left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )+\left (\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}\right )^{2}+b +k +\frac {1}{2}+\frac {\sqrt {1-4 b}}{2}}\right ] \end {array} \]

2.30.2.3 ✓ Mathematica. Time used: 0.065 (sec). Leaf size: 95

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

\begin{align*} y(x)&\to \sqrt {a} \sqrt {x} \left (c_1 \operatorname {Gamma}\left (1-\sqrt {1-4 b}\right ) \operatorname {BesselJ}\left (-\sqrt {1-4 b},2 \sqrt {a} \sqrt {x}\right )+c_2 \operatorname {Gamma}\left (\sqrt {1-4 b}+1\right ) \operatorname {BesselJ}\left (\sqrt {1-4 b},2 \sqrt {a} \sqrt {x}\right )\right ) \end{align*}

2.30.2.4 ✓ Sympy. Time used: 0.195 (sec). Leaf size: 54

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

\[ y{\left (x \right )} = \sqrt {x} \left (C_{1} J_{2 \sqrt {\frac {1}{4} - b}}\left (2 \sqrt {a} \sqrt {x}\right ) + C_{2} Y_{2 \sqrt {\frac {1}{4} - b}}\left (2 \sqrt {a} \sqrt {x}\right )\right ) \]

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)) 
 
('2nd_linear_bessel', '2nd_power_series_regular')