2.30.15 Problem 124

Internal problem ID [13785]
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 : 124
Date solved : Thursday, January 01, 2026 at 02:37:00 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.30.15.1 second order bessel ode

0.244 (sec)

\begin{align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-\left (\frac {1}{2}+n \right )^{2}\right ) y&=0 \\ \end{align*}

Entering second order bessel ode solverWriting the ode as

\begin{align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (x^{2}-\frac {1}{4}-n -n^{2}\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 &= 0\\ \beta &= 1\\ n &= -\frac {1}{2}-n\\ \gamma &= 1 \end{align*}

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

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

Summary of solutions found

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

2.30.15.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 19

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

\[ y = c_1 \operatorname {BesselJ}\left (n +\frac {1}{2}, x\right )+c_2 \operatorname {BesselY}\left (n +\frac {1}{2}, x\right ) \]

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 
   <- Bessel successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}\frac {d}{d x}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (x^{2}-\left (n +\frac {1}{2}\right )^{2}\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right )=\frac {\left (4 n^{2}-4 x^{2}+4 n +1\right ) y \left (x \right )}{4 x^{2}}-\frac {\frac {d}{d x}y \left (x \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}{d x}\frac {d}{d x}y \left (x \right )+\frac {\frac {d}{d x}y \left (x \right )}{x}-\frac {\left (4 n^{2}-4 x^{2}+4 n +1\right ) y \left (x \right )}{4 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 )=\frac {1}{x}, P_{3}\left (x \right )=-\frac {4 n^{2}-4 x^{2}+4 n +1}{4 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}}}=-\frac {1}{4}-n^{2}-n \\ {} & \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} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}\frac {d}{d x}y \left (x \right )\right )+4 x \left (\frac {d}{d x}y \left (x \right )\right )+\left (-4 n^{2}+4 x^{2}-4 n -1\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 \cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}\frac {d}{d x}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 (2 r +1+2 n \right ) \left (2 r -1-2 n \right ) x^{r}+a_{1} \left (2 r +3+2 n \right ) \left (2 r +1-2 n \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 r +1+2 n +2 k \right ) \left (2 r -1-2 n +2 k \right )+4 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (2 r +1+2 n \right ) \left (2 r -1-2 n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-n -\frac {1}{2}, n +\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2 r +3+2 n \right ) \left (2 r +1-2 n \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 r +1+2 n +2 k \right ) \left (2 r -1-2 n +2 k \right )+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (2 r +5+2 n +2 k \right ) \left (2 r +3-2 n +2 k \right )+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{\left (2 r +5+2 n +2 k \right ) \left (2 r +3-2 n +2 k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n -\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{\left (4+2 k \right ) \left (-4 n +2+2 k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n -\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n -\frac {1}{2}}, a_{k +2}=-\frac {4 a_{k}}{\left (4+2 k \right ) \left (-4 n +2+2 k \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =n +\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 a_{k}}{\left (4 n +6+2 k \right ) \left (4+2 k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =n +\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +n +\frac {1}{2}}, a_{k +2}=-\frac {4 a_{k}}{\left (4 n +6+2 k \right ) \left (4+2 k \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +n +\frac {1}{2}}\right ), a_{k +2}=-\frac {4 a_{k}}{\left (4+2 k \right ) \left (-4 n +2+2 k \right )}, a_{1}=0, b_{k +2}=-\frac {4 b_{k}}{\left (4 n +6+2 k \right ) \left (4+2 k \right )}, b_{1}=0\right ] \end {array} \]

2.30.15.3 ✓ Mathematica. Time used: 0.152 (sec). Leaf size: 26

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

\begin{align*} y(x)&\to c_1 \operatorname {BesselJ}\left (n+\frac {1}{2},x\right )+c_2 \operatorname {BesselY}\left (n+\frac {1}{2},x\right ) \end{align*}

2.30.15.4 ✓ Sympy. Time used: 0.178 (sec). Leaf size: 32

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

\[ y{\left (x \right )} = C_{1} J_{\frac {\sqrt {\left (2 n + 1\right )^{2}}}{2}}\left (x\right ) + C_{2} Y_{\frac {\sqrt {\left (2 n + 1\right )^{2}}}{2}}\left (x\right ) \]