2.30.8 Problem 117

Internal problem ID [13778]
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 : 117
Date solved : Friday, December 19, 2025 at 12:21:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} x^{2} y^{\prime \prime }-\left (a^{2} x^{4}+a \,x^{2} \left (2 b -1\right )+b \left (b +1\right )\right ) y&=0 \\ \end{align*}

2.30.8.1 ✓ Maple. Time used: 0.004 (sec). Leaf size: 38

ode:=x^2*diff(diff(y(x),x),x)-(a^2*x^4+a*(2*b-1)*x^2+b*(b+1))*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = {\mathrm e}^{-\frac {a \,x^{2}}{2}} x^{-b} \left (\Gamma \left (b +\frac {1}{2}\right ) c_2 -\Gamma \left (b +\frac {1}{2}, -a \,x^{2}\right ) c_2 +c_1 \right ) \]

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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
<- Kovacics algorithm 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^{2} x^{4}+a \left (2 b -1\right ) x^{2}+b \left (b +1\right )\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^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+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^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+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^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+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^{2}-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^{2} x^{4}-2 a \,x^{2} b +a \,x^{2}-b^{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 y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..4 \\ {} & {} & 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 +b \right ) \left (-r +1+b \right ) x^{r}-a_{1} \left (r +1+b \right ) \left (-r +b \right ) x^{1+r}+\left (-a_{2} \left (r +2+b \right ) \left (-r -1+b \right )-a_{0} a \left (2 b -1\right )\right ) x^{2+r}+\left (-a_{3} \left (r +3+b \right ) \left (-r -2+b \right )-a_{1} a \left (2 b -1\right )\right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (-a_{k} \left (r +k +b \right ) \left (-r +1-k +b \right )-a_{k -2} a \left (2 b -1\right )-a_{k -4} a^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (r +b \right ) \left (-r +1+b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-b , b +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 (r +1+b \right ) \left (-r +b \right )=0, -a_{2} \left (r +2+b \right ) \left (-r -1+b \right )-a_{0} a \left (2 b -1\right )=0, -a_{3} \left (r +3+b \right ) \left (-r -2+b \right )-a_{1} a \left (2 b -1\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{b^{2}-r^{2}+b -3 r -2}, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k} \left (r +k +b \right ) \left (-r +1-k +b \right )-a \left (a a_{k -4}+2 b a_{k -2}-a_{k -2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & -a_{k +4} \left (r +k +4+b \right ) \left (-r -3-k +b \right )-a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )}{\left (r +k +4+b \right ) \left (-r -3-k +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b \\ {} & {} & a_{k +4}=-\frac {a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b}, a_{k +4}=-\frac {a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )}, a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{-2+4 b}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =b +1 \\ {} & {} & a_{k +4}=-\frac {a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )}{\left (2 b +5+k \right ) \left (-k -4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =b +1 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +b +1}, a_{k +4}=-\frac {a \left (a a_{k}+2 b a_{k +2}-a_{k +2}\right )}{\left (2 b +5+k \right ) \left (-k -4\right )}, a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{b^{2}-\left (b +1\right )^{2}-2 b -5}, a_{3}=0\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 -b}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +b +1}\right ), c_{k +4}=-\frac {a \left (a c_{k}+2 b c_{k +2}-c_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )}, c_{1}=0, c_{2}=-\frac {c_{0} a \left (2 b -1\right )}{-2+4 b}, c_{3}=0, d_{k +4}=-\frac {a \left (a d_{k}+2 b d_{k +2}-d_{k +2}\right )}{\left (2 b +5+k \right ) \left (-k -4\right )}, d_{1}=0, d_{2}=-\frac {d_{0} a \left (2 b -1\right )}{b^{2}-\left (b +1\right )^{2}-2 b -5}, d_{3}=0\right ] \end {array} \]

2.30.8.2 ✓ Mathematica. Time used: 0.187 (sec). Leaf size: 66

ode=x^2*D[y[x],{x,2}]-(a^2*x^4+a*(2*b-1)*x^2+b*(b+1))*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {1}{2} e^{-\frac {a x^2}{2}} x^{-b} \left (a c_2 x^{2 b+3} \left (-a x^2\right )^{-b-\frac {3}{2}} \Gamma \left (b+\frac {1}{2},-a x^2\right )+2 c_1\right ) \end{align*}

2.30.8.3 ✗ Sympy

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**2*x**4 + a*x**2*(2*b - 1) + b*(b + 1))*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

ValueError : Expected Expr or iterable but got None
 

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_power_series_regular',)