2.30.6 Problem 115

Internal problem ID [13776]
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 : 115
Date solved : Friday, December 19, 2025 at 12:19:24 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.30.6.1 ✓ Maple. Time used: 0.029 (sec). Leaf size: 57

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

\[ y = c_1 \operatorname {WhittakerM}\left (-\frac {i b}{2 \sqrt {a}}, \frac {\sqrt {1-4 c}}{2}, 2 i \sqrt {a}\, x \right )+c_2 \operatorname {WhittakerW}\left (-\frac {i b}{2 \sqrt {a}}, \frac {\sqrt {1-4 c}}{2}, 2 i \sqrt {a}\, x \right ) \]

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 
   -> elliptic 
   -> Legendre 
   -> Whittaker 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
   <- Whittaker 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^{2}+b x +c \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^{2}+b x +c \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^{2}+b x +c \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^{2}+b x +c}{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}}}=c \\ {} & \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^{2}+b x +c \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^{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}+c -r \right ) x^{r}+\left (\left (r^{2}+c +r \right ) a_{1}+a_{0} b \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+c -k -r \right )+a_{k -1} b +a_{k -2} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+c -r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}, \frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (r^{2}+c +r \right ) a_{1}+a_{0} b =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} b}{r^{2}+c +r} \\ \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}+c -r \right ) a_{k}+a_{k -2} a +a_{k -1} b =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+\left (2 r -1\right ) \left (k +2\right )+r^{2}+c -r \right ) a_{k +2}+a_{k} a +a_{k +1} b =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+2 k r +r^{2}+c +3 k +3 r +2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+2 k \left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k -\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}, a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+2 k \left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k -\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}}, a_{1}=-\frac {a_{0} b}{\left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c -\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\sqrt {1-4 c}}{2}+\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+2 k \left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k +\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {\sqrt {1-4 c}}{2}+\frac {1}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}, a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+2 k \left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k +\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}}, a_{1}=-\frac {a_{0} b}{\left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}\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 -\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}\right ), d_{k +2}=-\frac {a d_{k}+b d_{k +1}}{k^{2}+2 k \left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k -\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}}, d_{1}=-\frac {d_{0} b}{\left (-\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c -\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}, e_{k +2}=-\frac {a e_{k}+b e_{k +1}}{k^{2}+2 k \left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )+\left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +3 k +\frac {3 \sqrt {1-4 c}}{2}+\frac {7}{2}}, e_{1}=-\frac {e_{0} b}{\left (\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}\right )^{2}+c +\frac {\sqrt {1-4 c}}{2}+\frac {1}{2}}\right ] \end {array} \]

2.30.6.2 ✓ Mathematica. Time used: 0.026 (sec). Leaf size: 88

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

\begin{align*} y(x)&\to c_1 M_{-\frac {i b}{2 \sqrt {a}},-\frac {1}{2} i \sqrt {4 c-1}}\left (2 i \sqrt {a} x\right )+c_2 W_{-\frac {i b}{2 \sqrt {a}},-\frac {1}{2} i \sqrt {4 c-1}}\left (2 i \sqrt {a} x\right ) \end{align*}

2.30.6.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq(x**2*Derivative(y(x), (x, 2)) + (a*x**2 + b*x + c)*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',)