2.27.6 Problem 6

Internal problem ID [13667]
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 Equations Containing Power Functions. page 213
Problem number : 6
Date solved : Friday, December 19, 2025 at 10:17:27 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.27.6.1 ✓ Maple. Time used: 0.007 (sec). Leaf size: 142

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

\[ y = {\mathrm e}^{-\frac {x \left (a x +b c \right )}{2 \sqrt {a}}} \left (2 \operatorname {hypergeom}\left (\left [-\frac {b^{2} c^{2}-12 a^{{3}/{2}}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {\left (2 a x +b c \right )^{2}}{4 a^{{3}/{2}}}\right ) c_2 a x +\operatorname {hypergeom}\left (\left [-\frac {b^{2} c^{2}-12 a^{{3}/{2}}}{16 a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {\left (2 a x +b c \right )^{2}}{4 a^{{3}/{2}}}\right ) c_2 b c +c_1 \operatorname {hypergeom}\left (\left [-\frac {b^{2} c^{2}-4 a^{{3}/{2}}}{16 a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {\left (2 a x +b c \right )^{2}}{4 a^{{3}/{2}}}\right )\right ) \]

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 
   -> hypergeometric 
      -> heuristic approach 
      -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      <- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Moebius`` i\ 
s resolved 
   <- hypergeometric successful 
<- special function solution successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \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 )=\left (a \,x^{2}+b x c \right ) y \left (x \right ) \\ \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 )+\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} \\ \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 =1..2 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d^{2}}{d x^{2}}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}+\left (6 a_{3}-a_{0} b c \right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-a_{k -1} b c -a_{k -2} a \right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}=0, 6 a_{3}-a_{0} b c =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=0, a_{3}=\frac {a_{0} b c}{6}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-a_{k -1} b c -a_{k -2} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}-a_{k +1} b c -a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=\frac {a_{k +1} b c +a_{k} a}{k^{2}+7 k +12}, a_{2}=0, a_{3}=\frac {a_{0} b c}{6}\right ] \end {array} \]

2.27.6.2 ✓ Mathematica. Time used: 0.031 (sec). Leaf size: 92

ode=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_2 \operatorname {ParabolicCylinderD}\left (-\frac {b^2 c^2}{8 a^{3/2}}-\frac {1}{2},\frac {i (b c+2 a x)}{\sqrt {2} a^{3/4}}\right )+c_1 \operatorname {ParabolicCylinderD}\left (\frac {1}{8} \left (\frac {b^2 c^2}{a^{3/2}}-4\right ),\frac {b c+2 a x}{\sqrt {2} a^{3/4}}\right ) \end{align*}

2.27.6.3 ✗ Sympy

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

False