2.28.3 Problem 13

Internal problem ID [13674]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-2
Problem number : 13
Date solved : Friday, December 19, 2025 at 10:23:51 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.28.3.1 ✓ Maple. Time used: 0.029 (sec). Leaf size: 74

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

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

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 
<- No Liouvillian solutions exists 
-> Trying a solution in terms of special functions: 
   -> Bessel 
   -> elliptic 
   -> Legendre 
   -> Kummer 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
   <- Kummer 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 )+a \left (\frac {d}{d x}y \left (x \right )\right )-\left (b \,x^{2}+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 )=-a \left (\frac {d}{d x}y \left (x \right )\right )+\left (b \,x^{2}+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 )+a \left (\frac {d}{d x}y \left (x \right )\right )+\left (-b \,x^{2}-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 =0..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}{d x}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k} k \,x^{k -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +1} \left (k +1\right ) 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} \\ {} & {} & a_{1} a -a_{0} c +2 a_{2}+\left (2 a a_{2}-a_{1} c +6 a_{3}\right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a a_{k +1} \left (k +1\right )-a_{k} c -a_{k -2} b \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}+a_{1} a -a_{0} c =0, 2 a a_{2}-a_{1} c +6 a_{3}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=-\frac {a_{1} a}{2}+\frac {a_{0} c}{2}, a_{3}=\frac {1}{6} a_{1} a^{2}-\frac {1}{6} a_{0} a c +\frac {1}{6} a_{1} c \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 a_{k +1} k +a a_{k +1}-a_{k -2} b -a_{k} c =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 a_{k +3} \left (k +2\right )+a a_{k +3}-a_{k} b -a_{k +2} c =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 a_{k +3}+3 a a_{k +3}-a_{k} b -a_{k +2} c}{k^{2}+7 k +12}, a_{2}=-\frac {a_{1} a}{2}+\frac {a_{0} c}{2}, a_{3}=\frac {1}{6} a_{1} a^{2}-\frac {1}{6} a_{0} a c +\frac {1}{6} a_{1} c \right ] \end {array} \]

2.28.3.2 ✓ Mathematica. Time used: 0.031 (sec). Leaf size: 96

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

\begin{align*} y(x)&\to e^{-\frac {1}{2} x \left (a+\sqrt {b} x\right )} \left (c_1 \operatorname {HermiteH}\left (\frac {-a^2-4 \left (c+\sqrt {b}\right )}{8 \sqrt {b}},\sqrt [4]{b} x\right )+c_2 \operatorname {Hypergeometric1F1}\left (\frac {a^2+4 \left (c+\sqrt {b}\right )}{16 \sqrt {b}},\frac {1}{2},\sqrt {b} x^2\right )\right ) \end{align*}

2.28.3.3 ✗ Sympy

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

False