2.28.18 Problem 28

Internal problem ID [13689]
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 : 28
Date solved : Friday, December 19, 2025 at 10:37:38 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.28.18.1 ✓ Maple. Time used: 0.006 (sec). Leaf size: 98

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

\[ y = {\mathrm e}^{-\frac {c x}{a}} \left (\operatorname {KummerU}\left (\frac {d \,a^{2}-a b c +c^{2}}{2 a^{3}}, \frac {1}{2}, -\frac {\left (x \,a^{2}+b a -2 c \right )^{2}}{2 a^{3}}\right ) c_2 +\operatorname {KummerM}\left (\frac {d \,a^{2}-a b c +c^{2}}{2 a^{3}}, \frac {1}{2}, -\frac {\left (x \,a^{2}+b a -2 c \right )^{2}}{2 a^{3}}\right ) c_1 \right ) \]

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 
   -> hypergeometric 
      -> heuristic approach 
      <- heuristic approach successful 
   <- 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 +b \right ) \left (\frac {d}{d x}y \left (x \right )\right )+\left (c x +d \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 {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..1 \\ {} & {} & 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} x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =\max \left (0, 1-m \right )}{\sum }}a_{k} k \,x^{k -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =\max \left (0, 1-m \right )+m -1}{\sum }}a_{k +1-m} \left (k +1-m \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} b +a_{0} d +2 a_{2}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k +1} \left (k +1\right ) b +a_{k} \left (a k +d \right )+a_{k -1} c \right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} b +a_{0} d +2 a_{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & k^{2} a_{k +2}+\left (a a_{k}+a_{k +1} b +3 a_{k +2}\right ) k +a_{k +1} b +a_{k -1} c +a_{k} d +2 a_{k +2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +1\right )^{2} a_{k +3}+\left (a a_{k +1}+a_{k +2} b +3 a_{k +3}\right ) \left (k +1\right )+a_{k +2} b +a_{k} c +a_{k +1} d +2 a_{k +3}=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 +3}=-\frac {a k a_{k +1}+b k a_{k +2}+a a_{k +1}+2 a_{k +2} b +a_{k} c +a_{k +1} d}{k^{2}+5 k +6}, a_{1} b +a_{0} d +2 a_{2}=0\right ] \end {array} \]

2.28.18.2 ✓ Mathematica. Time used: 0.045 (sec). Leaf size: 132

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

\begin{align*} y(x)&\to e^{\frac {c x}{a}-\frac {a x^2}{2}-b x} \left (c_2 \operatorname {Hypergeometric1F1}\left (\frac {a^3-d a^2+b c a-c^2}{2 a^3},\frac {1}{2},\frac {\left (x a^2+b a-2 c\right )^2}{2 a^3}\right )+c_1 \operatorname {HermiteH}\left (\frac {-a^3+d a^2-b c a+c^2}{a^3},\frac {x a^2+b a-2 c}{\sqrt {2} a^{3/2}}\right )\right ) \end{align*}

2.28.18.3 ✗ Sympy

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

False