2.35.5 Problem 5

Internal problem ID [13929]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with exponential functions
Problem number : 5
Date solved : Friday, December 19, 2025 at 08:50:21 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

2.35.5.1 ✓ Maple. Time used: 0.060 (sec). Leaf size: 73

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

\[ y = {\mathrm e}^{-\frac {\lambda x}{2}} \left (c_2 \operatorname {WhittakerW}\left (-\frac {b}{2 \lambda \sqrt {a}}, \frac {\sqrt {c}}{\lambda }, \frac {2 \sqrt {a}\, {\mathrm e}^{\lambda x}}{\lambda }\right )+c_1 \operatorname {WhittakerM}\left (-\frac {b}{2 \lambda \sqrt {a}}, \frac {\sqrt {c}}{\lambda }, \frac {2 \sqrt {a}\, {\mathrm e}^{\lambda x}}{\lambda }\right )\right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\ 
 dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   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 
   Change of variables used: 
      [x = ln(t)/lambda] 
   Linear ODE actually solved: 
      (-a*t^2-b*t-c)*u(t)+lambda^2*t*diff(u(t),t)+lambda^2*t^2*diff(diff(u(t),t\ 
),t) = 0 
<- change of variables successful
 

2.35.5.2 ✓ Mathematica. Time used: 0.311 (sec). Leaf size: 145

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

\begin{align*} y(x)&\to e^{-\frac {\sqrt {a} e^{\lambda x}}{\lambda }} \left (e^{\lambda x}\right )^{\frac {\sqrt {c}}{\lambda }} \left (c_1 \operatorname {HypergeometricU}\left (\frac {\frac {b}{\sqrt {a}}+\lambda +2 \sqrt {c}}{2 \lambda },\frac {2 \sqrt {c}}{\lambda }+1,\frac {2 \sqrt {a} e^{x \lambda }}{\lambda }\right )+c_2 L_{-\frac {\frac {b}{\sqrt {a}}+\lambda +2 \sqrt {c}}{2 \lambda }}^{\frac {2 \sqrt {c}}{\lambda }}\left (\frac {2 \sqrt {a} e^{x \lambda }}{\lambda }\right )\right ) \end{align*}

2.35.5.3 ✗ Sympy

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

False