Problem 8

2.3.8 Problem 8

2.3.8.1 Solved using ﬁrst_order_ode_riccati
2.3.8.2 ✓Maple
2.3.8.3 ✓Mathematica
2.3.8.4 ✗Sympy

Internal problem ID [13288]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3. Equations Containing Exponential Functions
Problem number : 8
Date solved : Sunday, January 18, 2026 at 07:08:31 PM
CAS classiﬁcation : [_Riccati]

2.3.8.1 Solved using ﬁrst_order_ode_riccati

1.089 (sec)

Entering ﬁrst order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = y^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simpliﬁcation) in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )+\left (a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2}\right ) u \left (x \right ) = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = c_1 \,{\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{2 \lambda x} a +b \right )}{4 \lambda \sqrt {a}}} \operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+c_2 \,{\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b +i {\mathrm e}^{4 \lambda x} a}{4 \lambda \sqrt {a}}} \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right ) \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \left (4 \lambda ^{2} \sqrt {a}+2 i \lambda \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{2 \lambda x} a +b \right )+2 i {\mathrm e}^{4 \lambda x} \lambda a \right ) {\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{2 \lambda x} a +b \right )}{4 \lambda \sqrt {a}}} \operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )}{4 \lambda \sqrt {a}}+\frac {i c_1 \,{\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{2 \lambda x} a +b \right )}{4 \lambda \sqrt {a}}} \left (8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}+1\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right ) \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right ) {\mathrm e}^{2 \lambda x}}{16 \lambda \,a^{2}}-\frac {c_2 \left (4 \lambda ^{2} \sqrt {a}+2 i \lambda \,{\mathrm e}^{2 \lambda x} b +4 i {\mathrm e}^{4 \lambda x} \lambda a \right ) {\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b +i {\mathrm e}^{4 \lambda x} a}{4 \lambda \sqrt {a}}} \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right ) \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )}{4 \lambda \sqrt {a}}+4 c_2 \,{\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b +i {\mathrm e}^{4 \lambda x} a}{4 \lambda \sqrt {a}}} \lambda \,{\mathrm e}^{2 \lambda x} a \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+\frac {i c_2 \,{\mathrm e}^{-\frac {4 \lambda ^{2} x \sqrt {a}+i {\mathrm e}^{2 \lambda x} b +i {\mathrm e}^{4 \lambda x} a}{4 \lambda \sqrt {a}}} \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2} \left (24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}+1\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right ) {\mathrm e}^{2 \lambda x}}{48 \lambda \,a^{2}} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= \text {Expression too large to display} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ \text {Expression too large to display} \]

Simplifying the above gives

\begin{align*} y &= \frac {-c_3 \left (\frac {b^{2} \left (i a^{2} \lambda +\frac {\sqrt {a}\, b^{2}}{24}-\frac {a^{{3}/{2}} c}{6}\right ) {\mathrm e}^{2 \lambda x}}{4}+\left (i a^{3} \lambda +\frac {a^{{3}/{2}} b^{2}}{24}-\frac {a^{{5}/{2}} c}{6}\right ) b \,{\mathrm e}^{4 \lambda x}+{\mathrm e}^{6 \lambda x} \left (i a^{4} \lambda +\frac {a^{{5}/{2}} b^{2}}{24}-\frac {a^{{7}/{2}} c}{6}\right )\right ) \operatorname {hypergeom}\left (\left [\frac {56 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+c_3 \left (\left (\frac {i a^{2} b^{2}}{4}-a^{{7}/{2}} \lambda \right ) {\mathrm e}^{2 \lambda x}+i {\mathrm e}^{6 \lambda x} a^{4}+i {\mathrm e}^{4 \lambda x} a^{3} b +\frac {a^{{5}/{2}} b \lambda }{2}\right ) \lambda \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+\left (-\frac {b \left (i a^{2} \lambda +\frac {\sqrt {a}\, b^{2}}{8}-\frac {a^{{3}/{2}} c}{2}\right ) {\mathrm e}^{2 \lambda x}}{4}-\frac {\left (i a^{3} \lambda +\frac {a^{{3}/{2}} b^{2}}{8}-\frac {a^{{5}/{2}} c}{2}\right ) {\mathrm e}^{4 \lambda x}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {40 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+\frac {\left (i {\mathrm e}^{4 \lambda x} a^{3}+\frac {i {\mathrm e}^{2 \lambda x} a^{2} b}{2}+a^{{5}/{2}} \lambda \right ) \operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right ) \lambda }{2}}{a^{{5}/{2}} \left (c_3 \left ({\mathrm e}^{2 \lambda x} a +\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {24 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {8 \lambda \,a^{{3}/{2}}+4 i a c -i b^{2}}{32 \lambda \,a^{{3}/{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{2 \lambda x} a +b \right )^{2}}{8 \lambda \,a^{{3}/{2}}}\right )}{2}\right ) \lambda } \\ \end{align*}

Summary of solutions found

2.3.8.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 578

ode:=diff(y(x),x) = y(x)^2+a*exp(8*lambda*x)+b*exp(6*lambda*x)+c*exp(4*lambda*x)-lambda^2; 
dsolve(ode,y(x), singsol=all);

\begin{align*} \text {Solution too large to show}\end{align*}

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying Riccati 
trying Riccati Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (-a*exp(8*lambda*x)- 
b*exp(6*lambda*x)-c*exp(4*lambda*x)+lambda^2)*y(x), y(x) 
      *** Sublevel 2 *** 
      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 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebi\ 
us 
               <- hyper3 successful: indirect Equivalence to 0F1 under ``^ @ Mo\ 
ebius`` is resolved 
            <- hypergeometric successful 
         <- special function solution successful 
         Change of variables used: 
            [x = ln(t)/lambda] 
         Linear ODE actually solved: 
            (a*t^8+b*t^6+c*t^4-lambda^2)*u(t)+lambda^2*t*diff(u(t),t)+lambda^2*\ 
t^2*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+a \,{\mathrm e}^{8 \lambda x}+b \,{\mathrm e}^{6 \lambda x}+c \,{\mathrm e}^{4 \lambda x}-\lambda ^{2} \end {array} \]

2.3.8.3 ✓ Mathematica. Time used: 1.737 (sec). Leaf size: 1515

ode=D[y[x],x]==y[x]^2+a*Exp[8*\[Lambda]*x]+b*Exp[6*\[Lambda]*x]+c*Exp[4*\[Lambda]*x]-\[Lambda]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} \text {Solution too large to show}\end{align*}

2.3.8.4 ✗ 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(8*lambda_*x) - b*exp(6*lambda_*x) - c*exp(4*lambda_*x) + lambda_**2 - y(x)**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

NotImplementedError : The given ODE -a*exp(8*lambda_*x) - b*exp(6*lambda_*x) - c*exp(4*lambda_*x) +

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)) 
 
('1st_power_series', 'lie_group')

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