2.3.7 Problem 7

Internal problem ID [13287]
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 : 7
Date solved : Friday, December 19, 2025 at 02:40:45 AM
CAS classification : [_Riccati]

\begin{align*} y^{\prime }&=y^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4} \\ \end{align*}

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = y^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4}\), \(f_1(x)=0\) 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}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{n}-\frac {\lambda ^{2}}{4} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) {\mathrm e}^{-\frac {\lambda x}{2}}+c_2 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \left ({\mathrm e}^{\frac {\lambda x}{2}}+b \,{\mathrm e}^{-\frac {\lambda x}{2}}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}+1\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2} {\mathrm e}^{\lambda x} {\mathrm e}^{-\frac {\lambda x}{2}}}{\left (n +1\right ) \lambda \left ({\mathrm e}^{\lambda x}+b \right )}-\frac {c_1 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \lambda \,{\mathrm e}^{-\frac {\lambda x}{2}}}{2}-\frac {c_2 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}+1\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2} {\mathrm e}^{\lambda x} \left ({\mathrm e}^{\frac {\lambda x}{2}}+b \,{\mathrm e}^{-\frac {\lambda x}{2}}\right )}{\left (n +3\right ) \lambda \left ({\mathrm e}^{\lambda x}+b \right )}+c_2 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \left (\frac {\lambda \,{\mathrm e}^{\frac {\lambda x}{2}}}{2}-\frac {b \lambda \,{\mathrm e}^{-\frac {\lambda x}{2}}}{2}\right ) \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

\[ y = -\frac {-\frac {\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}+1\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2} {\mathrm e}^{\lambda x} {\mathrm e}^{-\frac {\lambda x}{2}}}{\left (n +1\right ) \lambda \left ({\mathrm e}^{\lambda x}+b \right )}-\frac {\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \lambda \,{\mathrm e}^{-\frac {\lambda x}{2}}}{2}-\frac {c_3 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}+1\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2} {\mathrm e}^{\lambda x} \left ({\mathrm e}^{\frac {\lambda x}{2}}+b \,{\mathrm e}^{-\frac {\lambda x}{2}}\right )}{\left (n +3\right ) \lambda \left ({\mathrm e}^{\lambda x}+b \right )}+c_3 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \left (\frac {\lambda \,{\mathrm e}^{\frac {\lambda x}{2}}}{2}-\frac {b \lambda \,{\mathrm e}^{-\frac {\lambda x}{2}}}{2}\right )}{\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) {\mathrm e}^{-\frac {\lambda x}{2}}+c_3 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a \left ({\mathrm e}^{\lambda x}+b \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) \left ({\mathrm e}^{\frac {\lambda x}{2}}+b \,{\mathrm e}^{-\frac {\lambda x}{2}}\right )} \]

Simplifying the above gives

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The solution

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

was found not to satisfy the ode or the IC. Hence it is removed.

2.3.7.1 ✓ Maple. Time used: 0.004 (sec). Leaf size: 1514

ode:=diff(y(x),x) = y(x)^2+a*exp(2*lambda*x)*(exp(lambda*x)+b)^n-1/4*lambda^2; 
dsolve(ode,y(x), singsol=all);
 

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

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(2*lambda*x)* 
(exp(lambda*x)+b)^n+1/4*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 symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
            -> 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 0F1 ODE 
            <- Whittaker successful 
         <- special function solution successful 
         Change of variables used: 
            [x = ln(t)/lambda] 
         Linear ODE actually solved: 
            (4*a*t^2*(t+b)^n-lambda^2)*u(t)+4*lambda^2*t*diff(u(t),t)+4*lambda^\ 
2*t^2*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{13287}-\frac {\lambda ^{2}}{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+a \,{\mathrm e}^{2 \lambda x} \left ({\mathrm e}^{\lambda x}+b \right )^{13287}-\frac {\lambda ^{2}}{4} \end {array} \]

2.3.7.2 ✗ Mathematica

ode=D[y[x],x]==y[x]^2+a*Exp[2*\[Lambda]*x]*(Exp[\[Lambda]*x]+b)^n-1/4*\[Lambda]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.3.7.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
lambda_ = symbols("lambda_") 
n = symbols("n") 
y = Function("y") 
ode = Eq(-a*(b + exp(lambda_*x))**n*exp(2*lambda_*x) + lambda_**2/4 - y(x)**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -a*(b + exp(lambda_*x))**n*exp(2*lambda_*x) + lambda_**2/4 - y(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')