2.4.5 Problem 26

Internal problem ID [13305]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3-2. Equations with power and exponential functions
Problem number : 26
Date solved : Sunday, January 18, 2026 at 07:11:53 PM
CAS classification : [_Riccati]

2.4.5.1 Solved using first_order_ode_riccati

0.869 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=-\lambda \,{\mathrm e}^{\lambda x} y^{2}+a \,x^{n} y \,{\mathrm e}^{\lambda x}-x^{n} a \\ \end{align*}

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -\lambda \,{\mathrm e}^{\lambda x} y^{2}+a \,x^{n} y \,{\mathrm e}^{\lambda x}-x^{n} a \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-x^{n} a\), \(f_1(x)={\mathrm e}^{\lambda x} x^{n} a\) and \(f_2(x)=-\lambda \,{\mathrm e}^{\lambda x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u \lambda \,{\mathrm e}^{\lambda x}} \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' &=-\lambda ^{2} {\mathrm e}^{\lambda x}\\ f_1 f_2 &=-x^{n} {\mathrm e}^{2 \lambda x} a \lambda \\ f_2^2 f_0 &=-\lambda ^{2} {\mathrm e}^{2 \lambda x} x^{n} a \end{align*}

Substituting the above terms back in equation (2) gives

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

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}^{\lambda x}+c_2 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x {\mathrm e}^{\lambda x} \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = c_1 \lambda \,{\mathrm e}^{\lambda x}+\frac {c_2 \,{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x} {\mathrm e}^{\lambda x}}{\lambda }+c_2 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x \lambda \,{\mathrm e}^{\lambda x} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{-u \lambda \,{\mathrm e}^{\lambda x}} \\ y &= \frac {\left (c_1 \lambda \,{\mathrm e}^{\lambda x}+\frac {c_2 \,{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x} {\mathrm e}^{\lambda x}}{\lambda }+c_2 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x \lambda \,{\mathrm e}^{\lambda x}\right ) {\mathrm e}^{-\lambda x}}{\lambda \left (c_1 \,{\mathrm e}^{\lambda x}+c_2 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x {\mathrm e}^{\lambda x}\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {\left (\lambda \,{\mathrm e}^{\lambda x}+\frac {c_3 \,{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x} {\mathrm e}^{\lambda x}}{\lambda }+c_3 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x \lambda \,{\mathrm e}^{\lambda x}\right ) {\mathrm e}^{-\lambda x}}{\lambda \left ({\mathrm e}^{\lambda x}+c_3 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x {\mathrm e}^{\lambda x}\right )} \]

Summary of solutions found

\begin{align*} y &= \frac {\left (\lambda \,{\mathrm e}^{\lambda x}+\frac {c_3 \,{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x} {\mathrm e}^{\lambda x}}{\lambda }+c_3 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x \lambda \,{\mathrm e}^{\lambda x}\right ) {\mathrm e}^{-\lambda x}}{\lambda \left ({\mathrm e}^{\lambda x}+c_3 \int \frac {{\mathrm e}^{-\lambda x +a \int {\mathrm e}^{\lambda x} x^{n}d x}}{\lambda }d x {\mathrm e}^{\lambda x}\right )} \\ \end{align*}

2.4.5.2 ✓ Maple. Time used: 0.005 (sec). Leaf size: 88

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

\[ y = \frac {c_1 \,{\mathrm e}^{-2 \lambda x +a \int x^{n} {\mathrm e}^{\lambda x}d x}+{\mathrm e}^{-\lambda x} \left (\lambda \int {\mathrm e}^{-\lambda x +a \int x^{n} {\mathrm e}^{\lambda x}d x}d x c_1 +\lambda ^{2}\right )}{\lambda \left (\int {\mathrm e}^{-\lambda x +a \int x^{n} {\mathrm e}^{\lambda x}d x}d x c_1 +\lambda \right )} \]

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 sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (a*x^n*exp(lambda*x) 
+lambda)*diff(y(x),x)-lambda*exp(lambda*x)*a*x^n*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)] 
         <- linear_1 successful 
         Change of variables used: 
            [x = ln(t)/lambda] 
         Linear ODE actually solved: 
            a*(ln(t)/lambda)^n*u(t)-a*(ln(t)/lambda)^n*t*diff(u(t),t)+t*lambda*\ 
diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\lambda \,{\mathrm e}^{\lambda x} y \left (x \right )^{2}+a \,x^{13305} {\mathrm e}^{\lambda x} y \left (x \right )-a \,x^{13305} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\lambda \,{\mathrm e}^{\lambda x} y \left (x \right )^{2}+a \,x^{13305} {\mathrm e}^{\lambda x} y \left (x \right )-a \,x^{13305} \end {array} \]

2.4.5.3 ✓ Mathematica. Time used: 2.685 (sec). Leaf size: 158

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

\begin{align*} y(x)&\to \frac {e^{-2 \lambda x} \left (\exp \left (-\int _1^{e^{x \lambda }}-\frac {a \left (\frac {\log (K[1])}{\lambda }\right )^n}{\lambda }dK[1]\right )+e^{\lambda x} \int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a \left (\frac {\log (K[1])}{\lambda }\right )^n}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1 e^{\lambda x}\right )}{\int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a \left (\frac {\log (K[1])}{\lambda }\right )^n}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1}\\ y(x)&\to e^{\lambda (-x)} \end{align*}

2.4.5.4 ✗ Sympy

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

NotImplementedError : The given ODE -a*x**n*y(x)*exp(lambda_*x) + a*x**n + lambda_*y(x)**2*exp(lambd
 

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')