2.8.1 Problem 10

Internal problem ID [13353]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-2
Problem number : 10
Date solved : Sunday, January 18, 2026 at 07:28:59 PM
CAS classification : [_Riccati]

2.8.1.1 Solved using first_order_ode_riccati

0.913 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+a \ln \left (\beta x \right ) y-a b \ln \left (\beta x \right )-b^{2} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y^{2}+a \ln \left (\beta x \right ) y-a b \ln \left (\beta x \right )-b^{2} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = y^{2}+a \ln \left (\beta x \right ) y-a b \ln \left (\beta x \right )-b^{2} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-a b \ln \left (\beta x \right )-b^{2}\), \(f_1(x)=a \ln \left (\beta x \right )\) 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 &=a \ln \left (\beta x \right )\\ f_2^2 f_0 &=-a b \ln \left (\beta x \right )-b^{2} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )-a \ln \left (\beta x \right ) u^{\prime }\left (x \right )+\left (-a b \ln \left (\beta x \right )-b^{2}\right ) u \left (x \right ) = 0 \]

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 ) = \left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -c_1 \right ) {\mathrm e}^{-b x} c_2 \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{b x} c_2 -\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -c_1 \right ) {\mathrm e}^{-b x} c_2 b \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= -\frac {\left (-\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{b x} c_2 -\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -c_1 \right ) {\mathrm e}^{-b x} c_2 b \right ) {\mathrm e}^{b x}}{\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -c_1 \right ) c_2} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (-\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{b x} c_3 -\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -1\right ) {\mathrm e}^{-b x} c_3 b \right ) {\mathrm e}^{b x}}{\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -1\right ) c_3} \]

Summary of solutions found

\begin{align*} y &= -\frac {\left (-\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{b x} c_3 -\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -1\right ) {\mathrm e}^{-b x} c_3 b \right ) {\mathrm e}^{b x}}{\left (\int -\left (\beta x \right )^{a x} {\mathrm e}^{-a x} {\mathrm e}^{2 b x}d x -1\right ) c_3} \\ \end{align*}

2.8.1.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.048 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+a \ln \left (\beta x \right ) y-a b \ln \left (\beta x \right )-b^{2} \\ \end{align*}

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

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

Shows that

\begin{align*} f_0(x) & =-a b \ln \left (\beta x \right )-b^{2}\\ f_1(x) & =a \ln \left (\beta x \right )\\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = b \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

\[ y = b +\frac {{\mathrm e}^{\ln \left (\beta x \right ) a x -a x +2 b x}}{c_1 -\int {\mathrm e}^{\ln \left (\beta x \right ) a x -a x +2 b x}d x} \]

Summary of solutions found

\begin{align*} y &= b +\frac {{\mathrm e}^{\ln \left (\beta x \right ) a x -a x +2 b x}}{c_1 -\int {\mathrm e}^{\ln \left (\beta x \right ) a x -a x +2 b x}d x} \\ \end{align*}

2.8.1.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 48

ode:=diff(y(x),x) = y(x)^2+a*ln(beta*x)*y(x)-a*b*ln(beta*x)-b^2; 
dsolve(ode,y(x), singsol=all);
 

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

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: 
   <- Riccati particular case Kamke (b) successful
 

Maple step by step

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

2.8.1.4 ✓ Mathematica. Time used: 0.369 (sec). Leaf size: 187

ode=D[y[x],x]==y[x]^2+a*Log[\[Beta]*x]*y[x]-a*b*Log[\[Beta]*x]-b^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^x\frac {e^{2 b K[1]-a K[1]} (\beta K[1])^{a K[1]} (b+a \log (\beta K[1])+y(x))}{a (b-y(x))}dK[1]+\int _1^{y(x)}\left (\frac {e^{2 b x-a x} (x \beta )^{a x}}{a (K[2]-b)^2}-\int _1^x\left (\frac {e^{2 b K[1]-a K[1]} (b+K[2]+a \log (\beta K[1])) (\beta K[1])^{a K[1]}}{a (b-K[2])^2}+\frac {e^{2 b K[1]-a K[1]} (\beta K[1])^{a K[1]}}{a (b-K[2])}\right )dK[1]\right )dK[2]=c_1,y(x)\right ] \]

2.8.1.5 ✓ Sympy. Time used: 3.280 (sec). Leaf size: 85

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

\[ y{\left (x \right )} = \frac {\left (C_{1} b e^{a x} - b e^{a x} \int e^{- a x} e^{2 b x} e^{a x \log {\left (\beta x \right )}}\, dx + e^{x \left (a \log {\left (\beta x \right )} + 2 b\right )}\right ) e^{- a x}}{C_{1} - \int e^{- a x} e^{2 b x} e^{a x \log {\left (\beta x \right )}}\, dx} \]

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)) 
 
('lie_group',)