2.8.10 Problem 19

Internal problem ID [13362]
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 : 19
Date solved : Sunday, January 18, 2026 at 07:30:51 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

2.8.10.1 Solved using first_order_ode_riccati

1.196 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=\frac {a \,b^{2} x^{n} \ln \left (x \right )^{2}}{x}-\frac {b}{x}\), \(f_1(x)=\frac {2 b \ln \left (x \right ) a \,x^{n}}{x}\) and \(f_2(x)=\frac {a \,x^{n}}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a \,x^{n}}{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' &=\frac {a \,x^{n} n}{x^{2}}-\frac {a \,x^{n}}{x^{2}}\\ f_1 f_2 &=\frac {2 b \ln \left (x \right ) a^{2} x^{2 n}}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} x^{2 n} \left (\frac {a \,b^{2} x^{n} \ln \left (x \right )^{2}}{x}-\frac {b}{x}\right )}{x^{2}} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 \,x^{\frac {x^{n} a b}{n}} {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}+c_2 \,x^{\frac {n^{2}+a b \,x^{n}}{n}} {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}} \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = c_1 \,x^{\frac {x^{n} a b}{n}} \left (\frac {b \ln \left (x \right ) a \,x^{n}}{x}+\frac {a b \,x^{n}}{n x}\right ) {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}-\frac {c_1 \,x^{\frac {x^{n} a b}{n}} a b \,x^{n} {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}}{n x}+c_2 \,x^{\frac {n^{2}+a b \,x^{n}}{n}} \left (\frac {b \ln \left (x \right ) a \,x^{n}}{x}+\frac {n^{2}+a b \,x^{n}}{n x}\right ) {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}-\frac {c_2 \,x^{\frac {n^{2}+a b \,x^{n}}{n}} a b \,x^{n} {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}}{n x} \end{equation}

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

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

Doing change of constants, the above solution becomes

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

Simplifying the above gives

\begin{align*} y &= \frac {-\ln \left (x \right ) x^{n} c_3 a b -a b \ln \left (x \right )-c_3 n}{a \left (x^{n} c_3 +1\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-\ln \left (x \right ) x^{n} c_3 a b -a b \ln \left (x \right )-c_3 n}{a \left (x^{n} c_3 +1\right )} \\ \end{align*}

2.8.10.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.055 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime } x&=a \,x^{n} \left (y+b \ln \left (x \right )\right )^{2}-b \\ \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) & =\frac {a \,b^{2} x^{n} \ln \left (x \right )^{2}}{x}-\frac {b}{x}\\ f_1(x) & =\frac {2 b \ln \left (x \right ) a \,x^{n}}{x}\\ f_2(x) &=\frac {a \,x^{n}}{x} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -b \ln \left (x \right ) \]

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

\[ y = -b \ln \left (x \right )+\frac {n}{c_1 n -x^{n} a} \]

Summary of solutions found

\begin{align*} y &= -b \ln \left (x \right )+\frac {n}{c_1 n -x^{n} a} \\ \end{align*}

2.8.10.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 24

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

\[ y = -b \ln \left (x \right )+\frac {n}{c_1 n -a \,x^{n}} \]

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 (d) successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{13362} \left (y \left (x \right )+b \ln \left (x \right )\right )^{2}-b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{13362} \left (y \left (x \right )+b \ln \left (x \right )\right )^{2}-b}{x} \end {array} \]

2.8.10.4 ✓ Mathematica. Time used: 0.28 (sec). Leaf size: 35

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

\begin{align*} y(x)&\to -b \log (x)+\frac {n}{-a x^n+c_1 n}\\ y(x)&\to -b \log (x) \end{align*}

2.8.10.5 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*b**2*x**n*log(x)**2 + 2*a*b*x**n*y(x)*l
 

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)) 
 
('factorable', 'lie_group')