2.19.27 Problem 28

Internal problem ID [13475]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.8-1. Equations containing arbitrary functions (but not containing their derivatives).
Problem number : 28
Date solved : Sunday, January 18, 2026 at 08:25:47 PM
CAS classification : [_Riccati]

2.19.27.1 Solved using first_order_ode_riccati

2.111 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=-f \left (x \right )\), \(f_1(x)=a x f \left (x \right ) \ln \left (x \right )-f \left (x \right ) a x\) and \(f_2(x)=-a \ln \left (x \right )\). Let

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

Substituting the above terms back in equation (2) gives

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

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 x \left (-1+\ln \left (x \right )\right )+c_2 x \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = c_1 \left (-1+\ln \left (x \right )\right )+c_1 +c_2 \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_2 \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +\frac {c_2 \,{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x \left (-1+\ln \left (x \right )\right )} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{-u a \ln \left (x \right )} \\ y &= \frac {c_1 \left (-1+\ln \left (x \right )\right )+c_1 +c_2 \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_2 \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +\frac {c_2 \,{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x \left (-1+\ln \left (x \right )\right )}}{a \ln \left (x \right ) \left (c_1 x \left (-1+\ln \left (x \right )\right )+c_2 x \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {\ln \left (x \right )+c_3 \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +c_3 \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x +\frac {c_3 \,{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x \left (-1+\ln \left (x \right )\right )}}{a \ln \left (x \right ) \left (x \left (-1+\ln \left (x \right )\right )+c_3 x \left (-1+\ln \left (x \right )\right ) \int \frac {{\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}-f \left (x \right ) \ln \left (x \right ) a \,x^{2}+1}{x \ln \left (x \right )}d x}}{x^{2} \left (-1+\ln \left (x \right )\right )^{2}}d x \right )} \]

Summary of solutions found

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

2.19.27.2 ✓ Maple. Time used: 0.033 (sec). Leaf size: 227

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

\[ y = \frac {-x \left (\ln \left (x \right )-1\right ) {\mathrm e}^{\int \frac {f \left (x \right ) \ln \left (x \right )^{2} a \,x^{2}+\left (-2 f \left (x \right ) a \,x^{2}-2\right ) \ln \left (x \right )+f \left (x \right ) a \,x^{2}}{x \left (\ln \left (x \right )-1\right )}d x}+c_1 a -\int \ln \left (x \right ) {\mathrm e}^{a \int \frac {x f \left (x \right ) \ln \left (x \right )^{2}}{\ln \left (x \right )-1}d x -2 a \int \frac {x f \left (x \right ) \ln \left (x \right )}{\ln \left (x \right )-1}d x +a \int \frac {x f \left (x \right )}{\ln \left (x \right )-1}d x -2 \int \frac {\ln \left (x \right )}{x \left (\ln \left (x \right )-1\right )}d x}d x}{a x \left (\ln \left (x \right )-1\right ) \left (c_1 a -\int \ln \left (x \right ) {\mathrm e}^{a \int \frac {x f \left (x \right ) \ln \left (x \right )^{2}}{\ln \left (x \right )-1}d x -2 a \int \frac {x f \left (x \right ) \ln \left (x \right )}{\ln \left (x \right )-1}d x +a \int \frac {x f \left (x \right )}{\ln \left (x \right )-1}d x -2 \int \frac {\ln \left (x \right )}{x \left (\ln \left (x \right )-1\right )}d x}d x \right )} \]

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) = (f(x)*ln(x)^2*a*x^2- 
f(x)*ln(x)*a*x^2+1)/x/ln(x)*diff(y(x),x)-ln(x)*a*f(x)*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 
         -> 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 a symmetry of the form [xi=0, eta=F(x)] 
            trying 2nd order exact linear 
            trying symmetries linear in x and y(x) 
            trying to convert to a linear ODE with constant coefficients 
      <- unable to find a useful change of variables 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying 2nd order, integrating factor of the form mu(x,y) 
         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 
            -> Heun: Equivalence to the GHE or one of its 4 confluent cases und\ 
er a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = e\ 
xp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               trying a symmetry of the form [xi=0, eta=F(x)] 
               trying 2nd order exact linear 
               trying symmetries linear in x and y(x) 
               trying to convert to a linear ODE with constant coefficients 
         <- unable to find a useful change of variables 
            trying a symmetry of the form [xi=0, eta=F(x)] 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE, diff(y(x),x)-(-ln(x)*a*y(x)^2+y(x)+(f(x)*ln 
(x)*a*x-a*x*f(x))*y(x)*x-x^2*f(x))/x, y(x), explicit 
      *** Sublevel 2 *** 
      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 inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
trying inverse_Riccati 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
   -> Computing symmetries using: way = 4 
   -> Computing symmetries using: way = 2 
   -> Computing symmetries using: way = 6 
[0, exp(-Int((f(x)*ln(x)^2*a*x^2-2*f(x)*ln(x)*a*x^2+f(x)*a*x^2-2*ln(x))/x/(ln(x 
)-1),x))*(y-1/a/x/(ln(x)-1))^2] 
   <- successful computation of symmetries. 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful
 

Maple step by step

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

2.19.27.3 ✗ Mathematica

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

Not solved

2.19.27.4 ✗ Sympy

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

NotImplementedError : The given ODE -a*x*f(x)*y(x)*log(x) + a*x*f(x)*y(x) + a*y(x)**2*log(x) + f(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)) 
 
('factorable', 'lie_group')