2.20.2 Problem 35

Internal problem ID [13482]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.8-2. Equations containing arbitrary functions and their derivatives.
Problem number : 35
Date solved : Wednesday, December 31, 2025 at 09:49:52 PM
CAS classification : [_Riccati]

2.20.2.1 Solved using first_order_ode_riccati

12.539 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \textit {the\_rhs} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=g^{\prime }\left (x \right )\), \(f_1(x)=-f \left (x \right ) g \left (x \right )\) and \(f_2(x)=f \left (x \right )\). Let

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

Substituting the above terms back in equation (2) gives

\[ f \left (x \right ) u^{\prime \prime }\left (x \right )-\left (f^{\prime }\left (x \right )-f \left (x \right )^{2} g \left (x \right )\right ) u^{\prime }\left (x \right )+f \left (x \right )^{2} g^{\prime }\left (x \right ) u \left (x \right ) = 0 \]

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 \,{\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}+c_2 \int f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x}d x {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -c_1 f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}+c_2 f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x} {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}-c_2 \int f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x}d x f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} \end{equation}

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

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

Doing change of constants, the above solution becomes

\[ y = -\frac {-f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}+c_3 f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x} {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}-c_3 \int f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x}d x f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}}{f \left (x \right ) \left ({\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}+c_3 \int f \left (x \right ) {\mathrm e}^{-\int -f \left (x \right ) g \left (x \right )d x}d x {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x}\right )} \]

Summary of solutions found

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

2.20.2.2 ✗ Maple

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

\[ \text {No solution found} \]

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)^2*g(x)+diff(f 
(x),x))/f(x)*diff(y(x),x)-f(x)*diff(g(x),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 
      <- 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 
         <- 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)-(f(x)*y(x)^2+y(x)-f(x)*g(x)*y( 
x)*x+x^2*diff(g(x),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
 

Maple step by step

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

2.20.2.3 ✗ Mathematica

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

Not solved

2.20.2.4 ✗ Sympy

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

NotImplementedError : The given ODE f(x)*g(x)*y(x) - f(x)*y(x)**2 - Derivative(g(x), x) + Derivative(y(x), x) cannot be solved by the lie group method