2.20.7 Problem 40

Internal problem ID [13487]
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 : 40
Date solved : Sunday, January 18, 2026 at 08:29:19 PM
CAS classification : [_Riccati]

2.20.7.1 Solved using first_order_ode_riccati

2.354 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=f^{\prime }\left (x \right ) y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+{\mathrm e}^{\lambda x} a \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= f^{\prime }\left (x \right ) y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+{\mathrm e}^{\lambda x} a \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = f^{\prime }\left (x \right ) y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+{\mathrm e}^{\lambda x} a \]

With Riccati ODE standard form

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

Shows that \(f_0(x)={\mathrm e}^{\lambda x} a\), \(f_1(x)=a \,{\mathrm e}^{\lambda x} f \left (x \right )\) and \(f_2(x)=f^{\prime }\left (x \right )\). Let

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

Substituting the above terms back in equation (2) gives

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

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 f \left (x \right )+c_2 f \left (x \right ) \int \frac {{\mathrm e}^{\int \frac {f^{\prime \prime }\left (x \right )+{\mathrm e}^{\lambda x} f \left (x \right ) f^{\prime }\left (x \right ) a}{f^{\prime }\left (x \right )}d x}}{f \left (x \right )^{2}}d x \end{equation}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.20.7.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

5.650 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=f^{\prime }\left (x \right ) y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+{\mathrm e}^{\lambda x} a \\ \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) & ={\mathrm e}^{\lambda x} a\\ f_1(x) & =a \,{\mathrm e}^{\lambda x} f \left (x \right )\\ f_2(x) &=f^{\prime }\left (x \right ) \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -\frac {{\mathrm e}^{-\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \left (\ln \left (x \right ) x f^{\prime }\left (x \right )+f \left (x \right ) \textit {\_Z} \right )\right )} x^{\frac {\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \left (\ln \left (x \right ) x f^{\prime }\left (x \right )+f \left (x \right ) \textit {\_Z} \right )\right )}{\ln \left (x \right )}}}{f \left (x \right )} \]

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

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

Summary of solutions found

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

2.20.7.3 ✓ Maple. Time used: 0.013 (sec). Leaf size: 114

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

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

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) = (diff(f(x),x)*exp( 
lambda*x)*f(x)*a+diff(diff(f(x),x),x))/diff(f(x),x)*diff(y(x),x)-diff(f(x),x)*a 
*exp(lambda*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)-(diff(f(x),x)*y(x)^2+y(x)+a* 
exp(lambda*x)*f(x)*y(x)*x+x^2*a*exp(lambda*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((exp(lambda*x)*f(x)^2*a-2*diff(f(x),x))/f(x),x))*(y+1/f(x))^2] 
   <- successful computation of symmetries. 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful
 

Maple step by step

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

2.20.7.4 ✓ Mathematica. Time used: 42.574 (sec). Leaf size: 319

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

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

2.20.7.5 ✗ Sympy

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

NotImplementedError : The given ODE -a*f(x)*y(x)*exp(lambda_*x) - a*exp(lambda_*x) - y(x)**2*Derivat
 

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')