2.19.1 Problem 1

Internal problem ID [13449]
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 : 1
Date solved : Wednesday, December 31, 2025 at 09:38:21 PM
CAS classification : [_Riccati]

2.19.1.1 Solved using first_order_ode_riccati

11.152 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

Substituting the above terms back in equation (2) gives

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

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

Taking derivative gives

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.19.1.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.142 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+f \left (x \right ) y-a^{2}-a f \left (x \right ) \\ \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 f \left (x \right )-a^{2}\\ f_1(x) & =f \left (x \right )\\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = a \]

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

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

Summary of solutions found

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

2.19.1.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 36

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

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

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

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

2.19.1.4 ✓ Mathematica. Time used: 0.306 (sec). Leaf size: 166

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

\[ \text {Solve}\left [\int _1^x\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right ) (a+f(K[2])+y(x))}{a-y(x)}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x(-2 a-f(K[1]))dK[1]\right )}{(K[3]-a)^2}-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right ) (a+f(K[2])+K[3])}{(a-K[3])^2}+\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right )}{a-K[3]}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]

2.19.1.5 ✗ Sympy

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

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