2.20.4 Problem 37
Internal
problem
ID
[13484]
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
:
37
Date
solved
:
Wednesday, December 31, 2025 at 09:50:39 PM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
2.20.4.1 Solved using first_order_ode_riccati
16.357 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime }&=g \left (x \right ) \left (y-f \left (x \right )\right )^{2}+f^{\prime }\left (x \right ) \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= f \left (x \right )^{2} g \left (x \right )-2 f \left (x \right ) g \left (x \right ) y+g \left (x \right ) y^{2}+f^{\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 \(f_0(x)=f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\), \(f_1(x)=-2 f \left (x \right ) g \left (x \right )\) and \(f_2(x)=g \left (x \right )\). Let \begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u g \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' &=g^{\prime }\left (x \right )\\ f_1 f_2 &=-2 f \left (x \right ) g \left (x \right )^{2}\\ f_2^2 f_0 &=g \left (x \right )^{2} \left (f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\right ) \end{align*}
Substituting the above terms back in equation (2) gives
\[
g \left (x \right ) u^{\prime \prime }\left (x \right )-\left (g^{\prime }\left (x \right )-2 f \left (x \right ) g \left (x \right )^{2}\right ) u^{\prime }\left (x \right )+g \left (x \right )^{2} \left (f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\right ) u \left (x \right ) = 0
\]
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 (c_1 +\int g \left (x \right )d x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_2
\end{equation}
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_2 -\left (c_1 +\int g \left (x \right )d x \right ) f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_2
\end{equation}
Substituting equations (3,4) into (1) results in \begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{u g \left (x \right )} \\
y &= -\frac {\left (g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_2 -\left (c_1 +\int g \left (x \right )d x \right ) f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_2 \right ) {\mathrm e}^{\int f \left (x \right ) g \left (x \right )d x}}{g \left (x \right ) \left (c_1 +\int g \left (x \right )d x \right ) c_2} \\
\end{align*}
Doing
change of constants, the above solution becomes \[
y = -\frac {\left (g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_3 -\left (1+\int g \left (x \right )d x \right ) f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_3 \right ) {\mathrm e}^{\int f \left (x \right ) g \left (x \right )d x}}{g \left (x \right ) \left (1+\int g \left (x \right )d x \right ) c_3}
\]
Summary of solutions found
\begin{align*}
y &= -\frac {\left (g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_3 -\left (1+\int g \left (x \right )d x \right ) f \left (x \right ) g \left (x \right ) {\mathrm e}^{\int -f \left (x \right ) g \left (x \right )d x} c_3 \right ) {\mathrm e}^{\int f \left (x \right ) g \left (x \right )d x}}{g \left (x \right ) \left (1+\int g \left (x \right )d x \right ) c_3} \\
\end{align*}
2.20.4.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 17
ode:=diff(y(x),x) = g(x)*(y(x)-f(x))^2+diff(f(x),x);
dsolve(ode,y(x), singsol=all);
\[
y = f \left (x \right )+\frac {1}{c_1 -\int g \left (x \right )d x}
\]
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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=g \left (x \right ) \left (y \left (x \right )-f \left (x \right )\right )^{2}+\frac {d}{d x}f \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=g \left (x \right ) \left (y \left (x \right )-f \left (x \right )\right )^{2}+\frac {d}{d x}f \left (x \right ) \end {array} \]
2.20.4.3 ✓ Mathematica. Time used: 0.138 (sec). Leaf size: 31
ode=D[y[x],x]==g[x]*(y[x]-f[x])^2+D[ f[x],x];
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to f(x)+\frac {1}{-\int _1^xg(K[2])dK[2]+c_1}\\ y(x)&\to f(x) \end{align*}
2.20.4.4 ✗ Sympy
from sympy import *
x = symbols("x")
y = Function("y")
f = Function("f")
g = Function("g")
ode = Eq(-(-f(x) + y(x))**2*g(x) - Derivative(f(x), x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
IndexError : Index out of range: a[1]