2.19.13 Problem 13
Internal
problem
ID
[13461]
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
:
13
Date
solved
:
Wednesday, December 31, 2025 at 09:41:44 PM
CAS
classification
:
[_Riccati]
2.19.13.1 Solved using first_order_ode_riccati
12.559 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime }&=f \left (x \right ) y^{2}-a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+a \lambda \,{\mathrm e}^{\lambda x} \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= f \left (x \right ) y^{2}-a \,{\mathrm e}^{\lambda x} f \left (x \right ) y+a \lambda \,{\mathrm e}^{\lambda x} \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 \(f_0(x)=a \lambda \,{\mathrm e}^{\lambda x}\), \(f_1(x)=-a \,{\mathrm e}^{\lambda x} f \left (x \right )\) 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 &=-{\mathrm e}^{\lambda x} f \left (x \right )^{2} a\\ f_2^2 f_0 &=\lambda f \left (x \right )^{2} a \,{\mathrm e}^{\lambda x} \end{align*}
Substituting the above terms back in equation (2) gives
\[
f \left (x \right ) u^{\prime \prime }\left (x \right )-\left (-{\mathrm e}^{\lambda x} f \left (x \right )^{2} a +f^{\prime }\left (x \right )\right ) u^{\prime }\left (x \right )+\lambda f \left (x \right )^{2} a \,{\mathrm e}^{\lambda x} u \left (x \right ) = 0
\]
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 -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_2 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}
\end{equation}
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = -c_1 a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_2 f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x} {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}-c_2 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \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 a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_2 f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x} {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}-c_2 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}}{f \left (x \right ) \left (c_1 \,{\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_2 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}\right )} \\
\end{align*}
Doing
change of constants, the above solution becomes \[
y = -\frac {-a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_3 f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x} {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}-c_3 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}}{f \left (x \right ) \left ({\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_3 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}\right )}
\]
Summary of solutions found
\begin{align*}
y &= -\frac {-a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_3 f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x} {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}-c_3 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x a \,{\mathrm e}^{\lambda x} f \left (x \right ) {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}}{f \left (x \right ) \left ({\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}+c_3 \int f \left (x \right ) {\mathrm e}^{-\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}d x {\mathrm e}^{\int -a \,{\mathrm e}^{\lambda x} f \left (x \right )d x}\right )} \\
\end{align*}
2.19.13.2 ✗ Maple
ode:=diff(y(x),x) = f(x)*y(x)^2-a*exp(lambda*x)*f(x)*y(x)+a*lambda*exp(lambda*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*exp(lambda*
x)*a-diff(f(x),x))/f(x)*diff(y(x),x)-f(x)*a*lambda*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)-(f(x)*y(x)^2+y(x)-a*exp(lambda
*x)*f(x)*y(x)*x+x^2*a*lambda*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
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=f \left (x \right ) y \left (x \right )^{2}-a \,{\mathrm e}^{\lambda x} f \left (x \right ) y \left (x \right )+a \lambda \,{\mathrm e}^{\lambda x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=f \left (x \right ) y \left (x \right )^{2}-a \,{\mathrm e}^{\lambda x} f \left (x \right ) y \left (x \right )+a \lambda \,{\mathrm e}^{\lambda x} \end {array} \]
2.19.13.3 ✓ Mathematica. Time used: 42.397 (sec). Leaf size: 207
ode=D[y[x],x]==f[x]*y[x]^2-a*Exp[\[Lambda]*x]*f[x]*y[x]+a*\[Lambda]*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]+2 \lambda x\right ) \left (\int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1\right )}{\exp \left (\int _1^{e^{x \lambda }}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]+\lambda x\right ) \int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1 \exp \left (\int _1^{e^{x \lambda }}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]+\lambda x\right )+1}\\ y(x)&\to a e^{\lambda x} \end{align*}
2.19.13.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
lambda_ = symbols("lambda_")
y = Function("y")
f = Function("f")
ode = Eq(-a*lambda_*exp(lambda_*x) + a*f(x)*y(x)*exp(lambda_*x) - f(x)*y(x)**2 + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -a*lambda_*exp(lambda_*x) + a*f(x)*y(x)*exp(lambda_*x) - f(x)*y(x)**2 + Derivative(y(x), x) cannot be solved by the lie group method