2.2.1 Problem 1
Internal
problem
ID
[13207]
Book
:
Handbook
of
exact
solutions
for
ordinary
differential
equations.
By
Polyanin
and
Zaitsev.
Second
edition
Section
:
Chapter
1,
section
1.2.
Riccati
Equation.
1.2.2.
Equations
Containing
Power
Functions
Problem
number
:
1
Date
solved
:
Sunday, January 18, 2026 at 06:41:50 PM
CAS
classification
:
[_Riccati]
2.2.1.1 Solved using first_order_ode_riccati
0.121 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime }&=a y^{2}+b x +c \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= a y^{2}+b x +c \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[
y' = a y^{2}+b x +c
\]
With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows
that \(f_0(x)=b x +c\), \(f_1(x)=0\) and \(f_2(x)=a\). Let \begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u a} \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 &=0\\ f_2^2 f_0 &=a^{2} \left (b x +c \right ) \end{align*}
Substituting the above terms back in equation (2) gives
\[
a u^{\prime \prime }\left (x \right )+a^{2} \left (b x +c \right ) u \left (x \right ) = 0
\]
Entering second order Airy solverThis is
Airy ODE. It has the general form \[ a \frac {d^{2}u}{d x^{2}} + b \frac {d u}{d x} + c u x = F(x) \]
Where in this case \begin{align*} a &= a\\ b &= 0\\ c &= \frac {a^{2} \left (b x +c \right )}{x}\\ F &= 0 \end{align*}
Therefore the solution to the homogeneous Airy ODE becomes
\[
u = c_1 \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )+c_2 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )
\]
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = -c_1 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )-c_2 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )
\end{equation}
Substituting equations (3,4) into (1) results in \begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{u a} \\
y &= -\frac {-c_1 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )-c_2 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )}{a \left (c_1 \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )+c_2 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )} \\
\end{align*}
Doing change of constants, the above solution
becomes \[
y = -\frac {-\left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )-c_3 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )}
\]
Simplifying the above gives \begin{align*}
y &= \frac {\left (a b \right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right ) c_3 +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {\left (a b \right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right ) c_3 +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b x +c \right )}{b}\right )\right )} \\
\end{align*}
2.2.1.2 ✓ Maple. Time used: 0.008 (sec). Leaf size: 85
ode:=diff(y(x),x) = a*y(x)^2+b*x+c;
dsolve(ode,y(x), singsol=all);
\[
y = \frac {\left (\frac {b}{\sqrt {a}}\right )^{{1}/{3}} \left (\operatorname {AiryAi}\left (1, -\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{{2}/{3}}}\right ) c_1 +\operatorname {AiryBi}\left (1, -\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{{2}/{3}}}\right )\right )}{\sqrt {a}\, \left (c_1 \operatorname {AiryAi}\left (-\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{{2}/{3}}}\right )+\operatorname {AiryBi}\left (-\frac {b x +c}{\left (\frac {b}{\sqrt {a}}\right )^{{2}/{3}}}\right )\right )}
\]
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 Special
trying Riccati sub-methods:
<- Abel AIR successful: ODE belongs to the 0F1 0-parameter (Airy type) class
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a y \left (x \right )^{2}+b x +c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a y \left (x \right )^{2}+b x +c \end {array} \]
2.2.1.3 ✓ Mathematica. Time used: 0.131 (sec). Leaf size: 143
ode=D[y[x],x]==a*y[x]^2+b*x+c;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \frac {b \left (\operatorname {AiryBiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )\right )}{(-a b)^{2/3} \left (\operatorname {AiryBi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )\right )}\\ y(x)&\to \frac {b \operatorname {AiryAiPrime}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )}{(-a b)^{2/3} \operatorname {AiryAi}\left (-\frac {a (c+b x)}{(-a b)^{2/3}}\right )} \end{align*}
2.2.1.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
c = symbols("c")
y = Function("y")
ode = Eq(-a*y(x)**2 - b*x - c + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -a*y(x)**2 - b*x - c + Derivative(y(x), x) cannot be solved by t
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')