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2.7.2 Problem 2

2.7.2.1 Solved using first_order_ode_riccati
2.7.2.2 ✓Maple
2.7.2.3 ✓Mathematica
2.7.2.4 ✗Sympy

Internal problem ID [13347]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-1. Equations Containing Logarithmic Functions
Problem number : 2
Date solved : Wednesday, December 31, 2025 at 01:40:12 PM
CAS classification : [_Riccati]

2.7.2.1 Solved using first_order_ode_riccati

9.717 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime } x&=a y^{2}+b \ln \left (x \right )+c \\ \end{align*}
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= \frac {a y^{2}+b \ln \left (x \right )+c}{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 \(f_0(x)=\frac {b \ln \left (x \right )}{x}+\frac {c}{x}\), \(f_1(x)=0\) and \(f_2(x)=\frac {a}{x}\). Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a}{x}} \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' &=-\frac {a}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {a^{2} \left (\frac {b \ln \left (x \right )}{x}+\frac {c}{x}\right )}{x^{2}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {a u^{\prime \prime }\left (x \right )}{x}+\frac {a u^{\prime }\left (x \right )}{x^{2}}+\frac {a^{2} \left (\frac {b \ln \left (x \right )}{x}+\frac {c}{x}\right ) u \left (x \right )}{x^{2}} = 0 \]
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 \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_2 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) \end{equation}
Taking derivative gives
\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x}-\frac {c_2 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x} \end{equation}
Substituting equations (3,4) into (1) results in
\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u a}{x}} \\ y &= -\frac {\left (-\frac {c_1 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x}-\frac {c_2 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x}\right ) x}{a \left (c_1 \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_2 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \\ \end{align*}
Doing change of constants, the above solution becomes
\[ y = -\frac {\left (-\frac {\left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x}-\frac {c_3 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )}{x}\right ) x}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \]
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 \ln \left (x \right )+c \right )}{b}\right ) c_3 +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+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 \ln \left (x \right )+c \right )}{b}\right ) c_3 +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )}{a \left (\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_3 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \\ \end{align*}
2.7.2.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 91
ode:=x*diff(y(x),x) = a*y(x)^2+b*ln(x)+c; 
dsolve(ode,y(x), singsol=all);
\[ y = \frac {\left (a b \right )^{{1}/{3}} \left (\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_1 +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )}{a \left (c_1 \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{{1}/{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \]

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 sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -1/x*diff(y(x),x)-a* 
(b*ln(x)+c)/x^2*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 quadrature 
         checking if the LODE has constant coefficients 
         checking if the LODE is of Euler type 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying a Liouvillian solution using Kovacics algorithm 
         <- No Liouvillian solutions exists 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            <- Bessel successful 
         <- special function solution successful 
         Change of variables used: 
            [x = exp(t)] 
         Linear ODE actually solved: 
            (a*b*t+a*c)*u(t)+diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=a y \left (x \right )^{2}+b \ln \left (x \right )+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a y \left (x \right )^{2}+b \ln \left (x \right )+c}{x} \end {array} \]
2.7.2.3 ✓ Mathematica. Time used: 0.482 (sec). Leaf size: 149
ode=x*D[y[x],x]==a*y[x]^2+b*Log[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 \log (x))}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAiPrime}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )\right )}{(-a b)^{2/3} \left (\operatorname {AiryBi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )\right )}\\ y(x)&\to \frac {b \operatorname {AiryAiPrime}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )}{(-a b)^{2/3} \operatorname {AiryAi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )} \end{align*}
2.7.2.4 ✗ 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*log(x) - c + x*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (a*y(x)**2 + b*log(x) + c)/x cannot be solved by the factorable group method

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