2.7.7 Problem 9

Internal problem ID [13352]
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 : 9
Date solved : Wednesday, December 31, 2025 at 01:47:47 PM
CAS classification : [_Riccati]

2.7.7.1 Solved using first_order_ode_riccati

7.990 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {\ln \left (a x \right ) y^{2} x^{2}+1}{x^{2} \ln \left (a 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 \(f_0(x)=\frac {1}{x^{2} \ln \left (a x \right )}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=\frac {1}{x^{2} \ln \left (a x \right )} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 \ln \left (a x \right )+c_2 \left (\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) \ln \left (a x \right )+a x \right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1}{x}+\frac {c_2 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )}{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 {\frac {c_1}{x}+\frac {c_2 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )}{x}}{c_1 \ln \left (a x \right )+c_2 \left (\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) \ln \left (a x \right )+a x \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {1}{x}+\frac {c_3 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )}{x}}{\ln \left (a x \right )+c_3 \left (\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) \ln \left (a x \right )+a x \right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {-c_3 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )-1}{x \left (\ln \left (a x \right ) \operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) c_3 +c_3 a x +\ln \left (a x \right )\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-c_3 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )-1}{x \left (\ln \left (a x \right ) \operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) c_3 +c_3 a x +\ln \left (a x \right )\right )} \\ \end{align*}

2.7.7.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.209 (sec)

Entering first order ode riccati guess solver

\begin{align*} x^{2} \ln \left (a x \right ) \left (y^{\prime }-y^{2}\right )&=1 \\ \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) & =\frac {1}{x^{2} \ln \left (a x \right )}\\ f_1(x) & =0\\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -\frac {1}{x \ln \left (a x \right )} \]

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

\[ y = \frac {-c_1 a -\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )}{x \left (\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) \ln \left (a x \right )+a \left (c_1 \ln \left (a x \right )+x \right )\right )} \]

Summary of solutions found

\begin{align*} y &= \frac {-c_1 a -\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )}{x \left (\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right ) \ln \left (a x \right )+a \left (c_1 \ln \left (a x \right )+x \right )\right )} \\ \end{align*}

2.7.7.3 ✓ Maple. Time used: 0.001 (sec). Leaf size: 45

ode:=x^2*ln(a*x)*(diff(y(x),x)-y(x)^2) = 1; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {-c_1 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )-1}{x \left (\left (c_1 \,\operatorname {Ei}_{1}\left (-\ln \left (a x \right )\right )+1\right ) \ln \left (a x \right )+c_1 a x \right )} \]

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: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -1/x^2/ln(a*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 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)] 
         <- linear_1 successful 
         Change of variables used: 
            [x = exp(t)/a] 
         Linear ODE actually solved: 
            u(t)-t*diff(u(t),t)+t*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.7.7.4 ✓ Mathematica. Time used: 0.583 (sec). Leaf size: 102

ode=x^2*Log[a*x]*(D[y[x],x]-y[x]^2)==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to -\frac {c_1 \log (a x) \int _1^x\frac {1}{\log ^2(a K[1])}dK[1]+\log (a x)+c_1 x}{\log ^2(a x) \left (x+c_1 x \int _1^x\frac {1}{\log ^2(a K[1])}dK[1]\right )}\\ y(x)&\to -\frac {\frac {1}{\int _1^x\frac {1}{\log ^2(a K[1])}dK[1]}+\frac {\log (a x)}{x}}{\log ^2(a x)} \end{align*}

2.7.7.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
y = Function("y") 
ode = Eq(x**2*(-y(x)**2 + Derivative(y(x), x))*log(a*x) - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -y(x)**2 + Derivative(y(x), x) - 1/(x**2*log(a*x)) cannot be solved by the factorable group method