2.8.6 Problem 15

Internal problem ID [13358]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-2
Problem number : 15
Date solved : Sunday, January 18, 2026 at 07:29:27 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]

2.8.6.1 Solved using first_order_ode_riccati

2.628 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=a \ln \left (x \right )^{k} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{n -1} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c -2 \ln \left (x \right )^{k} x^{n} a b y+\ln \left (x \right )^{k} a \,c^{2}-2 \ln \left (x \right )^{k} a c y+\ln \left (x \right )^{k} a y^{2}+b n \,x^{n -1} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \ln \left (x \right )^{k} x^{2 n} a \,b^{2}-2 \ln \left (x \right )^{k} x^{n} a b y+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a y^{2}-2 \ln \left (x \right )^{k} a c y+\ln \left (x \right )^{k} a \,c^{2}+\frac {b \,x^{n} n}{x} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+\frac {b \,x^{n} n}{x}\), \(f_1(x)=-2 \ln \left (x \right )^{k} x^{n} a b -2 a c \ln \left (x \right )^{k}\) and \(f_2(x)=a \ln \left (x \right )^{k}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u a \ln \left (x \right )^{k}} \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 \ln \left (x \right )^{k} k}{x \ln \left (x \right )}\\ f_1 f_2 &=\left (-2 \ln \left (x \right )^{k} x^{n} a b -2 a c \ln \left (x \right )^{k}\right ) a \ln \left (x \right )^{k}\\ f_2^2 f_0 &=a^{2} \ln \left (x \right )^{2 k} \left (\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+\frac {b \,x^{n} n}{x}\right ) \end{align*}

Substituting the above terms back in equation (2) gives

\[ a \ln \left (x \right )^{k} u^{\prime \prime }\left (x \right )-\left (\frac {a \ln \left (x \right )^{k} k}{x \ln \left (x \right )}+\left (-2 \ln \left (x \right )^{k} x^{n} a b -2 a c \ln \left (x \right )^{k}\right ) a \ln \left (x \right )^{k}\right ) u^{\prime }\left (x \right )+a^{2} \ln \left (x \right )^{2 k} \left (\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+\frac {b \,x^{n} n}{x}\right ) u \left (x \right ) = 0 \]

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 ) = \frac {c_1 \ln \left (x \right )^{-\frac {k}{2}} {\mathrm e}^{-\frac {\int \frac {2 a x \ln \left (x \right )^{k +1} \left (b \,x^{n}+c \right )-k -\ln \left (x \right )}{\ln \left (x \right ) x}d x}{2}}}{\sqrt {x}}+\frac {c_2 \,{\mathrm e}^{-\frac {\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 a c \ln \left (x \right )^{k}-\frac {k}{x \ln \left (x \right )}-\frac {1}{x}\right )d x}{2}} \ln \left (x \right )^{\frac {k}{2}} \left (x +\left (-\ln \left (x \right )\right )^{-k} k \left (-\Gamma \left (k \right )+\Gamma \left (k , -\ln \left (x \right )\right )\right )\right )}{\sqrt {x}} \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \ln \left (x \right )^{-\frac {k}{2}} k \,{\mathrm e}^{-\frac {\int \frac {2 a x \ln \left (x \right )^{k +1} \left (b \,x^{n}+c \right )-k -\ln \left (x \right )}{\ln \left (x \right ) x}d x}{2}}}{2 x^{{3}/{2}} \ln \left (x \right )}-\frac {c_1 \ln \left (x \right )^{-\frac {k}{2}} {\mathrm e}^{-\frac {\int \frac {2 a x \ln \left (x \right )^{k +1} \left (b \,x^{n}+c \right )-k -\ln \left (x \right )}{\ln \left (x \right ) x}d x}{2}}}{2 x^{{3}/{2}}}-\frac {c_1 \ln \left (x \right )^{-\frac {k}{2}} \left (2 a x \ln \left (x \right )^{k +1} \left (b \,x^{n}+c \right )-k -\ln \left (x \right )\right ) {\mathrm e}^{-\frac {\int \frac {2 a x \ln \left (x \right )^{k +1} \left (b \,x^{n}+c \right )-k -\ln \left (x \right )}{\ln \left (x \right ) x}d x}{2}}}{2 x^{{3}/{2}} \ln \left (x \right )}+\frac {c_2 \left (-\ln \left (x \right )^{k} x^{n} a b -a c \ln \left (x \right )^{k}+\frac {k}{2 x \ln \left (x \right )}+\frac {1}{2 x}\right ) {\mathrm e}^{-\frac {\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 a c \ln \left (x \right )^{k}-\frac {k}{x \ln \left (x \right )}-\frac {1}{x}\right )d x}{2}} \ln \left (x \right )^{\frac {k}{2}} \left (x +\left (-\ln \left (x \right )\right )^{-k} k \left (-\Gamma \left (k \right )+\Gamma \left (k , -\ln \left (x \right )\right )\right )\right )}{\sqrt {x}}-\frac {c_2 \,{\mathrm e}^{-\frac {\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 a c \ln \left (x \right )^{k}-\frac {k}{x \ln \left (x \right )}-\frac {1}{x}\right )d x}{2}} \ln \left (x \right )^{\frac {k}{2}} \left (x +\left (-\ln \left (x \right )\right )^{-k} k \left (-\Gamma \left (k \right )+\Gamma \left (k , -\ln \left (x \right )\right )\right )\right )}{2 x^{{3}/{2}}}+\frac {c_2 \,{\mathrm e}^{-\frac {\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 a c \ln \left (x \right )^{k}-\frac {k}{x \ln \left (x \right )}-\frac {1}{x}\right )d x}{2}} \ln \left (x \right )^{\frac {k}{2}} k \left (x +\left (-\ln \left (x \right )\right )^{-k} k \left (-\Gamma \left (k \right )+\Gamma \left (k , -\ln \left (x \right )\right )\right )\right )}{2 x^{{3}/{2}} \ln \left (x \right )}+\frac {c_2 \,{\mathrm e}^{-\frac {\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 a c \ln \left (x \right )^{k}-\frac {k}{x \ln \left (x \right )}-\frac {1}{x}\right )d x}{2}} \ln \left (x \right )^{\frac {k}{2}} \left (1-\frac {\left (-\ln \left (x \right )\right )^{-k} k^{2} \left (-\Gamma \left (k \right )+\Gamma \left (k , -\ln \left (x \right )\right )\right )}{x \ln \left (x \right )}+\left (-\ln \left (x \right )\right )^{-k} k \left (-\ln \left (x \right )\right )^{k -1}\right )}{\sqrt {x}} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u a \ln \left (x \right )^{k}} \\ y &= \text {Expression too large to display} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ \text {Expression too large to display} \]

Summary of solutions found

\begin{align*} \text {Expression too large to display} \\ \end{align*}

2.8.6.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 24

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

\[ y = b \,x^{n}+c +\frac {1}{c_1 -a \int \ln \left (x \right )^{k}d x} \]

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \ln \left (x \right )^{k} \left (y \left (x \right )-b \,x^{13358}-c \right )^{2}+13358 b \,x^{13357} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \ln \left (x \right )^{k} \left (y \left (x \right )-b \,x^{13358}-c \right )^{2}+13358 b \,x^{13357} \end {array} \]

2.8.6.3 ✓ Mathematica. Time used: 0.617 (sec). Leaf size: 44

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

\begin{align*} y(x)&\to \frac {1}{-\int _1^xa \log ^k(K[2])dK[2]+c_1}+b x^n+c\\ y(x)&\to b x^n+c \end{align*}

2.8.6.4 ✗ Sympy

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

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0