2.2.40 Problem 43

Internal problem ID [13246]
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 : 43
Date solved : Sunday, January 18, 2026 at 06:51:25 PM
CAS classification : [_rational, _Riccati]

2.2.40.1 Solved using first_order_ode_riccati

0.737 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \frac {x^{2 n} a y^{2}}{x}+\frac {x^{n} b y}{x}-\frac {n y}{x}+\frac {c}{x} \]

With Riccati ODE standard form

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

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a \,x^{2 n}}{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 n}}{x^{2}}+\frac {2 a \,x^{2 n} n}{x^{2}}\\ f_1 f_2 &=\frac {\left (\frac {b \,x^{n}}{x}-\frac {n}{x}\right ) a \,x^{2 n}}{x}\\ f_2^2 f_0 &=\frac {a^{2} x^{4 n} c}{x^{3}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {a \,x^{2 n} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {a \,x^{2 n}}{x^{2}}+\frac {2 a \,x^{2 n} n}{x^{2}}+\frac {\left (\frac {b \,x^{n}}{x}-\frac {n}{x}\right ) a \,x^{2 n}}{x}\right ) u^{\prime }\left (x \right )+\frac {a^{2} x^{4 n} c u \left (x \right )}{x^{3}} = 0 \]

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}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+c_2 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_1 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_2 b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_2 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 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^{2 n}}{x}} \\ y &= -\frac {\left (\frac {c_1 b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_1 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_2 b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_2 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}\right ) x \,x^{-2 n}}{a \left (c_1 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+c_2 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (\frac {b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_3 b \,x^{n} {\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}+\frac {c_3 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} x^{n} n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\, \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )}{2 x}\right ) x \,x^{-2 n}}{a \left ({\mathrm e}^{\frac {b \,x^{n}}{2 n}} \sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+c_3 \,{\mathrm e}^{\frac {b \,x^{n}}{2 n}} \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {\left (\left (c_3 b +n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right ) \left (n c_3 \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}+b \right )\right ) x^{-n}}{2 a \left (\sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+c_3 \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= -\frac {\left (\left (c_3 b +n \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}\right ) \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+\sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right ) \left (n c_3 \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}+b \right )\right ) x^{-n}}{2 a \left (\sinh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )+c_3 \cosh \left (\frac {x^{n} \sqrt {\frac {-4 a c +b^{2}}{n^{2}}}}{2}\right )\right )} \\ \end{align*}

2.2.40.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.126 (sec)

Entering first order ode riccati guess solver

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

Using trial and error, the following particular solution was found

\[ y_p = \frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{-n}}{2 a} \]

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

\[ y = -\frac {\left (a \left (b +\sqrt {-4 a c +b^{2}}\right ) {\mathrm e}^{\frac {x^{n} \sqrt {-4 a c +b^{2}}}{n}}-4 c_1 \left (a c -\frac {b^{2}}{4}+\frac {b \sqrt {-4 a c +b^{2}}}{4}\right )\right ) x^{-n}}{2 a \left (a \,{\mathrm e}^{\frac {x^{n} \sqrt {-4 a c +b^{2}}}{n}}-c_1 \sqrt {-4 a c +b^{2}}\right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {\left (a \left (b +\sqrt {-4 a c +b^{2}}\right ) {\mathrm e}^{\frac {x^{n} \sqrt {-4 a c +b^{2}}}{n}}-4 c_1 \left (a c -\frac {b^{2}}{4}+\frac {b \sqrt {-4 a c +b^{2}}}{4}\right )\right ) x^{-n}}{2 a \left (a \,{\mathrm e}^{\frac {x^{n} \sqrt {-4 a c +b^{2}}}{n}}-c_1 \sqrt {-4 a c +b^{2}}\right )} \\ \end{align*}

2.2.40.3 ✓ Maple. Time used: 0.007 (sec). Leaf size: 72

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

\[ y = \frac {\left (\tan \left (\frac {\sqrt {4 b^{2} a c -b^{4}}\, \left (b \,x^{n}+c_1 n \right )}{2 b^{2} n}\right ) \sqrt {4 b^{2} a c -b^{4}}-b^{2}\right ) x^{-n}}{2 a b} \]

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 
<- Chini successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{26492} y \left (x \right )^{2}+\left (b \,x^{13246}-13246\right ) y \left (x \right )+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{26492} y \left (x \right )^{2}+\left (b \,x^{13246}-13246\right ) y \left (x \right )+c}{x} \end {array} \]

2.2.40.4 ✓ Mathematica. Time used: 0.477 (sec). Leaf size: 118

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

\begin{align*} y(x)&\to \frac {x^{-n} \left (-b+\frac {\sqrt {b^2-4 a c} \left (-e^{\frac {x^n \sqrt {b^2-4 a c}}{n}}+c_1\right )}{e^{\frac {x^n \sqrt {b^2-4 a c}}{n}}+c_1}\right )}{2 a}\\ y(x)&\to \frac {x^{-n} \left (\sqrt {b^2-4 a c}-b\right )}{2 a} \end{align*}

2.2.40.5 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*x**(2*n)*y(x)**2 + b*x**n*y(x) + c - n*
 

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)) 
 
('factorable', 'lie_group')