2.2.20 Problem 22

Internal problem ID [13226]
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 : 22
Date solved : Sunday, January 18, 2026 at 06:48:36 PM
CAS classification : [_rational, _Riccati]

2.2.20.1 Solved using first_order_ode_riccati

2.170 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=\frac {a \,x^{n}}{\left (x^{n} a +b \right ) x^{2}}\), \(f_1(x)=0\) and \(f_2(x)=\frac {b}{x^{n} a +b}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u b}{x^{n} a +b}} \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 {b a n \,x^{n}}{\left (x^{n} a +b \right )^{2} x}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {b^{2} a \,x^{n}}{\left (x^{n} a +b \right )^{3} x^{2}} \end{align*}

Substituting the above terms back in equation (2) gives

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

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 x \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+\frac {c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )}{x^{n} a +b}-\frac {2 c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1+\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n}}{b}+\frac {c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right )}{x \left (x^{n} a +b \right )}-\frac {c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [2, 1+\frac {1}{n}\right ], \left [\frac {n -1}{n}+1\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n} n}{\left (n -1\right ) b x} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u b}{x^{n} a +b}} \\ y &= -\frac {\left (c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+\frac {c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )}{x^{n} a +b}-\frac {2 c_1 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1+\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n}}{b}+\frac {c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right )}{x \left (x^{n} a +b \right )}-\frac {c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [2, 1+\frac {1}{n}\right ], \left [\frac {n -1}{n}+1\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n} n}{\left (n -1\right ) b x}\right ) \left (x^{n} a +b \right )}{b \left (c_1 x \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+c_2 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (\left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+\frac {\left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )}{x^{n} a +b}-\frac {2 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1+\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n}}{b}+\frac {c_3 \left (x^{n} a +b \right )^{\frac {1}{n}} a \,x^{n} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right )}{x \left (x^{n} a +b \right )}-\frac {c_3 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [2, 1+\frac {1}{n}\right ], \left [\frac {n -1}{n}+1\right ], -\frac {a \,x^{n}}{b}\right ) a \,x^{n} n}{\left (n -1\right ) b x}\right ) \left (x^{n} a +b \right )}{b \left (x \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [\frac {2}{n}\right ], \left [\right ], -\frac {a \,x^{n}}{b}\right )+c_3 \left (x^{n} a +b \right )^{\frac {1}{n}} \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {\left (a n c_3 \left (a \,x^{-1+2 n}+b \,x^{n -1}\right ) \operatorname {hypergeom}\left (\left [2, \frac {n +1}{n}\right ], \left [\frac {-1+2 n}{n}\right ], -\frac {a \,x^{n}}{b}\right )-b \left (n -1\right ) \left (a \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right ) c_3 \,x^{n -1}+b \left (\frac {x^{n} a +b}{b}\right )^{-\frac {2}{n}}\right )\right ) \left (\frac {x^{n} a +b}{b}\right )^{\frac {2}{n}}}{\left (n -1\right ) \left (\operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right ) c_3 \left (\frac {x^{n} a +b}{b}\right )^{\frac {2}{n}}+x \right ) b^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\left (a n c_3 \left (a \,x^{-1+2 n}+b \,x^{n -1}\right ) \operatorname {hypergeom}\left (\left [2, \frac {n +1}{n}\right ], \left [\frac {-1+2 n}{n}\right ], -\frac {a \,x^{n}}{b}\right )-b \left (n -1\right ) \left (a \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right ) c_3 \,x^{n -1}+b \left (\frac {x^{n} a +b}{b}\right )^{-\frac {2}{n}}\right )\right ) \left (\frac {x^{n} a +b}{b}\right )^{\frac {2}{n}}}{\left (n -1\right ) \left (\operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {n -1}{n}\right ], -\frac {a \,x^{n}}{b}\right ) c_3 \left (\frac {x^{n} a +b}{b}\right )^{\frac {2}{n}}+x \right ) b^{2}} \\ \end{align*}

2.2.20.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.148 (sec)

Entering first order ode riccati guess solver

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

Using trial and error, the following particular solution was found

\[ y_p = -\frac {1}{x} \]

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

\[ y = -\frac {1}{x}+\frac {{\mathrm e}^{\frac {2 \ln \left (x^{n} a +b \right )}{n}-\frac {2 \ln \left (x^{n}\right )}{n}}}{c_1 -\int \frac {{\mathrm e}^{\frac {2 \ln \left (x^{n} a +b \right )}{n}-\frac {2 \ln \left (x^{n}\right )}{n}} b}{x^{n} a +b}d x} \]

Summary of solutions found

\begin{align*} y &= -\frac {1}{x}+\frac {{\mathrm e}^{\frac {2 \ln \left (x^{n} a +b \right )}{n}-\frac {2 \ln \left (x^{n}\right )}{n}}}{c_1 -\int \frac {{\mathrm e}^{\frac {2 \ln \left (x^{n} a +b \right )}{n}-\frac {2 \ln \left (x^{n}\right )}{n}} b}{x^{n} a +b}d x} \\ \end{align*}

2.2.20.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 224

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

\[ y = \frac {\left (\frac {a \,x^{n}+b}{b}\right )^{\frac {2}{n}} \left (a n c_1 \left (a^{2} x^{3 n}+2 b a \,x^{2 n}+b^{2} x^{n}\right ) \operatorname {hypergeom}\left (\left [2, \frac {1+n}{n}\right ], \left [\frac {-1+2 n}{n}\right ], -\frac {a \,x^{n}}{b}\right )-\left (a c_1 \left (a \,x^{2 n}+x^{n} b \right ) \operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {-1+n}{n}\right ], -\frac {a \,x^{n}}{b}\right )+\left (\frac {a \,x^{n}+b}{b}\right )^{-\frac {2}{n}} b \left (x^{1+n} a +x b \right )\right ) \left (-1+n \right ) b \right )}{b^{2} \left (-1+n \right ) x \left (a \,x^{n}+b \right ) \left (x +\operatorname {hypergeom}\left (\left [1, \frac {1}{n}\right ], \left [\frac {-1+n}{n}\right ], -\frac {a \,x^{n}}{b}\right ) c_1 \left (\frac {a \,x^{n}+b}{b}\right )^{\frac {2}{n}}\right )} \]

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/(a*x^n+b)*a*x^n*n 
/x*diff(y(x),x)-1/(a*x^n+b)^2*b*a*x^(n-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 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
         -> Trying a Liouvillian solution using Kovacics algorithm 
            A Liouvillian solution exists 
            Reducible group (found an exponential solution) 
            Group is reducible, not completely reducible 
            Solution has integrals. Trying a special function solution free of \ 
integrals... 
            -> Trying a solution in terms of special functions: 
               -> Bessel 
               -> elliptic 
               -> Legendre 
               -> Whittaker 
                  -> hyper3: Equivalence to 1F1 under a power @ Moebius 
               -> hypergeometric 
                  -> heuristic approach 
                  <- heuristic approach successful 
               <- hypergeometric successful 
            <- special function solution successful 
               -> Trying to convert hypergeometric functions to elementary form\ 
... 
               <- elementary form could result into a too large expression - re\ 
turning special function form of solution, free of uncomputed integrals 
            <- Kovacics algorithm successful 
      <- Equivalence, under non-integer power transformations successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{13226}+b \right ) \left (\frac {d}{d x}y \left (x \right )\right )=b y \left (x \right )^{2}+a \,x^{13224} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {b y \left (x \right )^{2}+a \,x^{13224}}{a \,x^{13226}+b} \end {array} \]

2.2.20.4 ✓ Mathematica. Time used: 1.201 (sec). Leaf size: 289

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

\begin{align*} y(x)&\to \frac {-b^2 (-1)^{\frac {1}{n}} (n-1) \left (-\frac {a x^n}{b}\right )^{\frac {1}{n}}-a b c_1 (n-1) x^n \left (\frac {a x^n}{b}+1\right )^{2/n} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},\frac {n-1}{n},-\frac {a x^n}{b}\right )+a c_1 n x^n \left (a x^n+b\right ) \left (\frac {a x^n}{b}+1\right )^{2/n} \operatorname {Hypergeometric2F1}\left (2,1+\frac {1}{n},2-\frac {1}{n},-\frac {a x^n}{b}\right )}{b^2 (n-1) x \left ((-1)^{\frac {1}{n}} \left (-\frac {a x^n}{b}\right )^{\frac {1}{n}}+c_1 \left (\frac {a x^n}{b}+1\right )^{2/n} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},\frac {n-1}{n},-\frac {a x^n}{b}\right )\right )}\\ y(x)&\to \frac {a x^{n-1} \left (\frac {n \left (a x^n+b\right ) \operatorname {Hypergeometric2F1}\left (2,1+\frac {1}{n},2-\frac {1}{n},-\frac {a x^n}{b}\right )}{\operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},\frac {n-1}{n},-\frac {a x^n}{b}\right )}+b (-n)+b\right )}{b^2 (n-1)} \end{align*}

2.2.20.5 ✗ Sympy

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

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

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', '1st_power_series', 'lie_group')