Problem 21

2.2.19 Problem 21

2.2.19.1 Solved using ﬁrst_order_ode_riccati
2.2.19.2 ✓Maple
2.2.19.3 ✓Mathematica
2.2.19.4 ✗Sympy

Internal problem ID [13225]
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 : 21
Date solved : Sunday, January 18, 2026 at 06:47:04 PM
CAS classiﬁcation : [_Riccati]

2.2.19.1 Solved using ﬁrst_order_ode_riccati

0.709 (sec)

Entering ﬁrst order ode riccati solver

\begin{align*} x^{n +1} y^{\prime }&=x^{2 n} a y^{2}+c \,x^{m}+d \\ \end{align*}

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=\frac {x^{-n} c \,x^{m}}{x}+\frac {x^{-n} d}{x}\), \(f_1(x)=0\) and \(f_2(x)=\frac {a \,x^{n}}{x}\). Let

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

Substituting the above terms back in equation (2) gives

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

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 \,x^{\frac {n}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+c_2 \,x^{\frac {n}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \,x^{\frac {n}{2}} n \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {c_1 \,x^{\frac {n}{2}} \left (-\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{2}}}{x}+\frac {c_2 \,x^{\frac {n}{2}} n \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {c_2 \,x^{\frac {n}{2}} \left (-\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{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^{n}}{x}} \\ y &= -\frac {\left (\frac {c_1 \,x^{\frac {n}{2}} n \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {c_1 \,x^{\frac {n}{2}} \left (-\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{2}}}{x}+\frac {c_2 \,x^{\frac {n}{2}} n \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {c_2 \,x^{\frac {n}{2}} \left (-\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{2}}}{x}\right ) x^{-n} x}{a \left (c_1 \,x^{\frac {n}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+c_2 \,x^{\frac {n}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (\frac {x^{\frac {n}{2}} n \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {x^{\frac {n}{2}} \left (-\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{2}}}{x}+\frac {c_3 \,x^{\frac {n}{2}} n \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 x}+\frac {c_3 \,x^{\frac {n}{2}} \left (-\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+\frac {\sqrt {-4 a d +n^{2}}\, x^{-\frac {m}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )}{2 \sqrt {a c}}\right ) \sqrt {a c}\, x^{\frac {m}{2}}}{x}\right ) x^{-n} x}{a \left (x^{\frac {n}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )+c_3 \,x^{\frac {n}{2}} \operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {x^{-n} \left (\sqrt {a c}\, \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_3 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}-\frac {\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_3 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) \left (n +\sqrt {-4 a d +n^{2}}\right )}{2}\right )}{a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_3 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right )} \\ \end{align*}

Summary of solutions found

2.2.19.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 234

ode:=x^(n+1)*diff(y(x),x) = x^(2*n)*a*y(x)^2+c*x^m+d; 
dsolve(ode,y(x), singsol=all);

\[ y = \frac {x^{-n} \left (\sqrt {a c}\, \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_1 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}+1, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) x^{\frac {m}{2}}-\frac {\left (\operatorname {BesselY}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_1 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right ) \left (\sqrt {-4 d a +n^{2}}+n \right )}{2}\right )}{a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right ) c_1 +\operatorname {BesselJ}\left (\frac {\sqrt {-4 d a +n^{2}}}{m}, \frac {2 \sqrt {a c}\, x^{\frac {m}{2}}}{m}\right )\right )} \]

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) = (n-1)/x*diff(y(x),x) 
-a*x^(n-1)*(x^(-n-1+m)*c+d*x^(-n-1))*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 
         <- No Liouvillian solutions exists 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         <- Bessel successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{13226} \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{26450} y \left (x \right )^{2}+c \,x^{m}+d \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{26450} y \left (x \right )^{2}+c \,x^{m}+d}{x^{13226}} \end {array} \]

2.2.19.3 ✓ Mathematica. Time used: 0.689 (sec). Leaf size: 1890

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

\begin{align*} \text {Solution too large to show}\end{align*}

2.2.19.4 ✗ Sympy

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

NotImplementedError : The given ODE -a*x**(n - 1)*y(x)**2 - c*x**(m - n - 1) - d*x**(-n - 1) + Deriv

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')

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