2.2.14 Problem 15

Internal problem ID [13220]
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 : 15
Date solved : Sunday, January 18, 2026 at 06:45:20 PM
CAS classification : [_rational, _Riccati]

2.2.14.1 Solved using first_order_ode_riccati

0.306 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

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

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

Substituting the above terms back in equation (2) gives

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

Entering second order bessel ode solverWriting the ode as

\begin{align*} \left (\frac {d^{2}u}{d x^{2}}\right ) x^{2}+\left (a b \,x^{n}+a c \right ) u = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} \left (\frac {d^{2}u}{d x^{2}}\right ) x^{2}+\left (\frac {d u}{d x}\right ) x +\left (-n^{2}+x^{2}\right ) u = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} \left (\frac {d^{2}u}{d x^{2}}\right ) x^{2}+\left (1-2 \alpha \right ) x \left (\frac {d u}{d x}\right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) u = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} u&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= \frac {2 \sqrt {a b}}{n}\\ n &= \frac {\sqrt {-4 a c +1}}{n}\\ \gamma &= \frac {n}{2} \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} u = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {\sqrt {-4 a c +1}}{n}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}}}{n}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {\sqrt {-4 a c +1}}{n}, \frac {2 \sqrt {a b}\, x^{\frac {n}{2}}}{n}\right ) \end{align*}

Taking derivative gives

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

Doing change of constants, the above solution becomes

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

Simplifying the above gives

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

Summary of solutions found

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

2.2.14.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 220

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

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

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 Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -a*(b*x^(-2+n)*x^2+c 
)/x^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 
         <- 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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y \left (x \right )\right )=a \,x^{2} y \left (x \right )^{2}+b \,x^{13220}+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {a \,x^{2} y \left (x \right )^{2}+b \,x^{13220}+c}{x^{2}} \end {array} \]

2.2.14.3 ✓ Mathematica. Time used: 0.723 (sec). Leaf size: 1779

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

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

2.2.14.4 ✗ 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*y(x)**2 - b*x**n - c + x**2*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (a*x**2*y(x)**2 + b*x**n + c)/x**2 cannot
 

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')