2.2.34 Problem 37

Internal problem ID [13240]
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 : 37
Date solved : Wednesday, December 31, 2025 at 12:21:34 PM
CAS classification : [_rational, _Riccati]

2.2.34.1 Solved using first_order_ode_riccati

3.259 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \textit {the\_rhs} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=\frac {b \,x^{n}}{x}\), \(f_1(x)=\frac {a}{x}\) and \(f_2(x)=1\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \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 &=\frac {a}{x}\\ f_2^2 f_0 &=\frac {b \,x^{n}}{x} \end{align*}

Substituting the above terms back in equation (2) gives

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

Entering second order bessel ode solverWriting the ode as

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

Bessel ode has the form

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\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*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\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}+\frac {a}{2}\\ \beta &= \frac {2 \sqrt {b}}{n +1}\\ n &= -\frac {a +1}{n +1}\\ \gamma &= \frac {n}{2}+\frac {1}{2} \end{align*}

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

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

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Simplifying the above gives

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

Summary of solutions found

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

2.2.34.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 139

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

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

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/x*a*diff(y(x),x)-b 
*x^(-1+n)*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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )=x y \left (x \right )^{2}+a y \left (x \right )+b \,x^{13240} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x y \left (x \right )^{2}+a y \left (x \right )+b \,x^{13240}}{x} \end {array} \]

2.2.34.3 ✓ Mathematica. Time used: 0.448 (sec). Leaf size: 855

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

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

2.2.34.4 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*y(x) + b*x**n + x*y(x)**2)/x cannot be solved by the factorable group method