2.2.67 Problem 70

Internal problem ID [13273]
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 : 70
Date solved : Sunday, January 18, 2026 at 07:04:11 PM
CAS classification : [_rational, _Riccati]

2.2.67.1 Solved using first_order_ode_riccati

7.003 (sec)

Entering first order ode riccati solver

\begin{align*} x^{2} \left (x^{2}+a \right ) \left (y^{\prime }+\lambda y^{2}\right )+x \left (x^{2} b +c \right ) y+s&=0 \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {y^{2} \lambda \,x^{4}+a \,x^{2} \lambda y^{2}+y b \,x^{3}+y c x +s}{x^{2} \left (x^{2}+a \right )} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -\frac {x^{2} y^{2} \lambda }{x^{2}+a}-\frac {y^{2} a \lambda }{x^{2}+a}-\frac {x b y}{x^{2}+a}-\frac {c y}{x \left (x^{2}+a \right )}-\frac {s}{x^{2} \left (x^{2}+a \right )} \]

With Riccati ODE standard form

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

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

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

Substituting the above terms back in equation (2) gives

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

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 {a -c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}} \left (x^{2}+a \right )^{\frac {\left (-b +2\right ) a +c}{2 a}} \operatorname {hypergeom}\left (\left [\frac {3 a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}, \frac {-2 a b +5 a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}\right ], \left [\frac {2 a +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right )+c_2 \,x^{-\frac {-a +c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}} \left (x^{2}+a \right )^{\frac {\left (-b +2\right ) a +c}{2 a}} \operatorname {hypergeom}\left (\left [-\frac {-3 a -c +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}, \frac {-2 a b +5 a +c -\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{4 a}\right ], \left [-\frac {-2 a +\sqrt {a^{2}+\left (-4 \lambda s -2 c \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} \text {Expression too large to display} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u \left (-\frac {x^{2} \lambda }{x^{2}+a}-\frac {a \lambda }{x^{2}+a}\right )} \\ y &= \text {Expression too large to display} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ \text {Expression too large to display} \]

Simplifying the above gives

\begin{align*} \text {Expression too large to display} \\ \end{align*}

Summary of solutions found

\begin{align*} \text {Expression too large to display} \\ \end{align*}

2.2.67.2 ✓ Maple. Time used: 0.005 (sec). Leaf size: 1300

ode:=x^2*(x^2+a)*(diff(y(x),x)+lambda*y(x)^2)+x*(b*x^2+c)*y(x)+s = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ \text {Expression too large to display} \]

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 to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -(b*x^2+c)/x/(x^2+a) 
*diff(y(x),x)-lambda*s/x^2/(x^2+a)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      checking if the LODE has constant coefficients 
      checking if the LODE is of Euler type 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying a Liouvillian solution using Kovacics algorithm 
      <- No Liouvillian solutions exists 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         -> hypergeometric 
            -> heuristic approach 
            <- heuristic approach successful 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+a \right ) \left (\frac {d}{d x}y \left (x \right )+\lambda y \left (x \right )^{2}\right )+x \left (b \,x^{2}+c \right ) y \left (x \right )+s =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )^{2} \lambda \,x^{4}+y \left (x \right )^{2} a \lambda \,x^{2}+y \left (x \right ) b \,x^{3}+y \left (x \right ) c x +s}{x^{2} \left (x^{2}+a \right )} \end {array} \]

2.2.67.3 ✓ Mathematica. Time used: 3.037 (sec). Leaf size: 1908

ode=x^2*(x^2+a)*(D[y[x],x]+\[Lambda]*y[x]^2)+x*(b*x^2+c)*y[x]+s==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.2.67.4 ✗ Sympy

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

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0