2.2.42 Problem 45

Internal problem ID [13248]
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 : 45
Date solved : Sunday, January 18, 2026 at 06:52:04 PM
CAS classification : [_rational, _Riccati]

2.2.42.1 Solved using first_order_ode_riccati

12.709 (sec)

Entering first order ode riccati solver

\begin{align*} \left (a_{2} x +b_{2} \right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (a_{1} x +b_{1} \right ) y+a_{0} x +b_{0}&=0 \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {y^{2} a_{2} \lambda x +y^{2} b_{2} \lambda +y a_{1} x +y b_{1} +a_{0} x +b_{0}}{a_{2} x +b_{2}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -\frac {y^{2} a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {y^{2} b_{2} \lambda }{a_{2} x +b_{2}}-\frac {y a_{1} x}{a_{2} x +b_{2}}-\frac {y b_{1}}{a_{2} x +b_{2}}-\frac {a_{0} x}{a_{2} x +b_{2}}-\frac {b_{0}}{a_{2} x +b_{2}} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-\frac {a_{0} x}{a_{2} x +b_{2}}-\frac {b_{0}}{a_{2} x +b_{2}}\), \(f_1(x)=-\frac {a_{1} x}{a_{2} x +b_{2}}-\frac {b_{1}}{a_{2} x +b_{2}}\) and \(f_2(x)=-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u \left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\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 {a_{2}^{2} \lambda x}{\left (a_{2} x +b_{2} \right )^{2}}-\frac {a_{2} \lambda }{a_{2} x +b_{2}}+\frac {b_{2} \lambda a_{2}}{\left (a_{2} x +b_{2} \right )^{2}}\\ f_1 f_2 &=\left (-\frac {a_{1} x}{a_{2} x +b_{2}}-\frac {b_{1}}{a_{2} x +b_{2}}\right ) \left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\right )\\ f_2^2 f_0 &=\left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\right )^{2} \left (-\frac {a_{0} x}{a_{2} x +b_{2}}-\frac {b_{0}}{a_{2} x +b_{2}}\right ) \end{align*}

Substituting the above terms back in equation (2) gives

\[ \left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\right ) u^{\prime \prime }\left (x \right )-\left (\frac {a_{2}^{2} \lambda x}{\left (a_{2} x +b_{2} \right )^{2}}-\frac {a_{2} \lambda }{a_{2} x +b_{2}}+\frac {b_{2} \lambda a_{2}}{\left (a_{2} x +b_{2} \right )^{2}}+\left (-\frac {a_{1} x}{a_{2} x +b_{2}}-\frac {b_{1}}{a_{2} x +b_{2}}\right ) \left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\right )\right ) u^{\prime }\left (x \right )+\left (-\frac {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\right )^{2} \left (-\frac {a_{0} x}{a_{2} x +b_{2}}-\frac {b_{0}}{a_{2} x +b_{2}}\right ) u \left (x \right ) = 0 \]

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 \,{\mathrm e}^{-\frac {\left (a_{1} +\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\right ) x}{2 a_{2}}} \left (a_{2} x +b_{2} \right )^{\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}} \operatorname {KummerM}\left (\frac {\left (a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 \lambda b_{0} a_{2}^{2}+\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\right )+c_2 \,{\mathrm e}^{-\frac {\left (a_{1} +\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\right ) x}{2 a_{2}}} \left (a_{2} x +b_{2} \right )^{\frac {a_{1} b_{2} +a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}} \operatorname {KummerU}\left (\frac {\left (a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1} \right ) \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}-2 \lambda b_{0} a_{2}^{2}+\left (2 a_{0} \lambda b_{2} +a_{1} b_{1} \right ) a_{2} -a_{1}^{2} b_{2}}{2 \sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, a_{2}^{2}}, \frac {a_{1} b_{2} +2 a_{2}^{2}-a_{2} b_{1}}{a_{2}^{2}}, \frac {\sqrt {-4 a_{0} a_{2} \lambda +a_{1}^{2}}\, \left (a_{2} x +b_{2} \right )}{a_{2}^{2}}\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 {a_{2} \lambda x}{a_{2} x +b_{2}}-\frac {b_{2} \lambda }{a_{2} x +b_{2}}\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*}

2.2.42.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 827

ode:=(a__2*x+b__2)*(diff(y(x),x)+lambda*y(x)^2)+(a__1*x+b__1)*y(x)+a__0*x+b__0 = 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) = -(a__1*x+b__1)/(a__2 
*x+b__2)*diff(y(x),x)-lambda*(a__0*x+b__0)/(a__2*x+b__2)*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 
            <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
         <- Kummer successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.2.42.3 ✓ Mathematica. Time used: 4.28 (sec). Leaf size: 1432

ode=(a2*x+b2)*(D[y[x],x]+\[Lambda]*y[x]^2)+(a1*x+b1)*y[x]+a0*x+b0==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Too large to display

2.2.42.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a__0 = symbols("a__0") 
a__1 = symbols("a__1") 
a__2 = symbols("a__2") 
b__0 = symbols("b__0") 
b__1 = symbols("b__1") 
b__2 = symbols("b__2") 
lambda_ = symbols("lambda_") 
y = Function("y") 
ode = Eq(a__0*x + b__0 + (a__1*x + b__1)*y(x) + (a__2*x + b__2)*(lambda_*y(x)**2 + Derivative(y(x), 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