2.2.58 Problem 61

Internal problem ID [13264]
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 : 61
Date solved : Sunday, January 18, 2026 at 07:00:54 PM
CAS classification : [_rational, _Riccati]

2.2.58.1 Solved using first_order_ode_riccati

46.696 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

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

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

Substituting the above terms back in equation (2) gives

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

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} \text {Expression too large to display} \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'}{\frac {u}{a_{2} x^{2}+b_{2} x +c_{2}}} \\ y &= \text {Expression too large to display} \\ \end{align*}

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.2.58.2 ✓ Maple. Time used: 0.011 (sec). Leaf size: 545907

ode:=(a__2*x^2+b__2*x+c__2)*diff(y(x),x) = y(x)^2+(a__1*x+b__1)*y(x)-lambda*(lambda+a__1-a__2)*x^2+lambda*(b__2-b__1)*x+lambda*c__2; 
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-2*a__2*x+ 
b__1-b__2)/(a__2*x^2+b__2*x+c__2)*diff(y(x),x)+lambda*(a__1*x^2-a__2*x^2+lambda 
*x^2+b__1*x-b__2*x-c__2)/(a__2*x^2+b__2*x+c__2)^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 
         A Liouvillian solution exists 
         Reducible group (found an exponential solution) 
         Group is reducible, not completely reducible 
         Solution has integrals. Trying a special function solution free of int\ 
egrals... 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebi\ 
us 
               <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
            <- hypergeometric successful 
         <- special function solution successful 
            -> Trying to convert hypergeometric functions to elementary form... 
            <- elementary form is not straightforward to achieve - returning sp\ 
ecial function solution free of uncomputed integrals 
         <- Kovacics algorithm successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.2.58.3 ✓ Mathematica. Time used: 7.145 (sec). Leaf size: 250

ode=(a2*x^2+b2*x+c2)*D[y[x],x]==y[x]^2+(a1*x+b1)*y[x]-\[Lambda]*(\[Lambda]+a1-a2)*x^2+\[Lambda]*(b2-b1)*x+\[Lambda]*c2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {\text {c2} \left (-\exp \left (\int _1^x\frac {\text {b1}-\text {b2}+(\text {a1}-2 \text {a2}+2 \lambda ) K[1]}{\text {c2}+K[1] (\text {b2}+\text {a2} K[1])}dK[1]\right )\right )+\lambda x \int _1^x\exp \left (\int _1^{K[2]}\frac {\text {b1}-\text {b2}+(\text {a1}-2 \text {a2}+2 \lambda ) K[1]}{\text {c2}+K[1] (\text {b2}+\text {a2} K[1])}dK[1]\right )dK[2]-x \left (\text {b2} \exp \left (\int _1^x\frac {\text {b1}-\text {b2}+(\text {a1}-2 \text {a2}+2 \lambda ) K[1]}{\text {c2}+K[1] (\text {b2}+\text {a2} K[1])}dK[1]\right )+\text {a2} x \exp \left (\int _1^x\frac {\text {b1}-\text {b2}+(\text {a1}-2 \text {a2}+2 \lambda ) K[1]}{\text {c2}+K[1] (\text {b2}+\text {a2} K[1])}dK[1]\right )-c_1 \lambda \right )}{\int _1^x\exp \left (\int _1^{K[2]}\frac {\text {b1}-\text {b2}+(\text {a1}-2 \text {a2}+2 \lambda ) K[1]}{\text {c2}+K[1] (\text {b2}+\text {a2} K[1])}dK[1]\right )dK[2]+c_1}\\ y(x)&\to \lambda x \end{align*}

2.2.58.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a__1 = symbols("a__1") 
a__2 = symbols("a__2") 
b__1 = symbols("b__1") 
b__2 = symbols("b__2") 
c__2 = symbols("c__2") 
lambda_ = symbols("lambda_") 
y = Function("y") 
ode = Eq(-c__2*lambda_ + lambda_*x**2*(a__1 - a__2 + lambda_) - lambda_*x*(-b__1 + b__2) - (a__1*x + b__1)*y(x) + (a__2*x**2 + b__2*x + c__2)*Derivative(y(x), x) - y(x)**2,0) 
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