2.2.54 Problem 57

Internal problem ID [13260]
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 : 57
Date solved : Friday, December 19, 2025 at 02:17:42 AM
CAS classification : [_rational, _Riccati]

\begin{align*} \left (a \,x^{2}+b \right ) y^{\prime }+\alpha y^{2}+\beta x y+\gamma &=0 \\ \end{align*}

Entering first order ode riccati solver

\begin{align*} \left (a \,x^{2}+b \right ) y^{\prime }+\alpha y^{2}+\beta x y+\gamma &=0 \\ \end{align*}

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -\frac {\alpha \,y^{2}}{a \,x^{2}+b}-\frac {\beta x y}{a \,x^{2}+b}-\frac {\gamma }{a \,x^{2}+b} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-\frac {\gamma }{a \,x^{2}+b}\), \(f_1(x)=-\frac {\beta x}{a \,x^{2}+b}\) and \(f_2(x)=-\frac {\alpha }{a \,x^{2}+b}\). Let

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

Substituting the above terms back in equation (2) gives

\[ -\frac {\alpha u^{\prime \prime }\left (x \right )}{a \,x^{2}+b}-\left (\frac {2 \alpha x a}{\left (a \,x^{2}+b \right )^{2}}+\frac {\beta x \alpha }{\left (a \,x^{2}+b \right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\alpha ^{2} \gamma u \left (x \right )}{\left (a \,x^{2}+b \right )^{3}} = 0 \]

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 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )+c_2 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \beta x \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{2 \left (a \,x^{2}+b \right )}+\frac {c_1 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \left (\left (\frac {\beta }{2 a}-\frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}+1\right ) \operatorname {LegendreP}\left (1+\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )-\frac {\left (1+\frac {\beta }{2 a}\right ) a x \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{\sqrt {-a b}}\right ) a}{\sqrt {-a b}\, \left (-1-\frac {a \,x^{2}}{b}\right )}-\frac {c_2 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \beta x \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{2 \left (a \,x^{2}+b \right )}+\frac {c_2 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \left (\left (\frac {\beta }{2 a}-\frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}+1\right ) \operatorname {LegendreQ}\left (1+\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )-\frac {\left (1+\frac {\beta }{2 a}\right ) a x \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{\sqrt {-a b}}\right ) a}{\sqrt {-a b}\, \left (-1-\frac {a \,x^{2}}{b}\right )} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{-\frac {\alpha u}{a \,x^{2}+b}} \\ y &= -\frac {\left (\left (-2 a -\beta \right ) \sqrt {b}+\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}\right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_1 +\operatorname {LegendreQ}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_2 \right ) \sqrt {-a b}+2 a \sqrt {b}\, x \left (\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_1 +\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_2 \right ) \left (a +\beta \right )}{2 \sqrt {b}\, \left (\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_1 +\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_2 \right ) \alpha a} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {\left (-\frac {\left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \beta x \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{2 \left (a \,x^{2}+b \right )}+\frac {\left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \left (\left (\frac {\beta }{2 a}-\frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}+1\right ) \operatorname {LegendreP}\left (1+\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )-\frac {\left (1+\frac {\beta }{2 a}\right ) a x \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{\sqrt {-a b}}\right ) a}{\sqrt {-a b}\, \left (-1-\frac {a \,x^{2}}{b}\right )}-\frac {c_3 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \beta x \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{2 \left (a \,x^{2}+b \right )}+\frac {c_3 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \left (\left (\frac {\beta }{2 a}-\frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}+1\right ) \operatorname {LegendreQ}\left (1+\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )-\frac {\left (1+\frac {\beta }{2 a}\right ) a x \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )}{\sqrt {-a b}}\right ) a}{\sqrt {-a b}\, \left (-1-\frac {a \,x^{2}}{b}\right )}\right ) \left (a \,x^{2}+b \right )}{\alpha \left (\left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )+c_3 \left (a \,x^{2}+b \right )^{-\frac {\beta }{4 a}} \operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {\left (\left (-2 a -\beta \right ) \sqrt {b}+\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}\right ) \left (\operatorname {LegendreQ}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \sqrt {-a b}+2 a x \sqrt {b}\, \left (\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \left (a +\beta \right )}{2 \sqrt {b}\, \left (\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \alpha a} \\ \end{align*}

The solution

\[ y = -\frac {\left (\left (-2 a -\beta \right ) \sqrt {b}+\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}\right ) \left (\operatorname {LegendreQ}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {2 a +\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \sqrt {-a b}+2 a x \sqrt {b}\, \left (\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \left (a +\beta \right )}{2 \sqrt {b}\, \left (\operatorname {LegendreQ}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right ) c_3 +\operatorname {LegendreP}\left (\frac {\beta }{2 a}, \frac {\sqrt {4 a \alpha \gamma +b \,\beta ^{2}}}{2 a \sqrt {b}}, \frac {a x}{\sqrt {-a b}}\right )\right ) \alpha a} \]

was found not to satisfy the ode or the IC. Hence it is removed.

2.2.54.1 ✓ Maple. Time used: 0.028 (sec). Leaf size: 858

ode:=(a*x^2+b)*diff(y(x),x)+alpha*y(x)^2+beta*x*y(x)+gamma = 0; 
dsolve(ode,y(x), singsol=all);
 

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

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: 
   <- Abel AIR successful: ODE belongs to the 2F1 3-parameter class
 

Maple step by step

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

2.2.54.2 ✓ Mathematica. Time used: 0.409 (sec). Leaf size: 598

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

\begin{align*} y(x)&\to \frac {i \left (c_1 \left (\sqrt {4 a \alpha \gamma +b \beta ^2}-2 a \sqrt {b}-\sqrt {b} \beta \right ) P_{\frac {\beta }{2 a}+1}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )+2 i \sqrt {a} x (a+\beta ) Q_{\frac {\beta }{2 a}}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )+\left (\sqrt {4 a \alpha \gamma +b \beta ^2}-2 a \sqrt {b}-\sqrt {b} \beta \right ) Q_{\frac {\beta }{2 a}+1}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )\right )-2 \sqrt {a} c_1 x (a+\beta ) P_{\frac {\beta }{2 a}}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{2 \sqrt {a} \alpha \left (c_1 P_{\frac {\beta }{2 a}}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )+Q_{\frac {\beta }{2 a}}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )\right )}\\ y(x)&\to \frac {-2 x (a+\beta )+\frac {i \left (\sqrt {4 a \alpha \gamma +b \beta ^2}-2 a \sqrt {b}-\sqrt {b} \beta \right ) P_{\frac {\beta }{2 a}+1}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}{\sqrt {a} P_{\frac {\beta }{2 a}}^{\frac {\sqrt {b \beta ^2+4 a \alpha \gamma }}{2 a \sqrt {b}}}\left (\frac {i \sqrt {a} x}{\sqrt {b}}\right )}}{2 \alpha } \end{align*}

2.2.54.3 ✗ Sympy

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

Timed Out