2.2.57 Problem 60

Internal problem ID [13263]
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 : 60
Date solved : Wednesday, December 31, 2025 at 12:37:44 PM
CAS classification : [_rational, _Riccati]

2.2.57.1 Solved using first_order_ode_riccati

66.525 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {-\lambda ^{2} x^{2}+y a x +b \lambda x -\lambda \mu x +c \lambda +y \mu +y^{2}}{a \,x^{2}+b x +c} \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 \(f_0(x)=-\frac {\lambda ^{2} x^{2}}{a \,x^{2}+b x +c}+\frac {b \lambda x}{a \,x^{2}+b x +c}-\frac {\lambda \mu x}{a \,x^{2}+b x +c}+\frac {c \lambda }{a \,x^{2}+b x +c}\), \(f_1(x)=\frac {a x}{a \,x^{2}+b x +c}+\frac {\mu }{a \,x^{2}+b x +c}\) and \(f_2(x)=\frac {1}{a \,x^{2}+b x +c}\). Let

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

Substituting the above terms back in equation (2) gives

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

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 \,x^{2}+b x +c}} \\ 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*}

2.2.57.2 ✓ Maple. Time used: 0.011 (sec). Leaf size: 476004

ode:=(a*x^2+b*x+c)*diff(y(x),x) = y(x)^2+(a*x+mu)*y(x)-lambda^2*x^2+lambda*(b-mu)*x+lambda*c; 
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*x+b-mu)/(a*x^2+b 
*x+c)*diff(y(x),x)-lambda*(-lambda*x^2+b*x-mu*x+c)/(a*x^2+b*x+c)^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

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

2.2.57.3 ✓ Mathematica. Time used: 5.843 (sec). Leaf size: 245

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

\begin{align*} y(x)&\to \frac {c \left (-\exp \left (\int _1^x\frac {-b+\mu -a K[1]+2 \lambda K[1]}{c+K[1] (b+a K[1])}dK[1]\right )\right )+\lambda x \int _1^x\exp \left (\int _1^{K[2]}\frac {-b+\mu -a K[1]+2 \lambda K[1]}{c+K[1] (b+a K[1])}dK[1]\right )dK[2]-x \left (b \exp \left (\int _1^x\frac {-b+\mu -a K[1]+2 \lambda K[1]}{c+K[1] (b+a K[1])}dK[1]\right )+a x \exp \left (\int _1^x\frac {-b+\mu -a K[1]+2 \lambda K[1]}{c+K[1] (b+a K[1])}dK[1]\right )-c_1 \lambda \right )}{\int _1^x\exp \left (\int _1^{K[2]}\frac {-b+\mu -a K[1]+2 \lambda K[1]}{c+K[1] (b+a K[1])}dK[1]\right )dK[2]+c_1}\\ y(x)&\to \lambda x \end{align*}

2.2.57.4 ✗ Sympy

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

Timed Out