2.2.29 Problem 32

Internal problem ID [13235]
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 : 32
Date solved : Wednesday, December 31, 2025 at 12:18:50 PM
CAS classification : [_Riccati]

2.2.29.1 Solved using first_order_ode_riccati

13.356 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=-a n \,x^{n -1} y^{2}+c \,x^{m} \left (x^{n} a +b \right ) y-c \,x^{m} \\ \end{align*}

In canonical form the ODE is

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

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

Substituting the above terms back in equation (2) gives

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

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

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

\begin{equation} \tag{3} u \left (x \right ) = c_1 \left (x^{n} a +b \right )+c_2 \left (x^{n} a +b \right ) \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 a n \,x^{n}}{x}+\frac {c_2 a n \,x^{n} \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x}{x}+\frac {c_2 \,x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{x^{n} a +b} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{-\frac {u a n \,x^{n}}{x}} \\ y &= \frac {\left (\frac {c_1 a n \,x^{n}}{x}+\frac {c_2 a n \,x^{n} \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x}{x}+\frac {c_2 \,x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{x^{n} a +b}\right ) x^{-n} x}{a n \left (c_1 \left (x^{n} a +b \right )+c_2 \left (x^{n} a +b \right ) \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {\left (\frac {a n \,x^{n}}{x}+\frac {c_3 a n \,x^{n} \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x}{x}+\frac {c_3 \,x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{x^{n} a +b}\right ) x^{-n} x}{a n \left (x^{n} a +b +c_3 \left (x^{n} a +b \right ) \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x \right )} \]

Summary of solutions found

\begin{align*} y &= \frac {\left (\frac {a n \,x^{n}}{x}+\frac {c_3 a n \,x^{n} \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x}{x}+\frac {c_3 \,x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{x^{n} a +b}\right ) x^{-n} x}{a n \left (x^{n} a +b +c_3 \left (x^{n} a +b \right ) \int \frac {x^{n -1} {\mathrm e}^{\frac {\left (a \left (1+m \right ) x^{n}+b \left (m +n +1\right )\right ) c \,x^{1+m}}{\left (m +n +1\right ) \left (1+m \right )}}}{\left (x^{n} a +b \right )^{2}}d x \right )} \\ \end{align*}

2.2.29.2 ✓ Maple. Time used: 0.015 (sec). Leaf size: 199

ode:=diff(y(x),x) = -a*n*x^(n-1)*y(x)^2+c*x^m*(a*x^n+b)*y(x)-c*x^m; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {n a \left (a \,x^{n}+b \right ) \int \frac {x^{n -1} {\mathrm e}^{\frac {c \left (a \left (m +1\right ) x^{m +n +1}+b \,x^{m +1} \left (m +n +1\right )\right )}{\left (m +1\right ) \left (m +n +1\right )}}}{\left (a \,x^{n}+b \right )^{2}}d x -x^{n} c_1 a -c_1 b +{\mathrm e}^{\frac {c \left (a \left (m +1\right ) x^{m +n +1}+b \,x^{m +1} \left (m +n +1\right )\right )}{\left (m +1\right ) \left (m +n +1\right )}}}{\left (a \int \frac {x^{n -1} {\mathrm e}^{\frac {x^{m} c x \left (a \left (m +1\right ) x^{n}+b \left (m +n +1\right )\right )}{\left (m +1\right ) \left (m +n +1\right )}}}{\left (a \,x^{n}+b \right )^{2}}d x n -c_1 \right ) \left (a^{2} x^{2 n}+2 a \,x^{n} b +b^{2}\right )} \]

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_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (x^(m+n)*a*c*x+x^m*b 
*c*x+n-1)/x*diff(y(x),x)-a*n*x^(n-1)*c*x^m*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         -> hypergeometric 
            -> heuristic approach 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         -> Mathieu 
            -> Equivalence to the rational form of Mathieu ODE under a power @\ 
 Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\ 
(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
      <- unable to find a useful change of variables 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying 2nd order, integrating factor of the form mu(x,y) 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
         -> 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 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a powe\ 
r @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under \ 
a power @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under \ 
a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(\ 
int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
         -> Trying changes of variables to rationalize or make the ODE simpler 
         <- unable to find a useful change of variables 
            trying a symmetry of the form [xi=0, eta=F(x)] 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE, diff(y(x),x)-(-a*n*x^(n-1)*y(x)^2+y(x)+(x^( 
m+n)*a*c+x^m*b*c)*y(x)*x-x^2*c*x^m)/x, y(x), explicit 
      *** Sublevel 2 *** 
      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_symmetries 
      trying inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
trying inverse_Riccati 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
   -> Computing symmetries using: way = 4 
   -> Computing symmetries using: way = 2 
   -> Computing symmetries using: way = 6 
[0, exp(-b*c/(m+1)*x^m*x-a*c/(m+n+1)*x^m*x^n*x+2*ln(a*x^n+b))*(y-x^n/x/(x^(n-1) 
)/(a*x^n+b))^2] 
   <- successful computation of symmetries. 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-13235 a \,x^{13234} y \left (x \right )^{2}+c \,x^{m} \left (a \,x^{13235}+b \right ) y \left (x \right )-c \,x^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-13235 a \,x^{13234} y \left (x \right )^{2}+c \,x^{m} \left (a \,x^{13235}+b \right ) y \left (x \right )-c \,x^{m} \end {array} \]

2.2.29.3 ✓ Mathematica. Time used: 5.554 (sec). Leaf size: 304

ode=D[y[x],x]==-a*n*x^(n-1)*y[x]^2+c*x^m*(a*x^n+b)*y[x]-c*x^m; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.2.29.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
m = symbols("m") 
n = symbols("n") 
y = Function("y") 
ode = Eq(a*n*x**(n - 1)*y(x)**2 - c*x**m*(a*x**n + b)*y(x) + c*x**m + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -a*c*x**(m + n)*y(x) + a*n*x**(n - 1)*y(x)**2 - b*c*x**m*y(x) + c*x**m + Derivative(y(x), x) cannot be solved by the factorable group method