2.8.4 Problem 13

Internal problem ID [13356]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-2
Problem number : 13
Date solved : Wednesday, December 31, 2025 at 01:50:12 PM
CAS classification : [_Riccati]

2.8.4.1 Solved using first_order_ode_riccati

4.352 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

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

Substituting the above terms back in equation (2) gives

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

Entering second order change of variable on \(y\) method 2 solverIn normal form the ode

\begin{align*} \left (-n \,x^{n}-x^{n}\right ) \left (\frac {d^{2}u}{d x^{2}}\right )-\left (-\frac {n^{2} x^{n}}{x}-\frac {x^{n} n}{x}+a x \,x^{n} \ln \left (x \right )^{m} \left (-n \,x^{n}-x^{n}\right )\right ) \left (\frac {d u}{d x}\right )-\left (-n \,x^{n}-x^{n}\right )^{2} a \ln \left (x \right )^{m} u = 0\tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}u}{d x^{2}}+p \left (x \right ) \left (\frac {d u}{d x}\right )+q \left (x \right ) u&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-x^{n +1} \ln \left (x \right )^{m} a -\frac {n}{x}\\ q \left (x \right )&=\left (n +1\right ) x^{n} \ln \left (x \right )^{m} a \end{align*}

Applying change of variables on the depndent variable \(u = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(u\).

\begin{align*} \frac {d^{2}}{d x^{2}}v \left (x \right )+\left (\frac {2 n}{x}+p \right ) \left (\frac {d}{d x}v \left (x \right )\right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (-x^{n +1} \ln \left (x \right )^{m} a -\frac {n}{x}\right )}{x}+\left (n +1\right ) x^{n} \ln \left (x \right )^{m} a&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=n +1 \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} \frac {d^{2}}{d x^{2}}v \left (x \right )+\left (\frac {2 n +2}{x}-x^{n +1} \ln \left (x \right )^{m} a -\frac {n}{x}\right ) \left (\frac {d}{d x}v \left (x \right )\right )&=0 \\ \frac {d^{2}}{d x^{2}}v \left (x \right )+\frac {\left (n +2-a \ln \left (x \right )^{m} x^{n +2}\right ) \left (\frac {d}{d x}v \left (x \right )\right )}{x}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = \frac {d}{d x}v \left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} \frac {d}{d x}u \left (x \right )+\frac {\left (n +2-a \ln \left (x \right )^{m} x^{n +2}\right ) u \left (x \right )}{x} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} \frac {d}{d x}u \left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x}\\ &= {\mathrm e}^{\int -\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{\int -\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\int -\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\int -\frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x}\) gives the final solution

\[ u \left (x \right ) = {\mathrm e}^{\int \frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x} c_1 \]

Now that \(u \left (x \right )\) is known, then

\begin{align*} \frac {d}{d x}v \left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= \int {\mathrm e}^{\int \frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x} c_1 d x +c_2 \end{align*}

Hence

\begin{align*} u&= v \left (x \right ) x^{n}\\ &= \left (\int {\mathrm e}^{\int \frac {-n -2+a \ln \left (x \right )^{m} x^{n +2}}{x}d x} c_1 d x +c_2 \right ) x^{n +1}\\ &= x^{n +1} \left (c_1 \int {\mathrm e}^{\int \frac {x^{n +1} \ln \left (x \right )^{m} a x -n -2}{x}d x}d x +c_2 \right )\\ \end{align*}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.8.4.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.887 (sec)

Entering first order ode riccati guess solver

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

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

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

Shows that

\begin{align*} f_0(x) & =-a \ln \left (x \right )^{m}\\ f_1(x) & =a x \,x^{n} \ln \left (x \right )^{m}\\ f_2(x) &=-n \,x^{n}-x^{n} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = x^{-n -1} \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

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

Summary of solutions found

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

2.8.4.3 ✓ Maple. Time used: 0.010 (sec). Leaf size: 155

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

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

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) = (a*x^(n+1)*ln(x)^m*x 
+n)/x*diff(y(x),x)-a*ln(x)^m*x^n*(n+1)*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 
      -> 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 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> 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 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 
      <- 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 
         -> 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 
            trying a symmetry of the form [xi=0, eta=F(x)] 
            checking if the LODE is missing y 
            -> Heun: Equivalence to the GHE or one of its 4 confluent cases und\ 
er a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = e\ 
xp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               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 
         <- 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)-((-x^n*n-x^n)*y(x)^2+y(x)+a*x^ 
(n+1)*ln(x)^m*y(x)*x-x^2*a*ln(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(-Int((a*x^(n+2)*ln(x)^m-2*n-2)/x,x))*(y-1/(x^n)/x)^2] 
   <- successful computation of symmetries. 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-13357 x^{13356} y \left (x \right )^{2}+a \,x^{13357} \ln \left (x \right )^{m} y \left (x \right )-a \ln \left (x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-13357 x^{13356} y \left (x \right )^{2}+a \,x^{13357} \ln \left (x \right )^{m} y \left (x \right )-a \ln \left (x \right )^{m} \end {array} \]

2.8.4.4 ✓ Mathematica. Time used: 2.173 (sec). Leaf size: 248

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

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

2.8.4.5 ✗ Sympy

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

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