2.17.1 Problem 28

Internal problem ID [13442]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.7-4. Equations containing arccotangent.
Problem number : 28
Date solved : Wednesday, December 31, 2025 at 09:34:47 PM
CAS classification : [_Riccati]

2.17.1.1 Solved using first_order_ode_riccati

13.578 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+\lambda \operatorname {arccot}\left (x \right )^{n} y-a^{2}+a \lambda \operatorname {arccot}\left (x \right )^{n} \\ \end{align*}

In canonical form the ODE is

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \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' &=0\\ f_1 f_2 &=\operatorname {arccot}\left (x \right )^{n} \lambda \\ f_2^2 f_0 &=a \lambda \operatorname {arccot}\left (x \right )^{n}-a^{2} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )-\operatorname {arccot}\left (x \right )^{n} \lambda u^{\prime }\left (x \right )+\left (a \lambda \operatorname {arccot}\left (x \right )^{n}-a^{2}\right ) u \left (x \right ) = 0 \]

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 ) = {\mathrm e}^{\int \frac {\left (\int -{\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}d x {\mathrm e}^{\int \left (-\operatorname {arccot}\left (x \right )^{n} \lambda +2 a \right )d x} a +c_1 \,{\mathrm e}^{\int \left (-\operatorname {arccot}\left (x \right )^{n} \lambda +2 a \right )d x} a -1\right ) {\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}}{c_1 +\int -{\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}d x}d x} c_2 \end{equation}

Taking derivative gives

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

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

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

Summary of solutions found

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

2.17.1.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

6.743 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+\lambda \operatorname {arccot}\left (x \right )^{n} y-a^{2}+a \lambda \operatorname {arccot}\left (x \right )^{n} \\ \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 \lambda \operatorname {arccot}\left (x \right )^{n}-a^{2}\\ f_1(x) & =\operatorname {arccot}\left (x \right )^{n} \lambda \\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -a \]

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

\[ y = -a +\frac {{\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}}{c_1 -\int {\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}d x} \]

Summary of solutions found

\begin{align*} y &= -a +\frac {{\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}}{c_1 -\int {\mathrm e}^{\int \left (\operatorname {arccot}\left (x \right )^{n} \lambda -2 a \right )d x}d x} \\ \end{align*}

2.17.1.3 ✓ Maple. Time used: 0.068 (sec). Leaf size: 71

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

\[ y = \frac {-\int {\mathrm e}^{-\int \left (-\lambda \operatorname {arccot}\left (x \right )^{n}+2 a \right )d x}d x a -c_1 a -{\mathrm e}^{-\int \left (-\lambda \operatorname {arccot}\left (x \right )^{n}+2 a \right )d x}}{c_1 +\int {\mathrm e}^{-\int \left (-\lambda \operatorname {arccot}\left (x \right )^{n}+2 a \right )d x}d x} \]

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) = lambda*arccot(x)^n* 
diff(y(x),x)+(a^2-a*lambda*arccot(x)^n)*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 
            -> trying with_periodic_functions in the 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 
               -> trying with_periodic_functions in the 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)-(y(x)^2+y(x)+lambda*arccot(x)^ 
n*y(x)*x+x^2*(-a^2+a*lambda*arccot(x)^n))/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)] 
   <- symmetry pattern of the form [0, F(x)*G(y)] successful 
   <- Riccati with symmetry pattern of the form [0,F(x)*G(y)] successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+\lambda \mathrm {arccot}\left (x \right )^{13442} y \left (x \right )-a^{2}+a \lambda \mathrm {arccot}\left (x \right )^{13442} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=y \left (x \right )^{2}+\lambda \mathrm {arccot}\left (x \right )^{13442} y \left (x \right )-a^{2}+a \lambda \mathrm {arccot}\left (x \right )^{13442} \end {array} \]

2.17.1.4 ✓ Mathematica. Time used: 1.228 (sec). Leaf size: 210

ode=D[y[x],x]==y[x]^2+\[Lambda]*ArcCot[x]^n*y[x]-a^2+a*\[Lambda]*ArcCot[x]^n; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^x-\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \cot ^{-1}(K[1])^n\right )dK[1]\right ) \left (-\lambda \cot ^{-1}(K[2])^n+a-y(x)\right )}{n \lambda (a+y(x))}dK[2]+\int _1^{y(x)}\left (-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \cot ^{-1}(K[1])^n\right )dK[1]\right ) \left (-\lambda \cot ^{-1}(K[2])^n+a-K[3]\right )}{n \lambda (a+K[3])^2}+\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \cot ^{-1}(K[1])^n\right )dK[1]\right )}{n \lambda (a+K[3])}\right )dK[2]-\frac {\exp \left (-\int _1^x\left (2 a-\lambda \cot ^{-1}(K[1])^n\right )dK[1]\right )}{n \lambda (a+K[3])^2}\right )dK[3]=c_1,y(x)\right ] \]

2.17.1.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
lambda_ = symbols("lambda_") 
n = symbols("n") 
y = Function("y") 
ode = Eq(a**2 - a*lambda_*(-atan(x) + pi/2)**n - lambda_*(-atan(x) + pi/2)**n*y(x) - y(x)**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out