2.6.7 Problem 24

Internal problem ID [13342]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.4-2. Equations with hyperbolic tangent and cotangent.
Problem number : 24
Date solved : Wednesday, December 31, 2025 at 01:35:44 PM
CAS classification : [_Riccati]

2.6.7.1 Solved using first_order_ode_riccati

4.221 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}+a x \coth \left (b x \right )^{m} y+a \coth \left (b x \right )^{m} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= y^{2}+a x \coth \left (b x \right )^{m} y+a \coth \left (b 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 \(f_0(x)=a \coth \left (b x \right )^{m}\), \(f_1(x)=\coth \left (b x \right )^{m} a x\) 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 &=\coth \left (b x \right )^{m} a x\\ f_2^2 f_0 &=a \coth \left (b x \right )^{m} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )-\coth \left (b x \right )^{m} a x u^{\prime }\left (x \right )+a \coth \left (b x \right )^{m} u \left (x \right ) = 0 \]

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

\begin{align*} \frac {d^{2}u}{d x^{2}}-\coth \left (b x \right )^{m} a x \left (\frac {d u}{d x}\right )+a \coth \left (b 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 )&=-\coth \left (b x \right )^{m} a x\\ q \left (x \right )&=a \coth \left (b x \right )^{m} \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}}-n \coth \left (b x \right )^{m} a +a \coth \left (b x \right )^{m}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} 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}{x}-\coth \left (b x \right )^{m} a x \right ) \left (\frac {d}{d x}v \left (x \right )\right )&=0 \\ \frac {d^{2}}{d x^{2}}v \left (x \right )+\left (\frac {2}{x}-\coth \left (b x \right )^{m} a x \right ) \left (\frac {d}{d x}v \left (x \right )\right )&=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 )+\left (\frac {2}{x}-\coth \left (b x \right )^{m} a x \right ) u \left (x \right ) = 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 {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}d x}\\ &= {\mathrm e}^{\int -\frac {\coth \left (b x \right )^{m} a \,x^{2}-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 {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}d x}\right ) &= 0 \end{align*}

Integrating gives

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

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

\[ u \left (x \right ) = {\mathrm e}^{\int \frac {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_1 \]

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 {\coth \left (b x \right )^{m} a \,x^{2}-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 {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_1 d x +c_2 \right ) x\\ &= \left (c_1 \int {\mathrm e}^{\int \frac {\coth \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x +c_2 \right ) x\\ \end{align*}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.6.7.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

9.182 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=y^{2}+a x \coth \left (b x \right )^{m} y+a \coth \left (b 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 \coth \left (b x \right )^{m}\\ f_1(x) & =\coth \left (b x \right )^{m} a x\\ f_2(x) &=1 \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -\frac {1}{x} \]

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

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

Summary of solutions found

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

2.6.7.3 ✓ Maple. Time used: 0.007 (sec). Leaf size: 85

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

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

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 
found: 2 potential symmetries. Proceeding with integration step
 

Maple step by step

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

2.6.7.4 ✓ Mathematica. Time used: 1.668 (sec). Leaf size: 126

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

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

2.6.7.5 ✗ Sympy

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

Timed Out