2.6.1 Problem 18

Internal problem ID [13336]
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 : 18
Date solved : Friday, December 19, 2025 at 03:07:23 AM
CAS classification : [_Riccati]

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

Entering first order ode riccati solver

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

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -a^{2} \tanh \left (\lambda x \right )^{2}-a \tanh \left (\lambda x \right )^{2} \lambda +a \lambda +y^{2} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -a^{2} \tanh \left (\lambda x \right )^{2}-a \tanh \left (\lambda x \right )^{2} \lambda +a \lambda +y^{2} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-a^{2} \tanh \left (\lambda x \right )^{2}-a \tanh \left (\lambda x \right )^{2} \lambda +a \lambda \), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=-a^{2} \tanh \left (\lambda x \right )^{2}-a \tanh \left (\lambda x \right )^{2} \lambda +a \lambda \end{align*}

Substituting the above terms back in equation (2) gives

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

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 ) = c_1 \operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_2 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 \left (\operatorname {LegendreP}\left (1+\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )-\left (1+\frac {a}{\lambda }\right ) \tanh \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \lambda \left (1-\tanh \left (\lambda x \right )^{2}\right )}{\tanh \left (\lambda x \right )^{2}-1}+\frac {c_2 \left (\operatorname {LegendreQ}\left (1+\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )-\left (1+\frac {a}{\lambda }\right ) \tanh \left (\lambda x \right ) \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \lambda \left (1-\tanh \left (\lambda x \right )^{2}\right )}{\tanh \left (\lambda x \right )^{2}-1} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= \frac {\operatorname {LegendreP}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) c_1 \lambda +\operatorname {LegendreQ}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) c_2 \lambda -\tanh \left (\lambda x \right ) \left (c_1 \operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_2 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \left (a +\lambda \right )}{c_1 \operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_2 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {\left (\operatorname {LegendreP}\left (1+\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )-\left (1+\frac {a}{\lambda }\right ) \tanh \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \lambda \left (1-\tanh \left (\lambda x \right )^{2}\right )}{\tanh \left (\lambda x \right )^{2}-1}+\frac {c_3 \left (\operatorname {LegendreQ}\left (1+\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )-\left (1+\frac {a}{\lambda }\right ) \tanh \left (\lambda x \right ) \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \lambda \left (1-\tanh \left (\lambda x \right )^{2}\right )}{\tanh \left (\lambda x \right )^{2}-1}}{\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {\operatorname {LegendreP}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) \lambda +\operatorname {LegendreQ}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) c_3 \lambda -\tanh \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \left (a +\lambda \right )}{\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )} \\ \end{align*}

The solution

\[ y = \frac {\operatorname {LegendreP}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) \lambda +\operatorname {LegendreQ}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) c_3 \lambda -\tanh \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right ) \left (a +\lambda \right )}{\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )} \]

was found not to satisfy the ode or the IC. Hence it is removed.

2.6.1.1 ✓ Maple. Time used: 0.006 (sec). Leaf size: 122

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

\[ y = \frac {-\left (a +\lambda \right ) \tanh \left (\lambda x \right ) \left (c_1 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )\right )+c_1 \lambda \operatorname {LegendreQ}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+\operatorname {LegendreP}\left (\frac {a +\lambda }{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right ) \lambda }{c_1 \operatorname {LegendreQ}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )+\operatorname {LegendreP}\left (\frac {a}{\lambda }, \frac {a}{\lambda }, \tanh \left (\lambda x \right )\right )} \]

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 Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (a^2*tanh(lambda*x)^ 
2+a*tanh(lambda*x)^2*lambda-a*lambda)*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 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 \ 
integrals... 
            -> Trying a solution in terms of special functions: 
               -> Bessel 
               -> elliptic 
               -> Legendre 
               <- Legendre successful 
            <- special function solution successful 
               -> Trying to convert hypergeometric functions to elementary form\ 
... 
               <- elementary form could result into a too large expression - re\ 
turning special function form of solution, free of uncomputed integrals 
            <- Kovacics algorithm successful 
         Change of variables used: 
            [x = arctanh(t)/lambda] 
         Linear ODE actually solved: 
            (-a^2*t^2-a*lambda*t^2+a*lambda)*u(t)+(2*lambda^2*t^3-2*lambda^2*t)\ 
*diff(u(t),t)+(lambda^2*t^4-2*lambda^2*t^2+lambda^2)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.6.1.2 ✓ Mathematica. Time used: 3.391 (sec). Leaf size: 325

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

\begin{align*} y(x)&\to \frac {a \int -\frac {\lambda \left (2-2 e^{-2 \lambda x}\right )}{e^{2 \lambda x}+1} \, de^{2 \lambda x}}{2 \lambda }+\frac {-2 \lambda \exp \left (\int _1^{e^{2 x \lambda }}\frac {\frac {a (K[1]-1)}{\lambda (K[1]+1)}-1}{K[1]}dK[1]+2 \lambda x\right )-2 \lambda \exp \left (\int _1^{e^{2 x \lambda }}\frac {\frac {a (K[1]-1)}{\lambda (K[1]+1)}-1}{K[1]}dK[1]+4 \lambda x\right )+a \left (e^{2 \lambda x}-1\right ) \int _1^{e^{2 x \lambda }}\exp \left (\int _1^{K[2]}\frac {\frac {a (K[1]-1)}{\lambda (K[1]+1)}-1}{K[1]}dK[1]\right )dK[2]+a c_1 e^{2 \lambda x}-a c_1}{\left (e^{2 \lambda x}+1\right ) \left (\int _1^{e^{2 x \lambda }}\exp \left (\int _1^{K[2]}\frac {\frac {a (K[1]-1)}{\lambda (K[1]+1)}-1}{K[1]}dK[1]\right )dK[2]+c_1\right )}\\ y(x)&\to \frac {a \left (e^{2 \lambda x}-1\right )}{e^{2 \lambda x}+1}+\frac {a \int -\frac {\lambda \left (2-2 e^{-2 \lambda x}\right )}{e^{2 \lambda x}+1} \, de^{2 \lambda x}}{2 \lambda } \end{align*}

2.6.1.3 ✗ Sympy

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

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