2.11.1 Problem 27

Internal problem ID [13389]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.6-3. Equations with tangent.
Problem number : 27
Date solved : Friday, December 19, 2025 at 03:42:25 AM
CAS classification : [_Riccati]

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

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

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

Substituting the above terms back in equation (2) gives

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

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 \sqrt {\cos \left (\lambda x \right )}\, \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )+c_2 \sqrt {\cos \left (\lambda x \right )}\, \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda \sin \left (\lambda x \right )}{2 \sqrt {\cos \left (\lambda x \right )}}+\frac {c_1 \cos \left (\lambda x \right )^{{3}/{2}} \left (\operatorname {LegendreP}\left (1+\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )-\left (1+\frac {2 a -\lambda }{2 \lambda }\right ) \sin \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \lambda }{\sin \left (\lambda x \right )^{2}-1}-\frac {c_2 \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda \sin \left (\lambda x \right )}{2 \sqrt {\cos \left (\lambda x \right )}}+\frac {c_2 \cos \left (\lambda x \right )^{{3}/{2}} \left (\operatorname {LegendreQ}\left (1+\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )-\left (1+\frac {2 a -\lambda }{2 \lambda }\right ) \sin \left (\lambda x \right ) \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \lambda }{\sin \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 {\left (-\operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_1 a \sin \left (\lambda x \right )-\operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_2 a \sin \left (\lambda x \right )+\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_1 \lambda +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_2 \lambda \right ) \sec \left (\lambda x \right )}{c_1 \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )+c_2 \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {-\frac {\operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda \sin \left (\lambda x \right )}{2 \sqrt {\cos \left (\lambda x \right )}}+\frac {\cos \left (\lambda x \right )^{{3}/{2}} \left (\operatorname {LegendreP}\left (1+\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )-\left (1+\frac {2 a -\lambda }{2 \lambda }\right ) \sin \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \lambda }{\sin \left (\lambda x \right )^{2}-1}-\frac {c_3 \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda \sin \left (\lambda x \right )}{2 \sqrt {\cos \left (\lambda x \right )}}+\frac {c_3 \cos \left (\lambda x \right )^{{3}/{2}} \left (\operatorname {LegendreQ}\left (1+\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )-\left (1+\frac {2 a -\lambda }{2 \lambda }\right ) \sin \left (\lambda x \right ) \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \lambda }{\sin \left (\lambda x \right )^{2}-1}}{\sqrt {\cos \left (\lambda x \right )}\, \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )+c_3 \sqrt {\cos \left (\lambda x \right )}\, \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {\left (\sin \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) a +\operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_3 a \sin \left (\lambda x \right )-\lambda \left (\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_3 +\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right )\right ) \sec \left (\lambda x \right )}{\operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \\ \end{align*}

The solution

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

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

2.11.1.1 ✓ Maple. Time used: 0.003 (sec). Leaf size: 205

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

\[ y = -\frac {\sec \left (\lambda x \right ) \left (\sin \left (\lambda x \right ) \operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) a +\operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_1 a \sin \left (\lambda x \right )-\lambda \left (c_1 \operatorname {LegendreQ}\left (\frac {\lambda +2 a}{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )+\operatorname {LegendreP}\left (\frac {\lambda +2 a}{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right )\right )}{\operatorname {LegendreQ}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_1 +\operatorname {LegendreP}\left (\frac {2 a -\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \]

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*tan(lambda*x)^2 
-a*tan(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 
               -> Whittaker 
                  -> hyper3: Equivalence to 1F1 under a power @ Moebius 
               -> hypergeometric 
                  -> heuristic approach 
                  <- heuristic approach successful 
               <- hypergeometric successful 
            <- special function solution successful 
               -> Trying to convert hypergeometric functions to elementary form\ 
... 
               <- elementary form is not straightforward to achieve - returning\ 
 special function solution free of uncomputed integrals 
            <- Kovacics algorithm successful 
         Change of variables used: 
            [x = 1/lambda*arccos(t)] 
         Linear ODE actually solved: 
            (a^2*t^2-a^2+a*lambda)*u(t)-t^3*lambda^2*diff(u(t),t)+(-lambda^2*t^\ 
4+lambda^2*t^2)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.11.1.2 ✓ Mathematica. Time used: 1.335 (sec). Leaf size: 357

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

\begin{align*} y(x)&\to \frac {2 \left (a c_1 \sin ^2(\lambda x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )+(2 a-\lambda ) \sqrt {\sin ^2(\lambda x)} \left (a \sin (\lambda x) \cos ^{\frac {2 a}{\lambda }-1}(\lambda x)-c_1\right )\right )}{2 (2 a-\lambda ) \sqrt {\sin ^2(\lambda x)} \cos ^{\frac {2 a}{\lambda }}(\lambda x)+c_1 \sin (2 \lambda x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )}\\ y(x)&\to \frac {\tan (\lambda x) \left (a \sqrt {\sin ^2(\lambda x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )-2 a+\lambda \right )}{\sqrt {\sin ^2(\lambda x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )}\\ y(x)&\to \frac {\tan (\lambda x) \left (a \sqrt {\sin ^2(\lambda x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )-2 a+\lambda \right )}{\sqrt {\sin ^2(\lambda x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-\frac {a}{\lambda },\frac {3}{2}-\frac {a}{\lambda },\cos ^2(x \lambda )\right )} \end{align*}

2.11.1.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
lambda_ = symbols("lambda_") 
y = Function("y") 
ode = Eq(-a*lambda_ - a*(-a + lambda_)*tan(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*tan(lambda_*x)**2 - a*lambda_*tan(lambda_*x)**2 - a*lambda_
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', '1st_power_series', 'lie_group')