Problem 29

2.11.3 Problem 29

2.11.3.1 Solved using ﬁrst_order_ode_riccati
2.11.3.2 ✓Maple
2.11.3.3 ✓Mathematica
2.11.3.4 ✗Sympy

Internal problem ID [13391]
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 : 29
Date solved : Sunday, January 18, 2026 at 07:50:06 PM
CAS classiﬁcation : [_Riccati]

2.11.3.1 Solved using ﬁrst_order_ode_riccati

2.116 (sec)

Entering ﬁrst order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=c\), \(f_1(x)=b \tan \left (x \right )\) and \(f_2(x)=a\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u a} \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 &=b \tan \left (x \right ) a\\ f_2^2 f_0 &=a^{2} c \end{align*}

Substituting the above terms back in equation (2) gives

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

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 \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_2 \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) \end{equation}

Taking derivative gives

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

Doing change of constants, the above solution becomes

\[ y = -\frac {-\frac {\cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \left (\frac {1}{2}-\frac {b}{2}\right ) \sin \left (x \right ) \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )}{\cos \left (x \right )}+\frac {\cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \left (\left (\frac {\sqrt {4 a c +b^{2}}}{2}+1-\frac {b}{2}\right ) \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}\right ) \sin \left (x \right ) \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \cos \left (x \right )}{\sin \left (x \right )^{2}-1}-\frac {c_3 \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \left (\frac {1}{2}-\frac {b}{2}\right ) \sin \left (x \right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )}{\cos \left (x \right )}+\frac {c_3 \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \left (\left (\frac {\sqrt {4 a c +b^{2}}}{2}+1-\frac {b}{2}\right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}\right ) \sin \left (x \right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \cos \left (x \right )}{\sin \left (x \right )^{2}-1}}{a \left (\cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_3 \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {-\left (-\cos \left (x \right )^{2} \sqrt {4 a c +b^{2}}+\left (b -1\right ) \sin \left (x \right )^{2}-\cos \left (x \right )^{2}-b +1\right ) \sin \left (x \right ) \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-\left (-\cos \left (x \right )^{2} \sqrt {4 a c +b^{2}}+\left (b -1\right ) \sin \left (x \right )^{2}-\cos \left (x \right )^{2}-b +1\right ) c_3 \sin \left (x \right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\cos \left (x \right )^{2} \left (b -\sqrt {4 a c +b^{2}}-2\right ) \left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_3 +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )}{2 \cos \left (x \right ) \left (\sin \left (x \right )^{2}-1\right ) a \left (\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_3 \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )} \\ \end{align*}

Summary of solutions found

2.11.3.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 261

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

\[ y = \frac {-\left (-\cos \left (x \right )^{2} \sqrt {4 a c +b^{2}}+\left (b -1\right ) \sin \left (x \right )^{2}-\cos \left (x \right )^{2}-b +1\right ) \sin \left (x \right ) \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-c_1 \left (-\cos \left (x \right )^{2} \sqrt {4 a c +b^{2}}+\left (b -1\right ) \sin \left (x \right )^{2}-\cos \left (x \right )^{2}-b +1\right ) \sin \left (x \right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\left (-\sqrt {4 a c +b^{2}}+b -2\right ) \left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_1 +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \cos \left (x \right )^{2}}{2 \cos \left (x \right ) \left (\sin \left (x \right )^{2}-1\right ) a \left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_1 +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )} \]

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 sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = b*tan(x)*diff(y(x),x 
)-a*c*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 
         <- No Liouvillian solutions exists 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            <- Legendre successful 
         <- special function solution successful 
         Change of variables used: 
            [x = arcsin(t)] 
         Linear ODE actually solved: 
            a*c*u(t)+(-b*t-t)*diff(u(t),t)+(-t^2+1)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful

Maple step by step

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

2.11.3.3 ✓ Mathematica. Time used: 0.763 (sec). Leaf size: 488

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

\begin{align*} y(x)&\to \frac {\sin (x) \left (\left (-b^3+3 b^2+b-3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+2\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+2\right ),\frac {3-b}{2},\cos ^2(x)\right )+\cos (x) \left ((b+1) \cos (x) (a c+b-1) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+6\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+6\right ),\frac {5-b}{2},\cos ^2(x)\right )+a i^{b+1} (b-3) c c_1 \cos ^b(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}+4\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}+4\right ),\frac {b+3}{2},\cos ^2(x)\right )\right )\right )}{a (b-3) (b+1) \left (\cos (x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+2\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+2\right ),\frac {3-b}{2},\cos ^2(x)\right )-i i^b c_1 \cos ^b(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}\right ),\frac {b+1}{2},\cos ^2(x)\right )\right )}\\ y(x)&\to -\frac {c \sin (x) \cos (x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}+4\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}+4\right ),\frac {b+3}{2},\cos ^2(x)\right )}{(b+1) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}\right ),\frac {b+1}{2},\cos ^2(x)\right )} \end{align*}

2.11.3.4 ✗ Sympy

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

NotImplementedError : The given ODE -a*y(x)**2 - b*y(x)*tan(x) - c + Derivative(y(x), x) cannot be s

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)) 
 
('1st_power_series', 'lie_group')

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