Problem 26

2.10.11 Problem 26

2.10.11.1 Solved using ﬁrst_order_ode_riccati
2.10.11.2 ✓Maple
2.10.11.3 ✓Mathematica
2.10.11.4 ✗Sympy

Internal problem ID [13388]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.6-2. Equations with cosine.
Problem number : 26
Date solved : Sunday, January 18, 2026 at 07:48:21 PM
CAS classiﬁcation : [_Riccati]

2.10.11.1 Solved using ﬁrst_order_ode_riccati

1.487 (sec)

Entering ﬁrst order ode riccati solver

\begin{align*} \left (a \cos \left (\lambda x \right )+b \right ) \left (y^{\prime }-y^{2}\right )-a \,\lambda ^{2} \cos \left (\lambda x \right )&=0 \\ \end{align*}

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \frac {\cos \left (\lambda x \right ) a y^{2}}{a \cos \left (\lambda x \right )+b}+\frac {a \,\lambda ^{2} \cos \left (\lambda x \right )}{a \cos \left (\lambda x \right )+b}+\frac {b y^{2}}{a \cos \left (\lambda x \right )+b} \]

With Riccati ODE standard form

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

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

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )+\frac {a \,\lambda ^{2} \cos \left (\lambda x \right ) u \left (x \right )}{a \cos \left (\lambda x \right )+b} = 0 \]

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 ) = \left (-2 \cos \left (\frac {\lambda x}{2}\right )^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a b +\sin \left (\frac {\lambda x}{2}\right ) \cos \left (\frac {\lambda x}{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}\, a +\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a b -\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{2}\right ) c_1 +c_2 \left (a \cos \left (\lambda x \right )+b \right ) \end{equation}

Taking derivative gives

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

Doing change of constants, the above solution becomes

\[ y = -\frac {2 \cos \left (\frac {\lambda x}{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a b \sin \left (\frac {\lambda x}{2}\right ) \lambda -\frac {\cos \left (\frac {\lambda x}{2}\right )^{2} \left (a -b \right ) \lambda \left (1+\tan \left (\frac {\lambda x}{2}\right )^{2}\right ) a b}{\sqrt {\left (a -b \right ) \left (a +b \right )}\, \left (-\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )^{2}}{a +b}+1\right )}+\frac {\lambda \cos \left (\frac {\lambda x}{2}\right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}\, a}{2}-\frac {\sin \left (\frac {\lambda x}{2}\right )^{2} \lambda \sqrt {\left (a -b \right ) \left (a +b \right )}\, a}{2}+\frac {\left (a -b \right ) \lambda \left (1+\tan \left (\frac {\lambda x}{2}\right )^{2}\right ) a b}{2 \sqrt {\left (a -b \right ) \left (a +b \right )}\, \left (-\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )^{2}}{a +b}+1\right )}-\frac {\left (a -b \right ) \lambda \left (1+\tan \left (\frac {\lambda x}{2}\right )^{2}\right ) b^{2}}{2 \sqrt {\left (a -b \right ) \left (a +b \right )}\, \left (-\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )^{2}}{a +b}+1\right )}-c_3 a \lambda \sin \left (\lambda x \right )}{-2 \cos \left (\frac {\lambda x}{2}\right )^{2} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a b +\sin \left (\frac {\lambda x}{2}\right ) \cos \left (\frac {\lambda x}{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}\, a +\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a b -\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{2}+c_3 \left (a \cos \left (\lambda x \right )+b \right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {\left (2 \sqrt {a^{2}-b^{2}}\, \operatorname {arctanh}\left (\frac {\tan \left (\frac {\lambda x}{2}\right ) \left (a -b \right )}{\sqrt {a^{2}-b^{2}}}\right ) a b \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right )-2 \sqrt {a^{2}-b^{2}}\, c_3 a \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right )+\left (a -b \right ) \left (a +b \right ) \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}-\frac {b}{2}\right )\right ) \lambda }{-2 \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}+\frac {b}{2}\right ) b \sqrt {a^{2}-b^{2}}\, \operatorname {arctanh}\left (\frac {\tan \left (\frac {\lambda x}{2}\right ) \left (a -b \right )}{\sqrt {a^{2}-b^{2}}}\right )+2 \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}+\frac {b}{2}\right ) c_3 \sqrt {a^{2}-b^{2}}+a \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right )} \\ \end{align*}

Summary of solutions found

2.10.11.2 ✓ Maple. Time used: 0.006 (sec). Leaf size: 204

ode:=(cos(lambda*x)*a+b)*(diff(y(x),x)-y(x)^2)-a*lambda^2*cos(lambda*x) = 0; 
dsolve(ode,y(x), singsol=all);

\[ y = \frac {\left (2 \sqrt {a^{2}-b^{2}}\, \operatorname {arctanh}\left (\frac {\tan \left (\frac {\lambda x}{2}\right ) \left (a -b \right )}{\sqrt {a^{2}-b^{2}}}\right ) a b \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right )-2 \sqrt {a^{2}-b^{2}}\, c_1 a \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right )+\left (a -b \right ) \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}-\frac {b}{2}\right ) \left (a +b \right )\right ) \lambda }{\sqrt {a^{2}-b^{2}}\, \left (2 \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}+\frac {b}{2}\right ) b \,\operatorname {arctanh}\left (\frac {\tan \left (\frac {\lambda x}{2}\right ) \left (a -b \right )}{\sqrt {a^{2}-b^{2}}}\right )-\sqrt {a^{2}-b^{2}}\, a \cos \left (\frac {\lambda x}{2}\right ) \sin \left (\frac {\lambda x}{2}\right )-2 c_1 \left (a \cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {a}{2}+\frac {b}{2}\right )\right )} \]

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) = -1/(a*cos(lambda*x)+ 
b)*a*lambda^2*cos(lambda*x)*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)] 
      <- linear_1 successful 
   <- Riccati to 2nd Order successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \cos \left (\lambda x \right )+b \right ) \left (\frac {d}{d x}y \left (x \right )-y \left (x \right )^{2}\right )-a \,\lambda ^{2} \cos \left (\lambda x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\cos \left (\lambda x \right ) y \left (x \right )^{2} a +a \,\lambda ^{2} \cos \left (\lambda x \right )+y \left (x \right )^{2} b}{a \cos \left (\lambda x \right )+b} \end {array} \]

2.10.11.3 ✓ Mathematica. Time used: 1.786 (sec). Leaf size: 231

ode=(a*Cos[\[Lambda]*x]+b)*(D[y[x],x]-y[x]^2)-a*\[Lambda]^2*Cos[\[Lambda]*x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

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

2.10.11.4 ✗ Sympy

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

NotImplementedError : The given ODE Derivative(y(x), x) - (a*lambda_**2*cos(lambda_*x) + a*y(x)**2*c

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')

[prev] [prev-tail] [front] [up]