2.9.3 Problem 4

Internal problem ID [13369]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.6-1. Equations with sine
Problem number : 4
Date solved : Wednesday, December 31, 2025 at 02:00:41 PM
CAS classification : [_Riccati]

2.9.3.1 Solved using first_order_ode_riccati

14.849 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

Substituting the above terms back in equation (2) gives

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

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 \,{\mathrm e}^{b x}+c_2 \,{\mathrm e}^{b x} \int -i \beta \,{\mathrm e}^{\frac {-a \cos \left (\beta x \right )-2 b \beta x}{\beta }}d x \end{equation}

Taking derivative gives

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

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

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

Doing change of constants, the above solution becomes

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

Summary of solutions found

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

2.9.3.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.282 (sec)

Entering first order ode riccati guess solver

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

Using trial and error, the following particular solution was found

\[ y_p = -b \]

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

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

Summary of solutions found

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

2.9.3.3 ✓ Maple. Time used: 0.005 (sec). Leaf size: 76

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

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

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: 
   <- Riccati particular case Kamke (b) successful
 

Maple step by step

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

2.9.3.4 ✓ Mathematica. Time used: 2.074 (sec). Leaf size: 208

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

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

2.9.3.5 ✗ Sympy

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

NotImplementedError : The given ODE -a*b*sin(BETA*x) - a*y(x)*sin(BETA*x) + b**2 - y(x)**2 + Derivative(y(x), x) cannot be solved by the lie group method