2.9.9 Problem 11

Internal problem ID [13375]
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 : 11
Date solved : Wednesday, December 31, 2025 at 02:06:40 PM
CAS classification : [_Riccati]

2.9.9.1 Solved using first_order_ode_riccati

38.498 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime } x&=a \sin \left (\lambda x \right )^{m} y^{2}+k y+a \,b^{2} x^{2 k} \sin \left (\lambda x \right )^{m} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {a \sin \left (\lambda x \right )^{m} y^{2}+k y+a \,b^{2} x^{2 k} \sin \left (\lambda x \right )^{m}}{x} \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 \(f_0(x)=\frac {a \,b^{2} x^{2 k} \sin \left (\lambda x \right )^{m}}{x}\), \(f_1(x)=\frac {k}{x}\) and \(f_2(x)=\frac {a \sin \left (\lambda x \right )^{m}}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u a \sin \left (\lambda x \right )^{m}}{x}} \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' &=\frac {a \sin \left (\lambda x \right )^{m} m \lambda \cos \left (\lambda x \right )}{\sin \left (\lambda x \right ) x}-\frac {a \sin \left (\lambda x \right )^{m}}{x^{2}}\\ f_1 f_2 &=\frac {k a \sin \left (\lambda x \right )^{m}}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{3} \sin \left (\lambda x \right )^{3 m} b^{2} x^{2 k}}{x^{3}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {a \sin \left (\lambda x \right )^{m} u^{\prime \prime }\left (x \right )}{x}-\left (\frac {a \sin \left (\lambda x \right )^{m} m \lambda \cos \left (\lambda x \right )}{\sin \left (\lambda x \right ) x}-\frac {a \sin \left (\lambda x \right )^{m}}{x^{2}}+\frac {k a \sin \left (\lambda x \right )^{m}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{3} \sin \left (\lambda x \right )^{3 m} b^{2} x^{2 k} u \left (x \right )}{x^{3}} = 0 \]

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}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}+c_2 \,{\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x} \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = i c_1 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}-i c_2 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u a \sin \left (\lambda x \right )^{m}}{x}} \\ y &= -\frac {\left (i c_1 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}-i c_2 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right ) \sin \left (\lambda x \right )^{-m} x}{a \left (c_1 \,{\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}+c_2 \,{\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (i a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}-i c_3 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right ) \sin \left (\lambda x \right )^{-m} x}{a \left ({\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}+c_3 \,{\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {\left (i a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}-i c_3 a b \,x^{k -1} \sin \left (\lambda x \right )^{m} {\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right ) \sin \left (\lambda x \right )^{-m} x}{a \left ({\mathrm e}^{\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}+c_3 \,{\mathrm e}^{-\int i a b \,x^{k -1} \sin \left (\lambda x \right )^{m}d x}\right )} \\ \end{align*}

2.9.9.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

22.399 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime } x&=a \sin \left (\lambda x \right )^{m} y^{2}+k y+a \,b^{2} x^{2 k} \sin \left (\lambda x \right )^{m} \\ \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) & =\frac {a \,b^{2} x^{2 k} \sin \left (\lambda x \right )^{m}}{x}\\ f_1(x) & =\frac {k}{x}\\ f_2(x) &=\frac {a \sin \left (\lambda x \right )^{m}}{x} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -i b \,x^{k} \]

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

\[ y = -i b \,x^{k}+\frac {{\mathrm e}^{\int \left (-\frac {2 i b \,x^{k} a \sin \left (\lambda x \right )^{m}}{x}+\frac {k}{x}\right )d x}}{c_1 -\int \frac {{\mathrm e}^{\int \left (-\frac {2 i b \,x^{k} a \sin \left (\lambda x \right )^{m}}{x}+\frac {k}{x}\right )d x} a \sin \left (\lambda x \right )^{m}}{x}d x} \]

Summary of solutions found

\begin{align*} y &= -i b \,x^{k}+\frac {{\mathrm e}^{\int \left (-\frac {2 i b \,x^{k} a \sin \left (\lambda x \right )^{m}}{x}+\frac {k}{x}\right )d x}}{c_1 -\int \frac {{\mathrm e}^{\int \left (-\frac {2 i b \,x^{k} a \sin \left (\lambda x \right )^{m}}{x}+\frac {k}{x}\right )d x} a \sin \left (\lambda x \right )^{m}}{x}d x} \\ \end{align*}

2.9.9.3 ✓ Maple. Time used: 0.017 (sec). Leaf size: 31

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

\[ y = -\tan \left (-b a \int x^{-1+k} \sin \left (\lambda x \right )^{m}d x +c_1 \right ) b \,x^{k} \]

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 
<- Chini successful
 

Maple step by step

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

2.9.9.4 ✓ Mathematica. Time used: 0.464 (sec). Leaf size: 50

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

\begin{align*} y(x)&\to \sqrt {b^2} x^k \tan \left (\sqrt {b^2} \int _1^xa K[1]^{k-1} \sin ^m(\lambda K[1])dK[1]+c_1\right ) \end{align*}

2.9.9.5 ✗ Sympy

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

Timed Out