2.19.28 Problem 29

Internal problem ID [13476]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.8-1. Equations containing arbitrary functions (but not containing their derivatives).
Problem number : 29
Date solved : Sunday, January 18, 2026 at 08:26:11 PM
CAS classification : [_Riccati]

2.19.28.1 Solved using first_order_ode_riccati

8.156 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \lambda \sin \left (\lambda x \right ) y^{2}+f \left (x \right ) \cos \left (\lambda x \right ) y-f \left (x \right ) \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-f \left (x \right )\), \(f_1(x)=f \left (x \right ) \cos \left (\lambda x \right )\) and \(f_2(x)=\lambda \sin \left (\lambda x \right )\). Let

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

Substituting the above terms back in equation (2) gives

\[ \lambda \sin \left (\lambda x \right ) u^{\prime \prime }\left (x \right )-\left (\lambda ^{2} \cos \left (\lambda x \right )+f \left (x \right ) \cos \left (\lambda x \right ) \lambda \sin \left (\lambda x \right )\right ) u^{\prime }\left (x \right )-\lambda ^{2} \sin \left (\lambda x \right )^{2} f \left (x \right ) u \left (x \right ) = 0 \]

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 \cos \left (\lambda x \right )+c_2 \cos \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -c_1 \lambda \sin \left (\lambda x \right )-c_2 \lambda \sin \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x -c_2 \cos \left (\lambda x \right ) \lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \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 \lambda \sin \left (\lambda x \right )} \\ y &= -\frac {-c_1 \lambda \sin \left (\lambda x \right )-c_2 \lambda \sin \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x -c_2 \cos \left (\lambda x \right ) \lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )}{\lambda \sin \left (\lambda x \right ) \left (c_1 \cos \left (\lambda x \right )+c_2 \cos \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {-\lambda \sin \left (\lambda x \right )-c_3 \lambda \sin \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x -c_3 \cos \left (\lambda x \right ) \lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )}{\lambda \sin \left (\lambda x \right ) \left (\cos \left (\lambda x \right )+c_3 \cos \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {-\lambda \sin \left (\lambda x \right )-c_3 \lambda \sin \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x -c_3 \cos \left (\lambda x \right ) \lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )}{\lambda \sin \left (\lambda x \right ) \left (\cos \left (\lambda x \right )+c_3 \cos \left (\lambda x \right ) \int -\lambda \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )} \\ \end{align*}

2.19.28.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 62

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

\[ y = -\frac {c_1 \,{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}}{\lambda \int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x c_1 -1}+\sec \left (\lambda x \right ) \]

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) = cos(lambda*x)*(sin( 
lambda*x)*f(x)+lambda)/sin(lambda*x)*diff(y(x),x)+lambda*sin(lambda*x)*f(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)] 
      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 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 a symmetry of the form [xi=0, eta=F(x)] 
            trying 2nd order exact linear 
            trying symmetries linear in x and y(x) 
            <- linear symmetries successful 
         Change of variables used: 
            [x = arccos(t)/lambda] 
         Linear ODE actually solved: 
            (2*(-t^2+1)^(1/2)*f(arccos(t)/lambda)*t^2-2*(-t^2+1)^(1/2)*f(arccos\ 
(t)/lambda))*u(t)+(-2*f(arccos(t)/lambda)*(-t^2+1)^(1/2)*t^3+2*f(arccos(t)/lamb\ 
da)*(-t^2+1)^(1/2)*t)*diff(u(t),t)+(2*lambda*t^4-4*lambda*t^2+2*lambda)*diff(di\ 
ff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

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

2.19.28.3 ✗ Mathematica

ode=D[y[x],x]==\[Lambda]*Sin[\[Lambda]*x]*y[x]^2+f[x]*Cos[\[Lambda]*x]*y[x]-f[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.19.28.4 ✗ Sympy

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

NotImplementedError : The given ODE -lambda_*y(x)**2*sin(lambda_*x) - f(x)*y(x)*cos(lambda_*x) + f(x
 

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')