2.13.2 Problem 50
Internal
problem
ID
[13410]
Book
:
Handbook
of
exact
solutions
for
ordinary
differential
equations.
By
Polyanin
and
Zaitsev.
Second
edition
Section
:
Chapter
1,
section
1.2.
Riccati
Equation.
subsection
1.2.6-5.
Equations
containing
combinations
of
trigonometric
functions.
Problem
number
:
50
Date
solved
:
Friday, December 19, 2025 at 04:02:58 AM
CAS
classification
:
[_Riccati]
\begin{align*}
y^{\prime }&=a \cos \left (\lambda x \right ) y^{2}+b \cos \left (\lambda x \right ) \sin \left (\lambda x \right )^{n} \\
\end{align*}
Entering first order ode riccati solver\begin{align*}
y^{\prime }&=a \cos \left (\lambda x \right ) y^{2}+b \cos \left (\lambda x \right ) \sin \left (\lambda x \right )^{n} \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= \cos \left (\lambda x \right ) a \,y^{2}+b \cos \left (\lambda x \right ) \sin \left (\lambda x \right )^{n} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[
y' = \cos \left (\lambda x \right ) a \,y^{2}+b \cos \left (\lambda x \right ) \sin \left (\lambda x \right )^{n}
\]
With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows
that \(f_0(x)=b \cos \left (\lambda x \right ) \sin \left (\lambda x \right )^{n}\), \(f_1(x)=0\) and \(f_2(x)=\cos \left (\lambda x \right ) a\). Let \begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\cos \left (\lambda x \right ) a 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' &=-a \lambda \sin \left (\lambda x \right )\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\cos \left (\lambda x \right )^{3} a^{2} b \sin \left (\lambda x \right )^{n} \end{align*}
Substituting the above terms back in equation (2) gives
\[
\cos \left (\lambda x \right ) a u^{\prime \prime }\left (x \right )+a \lambda \sin \left (\lambda x \right ) u^{\prime }\left (x \right )+\cos \left (\lambda x \right )^{3} a^{2} b \sin \left (\lambda x \right )^{n} u \left (x \right ) = 0
\]
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 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )+c_2 \sin \left (\lambda x \right ) \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )
\end{equation}
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = -\frac {c_1 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}+1\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a b \sin \left (\lambda x \right )^{n +2} \cos \left (\lambda x \right )}{\left (n +1\right ) \lambda \sin \left (\lambda x \right )}+c_2 \lambda \cos \left (\lambda x \right ) \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )-\frac {c_2 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}+1\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a b \sin \left (\lambda x \right )^{n +2} \cos \left (\lambda x \right )}{\left (n +3\right ) \lambda }
\end{equation}
Substituting equations (3,4) into (1) results in \begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{\cos \left (\lambda x \right ) a u} \\
y &= \text {Expression too large to display} \\
\end{align*}
Doing
change of constants, the above solution becomes \[
y = -\frac {-\frac {\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}+1\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a b \sin \left (\lambda x \right )^{n +2} \cos \left (\lambda x \right )}{\left (n +1\right ) \lambda \sin \left (\lambda x \right )}+c_3 \lambda \cos \left (\lambda x \right ) \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )-\frac {c_3 \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}+1\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right ) a b \sin \left (\lambda x \right )^{n +2} \cos \left (\lambda x \right )}{\left (n +3\right ) \lambda }}{\cos \left (\lambda x \right ) a \left (\operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +1}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )+c_3 \sin \left (\lambda x \right ) \operatorname {hypergeom}\left (\left [\right ], \left [\frac {n +3}{n +2}\right ], -\frac {a b \sin \left (\lambda x \right )^{n +2}}{\lambda ^{2} \left (n +2\right )^{2}}\right )\right )}
\]
Simplifying the above gives \begin{align*}
\text {Expression too large to display} \\
\end{align*}
The solution \[
\text {Expression too large to display}
\]
was
found not to satisfy the ode or the IC. Hence it is removed.
2.13.2.1 ✓ Maple. Time used: 0.003 (sec). Leaf size: 1039
ode:=diff(y(x),x) = cos(lambda*x)*a*y(x)^2+b*cos(lambda*x)*sin(lambda*x)^n;
dsolve(ode,y(x), singsol=all);
\[
\text {Expression too large to display}
\]
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) = -lambda*sin(lambda*x
)/cos(lambda*x)*diff(y(x),x)-a*cos(lambda*x)^2*b*sin(lambda*x)^n*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
-> Trying an equivalence, under non-integer power transformations,
to LODEs admitting Liouvillian solutions.
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 0F1 ODE
<- Kummer successful
<- special function solution successful
Change of variables used:
[x = arccos(t)/lambda]
Linear ODE actually solved:
4*a*b*(-t^2+1)^(1/2*n)*t^3*u(t)-4*lambda^2*diff(u(t),t)+(-4*lambda^\
2*t^3+4*lambda^2*t)*diff(diff(u(t),t),t) = 0
<- change of variables successful
<- Riccati to 2nd Order successful
2.13.2.2 ✓ Mathematica. Time used: 0.426 (sec). Leaf size: 633
ode=D[y[x],x]==a*Cos[\[Lambda]*x]*y[x]^2+b*Cos[\[Lambda]*x]*Sin[\[Lambda]*x]^n;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} \text {Solution too large to show}\end{align*}
2.13.2.3 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
lambda_ = symbols("lambda_")
n = symbols("n")
y = Function("y")
ode = Eq(-a*y(x)**2*cos(lambda_*x) - b*sin(lambda_*x)**n*cos(lambda_*x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -(a*y(x)**2 + b*sin(lambda_*x)**n)*cos(lambda_*x) + Derivative(y(x), x) cannot be solved by the factorable group method