2.10.6 Problem 21
Internal
problem
ID
[13383]
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
:
21
Date
solved
:
Wednesday, December 31, 2025 at 02:13:33 PM
CAS
classification
:
[_Riccati]
2.10.6.1 Solved using first_order_ode_riccati
41.705 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime }&=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y^{2}+\lambda -a +a \cos \left (\lambda x \right )^{2} \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= y^{2} \cos \left (\lambda x \right )^{2} a +\lambda y^{2}+a \cos \left (\lambda x \right )^{2}-a +\lambda \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 \(f_0(x)=a \cos \left (\lambda x \right )^{2}-a +\lambda \), \(f_1(x)=0\) and \(f_2(x)=\lambda +a \cos \left (\lambda x \right )^{2}\). Let \begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u \left (\lambda +a \cos \left (\lambda x \right )^{2}\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' &=-2 \sin \left (\lambda x \right ) a \lambda \cos \left (\lambda x \right )\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right )^{2} \left (a \cos \left (\lambda x \right )^{2}-a +\lambda \right ) \end{align*}
Substituting the above terms back in equation (2) gives
\[
\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) u^{\prime \prime }\left (x \right )+2 \sin \left (\lambda x \right ) a \lambda \cos \left (\lambda x \right ) u^{\prime }\left (x \right )+\left (\lambda +a \cos \left (\lambda x \right )^{2}\right )^{2} \left (a \cos \left (\lambda x \right )^{2}-a +\lambda \right ) u \left (x \right ) = 0
\]
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 ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}+c_2 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \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 ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}-\frac {c_1 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}}{2}-c_2 \lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x -\frac {c_2 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x}{2}+2 i c_2 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )
\end{equation}
Substituting equations (3,4) into (1) results in \begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{u \left (\lambda +a \cos \left (\lambda x \right )^{2}\right )} \\
y &= -\frac {-c_1 \lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}-\frac {c_1 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}}{2}-c_2 \lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x -\frac {c_2 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x}{2}+2 i c_2 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )}{\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) \left (c_1 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}+c_2 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x \right )} \\
\end{align*}
Doing
change of constants, the above solution becomes \[
y = -\frac {-\lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}-\frac {\cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}}{2}-c_3 \lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x -\frac {c_3 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x}{2}+2 i c_3 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )}{\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) \left (\cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}+c_3 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x \right )}
\]
Summary of solutions found
\begin{align*}
y &= -\frac {-\lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}-\frac {\cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}}{2}-c_3 \lambda \sin \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x -\frac {c_3 \cos \left (\lambda x \right ) \sin \left (2 \lambda x \right ) a \,{\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x}{2}+2 i c_3 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )}{\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) \left (\cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }}+c_3 \cos \left (\lambda x \right ) {\mathrm e}^{\frac {a \cos \left (2 \lambda x \right )}{4 \lambda }} \int 2 i \lambda \,{\mathrm e}^{-\frac {a \cos \left (2 \lambda x \right )}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x \right )} \\
\end{align*}
2.10.6.2 ✓ Maple. Time used: 0.005 (sec). Leaf size: 102
ode:=diff(y(x),x) = (lambda+a*cos(lambda*x)^2)*y(x)^2+lambda-a+a*cos(lambda*x)^2;
dsolve(ode,y(x), singsol=all);
\[
y = \frac {-2 \lambda \int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x c_1 \tan \left (\lambda x \right )+i \tan \left (\lambda x \right )+2 \,{\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} c_1 \lambda \sec \left (\lambda x \right )^{2}}{-2 \lambda \int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\lambda \sec \left (\lambda x \right )^{2}+a \right )d x c_1 +i}
\]
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) = -2*a*cos(lambda*x)*
lambda*sin(lambda*x)/(lambda+a*cos(lambda*x)^2)*diff(y(x),x)-(lambda+a*cos(
lambda*x)^2)*(a*cos(lambda*x)^2-a+lambda)*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 quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
Solution has integrals. Trying a special function solution free of \
integrals...
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
-> hypergeometric
-> heuristic approach
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Mo\
ebius
-> Mathieu
-> Equivalence to the rational form of Mathieu ODE under a p\
ower @ Moebius
-> Heun: Equivalence to the GHE or one of its 4 confluent cases \
under a power @ Moebius
No special function solution was found.
<- Kovacics algorithm successful
Change of variables used:
[x = 1/2*arccos(t)/lambda]
Linear ODE actually solved:
(4*a^3*t^3+4*a^3*t^2+24*a^2*lambda*t^2-4*a^3*t+16*a^2*lambda*t+48*a\
*lambda^2*t-4*a^3-8*a^2*lambda+16*a*lambda^2+32*lambda^3)*u(t)+(-64*a*lambda^2*\
t-128*lambda^3*t-64*a*lambda^2)*diff(u(t),t)+(-64*a*lambda^2*t^3-64*a*lambda^2*\
t^2-128*lambda^3*t^2+64*a*lambda^2*t+64*a*lambda^2+128*lambda^3)*diff(diff(u(t)\
,t),t) = 0
<- change of variables successful
<- Riccati to 2nd Order successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y \left (x \right )^{2}+\lambda -a +a \cos \left (\lambda x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y \left (x \right )^{2}+\lambda -a +a \cos \left (\lambda x \right )^{2} \end {array} \]
2.10.6.3 ✓ Mathematica. Time used: 7.995 (sec). Leaf size: 191
ode=D[y[x],x]==(\[Lambda]+a*Cos[\[Lambda]*x]^2)*y[x]^2+\[Lambda]-a+a*Cos[\[Lambda]*x]^2;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to \frac {2 \left (c_1 \tan (\lambda x) \int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]+c_1 \sec ^2(\lambda x) \left (-e^{-\frac {a \cos ^2(\lambda x)}{\lambda }}\right )+\tan (\lambda x)\right )}{2+2 c_1 \int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]}\\ y(x)&\to \frac {1}{2} \sec ^2(\lambda x) \left (\sin (2 \lambda x)-\frac {2 e^{-\frac {a \cos ^2(\lambda x)}{\lambda }}}{\int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]}\right ) \end{align*}
2.10.6.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
lambda_ = symbols("lambda_")
y = Function("y")
ode = Eq(-a*cos(lambda_*x)**2 + a - lambda_ - (a*cos(lambda_*x)**2 + lambda_)*y(x)**2 + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -a*y(x)**2*cos(lambda_*x)**2 - a*cos(lambda_*x)**2 + a - lambda_*y(x)**2 - lambda_ + Derivative(y(x), x) cannot be solved by the factorable group method