2.13.10 Problem 59
Internal
problem
ID
[13418]
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
:
59
Date
solved
:
Wednesday, December 31, 2025 at 09:21:49 PM
CAS
classification
:
[_Riccati]
2.13.10.1 Solved using first_order_ode_riccati
23.607 (sec)
Entering first order ode riccati solver
\begin{align*}
y^{\prime }&=\lambda \sin \left (\lambda x \right ) y^{2}+a \sin \left (\lambda x \right ) y-a \tan \left (\lambda x \right ) \\
\end{align*}
In canonical form the ODE is \begin{align*} y' &= F(x,y)\\ &= \lambda \sin \left (\lambda x \right ) y^{2}+a \sin \left (\lambda x \right ) y-a \tan \left (\lambda x \right ) \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 \(f_0(x)=-a \tan \left (\lambda x \right )\), \(f_1(x)=\sin \left (\lambda x \right ) a\) 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 &=\sin \left (\lambda x \right )^{2} a \lambda \\ f_2^2 f_0 &=-\lambda ^{2} \sin \left (\lambda x \right )^{2} a \tan \left (\lambda x \right ) \end{align*}
Substituting the above terms back in equation (2) gives
\[
\lambda \sin \left (\lambda x \right ) u^{\prime \prime }\left (x \right )-\left (\lambda ^{2} \cos \left (\lambda x \right )+\sin \left (\lambda x \right )^{2} a \lambda \right ) u^{\prime }\left (x \right )-\lambda ^{2} \sin \left (\lambda x \right )^{2} a \tan \left (\lambda x \right ) u \left (x \right ) = 0
\]
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 \left (\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \cos \left (\lambda x \right ) a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda \right )
\end{equation}
Taking derivative gives \begin{equation}
\tag{4} u^{\prime }\left (x \right ) = -c_1 \lambda \sin \left (\lambda x \right )-c_2 \,\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \lambda \sin \left (\lambda x \right ) a
\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 \,\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \lambda \sin \left (\lambda x \right ) a}{\lambda \sin \left (\lambda x \right ) \left (c_1 \cos \left (\lambda x \right )+c_2 \left (\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \cos \left (\lambda x \right ) a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda \right )\right )} \\
\end{align*}
Doing change of constants, the above solution becomes \[
y = -\frac {-\lambda \sin \left (\lambda x \right )-c_3 \,\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \lambda \sin \left (\lambda x \right ) a}{\lambda \sin \left (\lambda x \right ) \left (\cos \left (\lambda x \right )+c_3 \left (\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \cos \left (\lambda x \right ) a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda \right )\right )}
\]
Simplifying the above gives
\begin{align*}
y &= \frac {1+c_3 a \,\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )}{\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \cos \left (\lambda x \right ) c_3 a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} c_3 \lambda +\cos \left (\lambda x \right )} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= \frac {1+c_3 a \,\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )}{\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) \cos \left (\lambda x \right ) c_3 a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} c_3 \lambda +\cos \left (\lambda x \right )} \\
\end{align*}
2.13.10.2 ✓ Maple. Time used: 0.003 (sec). Leaf size: 61
ode:=diff(y(x),x) = lambda*sin(lambda*x)*y(x)^2+a*sin(lambda*x)*y(x)-a*tan(lambda*x);
dsolve(ode,y(x), singsol=all);
\[
y = \frac {\operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) c_1 a +1}{\cos \left (\lambda x \right ) \operatorname {Ei}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right ) c_1 a -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} c_1 \lambda +\cos \left (\lambda x \right )}
\]
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) = (a*sin(lambda*x)^2+
lambda*cos(lambda*x))/sin(lambda*x)*diff(y(x),x)+lambda*sin(lambda*x)*a*tan(
lambda*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 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)]
<- linear_1 successful
Change of variables used:
[x = arccos(t)/lambda]
Linear ODE actually solved:
(-2*a*t^4+4*a*t^2-2*a)*u(t)+(2*a*t^5-4*a*t^3+2*a*t)*diff(u(t),t)+(2\
*lambda*t^5-4*lambda*t^3+2*lambda*t)*diff(diff(u(t),t),t) = 0
<- change of variables successful
<- Riccati to 2nd Order successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\lambda \sin \left (\lambda x \right ) y \left (x \right )^{2}+a \sin \left (\lambda x \right ) y \left (x \right )-a \tan \left (\lambda x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\lambda \sin \left (\lambda x \right ) y \left (x \right )^{2}+a \sin \left (\lambda x \right ) y \left (x \right )-a \tan \left (\lambda x \right ) \end {array} \]
2.13.10.3 ✗ Mathematica
ode=D[y[x],x]==\[Lambda]*Sin[\[Lambda]*x]*y[x]^2+a*Sin[\[Lambda]*x]*y[x]-a*Tan[\[Lambda]*x];
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
Not solved
2.13.10.4 ✗ Sympy
from sympy import *
x = symbols("x")
a = symbols("a")
lambda_ = symbols("lambda_")
y = Function("y")
ode = Eq(-a*y(x)*sin(lambda_*x) + a*tan(lambda_*x) - lambda_*y(x)**2*sin(lambda_*x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
NotImplementedError : The given ODE -a*y(x)*sin(lambda_*x) + a*tan(lambda_*x) - lambda_*y(x)**2*sin(lambda_*x) + Derivative(y(x), x) cannot be solved by the lie group method