Problem 8

2.5.8 Problem 8

2.5.8.1 Solved using ﬁrst_order_ode_riccati
2.5.8.2 ✓Maple
2.5.8.3 ✓Mathematica
2.5.8.4 ✗Sympy

Internal problem ID [13327]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.4-1. Equations with hyperbolic sine and cosine
Problem number : 8
Date solved : Wednesday, December 31, 2025 at 01:24:46 PM
CAS classiﬁcation : [_Riccati]

2.5.8.1 Solved using ﬁrst_order_ode_riccati

8.483 (sec)

Entering ﬁrst order ode riccati solver

\begin{align*} y^{\prime }&=\alpha y^{2}+\beta +\gamma \cosh \left (x \right ) \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \alpha y^{2}+\beta +\gamma \cosh \left (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 \(f_0(x)=\gamma \cosh \left (x \right )+\beta \), \(f_1(x)=0\) and \(f_2(x)=\alpha \). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u \alpha } \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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\alpha ^{2} \left (\gamma \cosh \left (x \right )+\beta \right ) \end{align*}

Substituting the above terms back in equation (2) gives

\[ \alpha u^{\prime \prime }\left (x \right )+\alpha ^{2} \left (\gamma \cosh \left (x \right )+\beta \right ) u \left (x \right ) = 0 \]

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 \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_2 \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {i c_1 \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2}+\frac {i c_2 \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u \alpha } \\ y &= -\frac {\frac {i c_1 \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2}+\frac {i c_2 \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2}}{\alpha \left (c_1 \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_2 \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {i \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2}+\frac {i c_3 \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )}{2}}{\alpha \left (\operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_3 \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \]

Simplifying the above gives

\begin{align*} y &= -\frac {i \left (c_3 \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (\operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_3 \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \\ \end{align*}

Summary of solutions found

2.5.8.2 ✓ Maple. Time used: 0.002 (sec). Leaf size: 70

ode:=diff(y(x),x) = alpha*y(x)^2+beta+gamma*cosh(x); 
dsolve(ode,y(x), singsol=all);

\[ y = -\frac {i \left (c_1 \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_1 \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \]

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 Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = -alpha*(beta+gamma* 
cosh(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)] 
         checking if the LODE is missing y 
         -> 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 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebi\ 
us 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a powe\ 
r @ Moebius 
               Equivalence transformation and function parameters: {z = 1/2*t+1\ 
/2}, {kappa = -16*alpha*beta+16*alpha*gamma-4, mu = 32*alpha*gamma} 
               <- Equivalence to the rational form of Mathieu ODE successful 
            <- Mathieu successful 
         <- special function solution successful 
         Change of variables used: 
            [x = arccosh(t)] 
         Linear ODE actually solved: 
            (alpha*gamma*t+alpha*beta)*u(t)+t*diff(u(t),t)+(t^2-1)*diff(diff(u(\ 
t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful

Maple step by step

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

2.5.8.3 ✓ Mathematica. Time used: 0.162 (sec). Leaf size: 140

ode=D[y[x],x]==\[Alpha]*y[x]^2+\[Beta]+\[Gamma]*Cosh[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)&\to -\frac {i c_1 \text {MathieuCPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]-i \text {MathieuSPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]}{2 \alpha c_1 \text {MathieuC}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]-2 \alpha \text {MathieuS}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]}\\ y(x)&\to -\frac {i \text {MathieuCPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]}{2 \alpha \text {MathieuC}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]} \end{align*}

2.5.8.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
Alpha = symbols("Alpha") 
BETA = symbols("BETA") 
Gamma = symbols("Gamma") 
y = Function("y") 
ode = Eq(-Alpha*y(x)**2 - BETA - Gamma*cosh(x) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

NotImplementedError : The given ODE -Alpha*y(x)**2 - BETA - Gamma*cosh(x) + Derivative(y(x), x) cannot be solved by the lie group method

[next] [prev] [prev-tail] [front] [up]