5.13 problem 13

5.13.1 Solving as riccati ode
5.13.2 Maple step by step solution

Internal problem ID [10460]
Internal file name [OUTPUT/9408_Monday_June_06_2022_02_24_42_PM_90381490/index.tex]

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: 13.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "riccati"

Maple gives the following as the ode type

[_Riccati]

\[ \boxed {y^{\prime }-y^{2}=-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4}} \]

5.13.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}-\lambda ^{2}+\frac {a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n}}{\sinh \left (\lambda x \right )^{4}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4}\right ) u \left (x \right ) &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}

Verification of solutions

\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\sinh \left (\lambda x \right )^{-n -4} \textit {\_Y} \left (x \right ) \cosh \left (\lambda x \right )^{n} a -\textit {\_Y} \left (x \right ) \lambda ^{2}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.

5.13.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-\lambda ^{2}+a \cosh \left (\lambda x \right )^{n} \sinh \left (\lambda x \right )^{-n -4} \end {array} \]

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) = (lambda^2-a*cosh(lambda*x)^n*sinh(lambda*x)^(-n-4))*y(x), y(x)`      * 
      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 a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> 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) 
            trying to convert to a linear ODE with constant coefficients 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> 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) 
            trying to convert to a linear ODE with constant coefficients 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> 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) 
            trying to convert to a linear ODE with constant coefficients 
      <- unable to find a useful change of variables 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE`, diff(y(x), x)-(y(x)^2+y(x)+x^2*(-lambda^2+a*cosh(lambda*x)^n*sinh(lambda*x)^(-n-4)))/x, y(x), e 
      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 inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
trying inverse_Riccati 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
`, `-> Computing symmetries using: way = 4 
`, `-> Computing symmetries using: way = 2 
`, `-> Computing symmetries using: way = 6`
 

Solution by Maple

dsolve(diff(y(x),x)=y(x)^2-lambda^2+a*cosh(lambda*x)^n*sinh(lambda*x)^(-n-4),y(x), singsol=all)
 

\[ \text {No solution found} \]

Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[y'[x]==y[x]^2-\[Lambda]^2+a*Cosh[\[Lambda]*x]^n*Sinh[\[Lambda]*x]^(-n-4),y[x],x,IncludeSingularSolutions -> True]
 

Not solved