5.11 problem 11

5.11.1 Solving as riccati ode
5.11.2 Maple step by step solution

Internal problem ID [10458]
Internal file name [OUTPUT/9406_Monday_June_06_2022_02_24_34_PM_92845045/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: 11.
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]

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

5.11.1 Solving as riccati ode

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

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

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

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

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

\[ y = \frac {-2 \,{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \operatorname {sech}\left (\lambda x \right )^{2} \lambda -2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right ) \tanh \left (\lambda x \right )+c_{3} \tanh \left (\lambda x \right )}{-2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right )+c_{3}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-2 \,{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \operatorname {sech}\left (\lambda x \right )^{2} \lambda -2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right ) \tanh \left (\lambda x \right )+c_{3} \tanh \left (\lambda x \right )}{-2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right )+c_{3}} \\ \end{align*}

Verification of solutions

\[ y = \frac {-2 \,{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \operatorname {sech}\left (\lambda x \right )^{2} \lambda -2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right ) \tanh \left (\lambda x \right )+c_{3} \tanh \left (\lambda x \right )}{-2 \lambda \left (\int -{\mathrm e}^{\frac {\cosh \left (2 \lambda x \right ) a}{2 \lambda }} \left (a -\operatorname {sech}\left (\lambda x \right )^{2} \lambda \right )d x \right )+c_{3}} \] Verified OK.

5.11.2 Maple step by step solution

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

Maple trace Kovacic algorithm successful

`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*cosh(lambda*x)*lambda*sinh(lambda*x)*(diff(y(x), x))/(a*cosh(lambd 
      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 @ 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 
            No special function solution was found. 
         <- Kovacics algorithm successful 
         Change of variables used: 
            [x = 1/2*arccosh(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*lamb 
      <- change of variables successful 
   <- Riccati to 2nd Order successful`
 

Solution by Maple

Time used: 0.016 (sec). Leaf size: 104

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

\[ y \left (x \right ) = \frac {2 \tanh \left (x \lambda \right ) \lambda \left (\int -{\mathrm e}^{\frac {a \cosh \left (2 x \lambda \right )}{2 \lambda }} \left (a -\operatorname {sech}\left (x \lambda \right )^{2} \lambda \right )d x \right ) c_{1} +2 \operatorname {sech}\left (x \lambda \right )^{2} {\mathrm e}^{\frac {a \cosh \left (2 x \lambda \right )}{2 \lambda }} c_{1} \lambda -\tanh \left (x \lambda \right )}{2 \lambda \left (\int -{\mathrm e}^{\frac {a \cosh \left (2 x \lambda \right )}{2 \lambda }} \left (a -\operatorname {sech}\left (x \lambda \right )^{2} \lambda \right )d x \right ) c_{1} -1} \]

Solution by Mathematica

Time used: 49.81 (sec). Leaf size: 211

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

\begin{align*} y(x)\to \frac {\text {sech}^2(\lambda x) \left (c_1 \sinh (2 \lambda x) \int _1^xe^{\frac {a \cosh ^2(\lambda K[1])}{\lambda }} \left (\lambda -a \cosh ^2(\lambda K[1])\right ) \text {sech}^2(\lambda K[1])dK[1]+2 c_1 e^{\frac {a \cosh ^2(\lambda x)}{\lambda }}+\sinh (2 \lambda x)\right )}{2+2 c_1 \int _1^xe^{\frac {a \cosh ^2(\lambda K[1])}{\lambda }} \left (\lambda -a \cosh ^2(\lambda K[1])\right ) \text {sech}^2(\lambda K[1])dK[1]} \\ y(x)\to \frac {1}{2} \text {sech}^2(\lambda x) \left (\frac {2 e^{\frac {a \cosh ^2(\lambda x)}{\lambda }}}{\int _1^xe^{\frac {a \cosh ^2(\lambda K[1])}{\lambda }} \left (\lambda -a \cosh ^2(\lambda K[1])\right ) \text {sech}^2(\lambda K[1])dK[1]}+\sinh (2 \lambda x)\right ) \\ \end{align*}