problem 26

Internal problem ID [10473]
Internal ﬁle name [OUTPUT/9421_Monday_June_06_2022_02_31_39_PM_41931427/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 1.2.4-2. Equations with hyperbolic tangent and cotangent.
Problem number: 26.
ODE order: 1.
ODE degree: 1.

\[ \boxed {y^{\prime }-y^{2}=-2 \lambda ^{2} \tanh \left (\lambda x \right )^{2}-2 \coth \left (\lambda x \right )^{2} \lambda ^{2}} \]

6.9.1 Solving as riccati ode

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

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

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

\[ u \left (x \right ) = \operatorname {sech}\left (\lambda x \right ) \operatorname {csch}\left (\lambda x \right ) \left (-c_{2} \ln \left (\coth \left (\lambda x \right )-1\right )+c_{2} \ln \left (\coth \left (\lambda x \right )+1\right )+c_{1} -2 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right ) \left (2 \cosh \left (\lambda x \right )^{2}-1\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = 2 \left (c_{2} \left (\cosh \left (\lambda x \right )^{2}-\frac {1}{2}\right ) \ln \left (\coth \left (\lambda x \right )-1\right )+c_{2} \left (-\cosh \left (\lambda x \right )^{2}+\frac {1}{2}\right ) \ln \left (\coth \left (\lambda x \right )+1\right )-4 \cosh \left (\lambda x \right )^{5} \sinh \left (\lambda x \right ) c_{2} +4 c_{2} \cosh \left (\lambda x \right )^{3} \sinh \left (\lambda x \right )-\cosh \left (\lambda x \right )^{2} c_{1} +c_{2} \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )+\frac {c_{1}}{2}\right ) \operatorname {sech}\left (\lambda x \right )^{2} \operatorname {csch}\left (\lambda x \right )^{2} \lambda \] Using the above in (1) gives the solution \[ y = -\frac {2 \left (c_{2} \left (\cosh \left (\lambda x \right )^{2}-\frac {1}{2}\right ) \ln \left (\coth \left (\lambda x \right )-1\right )+c_{2} \left (-\cosh \left (\lambda x \right )^{2}+\frac {1}{2}\right ) \ln \left (\coth \left (\lambda x \right )+1\right )-4 \cosh \left (\lambda x \right )^{5} \sinh \left (\lambda x \right ) c_{2} +4 c_{2} \cosh \left (\lambda x \right )^{3} \sinh \left (\lambda x \right )-\cosh \left (\lambda x \right )^{2} c_{1} +c_{2} \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )+\frac {c_{1}}{2}\right ) \operatorname {sech}\left (\lambda x \right ) \operatorname {csch}\left (\lambda x \right ) \lambda }{-c_{2} \ln \left (\coth \left (\lambda x \right )-1\right )+c_{2} \ln \left (\coth \left (\lambda x \right )+1\right )+c_{1} -2 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right ) \left (2 \cosh \left (\lambda x \right )^{2}-1\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {\operatorname {sech}\left (\lambda x \right ) \operatorname {csch}\left (\lambda x \right ) \left (8 \cosh \left (\lambda x \right )^{5} \sinh \left (\lambda x \right )-8 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right )^{3}-2 \cosh \left (\lambda x \right )^{2} \ln \left (\coth \left (\lambda x \right )-1\right )+2 \cosh \left (\lambda x \right )^{2} \ln \left (\coth \left (\lambda x \right )+1\right )+2 \cosh \left (\lambda x \right )^{2} c_{3} -2 \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )+\ln \left (\coth \left (\lambda x \right )-1\right )-\ln \left (\coth \left (\lambda x \right )+1\right )-c_{3} \right ) \lambda }{-4 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right )^{3}+2 \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )-\ln \left (\coth \left (\lambda x \right )-1\right )+\ln \left (\coth \left (\lambda x \right )+1\right )+c_{3}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {sech}\left (\lambda x \right ) \operatorname {csch}\left (\lambda x \right ) \left (8 \cosh \left (\lambda x \right )^{5} \sinh \left (\lambda x \right )-8 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right )^{3}-2 \cosh \left (\lambda x \right )^{2} \ln \left (\coth \left (\lambda x \right )-1\right )+2 \cosh \left (\lambda x \right )^{2} \ln \left (\coth \left (\lambda x \right )+1\right )+2 \cosh \left (\lambda x \right )^{2} c_{3} -2 \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )+\ln \left (\coth \left (\lambda x \right )-1\right )-\ln \left (\coth \left (\lambda x \right )+1\right )-c_{3} \right ) \lambda }{-4 \sinh \left (\lambda x \right ) \cosh \left (\lambda x \right )^{3}+2 \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )-\ln \left (\coth \left (\lambda x \right )-1\right )+\ln \left (\coth \left (\lambda x \right )+1\right )+c_{3}} \\ \end{align*}

Veriﬁcation of solutions

6.9.2 Maple step by step solution

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

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 Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (2*lambda^2*tanh(lambda*x)^2+2*lambda^2*coth(lambda*x)^2)*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 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 
         <- Kovacics algorithm successful 
         Change of variables used: 
            [x = arccoth(t)/lambda] 
         Linear ODE actually solved: 
            (-2*t^4-2)*u(t)+(2*t^5-2*t^3)*diff(u(t),t)+(t^6-2*t^4+t^2)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful`

\[ y \left (x \right ) = \frac {2 \left (-\frac {1}{2}+c_{1} \left (-\cosh \left (x \lambda \right )^{2}+\frac {1}{2}\right ) \ln \left (\coth \left (x \lambda \right )-1\right )+c_{1} \left (\cosh \left (x \lambda \right )^{2}-\frac {1}{2}\right ) \ln \left (\coth \left (x \lambda \right )+1\right )+4 \cosh \left (x \lambda \right )^{5} c_{1} \sinh \left (x \lambda \right )-4 \cosh \left (x \lambda \right )^{3} c_{1} \sinh \left (x \lambda \right )-\sinh \left (x \lambda \right ) \cosh \left (x \lambda \right ) c_{1} +\cosh \left (x \lambda \right )^{2}\right ) \lambda \,\operatorname {csch}\left (x \lambda \right ) \operatorname {sech}\left (x \lambda \right )}{-4 \cosh \left (x \lambda \right )^{3} c_{1} \sinh \left (x \lambda \right )+2 \sinh \left (x \lambda \right ) \cosh \left (x \lambda \right ) c_{1} +\ln \left (\coth \left (x \lambda \right )+1\right ) c_{1} -\ln \left (\coth \left (x \lambda \right )-1\right ) c_{1} +1} \]

\begin{align*} y(x)\to -\frac {2 \lambda \left (e^{12 \lambda x}+2 e^{4 \lambda x} \left (e^{4 \lambda x}+1\right ) \log \left (e^{4 \lambda x}\right )+(-7+c_1) \left (-e^{4 \lambda x}\right )-(7+c_1) e^{8 \lambda x}-1\right )}{\left (e^{4 \lambda x}-1\right ) \left (e^{8 \lambda x}-2 e^{4 \lambda x} \log \left (e^{4 \lambda x}\right )+c_1 e^{4 \lambda x}-1\right )} \\ y(x)\to \frac {2 \lambda \left (e^{4 \lambda x}+1\right )}{e^{4 \lambda x}-1} \\ \end{align*}

6.9 problem 26

6.9.1 Solving as riccati ode

6.9.2 Maple step by step solution