"Assignment Partial Differential EquationQ1. Explain in detail about Tanh-Coth Method. How it can be applied to nonlinear PartialDifferential Equations. Draw Graph and compare solution with exact Solution. Note: Use MAPLE for graphs. Tanh - Coth MethodThe Tanh-Coth method has been applied for a wide variety of nonlinear problems and will beused in this text for single travelling wave solution, which arises in various branches ofphysics, engineering and applied sciences. This method is proved to be powerful, reliable andeffective in handling a large number of dispersive and dissipative equations. Numericalresults clearly indicate the reliability and efficiency of this technique.Example 2.1Consider the Caudrey-Dodd-Gibbon (CDG) Equation ? 3? + (? + 30? ? + 60? ) = 0. (2.1)? ? ? ? ? ? ? ? ? Using the wave variable ? = ? -? ? carries out eq. (2.1) to the ODE' ' ? ? '' 3-? ? + ? + 30? ? + 60? = 0. (2.2)Integrating eq. (2.2), setting constant of integration to zero, we obtain? ? '' 3 -? ? +? + 30? ? + 60? = 0. (2.3)'' 3 Balancing ? ? with ? , we find3? = 2? + 2,so that ? = 2. (2.4)The tanh–coth method introduces the finite expansion2 2 ? -?? ? ,? = ? ? = ? ? + ? ? ,(2.5)? ? ? =0 ? =1? = tanh ? ? . (2.6)Substituting (2.5) into (2.3), collecting the coefficients of each power of Y, and using Mapleto solve the resulting system of algebraic equations we obtain the following sets of distinctsolutions of parameters:2 2 2 ? = ? ,? = 0,? = -? ,? = ? = 0,? = 16? ,0 1 2 1 2 2 2 4 ? = ? ,? = ? = 0,? = 0,? = -? ,? = 16? ,0 1 2 1 215 ± 105 2 2 4 ? = ? ,? = 0,? = -? ,? = ? = 0,? = (22± 2 105)? ,0 1 2 1 2 30 15 ± 105 2 2 4 ? = ? ,? = ? = 0,? = 0,? = -? ,? = (22± 2 105)? ,0 1 2 1 2 30 where ? is selected as a free parameter. Consequently, we obtain the soliton solutions.2 2 4? ? ,? = ? sech ? ? - 16? ? , (2.7)1 15± 105 2 2 4? ? ,? = ? { - tanh [? ? - 2? 11± 105 ? ]}, (2.8)2 30 2 2 4 ? ? ,? = -? ? ? sech ? ? - 16? ? ,(2.9) 3 15± 105 2 2 4? ? ,? = ? { - coth [? ? - 2? 11± 105 ? ]}. (2.10)4 30 (a) (b)(c)(d)Figure 2.1 The graphical representation of? ? ,?in 3 dimensions for (2.7)-(2.10). (a)(b) (c) (d) Figure 2.2 The graphical representation of? ? ,? in 2 dimensions for (2.7)-(2.10). Example 2.2 Consider Huxley equation ? = ? +? ? -? ? - 1 , ? ? 0. `(2.11)? ? ? Using the wave variable ? = ? -? ? carries out eq. (2.11) to the ODE' ''? ? +? +? ? -? 1-? = 0, (2.12)'' 3 balancing ? with ? , we find 3? = ? + 2, ? = 1.(2.13)The tanh–coth method introduces the finite expansion1 1 ? -?? ? ,? = ? ? = ? ? + ? ? ,(2.14)? ? ? =0 ? =1 where ? = tanh ? ? .-1 ? ? = ? +? ? +? ? . (2.15)0 1 1 ? Substituting (2.15) into (2.12), we obtain an equation that contains ? , 0 = ? = 6. Collecting? the coefficients of ? , set it to zero, and solving the resulting system we find we obtain thefollowing sets of solutions:1 1 1 2? - 1 ? = ,? = ,? = 0,? = ,? = , 0 1 1 2 2 2 2 2? ? ? ? - 2 ? = ,? = - ,? = 0,? = ,? = ,0 1 1 2 2 2 2 21 1 1 2? - 1 ? = ,? = 0,? = ,? = ,? = , 0 1 1 2 2 2 2 2? ? ? ? - 2 ? = ,? = 0,? = - ,? = ,? = ,0 1 1 2 2 2 2 2This in turn gives the solution1 1 2? -1? ? ,? = (1 + tanh? - ?, (2.16)1 2 2 2 2 ? ? ? -2 ? ? ,? = (1- tanh? - ?, (2.17)2 2 2 2 21 1 2? -1 ? ? ,? = (1 + coth? - ?,(2.18)3 2 2 2 2 ? ? ? -2? ? ,? = (1- coth? - ?. (2.19)4 2 2 2 2 "