**2. Instructions for the Methods**

Assume a nonlinear partial differential equation (NLPDE) as follows:

$$P\left(\mathcal{U}\_{\prime}\mathcal{U}\_{\mathbf{x}\prime}\mathcal{U}\_{\prime\prime}\mathcal{U}\_{\prime\prime}\mathcal{U}\_{\mathbf{x}\prime\prime}\mathcal{U}\_{\mathrm{tt}\prime}\mathcal{U}\_{\mathrm{tx}\prime}\dots\right) = 0,\tag{5}$$

and define the traveling wave transformation as follows,

$$
\mathcal{U}\left(\mathbf{x}, \mathbf{y}, t\right) = \boldsymbol{\Phi}\left(\mathbb{\zeta}\right), \quad \mathbb{\zeta} = \mathbf{x} - \nu t. \tag{6}
$$

Putting Equation (6) into Equation (5), the outcome is:

$$N\left(\phi, \phi', \phi'', \dots\right) = 0.\tag{7}$$

For the *m* + *<sup>G</sup>*- (*ζ*) *<sup>G</sup>*(*ζ*) expansion method, we take the trial solution for Equation (7) as follows:

$$\phi\left(\zeta\right) = \sum\_{i=-n}^{n} a\_i (m+F)^i = a\_{-n}(m+F)^{-n} + \dots + m \, a\_0 + a\_1 \left(m+F\right) + \dots + a\_n (m+F)^n,\tag{8}$$

where *ai*, *i* = 0, 1, ... , *n* and *m* are nonzero constants. According to the principles of balance, we find the value of *n*. In this manuscript, we define *F* to be a function as:

$$F = \frac{G'\left(\zeta\right)}{G\left(\zeta\right)},\tag{9}$$

where *G* (*ζ*) satisfy *G*-- + (*λ* + 2*m*) *G*-+ *μG* = 0.

Putting Equation (8) into Equation (7) by using Equation (9), then collecting all terms with the same order of (*m* + *F*) *<sup>n</sup>*, we get the system of algebraic equations for *ν*, *an*, *n* = 0, 1, ... , *n*, *λ*, and *μ*. As a result, solving the obtained system, we get the explicit and exact solutions of Equation (5).

For the (exp − *ϕ* (*ξ*)) expansion method, we use the trial solution as follows:

$$\phi\left(\xi\right) = \sum\_{i=0}^{n} b\_i \left(\exp\left(-\varphi\left(\xi\right)\right)\right)^i, \quad i = 1, 2, \dots, n \tag{10}$$

where *bi* are non-zero constants. The auxiliary ODE *ϕ* (*ξ*) is defined as follows:

$$\varphi'\left(\xi\right) = \exp\left(-\varphi\left(\xi\right)\right) + \mu \exp\left(\varphi\left(\xi\right)\right) + \lambda. \tag{11}$$

Solving Equation (11), we have:

Case 1. When Δ > 0 and *μ* = 0, we get the hyperbolic function solution:

$$\varphi\left(\xi\right) = \ln\left(\frac{-\lambda - \sqrt{\Delta}\tanh\left(\frac{1}{2}\sqrt{\Delta}\left(\xi + c\right)\right)}{2\mu}\right).\tag{12}$$

Case 2. When Δ < 0 and *μ* = 0, we get the trigonometric function solution:

$$\varphi\left(\xi\right) = \ln\left(\frac{-\lambda + \sqrt{-\Delta}\tan\left(\frac{1}{2}\sqrt{-\Delta}\left(\xi + c\right)\right)}{2\mu}\right).\tag{13}$$

Case 3. When Δ > 0, *μ* = 0, and *λ* = 0, we get hyperbolic function solution

$$\varphi\left(\vec{\xi}\right) = -\ln\left(\frac{\lambda}{-1 + \cosh\left(\lambda\left(\vec{\xi} + c\right)\right) + \sinh\left(\lambda\left(\vec{\xi} + c\right)\right)}\right).\tag{14}$$

Case 4. When Δ = 0, *μ* = 0 and *λ* = 0, we get the rational function solution:

$$\varphi\left(\xi\right) = \ln\left(\frac{-2 - 2\lambda\left(\xi + c\right)}{\lambda^2 \left(\xi + c\right)}\right).\tag{15}$$

Case 5. When Δ = 0, *μ* = 0, and *λ* = 0, we get:

$$\varphi\left(\xi\right) = \ln\left(\xi + c\right),\tag{16}$$

where *<sup>c</sup>* is the non-zero constant of integration and <sup>Δ</sup> = *<sup>λ</sup>*<sup>2</sup> − <sup>4</sup>*μ*.

#### **3. Application to the** - *m* + *<sup>G</sup>*- *G* **Expansion Method**

In this section, we use the *m* + *<sup>G</sup>*- *G* expansion method for the cubic-quartic nonlinear Schrödinger and cubic-quartic resonant nonlinear Schrödinger equations.
