## 2. Instructions for the Methods
### The Cubic-Quartic Nonlinear Schrodinger Equation  
To apply this method on the cubic-quartic nonlinear Schrodinger equation, Equation (3), we utilize the same wave transformation of Equation (17). As a result, we get Equation (22). Finding the balance, we gain \(n=1\). By inserting the value of the balance into Equation (10), we get:  
\[U\left(\xi\right)=b_{0}+b_{1}e^{-\varphi\left(\xi\right)}  
\[u\left(x,t\right)=\mathrm{e}^{\mathrm{i}\left(\frac{\sqrt{\lambda^{2}-4 \mu}}{\sqrt{6}}x+\frac{5}{12}a_{2}\left(\lambda^{2}-4\mu\right)^{2}t\right)} \tag{37}\] \[\left(\frac{\lambda b_{1}}{2}+\frac{2\mu b_{1}}{-\lambda-\sqrt{ \lambda^{2}-4\mu}\tanh\left[\frac{1}{2}\left(c+x-\frac{2}{3}\sqrt{\frac{2}{3}} a_{2}\left(\lambda^{2}-4\mu\right)^{3/2}t\right)\sqrt{\lambda^{2}-4\mu}\right]} \right).\]  
This is a dark soliton solution, as shown in Figure 8.  
Solution 2. When \(\lambda^{2}-4\mu>0\), \(\mu=0\), and \(\lambda\
eq 0\), we have hyperbolic function solutions:  
\[\begin{split}\mathrm{u}\left(x,t\right)=&\mathrm{e}^ {\mathrm{i}\left(\frac{5}{12}\beta\lambda^{4}t+\frac{\sqrt{\lambda^{2}}}{\sqrt {6}}x\right)}\\ &\left(\frac{\lambda b_{1}}{2}+\frac{\lambda b_{1}}{-1+\cosh \left(\lambda\left(c+x-\frac{2}{3}\sqrt{\frac{2}{3}}\beta{\left(\lambda^{2} \right)}^{3/2}t\right)\right)+\sinh\left(\lambda\left(c+x-\frac{2}{3}\sqrt{ \frac{2}{3}}\beta{\left(\lambda^{2}\right)}^{3/2}t\right)\right)}\right). \end{split} \tag{38}\]  
This is a bright singular combo soliton solution, as shown in Figure 9.  
Figure 8: 3D surface of Equation (37), which is a bright singular combo soliton solution plotted when \(b_{1}=0.2,\beta=0.2,c=1,\lambda=3,\mu=1\), and \(t=2\) for 2D.  
Case 2. When \(b_{0}=\frac{\sqrt{3}\sqrt{\zeta_{1}}\lambda}{2\sqrt{-c_{2}(\lambda^{2}-4\mu)}}\), \(b_{1}=\frac{\sqrt{3}\sqrt{\zeta_{1}}}{\sqrt{-c_{2}(\lambda^{2}-4\mu)}}\), \(\kappa=-\frac{\sqrt{\lambda^{2}-4\mu}}{\sqrt{6}}\), \(\omega=-\frac{5c_{1}^{2}}{32c_{2}}\), and \(\beta=-\frac{3c_{1}^{2}}{8c_{2}(\lambda^{2}-4\mu)^{2}}\), we get the following solutions:  
Solution 1. When \(\lambda^{2}-4\mu>0\) and \(\mu\
eq 0\), we get a dark solution, as shown in Figure 10:  
\[\begin{split}\mathfrak{u}\left(x,t\right)=& \mathrm{e}^{\mathrm{i}\left(-\frac{5\lambda^{2}}{32c_{2}}t+\frac{\sqrt{ \lambda^{2}-4\mu}}{\sqrt{6}}x\right)}\\ &\left(\frac{\sqrt{3}\sqrt{c_{1}}\left(\lambda^{2}-4\mu+\lambda \sqrt{\lambda^{2}-4\mu}\tanh\left(\frac{c_{1}^{2}t}{4\sqrt{6}c_{2}}+\frac{1}{2 }\left(c+x\right)\sqrt{\lambda^{2}-4\mu}\right)\right)}{2\sqrt{-c_{2}\left( \lambda^{2}-4\mu\right)}\left(\lambda+\sqrt{\lambda^{2}-4\mu}\tanh\left(\frac{ c_{1}^{2}t}{4\sqrt{6}c_{2}}+\frac{1}{2}\left(c+x\right)\sqrt{\lambda^{2}-4\mu} \right)\right)}\right).\end{split} \tag{39}\]  
Figure 10: 3D surface of Equation (39), which is a dark soliton solution plotted when \(b_{1}=0.2,\beta=0.2,c_{1}=0.1,c_{2}=-0.1,c=1,\lambda=3,\mu=1\), and \(t=2\) for 2D.  
Figure 9: 3D surface of Equation (38), which is a bright singular combo soliton solution plotted when \(b_{1}=0.04\), \(\beta=0.2,c=0.2,\lambda=1,\mu=0\), and \(t=2\) for 2D.  
Solution 2. When \(\lambda^{2}-4\mu>0\) and \(\mu=0\), we get hyperbolic function solution:  
\[\mathrm{u}\left(x,t\right)= \mathrm{e}^{\mathrm{i}\left(-\frac{\sqrt{\lambda^{2}-4\mu}}{3 \lambda^{2}}t+\frac{\sqrt{\lambda^{2}-4\mu}}{\sqrt{6}}x\right)}\left(\frac{ \sqrt{3}\sqrt{c_{1}}\lambda\coth\left(\frac{1}{24}\lambda\left(12\left(c+x \right)+\frac{\sqrt{6}c_{1}^{2}}{c_{2}\sqrt{\lambda^{2}}}t\right)\right)}{2 \sqrt{-c_{2}\lambda^{2}}}\right). \tag{40}\]  
This is a singular soliton solution, as shown in Figure 11.