## 4 Result and Simulations
\begin{array}{l}\hat{u}_{1}(t)=0.7360(2\tan^{-1}e^{(1.5572t+0.2711)}-0.5 \pi)-1.3828(2\tan^{-1}e^{(0.0176t+1.1785)}-0.5\pi)+\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 3.1285(2\tan^{-1}e^{(-0. 320t-0.1422)}-0.5\pi)-0.5492(2\tan^{-1}e^{(-0.0149t-0.104)}-0.5\pi)-\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 0.3260(2\tan^{-1}e^{(-0. 261t-1.2335)}-0.5\pi)-0.8072(2\tan^{-1}e^{(0.6869t-0.8473)}-0.5\pi)+\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 1.8626(2\tan^{-1}e^{(1. 0913t+0.6863)}-0.5\pi)-1.2435(2\tan^{-1}e^{(1.3372t-0.1917)}-0.5\pi)-\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 0.2982(2\tan^{-1}e^{(-1. 018t+1.7279)}-0.5\pi)+0.445(2\tan^{-1}e^{(0.3368t+0.5995)}-0.5\pi),\end{array}  \tag{16}\]  
\[ \begin{array}{l}\hat{u}_{2}(t)=-5.627(2\tan^{-1}e^{(4.107t+13.776)}-0.5\pi)+ 19.796(2\tan^{-1}e^{(-1.1441t+2.700)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad -16.9872(2\tan^{-1}e^{(-1.825t-5.530)}-0.5\pi)+18.171(2\tan^{-1}e^{(1. 1842t-2.7152)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad -1.1341(2\tan^{-1}e^{(10.9932t+17.9253)}-0.5\pi)+9.3739(2\tan^{-1}e^{(0. 2126t-13.113)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +14.154(2\tan^{-1}e^{(6.1138t+17.6261)}-0.5\pi)+8.7317(2\tan^{-1}e^{(-6.7862 t-16.019)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +5.2472(2\tan^{-1}e^{(-4.4685t-6.8103)}-0.5\pi)+1.7752(2\tan^{-1}e^{(9.4095 t-19.1893)}-0.5\pi),\end{array}  \tag{17}\]  
\[ \begin{array}{l}\hat{u}_{2}(t)=0.7499(2\tan^{-1}e^{(-0.5392t+0.7869)}-0.5\pi)+ 0.435(2\tan^{-1}e^{(0.2239t-0.3881)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +1.3708(2\tan^{-1}e^{(-0.3328t+0.3684)}-0.5\pi)+0.9967(2\tan^{-1}e^{(-0. 034t-0.269)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +1.3307(2\tan^{-1}e^{(0.0317t-1.9519)}-0.5\pi)-0.2491(2\tan^{-1}e^{(-1. 0871t-0.2457)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +0.0744(2\tan^{-1}e^{(0.0144t-18.028)}-0.5\pi)+2.1415(2\tan^{-1}e^{(-0. 1161t-0.1017)}-0.5\pi)\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad -1.6038(2\tan^{-1}e^{(-0.8301t-1.3457)}-0.5\pi)+0.0044(2\tan^{-1}e^{(-0. 445t+0.4301)}-0.5\pi),\end{array}  \tag{18}\]  
The best weights set for 10 neurons and comparison of the mean, exact, and the best results for all the problems of the second kind of LE-NSM is shown in Figure 1. The set of best weight is drawn by using Equations 16 to 18. It is seen that the overlapping of the mean, exact, and best results are performed for all the problems of the second kind of LE-NSM. These results assessment shown in Figure 1 indicates the correctness and exactness of the suggested GNN-GAASM. Figure 2 shows the performance investigations based on the TIC, ENSE, and EVAF operator together with the best, worst, and mean values of AE for all variants of the second kind of LE-NSM. In order to evaluate the performance measures for Problem I, the calculations of the best Fit, ENSE, and EVAF values lie \(10^{-10}\)-\(10^{-12}\), and the mean values of the Fit, ENSE and EVAF lie \(10^{-4}\)-\(10^{-6}\). The performance of the best Fit, ENSE, and EVAF values for Problem 2 and 3 lie \(10^{-10}\)-\(10^{-15}\), \(10^{-6}\)-\(10^{-10}\), and \(10^{-7}\)-\(10^{-10}\), respectively, while the mean Fit, ENSE, and EVAF are close to \(10^{-5}\). In order to measure the absolute error (AE), the best values have been calculated around \(10^{-6}\) to \(10^{-7}\) for ProblemI, while for another two problems, the best AE lie \(10^{-5}\)-\(10^{-6}\). The mean AE values have also been noticed in suitable measures for all variants of the second kind of LE-NSM.