Theory and Modern Applications
Table 3 Comparing the results of Lu et al. [25] with our results
Obtained results | Results of Lu et al. |
|---|---|
If \(H_{1}=\frac{a_{1} ( V-2k ) }{b_{1}\sqrt{V ( V-2k ) }}\), a = 0 and V = 5 then the solution is \(u_{19}=H_{1}\coth ( \frac{5}{2}-x ) \) | If \(H_{1}=\frac{k\sqrt{6\beta }}{\beta _{2}\sqrt{-\delta ( \beta _{1}k^{2}+2 ) }}\), ϵ = 1, k = −1, ω = 1, \(\beta _{1}=1\), \(\sqrt{\beta _{1}}=5\), \(\sqrt{\beta _{3}}=\frac{1}{2}\) and \(\zeta _{0}=0\) then the solution is \(u_{4}=H_{1}\coth ( \frac{5}{2}-x ) \) |
If \(H_{1}=\frac{a_{1} ( V-2k- ) }{b_{1}\sqrt{V ( V-2k ) }} \), a = 0 and V = 5 then the solution is \(u_{19}=H_{1}\coth ( \frac{5}{2}-x ) \) | If \(H_{1}=\frac{2\sqrt{3}k\beta _{1}\sqrt{\beta _{3}\beta }}{\beta _{2}\sqrt{-\delta ( \beta _{1}k^{2}+2 ) }}\), ϵ = 1, k = −1, ω = 1, \(\beta _{1}=1\), \(\sqrt{\beta _{1}}=5\), \(\sqrt{\beta _{3}}=\frac{1}{2}\) and \(\zeta _{0}=0\), then the solution is \(u_{7}=H_{1}\coth ( \frac{5}{2}-x ) \) |