## 4 Applications and Discussion
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]1 Represents the analytical solution of Example 1 at \(\beta=2\). _In Table 2, the solutions of Shehu transform decomposition method (SDM) and Adomian decomposition method (ADM) are compared with each other. The comparison has shown that the solutions of proposed method are in strong agreement with the solution of ADM._  
**Example 2**. : _Consider the (2+time fractional-order) dimensional Heat model:_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]2  
_with initial condition_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]3  
_If \(\beta=1\), then the exact solution of Equation (32) is_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]4  
_Taking Shehu transform of Equation (32)_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]5  
_Simplifying Equation (35), we get as_  
\[u(\Im,\Re,\tau)=\Im^{4}\sinh(t)+\Re^{4}\cosh(t), \tag{21}\]6  
_Applying inverse Shehu transform, we get_ \begin{table}
\begin{tabular}{c c c c c c c} \hline  & & **SDM (m = 5)** & **SDM (m = 3)** & **SDM (m = 5)** & **ADM (m = 5)** & **AE of SDM** \\ \hline \(\Im\) & \(\Re\) & \(\beta=1.75\) & \(\beta=2\) & \(\beta=2\) & \(\beta=2\) & \(\beta=2\) \\ \hline
1 & 1 & 1.111568974 & 1.105195833 & 1.10519608 & 1.10519609 & 2.51 \(\times 10^{-5}\) \\
2 & 2 & 17.78510358 & 17.68313333 & 17.6831373 & 17.6831374 & 4.02 \(\times 10^{-4}\) \\
3 & 3 & 90.03708688 & 89.52086250 & 89.5208829 & 89.5208828 & 2.03 \(\times 10^{-3}\) \\
4 & 4 & 284.5616573 & 282.9301334 & 282.930198 & 282.930199 & 6.44 \(\times 10^{-3}\) \\
5 & 5 & 694.7306086 & 690.7473959 & 690.747553 & 690.747552 & 1.57 \(\times 10^{-2}\) \\ \hline \end{tabular}
\end{table} 
Table 2: Comparison of SDM and ADM [42] of Example 1 at \(\tau=0.1\). Figure 4: Represents the solution at different fractional order of Example 1. \[u(\Im,\Re,\tau)=u(0)+S^{-1}\left\{\frac{\mu^{\beta}}{s^{\beta}}S\left\{u_{\Im \Im}(\Im,\Re,\tau)+u_{\Re\Re}(\Im,\Re,\tau)\right\}\right\}. \tag{37}\]  
_Thus we get the following recursive scheme_  
\[u_{0}(\Im,\Re,\tau)=u(0)=\sin(\Im)\cos(\Re), \tag{38}\]  
\[u_{m+1}(\Im,\Re,\tau)=S^{-1}\left\{\frac{\mu^{\beta}}{s^{\beta}}S\left\{u_{m \Im\Im}(\Im,\Re,\tau)+u_{m\Re\Re}(\Im,\Re,\tau)\right\}\right\}, \tag{39}\]  
_Using Equation (39), for \(m=0,1,2,3,\cdots\) we get the following values_  
\[ \begin{split}& u_{1}(\Im,\Re,\tau)=-2\sin(\Im)\cos(\Re)\frac{ \tau^{\beta}}{(\beta)!},\\ & u_{2}(\Im,\Re,\tau)=4\sin(\Im)\cos(\Re)\frac{\tau^{2\beta}}{(2 \beta)!},\\ & u_{3}(\Im,\Re,\tau)=-8\sin(\Im)\cos(\Re)\frac{\tau^{3\beta}}{(3 \beta)!},\\ & u_{4}(\Im,\Re,\tau)=16\sin(\Im)\cos(\Re)\frac{\tau^{4\beta}}{(4 \beta)!},\\ \end{split}  \tag{40}\]  
_Now using the values of \(u_{0},u_{1},u_{2},u_{3},\cdots\), we get Shehu transformation solution for example 2_  
\[ \begin{split}& u(\Im,\Re,\tau)=\sin(\Im)\cos(\Re)-2\sin(\Im) \cos(\Re)\frac{\tau^{\beta}}{(\beta)!}+4\sin(\Im)\cos(\Re)\frac{\tau^{2\beta}} {(2\beta)!}+\\ &-8\sin(\Im)\cos(\Re)\frac{\tau^{3\beta}}{(3\beta)!}+16\sin(\Im) \cos(\Re)\frac{\tau^{4\beta}}{(4\beta)!}+\cdots\\ \end{split}  \tag{41}\]  
_After simplification, we get_  
\[u(\Im,\Re,\tau)=\sin(\Im)\cos(\Re)\left\{1-2\frac{\tau^{\beta}}{(\beta)!}+4 \frac{\tau^{2\beta}}{(2\beta)!}+-8\frac{\tau^{3\beta}}{(3\beta)!}+16\frac{ \tau^{4\beta}}{(4\beta)!}+\cdots\right\}, \tag{42}\]  
_which converge to the solution_  
\[u(\Im,\Re,\tau)=\sin(\Im)\cos(\Re)E_{\beta}(-2\tau^{\beta}), \tag{43}\]  
_For particular case \(\beta\to 1\), the Shehu transform solution become as_  
\[u(\Im,\Re,\tau)=\sin(\Im)\cos(\Re)e^{-2\tau}. \tag{44}\]Figure 5: Represents the exact solution of Example 2 at \(\beta=1\). Figure 6: Represents the analytical solution of Example 2 at \(\beta=1\). Figure 7: \(u(\Im,\Re,\tau)\) Represents the solution at different fractional order of Example 2. **Example 3**.