## 4 Solution in the Transformed Domain
Likewise, we may obtain the following DE by removing \(\bar{w}\) between Equations (31) and (32) as:  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]3  
where \(A_{10}=\frac{A_{2}}{\delta^{4}}\big{(}1+\delta^{2}A_{0}+\delta^{4}A_{1}\big{)}\) and its solution may be written as:  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]4  
where  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]5  
It is possible to create a solution for the functions \(\overline{M}\), \(\overline{u}\), \(\overline{e}\), and \(\overline{\sigma}_{x}\) in the Laplace field with the assistance of Equations (40) and (42) by writing the solution as follows:  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]6  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]7  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]8  
\[\left(\frac{d^{4}}{dx^{4}}+A_{0}\frac{d^{2}}{dx^{2}}+A_{1}\right)\bar{w}=-A_{ 2}\frac{d^{2}\Theta}{dx^{2}} \tag{31}\]9  
The energy that is put into a material in the form of work is referred to as the material stress energy. One may refer to the internal work that is done by the stress products as they travel through related distortions as stress energy. The material is expected to maintain its flexibility while working on it, ensuring that no energy is wasted, and no permanent deformation is caused by the material's production. The total stress energy or work done by the microbeam can be represented as:  
\[-A_{4}\frac{d^{2}\bar{w}}{dx^{2}}=\left(\frac{d^{2}\bar{w}}{dx^{2}}-A_{3} \right)\tilde{\Theta}+A_{7}e^{-x/\delta} \tag{32}\]0  
In the domain of Laplace transformation, the conditions (27)-(29) will have the form:  
\[ \begin{array}{l}\bar{w}(0,s)=\bar{w}(L,s)=0,\frac{\partial^{2}\bar{w}(0,s)}{ \partial x^{2}}=\frac{\partial^{2}\bar{w}(L,s)}{\partial x^{2}}=0\\ \frac{\partial\bar{\Theta}(0,s)}{\partial x}=\Theta_{0}\Big{(}\frac{s}{s^{2}+ \omega^{2}}\Big{)}=\bar{G}(s),\frac{\partial\bar{\Theta}(L,s)}{\partial x}=0 \end{array}  \tag{49}\]  
Substituting Equations (40) and (42) into the boundary conditions (49) yields the following results:  
\[ \begin{array}{l}\sum\limits_{n=1}^{3}m_{n}^{2}(C_{n}+C_{n+1})=-A_{9},\\ \sum\limits_{n=1}^{3}m_{n}^{2}\big{(}C_{n}e^{-m_{n}L}+C_{n+1}e^{m_{n}L}\big{)}= -A_{9}e^{-L/\delta},\end{array}  \tag{50}\]\[ \begin{array}{c}\sum\limits_{n=1}^{3}m_{n}^{2}(C_{n}+C_{n+1})=-\frac{A_{B}}{ \delta^{2}},\\ \sum\limits_{n=1}^{3}m_{n}^{2}(C_{n}e^{-m_{n}L}+C_{n+1}e^{m_{n}L})=-\frac{A_{B}} {\delta^{2}}e^{-L/\delta},\\ \sum\limits_{n=1}^{3}\beta_{n}(C_{n}+C_{n+1})=A_{11}+G(s),\\ \sum\limits_{n=1}^{3}m_{n}\beta_{n}\big{(}C_{n}e^{-m_{n}L}-C_{n+1}e^{m_{n}L} \big{)}=-\frac{A_{11}}{\delta}e^{-L/\delta},\end{array}  \tag{51}\]  
As a result of the governing equation being solved, we may deduce the indeterminate constants \(C_{n}\), \((n=1,2,\ldots,6)\). The Riemann sum approximation technique is employed to get the numerical values of the studied field variables [42]. Many algorithms are used for this purpose. In the current study, the following relationship will be utilized to convert the functions from the Laplace field to the time realm:  
\[g(x,t)=\frac{et}{t}\Bigg{[}\frac{1}{2}\;\delta(x,\epsilon)+\text{Re}\Bigg{\{} \sum\limits_{k=1}^{M}\delta\bigg{(}x,\epsilon+\frac{ik\pi}{t}\bigg{)}(-1)^{k} \Bigg{\}}\Bigg{]},\quad 0\leq t\leq t_{1} \tag{53}\]  
In the above relation, the abbreviation \(i=\sqrt{-1}\) stands for the imaginary number, Re stands for the real component, and \(M\) stands for the number of truncated terms that were used in the numerical solution.