## 3 Solution of the Problem
### Exact Solution of Velocity Profile
Equation (29) can be rearranged in another form as \[\bar{w}(y,s)=F(s)\bar{A}_{0}(y,s)-\frac{\xi Gr}{a_{1}}\frac{1}{(\sqrt{s}+\xi)} \bigg{[}\frac{a_{5}}{s}+\frac{a_{6}}{s+m_{1}}+\frac{a_{7}}{s+n_{1}}\bigg{]}[\bar {A}_{0}(y,s)+\bar{w}_{4}(y,s)-\bar{w}_{5}(y,s)], \tag{30}\]  
\[\bar{w}(y,s)=F(s)\bar{A}_{0}(y,s)-\frac{\xi Gr}{a_{1}}[a_{5}\bar{w}_{1}(y,s)+a_ {6}\bar{w}_{2}(y,s)+a_{7}\bar{w}_{3}(y,s)][\bar{A}_{0}(y,s)+\bar{w}_{4}(y,s)- \bar{w}_{5}(y,s)], \tag{31}\]  
where:  
\[\bar{A}_{0}(y,s)= e^{-y\sqrt{\frac{\lambda_{1}p^{2}+as+M}{1+\lambda_{2}s}}}. \frac{1}{1+\gamma\sqrt{\frac{\lambda_{1}s^{2}+as+M}{1+\lambda_{2}s}}},\] \[= \Bigg{[}\sum_{a=0}^{\infty}\frac{(-y)^{a}}{\alpha!}\bigg{(}\frac {\lambda_{1}s^{2}+as+M}{1+\lambda_{2}s}\bigg{)}^{\frac{\alpha}{2}}\Bigg{]} \Bigg{[}\sum_{\beta=0}^{\infty}(-1)^{\beta}(\gamma)^{\beta}\bigg{(}\frac{ \lambda_{1}s^{2}+as+M}{1+\lambda_{2}s}\bigg{)}^{\frac{\alpha}{2}}\Bigg{]},\]  
By applying the discrete convolution known as the Cauchy product, each of which have m terms with two truncated series, yields:  
\[\bar{A}_{0}(y,s)= \sum_{a=0}^{m}\sum_{\beta=0}^{m}\frac{(-y)^{a}(-1)^{m-\beta}( \gamma)^{m-\beta}}{\alpha!}\bigg{(}\frac{\lambda_{1}s^{2}+as+M}{1+\lambda_{2} s}\bigg{)}^{m+\frac{\alpha}{2}-\frac{\beta}  
\[f(t)=\mathcal{L}^{-1}\bigg{(}\frac{1}{\sqrt{s}+\xi}\bigg{)}=\frac{1} {\sqrt{\pi t}}-\xi e^{\xi^{2}t}erfc(\xi\sqrt{t}),\quad f_{1}(t)=\mathcal{L}^{-1} \bigg{(}\frac{1}{s}\bigg{)}=1,\] \[f_{2}(t)=\mathcal{L}^{-1}\bigg{(}\frac{1}{s+m_{1}}\bigg{)}=e^{-m _{1}t},\quad f_{3}(t)=\mathcal{L}^{-1}\bigg{(}\frac{1}{s+n_{1}}\bigg{)}=e^{-n_ {1}t},\] \[\tilde{w}(y,s)=c_{3}e^{y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}+c_{4}e ^{-y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}-\bigg{(}\frac{Gr(1+\lambda _{1}s)e^{-y\sqrt{Prs}}}{s\big{(}\sqrt{Prs}-1\big{)}(Prs(1+\lambda_{2}s)-( \lambda_{1}s^{2}+as+M)\bigg{)}}\bigg{)}. \tag{28}\]0  
Finally, the solution which requires employing the inverse Laplace transformation for the momentum equation with the convolution product on Equation (31) has the following form:  
\[\tilde{w}(y,s)=c_{3}e^{y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}+c_{4}e ^{-y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}-\bigg{(}\frac{Gr(1+\lambda _{1}s)e^{-y\sqrt{Prs}}}{s\big{(}\sqrt{Prs}-1\big{)}(Prs(1+\lambda_{2}s)-( \lambda_{1}s^{2}+as+M)\bigg{)}}\bigg{)}. \tag{28}\]1 \[\tilde{w}(y,s)=c_{3}e^{y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}+c_{4}e ^{-y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}-\bigg{(}\frac{Gr(1+\lambda _{1}s)e^{-y\sqrt{Prs}}}{s\big{(}\sqrt{Prs}-1\big{)}(Prs(1+\lambda_{2}s)-( \lambda_{1}s^{2}+as+M)\bigg{)}}\bigg{)}. \tag{28}\]2 \[\tilde{w}(y,s)=c_{3}e^{y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}+c_{4}e ^{-y\sqrt{\frac{\lambda_{1}s^{2}+ss+M}{1+2s^{2}}}}-\bigg{(}\frac{Gr(1+\lambda _{1}s)e^{-y\sqrt{Prs}}}{s\big{(}\sqrt{Prs}-1\big{)}(Prs(1+\lambda_{2}s)-( \lambda_{1}s^{2}+as+M)\bigg{)}}\bigg{)}. \tag{28}\]3  
We recover the Maxwell model by considering time retardation parameter of value zero, i.e., \(\lambda_{2}=0\), as acquired by [PERSON] et al. [59]. Furthermore, the velocity field solution for the second grade fluid is traced out by considering \(\lambda_{1}=0\) in Equation (33), as derived by [PERSON] et al. [34], which validates our current results with the previous literature.