To acquire the exact outcomes of the above equations (16) and (17) with fractional derivative, we utilized the technique of Laplace transform. Firstly, we execute the Laplace transform method on Eq. (17) and then obtain the solution of Eq. (16). The Laplace transforms of velocity, temperature, and Caputo-Fabrizio time fractional derivative are defined in the following equation [32]

$$\stackrel{-}{{U}^{*}}\left({Y}^{*},s\right)=\mathcal{L}\left[{U}^{*}\left({Y}^{*},{t}^{*}\right)\right]={\int }_{0}^{\infty }{U}^{*}\left({Y}^{*},{t}^{*}\right){e}^{-s{t}^{*}}d{t}^{*}$$

$$\stackrel{-}{{T}^{*}}\left({Y}^{*},s\right)=\mathcal{L}\left[{T}^{*}\left({Y}^{*},{t}^{*}\right)\right]={\int }_{0}^{\infty }{T}^{*}\left({Y}^{*},{t}^{*}\right){e}^{-s{t}^{*}}d{t}^{*}$$

$$\mathcal{L}\left[{D}_{{t}^{*}}^{\alpha }{U}^{*}\left({Y}^{*},{t}^{*}\right)\right]=\frac{s\mathcal{L}\left[{U}^{*}\left({Y}^{*},{t}^{*}\right)\right]-{U}^{*}\left({Y}^{*},0\right)}{\alpha +\left(1-\alpha \right)s}$$

18

By using Eq. (18), Eq. (17) has the following form

$$\frac{{d}^{2}\stackrel{-}{{T}^{*}}}{d{{Y}^{*}}^{2}}-{Pr}_{eff}\frac{s\gamma }{s+\alpha \gamma }\stackrel{-}{{T}^{*}}=0,$$

19

where\(\gamma =\frac{1}{1-\alpha }.\)

The transformed boundary conditions of temperature are given as

$$\stackrel{-}{{T}^{*}}\left(0,s\right)=\frac{1}{s}, \stackrel{-}{{T}^{*}}\left(\infty ,s\right)=0.$$

20

Eq. (19) has the following solution corresponding to Eq. (20)

$$\stackrel{-}{{T}^{*}}\left({Y}^{*},s\right)=Exp\left(\sqrt{\frac{{Pr}_{eff}s\gamma }{s+\alpha \gamma }}{Y}^{*}\right){c}_{1}+Exp\left(-\sqrt{\frac{{Pr}_{eff}s\gamma }{s+\alpha \gamma }}{Y}^{*}\right){c}_{2}$$

21

$$\text{w}\text{h}\text{e}\text{r}\text{e} {c}_{1}=0 \text{a}\text{n}\text{d} {c}_{2}=\frac{1}{s}.$$

Another form of Eq. (21) can be written as follow

$$\stackrel{-}{{T}^{*}}\left({Y}^{*},s\right)=\psi ({Y}^{*}, s,{Pr}_{eff}\gamma ,\alpha \gamma )$$

22

As the inverse Laplace transform of the function \({\psi }_{1}\left(y, s,g,h\right)=\frac{1}{s}Exp\left(-\sqrt{\frac{gs}{s+h}}y\right)\)is [29]

$${\psi }_{2}\left(y,t,g,h\right)={L}^{-1}{\psi }_{1}\left(y, s,g,h\right)=1-\frac{2g}{\pi }{\int }_{0}^{\infty }\frac{\text{sin}\left(yx\right)}{x\left(g+{x}^{2}\right)}Exp\left(\frac{-ht{x}^{2}}{g+{x}^{2}}\right)dx$$

23

According to Eq. (23), the inverse Laplace transform of Eq. (22) has the following expression

$${T}^{*}\left({Y}^{*},{t}^{*}\right)=1-\frac{2{Pr}_{eff}\gamma }{\pi }{\int }_{0}^{\infty }\frac{\text{sin}\left({Y}^{*}x\right)}{x\left({Pr}_{eff}\gamma +{x}^{2}\right)}Exp\left(\frac{-\alpha \gamma {t}^{*}{x}^{2}}{{Pr}_{eff}\gamma +{x}^{2}}\right)dx$$

24

To evaluate the above integral, we use the series representation of an exponential function. So, we have

$${\int }_{0}^{\infty }\frac{\text{sin}\left({Y}^{*}x\right)}{x\left({Pr}_{eff}\gamma +{x}^{2}\right)}Exp\left(\frac{-\alpha \gamma {t}^{*}{x}^{2}}{{Pr}_{eff}\gamma +{x}^{2}}\right)dx=\sum _{n=0}^{\infty }\frac{{(-1)}^{n}{\left({t}^{*}\alpha \gamma \right)}^{n}}{n!}{\int }_{0}^{\infty }\frac{\text{sin}\left({Y}^{*}x\right)}{x}{\left({Pr}_{eff}\gamma +{x}^{2}\right)}^{-1-n}({{x}^{2})}^{n}dx$$

25

After evaluating the above integral, Eq. (25) can be written as

We get the exact solution of the temperature distribution after using Eqs. (25) and 26) as follows

Now to acquire the solution of the velocity distribution given in Eq. (16), we apply the Laplace transform technique and get the following form

$$\left(1+N\right)\frac{{d}^{2}\stackrel{-}{{U}^{*}}}{d{{Y}^{*}}^{2}}-\frac{s\gamma }{s+\alpha \gamma }\stackrel{-}{{U}^{*}}+Gr\stackrel{-}{{T}^{*}}=0,$$

28

The transformed conditions are

$$\stackrel{-}{{U}^{*}}\left(0,s\right)=0, \stackrel{-}{{U}^{*}}\left(\infty ,s\right)=0.$$

29

Eq. (28) has the following solution

$$\stackrel{-}{{U}^{*}}\left({Y}^{*},s\right)=\frac{Gr(s+\alpha \gamma )}{{s}^{2}\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)} Exp\left(-\sqrt{\frac{{Pr}_{eff}s\gamma }{(s+\alpha \gamma )}}{Y}^{*}\right)+{c}_{3}Exp\left(\sqrt{\frac{s\gamma }{(1+N)(s+\alpha \gamma )}}{Y}^{*}\right){+c}_{4}Exp\left(-\sqrt{\frac{s\gamma }{(1+N)(s+\alpha \gamma )}}{Y}^{*}\right)$$

30

Using Eq. (29), we have the values of constants\({c}_{3}=0, {c}_{4}=-\frac{Gr(s+\alpha \gamma )}{{s}^{2}\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)}.\)

Eq. (30) can be written as

$$\stackrel{-}{{U}^{*}}\left({Y}^{*},s\right)=\frac{Gr(s+\alpha \gamma )}{{s}^{2}\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)} Exp\left(-\sqrt{\frac{{Pr}_{eff}s\gamma }{(s+\alpha \gamma )}}{Y}^{*}\right)-\frac{Gr(s+\alpha \gamma )}{{s}^{2}\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)}Exp\left(-\sqrt{\frac{s\gamma }{(1+N)(s+\alpha \gamma )}}{Y}^{*}\right)$$

31

Eq. (31) can also be written as follows

$$\stackrel{-}{{U}^{*}}\left({Y}^{*},s\right)=\frac{{\beta }_{0}}{s} Exp\left(-\sqrt{\frac{{\delta }_{0}s}{({\delta }_{1}+s)}}{Y}^{*}\right)+\frac{{\beta }_{1}}{{s}^{2}} Exp\left(-\sqrt{\frac{{\delta }_{0}s}{({\delta }_{1}+s)}}{Y}^{*}\right)-\frac{{\beta }_{0}}{s} Exp\left(-\sqrt{\frac{{\delta }_{2}s}{\left({\delta }_{1}+s\right)}}{Y}^{*}\right)- \frac{{\beta }_{1}}{{s}^{2}} Exp\left(-\sqrt{\frac{{\delta }_{2}s}{({\delta }_{1}+s)}}{Y}^{*}\right)$$

32

where

$${\beta }_{0}=\frac{Gr}{\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)}, {\beta }_{1}=\frac{Gr\alpha \gamma }{\left(\gamma -\left(1+N\right){Pr}_{eff}\gamma \right)}, {\delta }_{0}={Pr}_{eff}\gamma , {\delta }_{1}=\alpha \gamma ,$$

$${\delta }_{2}=\frac{\gamma }{(1+N)}.$$

33

The more suitable expression of Eq. (31) can be written as follows

$$\stackrel{-}{{U}^{*}}\left({Y}^{*},s\right)={\beta }_{0}{\stackrel{-}{\phi }}_{1}({Y}^{*},s,{\delta }_{0},{\delta }_{1})+{\beta }_{1}{\stackrel{-}{\phi }}_{2}({Y}^{*},s,{\delta }_{0},{\delta }_{1})-{\beta }_{0}{\stackrel{-}{\phi }}_{3}({Y}^{*},s,{\delta }_{1},{\delta }_{2})-{\beta }_{1}{\stackrel{-}{\phi }}_{4}({Y}^{*},s,{\delta }_{1},{\delta }_{2})$$

34

The solution of velocity distribution is achieved after applying the inverse Laplace transformation on Eq. (34) as follows

$${U}^{*}\left({Y}^{*},{t}^{*}\right)={\beta }_{0}{\phi }_{1}({Y}^{*},{t}^{*},{\delta }_{0},{\delta }_{1})+{\beta }_{1}{\phi }_{2}({Y}^{*},{t}^{*},{\delta }_{0},{\delta }_{1})-{\beta }_{0}{\phi }_{3}({Y}^{*},{t}^{*},{\delta }_{1},{\delta }_{2})-{\beta }_{1}{\phi }_{4}({Y}^{*},{t}^{*},{\delta }_{1},{\delta }_{2})$$

35

where \({\phi }_{1},{\phi }_{2},{\phi }_{3}, {\phi }_{4}\) can be evaluated in a same manner as \({\psi }_{2}(y, t,g,h)\) in Eq. (23).