**3. Method of solution**

To find the numerical solution of a nonlinear system (7) and (8), the set of firstorder linear equations is obtained by considering the following assumptions. By putting these assumptions in the above equations, we get first-order linear equations, which are then used in MATLAB by using BVP4C scheme to get numerical and graphical results.

$$\mathcal{Y}\_1 = f,\tag{13}$$

$$
\mathcal{Y}\_1' = \mathcal{Y}\_2,\tag{14}
$$

$$\mathcal{Y}\_2' = \mathcal{Y}\_3,\tag{15}$$

$$\mathbf{y}'\_3 = \mathbf{g}\_1,\tag{16}$$

$$
\theta = \mathcal{y}\_{\text{4}},
\tag{17}
$$

$$\mathbf{y}'\_4 = \mathbf{g}\_1,\tag{18}$$

$$\mathbf{y}'\_{\mathfrak{F}} = \mathbf{g}\_2,\tag{19}$$

$$y\_1(\mathbf{0}) = \mathbf{0}, y\_2(\mathbf{0}) = \mathbf{1}, y\_4(\mathbf{0}) - \mathbf{1} = \mathbf{0} \text{ at } \boldsymbol{\eta} = \mathbf{0} \tag{20}$$

$$\mathcal{Y}\_2(\eta) = \mathbf{0}, \mathcal{Y}\_4(\eta) = \mathbf{0}. \text{as } \eta \to \infty \tag{21}$$


## **Table 3.**

*Nusselt number of platelet shape nanoparticle.*

where

$$\mathbf{g}\_1 = \frac{1}{c\_1(1+2\eta \mathbf{C})} \left( c\mathbf{y}\mathbf{M}\mathbf{y}\_2 - 2\varepsilon \mathbf{c}\_1 \mathbf{C} \mathbf{y}\_3 - \lambda \mathbf{y}\_4 - \left( \mathbf{y}\_1 \mathbf{y}\_3 - \mathbf{y}\_2^2 - \mathbf{S} \left( \mathbf{y}\_2 + \frac{\eta}{2} \mathbf{y}\_3 \right) \right) \right), \tag{22}$$

and

$$\begin{aligned} g\_2 &= \frac{1}{(1+2\eta C)e\_3} - \beta\_1 \left(\frac{(S\eta)^2}{4} - S\eta\_1 + \boldsymbol{\chi}\_1^2\right) \\ &- (1+2\eta C)e\_1E\boldsymbol{\chi}\_3^2 - \beta\_1 \left(e\_1Ec(1+2\eta C)\left(3\boldsymbol{\mathcal{S}}\boldsymbol{\chi}\_3^2\right) + \boldsymbol{\mathcal{S}}\boldsymbol{\mathfrak{s}}\_2\boldsymbol{E}C\boldsymbol{\mathcal{S}}\boldsymbol{\chi}\_3\boldsymbol{\chi}\_3^2 - 2\boldsymbol{\chi}\_1\boldsymbol{\chi}\_3\boldsymbol{\chi}\_3'\right) \\ &- \left(S^2\left(6\boldsymbol{\mathcal{y}}\_4 + \frac{11}{4}\eta\boldsymbol{\mathcal{y}}\_5\right) - S\left(5\boldsymbol{\mathcal{y}}\_2\boldsymbol{\chi}\_4 - \frac{11}{2}\boldsymbol{\mathcal{y}}\_1\boldsymbol{\mathcal{y}}\_5\right) + \left(\eta - \frac{1}{2}\boldsymbol{\mathcal{y}}\_2\boldsymbol{\mathcal{y}}\_5 + \frac{\eta}{2}\boldsymbol{\mathcal{y}}\_3\boldsymbol{\mathcal{y}}\_5 + \boldsymbol{\chi}\_3^2\boldsymbol{\mathcal{y}}\_4\right)\right)\right). \end{aligned} \tag{23}$$

*g*<sup>1</sup> and *g*<sup>2</sup> are the obtained first-order linear equations.
