\[C_f=-\frac{\theta_r}{\theta-\theta_r}R_e^\frac{1}{2}f''(0)\]
\[Nu=-R_e^{\frac{1}{2}}\theta'(0)\]

Numerical Method for solution

To find an exact solution of such kind of problems in fluid dynamics several numerical techniques have been proposed. Among these bvp4c function technique is most useful technique to solve higher non-linear differential equation. For this we need to convert the equations (12) and (13) with suitable boundary conditions (14) are in first order differential equation as assumed by new variables such as : \(f=y_1,f'=y_2, f''=y_3, \theta=y_4, \theta'=y_5\) . All these process are simplified in MATLAB software. So after introducing new variables in equations (12), (13) and (14) we get the following form.
\[\begin{equation}\label{eq:1} \begin{aligned} f'=y_2\\ f''=y_2'=y_3\\ f'''=y_3'=\frac{y_3y_5}{y_4-\theta_r}+\frac{y_4-\theta_r}{\theta_r}y_1y_3\\ +\frac{y_4-\theta_r}{\theta_r}y_2^2-+\frac{y_4-\theta_r}{\theta_r}\left(\frac{2\beta}{(\eta+\alpha)^4}\right)y_4-\frac{y_4-\theta_r}{\theta_r}My_2\\ \theta'=y_5\\ \theta''=y_5'=-\frac{ay_5^2}{1+ay_4}-Pr\frac{y_1y_5}{1+ay_4}+\frac{2\beta\lambda(y_4+\epsilon)y_1}{(1+ay_4)(\eta+\alpha)^3} \end{aligned} \end{equation} \]
Boundary conditions are:
\[y_1(0)=s,\ y_2(0)=1, \ y_4(0)=1, \ y_2(\infty)=0,\ y_4(\infty)=0\]
Set of equation (19) as well as boundary condition (20) are integrated numerically as an initial value problem to a given terminal point.