\[\frac{1}{\mu}=\frac{1}{\mu_\infty}\left(1+\gamma(\left(T-T_\infty\right) \right)\]
or \(\frac{1}{\mu}=b\left (T-T_\infty\right)\) here \(b=\frac{\gamma}{\mu_\infty}, \ T_r=T_\infty-\frac{1}{\gamma}\)
Thermal conductivity of fluid as described by \cite{dada2016} in following way:
\(k=k_\infty\left(1+a\theta\right)\) here \(a=\frac{K_w-K_\infty}{K_\infty}\), Where \(a\) presents thermal conductivity parameter, \(b, c, T_r\) are means constants. Where, numerical calculations for liquid are obtained when \(b>0\) and \(b<0\) for gases.

Solution Procedures

Following similarity transformations of   \cite{radiation2015} ,we convert the partial differential equation with associated boundary conditions are in ordinary differential equations:
\[ \begin{equation}\label{eq:1} \begin{aligned} u=cxf',\ v=-\sqrt{c}\nu_\infty f\left(\eta\right),\ \eta\left(y\right)\\ =-\sqrt\frac{{c}}{\nu_\infty}y, \\ \theta(\eta)=\frac{T-T_\infty}{T_w-T_\infty}, \ \ \ \ \ \ \ \ \ \xi(x)=\sqrt\frac{c}{\nu_\infty}x\end{aligned} \end{equation} \]The continuity equation satisfied after defining the velocity component as follows  \(u=\frac{\delta \psi}{\delta y}\) and \(v=\frac{\delta \psi}{\delta x}\)    and the remaining momentum and energy equations are in following form:
\[\begin{equation}\label{eq:1} \begin{aligned} f'''-\frac{\theta'}{\theta-\theta_r}f''-\frac{\theta-\theta'}{\theta_r}ff'\\ +\frac{\theta-\theta'}{\theta_r}f'^2+\frac{\theta-\theta'}{\theta_r}Mf'+\frac{\theta-\theta'}{\theta_r}\left(\frac{2\beta}{(\eta+\alpha)^4}-\lambda_1\right)\theta\\ =0\end{aligned} \end{equation} \]
\[\left(1+a\theta\right)\theta''+a\theta'^2+\Pr f\theta'-2\beta\lambda(\epsilon+\theta)\frac{f}{(\eta+\alpha)^3}=0\]