Now we are going to apply our procedure to obtain approximate solutions of Eqs. (5) and (6) with the initial / boundary conditions Eq. (7).

For this purpose, in the case of the nonlinear equation Eq. (5), we choose the linear operator of the form:
$$\begin{array}{}{\displaystyle {L}_{f}\left(\eta \right)={f}^{\u2034}(\eta )+\frac{3K}{K\eta +1}{f}^{\u2033}(\eta ),}\end{array}$$(22)

where *K* is an unknown positive parameter and will be determined later.

As in Marinca and Herisanu [12], it is easy to show that the linear operator is not unique.

The initial approximation *f*_{0}(*η*) can be obtained from the following problem:
$$\begin{array}{}{\displaystyle {f}_{0}^{\u2034}(\eta )+\frac{3K}{K\eta +1}{f}_{0}^{\u2033}(\eta )=0,}\\ \\ {f}_{0}(0)=S,\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{0}^{\prime}(0)=1+\lambda {f}_{0}^{\u2033}(0),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{f}_{0}^{\prime}(\mathrm{\infty})=0,\end{array}$$(23)

which has the solution
$$\begin{array}{}{\displaystyle {f}_{0}(\eta )=S+\frac{1}{K(2\lambda K+1)}-\frac{1}{K(2\lambda K+1)}\cdot \frac{1}{K\eta +1}.}\end{array}$$(24)

The nonlinear operator *N*_{f}(*η*), corresponding to nonlinear differential Eq. (5), is defined by:
$$\begin{array}{}{\displaystyle {N}_{f}(\eta )=-\frac{3K}{K\eta +1}{f}^{\u2033}+f{f}^{\u2033}-2({f}^{\prime}{)}^{2}.}\end{array}$$(25)

For the initial approximation *f*_{0}(*η*) given by Eq. (24), the nonlinear operator Eq. (25) becomes:
$$\begin{array}{}{\displaystyle {N}_{{f}_{0}}(\eta )=-\frac{3K}{K\eta +1}{f}_{0}^{\u2033}+{f}_{0}{f}_{0}^{\u2033}-2({f}_{0}^{\prime}{)}^{2}=}\\ \\ {\displaystyle =-\frac{2(KS+1)}{(2K\lambda +1{)}^{2}}\cdot \frac{1}{(K\eta +1{)}^{3}}+\frac{6{K}^{2}}{2K\lambda +1}\cdot \frac{1}{(K\eta +1{)}^{4}}.}\end{array}$$(26)

Comparing Eqs. (26) and (18), one can write:
$$\begin{array}{}{\displaystyle {h}_{1}(\eta )=-\frac{2(KS+1)}{(2K\lambda +1{)}^{2}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{1}(\eta )=\frac{1}{(K\eta +1{)}^{3}},}\\ \\ {\displaystyle {h}_{2}(\eta )=\frac{6{K}^{2}}{2K\lambda +1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{2}(\eta )=\frac{1}{(K\eta +1{)}^{4}}.}\end{array}$$(27)

The function *f*_{1}(*η*) given by Eq. (19) becomes:
$$\begin{array}{}{\displaystyle {f}_{1}(\eta ,{C}_{i})={H}_{1}(\eta ,{C}_{i})\frac{1}{(K\eta +1{)}^{3}}+{H}_{2}(\eta ,{C}_{i})\frac{1}{(K\eta +1{)}^{4}},}\end{array}$$(28)

where we have freedom to choose a lot of possibilities for the unknown functions *H*_{i}, *i* = 1, 2, as follows (see Marinca and Herisanu [12]):
$$\begin{array}{}{H}_{1}(\eta ,{C}_{i})={A}_{1}(K\eta +1{)}^{3}-\frac{{A}_{2}}{K(2K\lambda +1)}(K\eta +1{)}^{2}+\\ \\ +{C}_{1}(K\eta +1)+{C}_{2},\\ \\ {H}_{2}(\eta ,{C}_{i})={C}_{3}+\sum _{i=1}^{7}\frac{{C}_{i+3}}{(K\eta +1{)}^{i}}+\frac{{C}_{11}{\eta}^{2}}{(K\eta +1{)}^{10}}.\end{array}$$(29)

Substituting Eq. (29) into Eq. (28) we have:
$$\begin{array}{}{f}_{1}(\eta ,{C}_{i})={A}_{1}-\frac{1}{K(2K\lambda +1)}\frac{{A}_{2}}{K\eta +1}+\\ \\ +\sum _{i=1}^{10}\frac{{C}_{i}}{(K\eta +1{)}^{i+1}}+\frac{{C}_{11}{\eta}^{2}}{(K\eta +1{)}^{14}},\end{array}$$(30)

where
$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{A}_{1}=\frac{{A}_{2}}{K(2K\lambda +1)}-\sum _{i=1}^{10}{C}_{i},}\\ {\displaystyle {A}_{2}=K\cdot \sum _{i=1}^{10}(i+1){C}_{i}+\lambda {K}^{2}\cdot \sum _{i=1}^{10}(i+1)(i+2){C}_{i}+2\lambda {C}_{11}.}\end{array}$$

The first-order approximate solution given by Eq. (21) is obtained from Eqs. (24) and (30):
$$\begin{array}{}\overline{f}(\eta ,{C}_{i})={f}_{0}(\eta )+{f}_{1}(\eta ,{C}_{i})=S+{A}_{1}+\frac{1}{K(2K\lambda +1)}-\\ \\ -\frac{1}{K(2K\lambda +1)}\frac{1+{A}_{2}}{K\eta +1}+\sum _{i=1}^{10}\frac{{C}_{i}}{(K\eta +1{)}^{i+1}}+\frac{{C}_{11}{\eta}^{2}}{(K\eta +1{)}^{14}}.\end{array}$$(31)

In this way, we can find other solutions as well.

For Eq. (6) with initial/boundary condition given by Eq. (7) (for the unknown function *φ*), the expression for the linear operator *L*_{φ}(*η*) is chosen as:
$$\begin{array}{}{\displaystyle {L}_{\phi}(\eta )={\phi}^{\u2033}+\frac{2}{{K}_{1}\eta +1}{\phi}^{\prime},}\end{array}$$(32)

where *K*_{1} > 0 is an unknown parameter at this moment.

Eq. (16) can be written in the form:
$$\begin{array}{}{\displaystyle {\phi}_{0}^{\u2033}+\frac{2}{{K}_{1}\eta +1}{\phi}_{0}^{\prime}=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\phi}_{0}(0)=1,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\phi}_{0}(\mathrm{\infty})=0}\end{array}$$(33)
and has the solution
$$\begin{array}{}{\displaystyle {\phi}_{0}(\eta )=\frac{1}{{K}_{1}\eta +1}.}\end{array}$$(34)

The nonlinear operator *N*_{φ} corresponding to the unknown function *φ* is obtained from the expression Eq. (6) in the form:
$$\begin{array}{}{\displaystyle {N}_{\phi}(\eta )=-\frac{2}{{K}_{1}\eta +1}{\phi}^{\prime}+Sc\left(f{\phi}^{\prime}-{f}^{\prime}\phi -\beta \phi \right).}\end{array}$$(35)

For the initial approximation *φ*_{0}(*η*) given by Eq. (34), the nonlinear operator Eq. (35) becomes:
$$\begin{array}{}{N}_{{\phi}_{0}}(\eta )=-\frac{Sc\beta}{{K}_{1}\eta +1}-{K}_{1}Sc\left[S+\frac{1}{K(2K\lambda +1)}\right]\times \\ \\ \times \frac{1}{({K}_{1}\eta +1{)}^{2}}+\frac{2{K}_{1}}{({K}_{1}\eta +1{)}^{3}}-\frac{Sc}{2\lambda K+1}\frac{1}{(K\eta +1{)}^{2}}\frac{1}{{K}_{1}\eta +1}+\\ \\ +\frac{Sc{K}_{1}}{K(2\lambda K+1)}\frac{1}{K\eta +1}\frac{1}{({K}_{1}\eta +1{)}^{2}}.\end{array}$$(36)

By comparing the Eqs. (18) and (36) one can get:
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}{h}_{1}^{\ast}(\eta )=-Sc\beta ,{g}_{1}^{\ast}(\eta )=\frac{1}{{K}_{1}\eta +1},\\ {h}_{2}^{\ast}(\eta )=-{K}_{1}Sc\left[S+\frac{1}{K(2K\lambda +1)}\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{g}_{2}^{\ast}(\eta )=\frac{1}{({K}_{1}\eta +1{)}^{2}},\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{h}_{3}^{\ast}(\eta )=2{K}_{1},{g}_{3}^{\ast}(\eta )=\frac{1}{({K}_{1}\eta +1{)}^{3}},\\ {h}_{4}^{\ast}(\eta )=\frac{Sc{K}_{1}}{K(2\lambda K+1)},{g}_{4}^{\ast}(\eta )=\frac{1}{K\eta +1}\frac{1}{({K}_{1}\eta +1{)}^{2}},\\ \phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{h}_{5}^{\ast}(\eta )=-\frac{Sc}{2\lambda K+1},{g}_{5}^{\ast}(\eta )=\frac{1}{(K\eta +1{)}^{2}}\frac{1}{{K}_{1}\eta +1}.\end{array}$$(37)

The first approximation *φ*_{1}(*η*, *D*_{i}), given by Eq. (19), becomes:
$$\begin{array}{}{\phi}_{1}(\eta ,{D}_{i})={H}_{1}^{\ast}(\eta ,{D}_{i})\frac{1}{{K}_{1}\eta +1}+{H}_{2}^{\ast}(\eta ,{D}_{i})\frac{1}{({K}_{1}\eta +1{)}^{2}}+\\ \\ +{H}_{3}^{\ast}(\eta ,{D}_{i})\frac{1}{({K}_{1}\eta +1{)}^{3}}+{H}_{4}^{\ast}(\eta ,{D}_{i})\frac{1}{K\eta +1}\frac{1}{({K}_{1}\eta +1{)}^{2}}+\\ \\ +{H}_{5}^{\ast}(\eta ,{D}_{i})\frac{1}{(K\eta +1{)}^{2}}\frac{1}{{K}_{1}\eta +1},\end{array}$$(38)

where *D*_{i} are unknown parameters, and the unknown auxiliary functions $\begin{array}{}{H}_{1}^{\ast}(\eta ,{D}_{i}),...,{H}_{5}^{\ast}(\eta ,{D}_{i})\end{array}$ can be chosen in the form:
$$\begin{array}{}{H}_{1}^{\ast}(\eta ,{D}_{i})={M}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{H}_{2}^{\ast}(\eta ,{D}_{i})={D}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{H}_{3}^{\ast}(\eta ,{D}_{i})={D}_{2},\\ \\ {H}_{4}^{\ast}(\eta ,{D}_{i})={D}_{4}({K}_{1}\eta +1{)}^{2}+\frac{{D}_{5}}{({K}_{1}\eta +1{)}^{3}}+\frac{{D}_{6}}{({K}_{1}\eta +1{)}^{4}}+\\ \\ +\frac{{D}_{7}}{({K}_{1}\eta +1{)}^{5}}+\frac{{D}_{8}{\eta}^{2}}{({K}_{1}\eta +1{)}^{6}},\\ \\ {\displaystyle {H}_{5}^{\ast}(\eta ,{D}_{i})={D}_{9}({K}_{1}\eta +1)+\frac{{D}_{10}}{({K}_{1}\eta +1{)}^{8}}+}\\ \\ +\frac{{D}_{11}}{({K}_{1}\eta +1{)}^{9}}+\frac{{D}_{12}{\eta}^{2}}{({K}_{1}\eta +1{)}^{10}},\end{array}$$(39)

where $\begin{array}{}{M}_{1}=-\sum _{i=1}^{7}{D}_{i}-\sum _{i=9}^{11}{D}_{i}.\end{array}$

Substituting Eq. (39) into Eq. (38) one can get:
$$\begin{array}{}{\displaystyle {\phi}_{1}(\eta ,{D}_{i})=\frac{{M}_{1}}{{K}_{1}\eta +1}+\sum _{i=1}^{3}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i+1}}+}\\ \\ {\displaystyle +\left[{D}_{4}+\sum _{i=5}^{7}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i}}+\frac{{D}_{8}{\eta}^{2}}{({K}_{1}\eta +1{)}^{8}}\right]\cdot \frac{1}{K\eta +1}+}\\ \\ {\displaystyle +\left[{D}_{9}+\sum _{i=10}^{11}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i-1}}+\frac{{D}_{12}{\eta}^{2}}{({K}_{1}\eta +1{)}^{11}}\right]\cdot \frac{1}{(K\eta +1{)}^{2}}.}\end{array}$$(40)

The first-order approximate solution given by Eq. (21) is obtained from Eqs. (34) and (40):
$$\begin{array}{}{\displaystyle \overline{\phi}(\eta ,{D}_{i})={\phi}_{0}(\eta )+{\phi}_{1}(\eta ,{D}_{i})=}\\ \\ {\displaystyle =\frac{{M}_{1}+1}{{K}_{1}\eta +1}+\sum _{i=1}^{3}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i+1}}+}\\ \\ +\left[{D}_{4}+\sum _{i=5}^{7}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i}}+\frac{{D}_{8}{\eta}^{2}}{({K}_{1}\eta +1{)}^{8}}\right]\cdot \frac{1}{K\eta +1}+\\ \\ {\displaystyle +\left[{D}_{9}+\sum _{i=10}^{11}\frac{{D}_{i}}{({K}_{1}\eta +1{)}^{i-1}}+\frac{{D}_{12}{\eta}^{2}}{({K}_{1}\eta +1{)}^{11}}\right]\cdot \frac{1}{(K\eta +1{)}^{2}}.}\end{array}$$(41)

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