Table 1 Comparison between Ma et al. ( 2010 ) solutions and our solutions

i. If  b₀ = 0 and b₁ = 1, solutions Eq. (3.5) becomes:$\mathit{u}=\frac{{\mathit{k}}_0+{\mathit{b}}_1{\mathit{e}}^{\mathit{\xi }}}{{\mathit{e}}^{\mathit{\xi }}}$

i. If a₀ = 0, k₀ = − 2μ, a₁ = − 2μλ, f_1 i(ξ) = e^ξ then the solution of equation (20) is$\mathit{u}=\frac{{\mathit{k}}_0+{\mathit{b}}_1{\mathit{e}}^{\mathit{\xi }}}{{\mathit{e}}^{\mathit{\xi }}}$