conditions will give $$\begin{aligned}
C_1 =& \frac{R_0^{1-q}}{2(q-r)} \bigg[4r - \alpha(3+3h-2r)~R_0^{\frac{3h+1}{2}} - 2(2\beta +2h\beta - r\beta +\gamma +h\gamma)~R_0^{2h+1} \nonumber \\
& -2\delta (1 + h -2r)~R_0^h - 2\gamma (2+2h-r)R_0^{2h+1}\ln ~[R_0^{h+1}]\bigg]\end{aligned}$$ and $$\begin{aligned}
C_2 =& \frac{R_0^{1-r}}{2(r-q)} \bigg[4q - \alpha(3+3h-2q)~R_0^{\frac{3h+1}{2}} - 2(2\beta +2h\beta - q\beta +\gamma +h\gamma)~R_0^{2h+1} \nonumber \\
& -2\delta (1 + h -2q)~R_0^h - 2\gamma (2+2h-q)R_0^{2h+1}\ln ~[R_0^{h+1}]\bigg]\end{aligned}$$ In general the equation of state of eqn.(\[eos\]) will be a function of $H$ and hence time in this case and so it can explain the transition from quintessence ($\omega_R > -1$) to phantom dominated regime ($\omega_R < -1$) as predicted by recent observations [@f100; @f101; @f102]. For the quintessence scale factor (\[mGB2\]) the conformal time can be evaluated with eqns.(\[quin1\]) and (\[quin2\]) as $$\eta = \int_0^{t}~\frac{dt}{a} = \frac{1}{a_0(1-h)} \left[\frac{6h(2h-1)}{R}\right]^{\frac{1-h}{2}} ~~,~~~~~~~~~~~~\frac{1}{2}<h<1 ~~.$$ For a real finite conformal time it is necessary to have $\frac{1}{2}<h<1$. Substituting this in eqn.(\[rho2\]) we get $$\begin{aligned}
\label{rhaq}
\frac{\rho_{\Lambda}}{m_p^2} &= \frac{3~n^2 ~a_0^2 (1-h)^2}{(6h)^{1-h}(2h-1)^{1-h}}~R^{1-h} + \frac{a ~a_0^3 (1-h)^3}{m_p~(6h)^{\frac{3-3h}{2}}(2h-1)^{\frac{3-3h}{2}}}~R^{\frac{3-3h}{2}}  \nonumber \\
& + \frac{ a_0^4 (1-h)^4}{m_p^2~(6h)^{2-2h}(2h-1)^{2-2h}}\left[b~ \ln \left\{\frac{(6h)^{1-h}(2h-1)^{1-h}m_p^2}{a_0^2(1-h)^2}\right\}
+ c \right]R^{2-2h}  \nonumber \\
& - \frac{b~a_0^4 (1-h)^4}{m_p^2~(6h)^{2-2h}(2h-1)^{2-2h}} ~R^{2-2h} ~\ln ~(R^{1-h}) ~~.\end{aligned}$$ Solving the inhomogeneous Euler differential equation (\[maineq\]) with eqns.(\[rhaq\]), (\[quin1\]) and (\[quin2\]) we get the form of $f(R)$ as $$f(R) = 