I am trying to understand the way Smith demonstrates that the general solution to his equation (12) is (13) (see page 6).
(12) \begin{eqnarray} \dot{z} &=& (1- \alpha)\left[1-\left(\delta + \frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z\right] \end{eqnarray}
In the Appendix I demonstrate that the general solution to Equation (12) is:
(13)\begin{eqnarray} z &=& \frac{1}{\bar{x}+\delta} 2F1(1-\alpha,1,d;\omega)+B\bar{x}^{\alpha-1}e^{-(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{1-\alpha} \end{eqnarray}
The Appendix: A.1 \begin{eqnarray} \dot{z} &=& -(1- \alpha)\left(\delta + \frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z \end{eqnarray} This can be integrated to find the complementary solution:(A.2) \begin{eqnarray} z_c &=& \bar{x}^{\alpha-1}e^{-(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{1-\alpha} \end{eqnarray}
To find the particular solution to equation (12), I will use the method of variation of parameters. Conjecture that the particular solution is $z_p$ $=$ $z_c$$\Psi$, where $\Psi$ is an unknown function of time. Substituting this conjecture into equation (12), it follows that: (A.3) \begin{eqnarray} \dot{\Psi} &=& \frac{1-\alpha}{z_c} &=& (1-\alpha)\bar{x}^{1-\alpha}e^{(1-\alpha)(\bar{x}+\delta)t}(1+\bar{x}Ae^{\bar{x}t})^{\alpha-1} \end{eqnarray}
First, is (A.2) then just an integrated version of (A.1) or are there any other steps involved?
Second, I really do not see how he substituted the conjecture into 12 and how can he obtain (A.3) from this substitution? Especially here I am really lost and am lacking the imagination how he came up with $\dot{\Psi}$. Did he substitute $z_p$ for $\dot{z}$ or $z$ in (12)?
Citation: [Smith, William. (2006). A Closed Form Solution to the Ramsey Model. Contributions to Macroeconomics. 6.]