Solution method in Smith (2006) A Closed Form Solution to the Ramsey Model 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} &amp;=&amp; (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 &amp;=&amp; \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} &amp;=&amp; -(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 &amp;=&amp; \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} &amp;=&amp; \frac{1-\alpha}{z_c} &amp;=&amp; (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.]
Let's solve the differential equation (12).

As a first step, we look for a "simpler" differential equation, namely (A.1).
(A.1) can be written
$$\frac{\dot{z}}{z}=-(1-\alpha)\left(\delta+\frac{\bar{x}e^{-\bar{x}t}}{e^{-\bar{x}t}+\bar{x}A}\right)$$
On the left-hand side, you have the derivative of $\ln(z)$. You can integrate to obtain the form of any solution $z_c$: 
$$\ln(z_c)=-(1-\alpha)\left(\delta t-\ln(e^{-\bar{x}t}+\bar{x}A)\right)+ constant$$
You obtain $z_c=\kappa e^{-(1-\alpha)\delta t}\left(e^{-\bar{x}t}+\bar{x}A\right)^{1-\alpha}$, where $\kappa$ is the exponential of the constant in the previous equation.
This equation is equivalent to (A.2) for a particular constant, such that $\kappa=\bar{x}^{\alpha-1}$. This choice of the constant comes from some boundary conditions on $z$ (which should be stated in the paper).

As a second step, we use the method of variation of parameters, meaning we are looking for a solution of (12), $z_p$, that has a particular form, $z_p=z_c.\Psi$. If we find an expression for $\Psi$, then we have found a solution of (12).
We substitute $z_p$ in (12):
$$\dot{z_c}.\Psi+z_c.\dot{\Psi}=(1-\alpha)\left[1-\left(\delta+\frac{\bar{x}}{1+\bar{x}Ae^{\bar{x}t}}\right)z_c.\Psi\right]$$
This expression simplifies since $z_c$ satisfies (A.1):
$$z_c.\dot{\Psi}=(1-\alpha)$$
This is (A.3).
Then, I guess you we have to find a solution $\Psi$ of this equation, and we are done.