Solution method in Smith (2006) A Closed Form Solution to the Ramsey Model
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.