Bernoulli transformation in " A Close Form Solution to the Ramsey model" by Smith (2006) I am trying to understand the way Smith arrived from (1) and (2) to (6) and (8). 

Could anyone give me a hint? Thank you!


  It is well known that the dynamics per capita consumption $c$ and capital $k$ are determined by the following pair of differential equations:
  \begin{eqnarray}
\dot{k} &amp;=&amp; k^\alpha - \delta k - c \tag{1} \\
\frac{{\dot c}}{c} &amp;=&amp; \sigma(\alpha k^{\alpha - 1} - \delta - \rho) \tag{2}
\end{eqnarray}
  A Solution
  
  The system of equations (1)-(2) is dauntingly nonlinear. To approach a solution, define the capital-output ratio (a Bernoulli transformation) by $z = k^{1-\alpha}$ and the consumption-capital ratio by $x = c/k$. Using these  transformations the system can be rewritten as 
  \begin{eqnarray}
\dot{z} &amp;=&amp; (1-\alpha)[1 - (\delta + x)z] \tag{6} \\
\frac{{\dot x}}{x} &amp;=&amp;  \frac{\sigma \alpha - 1}{z} + \delta (1 - \sigma) - \rho\sigma + x \tag{7}
\end{eqnarray}
  The Bernoulli transformation converts the capital accumulation equation [Equation (1)] into the linear, albeit non-autonomous, differential equation in Equation (6). Despite this simplification, the system still does not allow an analytical solution. Equation (7) is still non-linear.
  
  Suppose, however, that $\sigma = 1/\alpha$. In that case $z$ disappears from Equation (7), and the system becomes recursive: Equation (7) reduces to
  $$
\frac{\dot{x}}{x} = - \frac{\delta(1- \alpha) \rho}{\alpha} + x \tag{8}
$$
  while $z$ evolves according to Equation (6), given the forcing process $x$ determined in Equation (8). This is a simple, autonomous logistic equation.


Citation: Smith, William. (2006). A Closed Form Solution to the Ramsey Model. Contributions to Macroeconomics. 6.
If $z = k^{1-\alpha}$ then

\begin{eqnarray}
\dot{z} &amp;=&amp; (1-\alpha)k^{1-\alpha-1}\dot{k} = (1-\alpha)k^{-\alpha} \dot{k} \\ &amp;=&amp; (1-\alpha) k^{-\alpha} [k^{\alpha} - \delta k - c] \\
&amp;=&amp; (1-\alpha) [1 - \delta k^{1-\alpha} - c k^{-\alpha}\color{blue}{(k/k)}] \\
&amp;=&amp; (1-\alpha) [1 - \delta z - c(k^{1-\alpha})/k] \\
&amp;=&amp; (1-\alpha) [1 - \delta z - x z]
\end{eqnarray}

I will leave $\dot{c}/c$ for you to work out