Analytically tractable Ramsey model: how to solve ODE for optimal trajectories In Brunner and Strulik (2002) the authors claim, that the solution of
\begin{align}
\dot c &amp;= \frac{c}{\sigma}(\alpha k^{\alpha-1} - \delta - \rho)\\
\dot k &amp;= k^\alpha - \delta k - c
\end{align}

is given by (see eq. 27)
\begin{align}
c(t) &amp;= \left(1 - \frac{1}{\sigma}\right) k(t)^\alpha\\
k(t) &amp;= \left[\frac{1}{\delta\sigma} +\left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp(-\delta(1-\alpha)t) \right]^{\frac{1}{1-\alpha}}
\end{align}
if $\alpha\delta\sigma = \delta + \rho$.
I can verify the solution for $c(t)$ (see eg. this thread). However I'm not sure how to solve for $k(t)$. We can plug in $c(t)$ into $\dot k$, which gives
\begin{align}
\dot k &amp;=  \frac{1}{\sigma} k^\alpha - \delta k. 
\end{align}


How would one proceed?


Solution

With respect to Alecos answer we may solve the following ODE

\begin{align}
\dot z + (1-\alpha)\delta z = \frac{1-\alpha}{\sigma}.
\end{align}

The genereal solution is given by
\begin{align}
z(t) &amp;= \frac{1}{\exp(\int (1-\alpha)\delta dt)}\left[\int \exp\left(\int (1-\alpha)\delta dt\right) \frac{1-\alpha}{\sigma} dt + C \right]\\
&amp;= \frac{1}{\delta\sigma} + C\exp((\alpha-1)\delta t).
\end{align}

With $z(0)$ given we pin down $C$
\begin{align}
C = z(0) - \frac{1}{\delta\sigma}
\end{align}

which yields the desired result
\begin{align}
z(t) &amp;= \frac{1}{\delta\sigma} + \left(z(0) - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t) \\
\Longleftrightarrow \quad k(t)^{1-\alpha} &amp;= \frac{1}{\delta\sigma} + \left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t)\\
\Longleftrightarrow \quad k(t) &amp;= \left[\frac{1}{\delta\sigma} + \left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t)\right]^{\frac{1}{1-\alpha}}.
\end{align}
The differential equation 

$$\dot k =  \frac{1}{\sigma} k^\alpha - \delta k$$

has the structure of a Bernoulli equation. We solve it by the following transformation steps:

1) Mulitply throughout by $k^{-\alpha}$:

$$k^{-\alpha}\dot k =  \frac{1}{\sigma}  - \delta k^{1-\alpha} \tag{1}$$

2) Define the variable 
$$z \equiv k^{1-\alpha} \implies \dot z = (1-\alpha)k^{-a}\dot k \tag{2}$$

3) Combine to get

$$(1),(2) \implies \frac {1}{1-\alpha}\dot z = \frac{1}{\sigma} - \delta z $$

$$\implies \dot z + (1-\alpha)\delta z = \frac {1-\alpha}{\sigma}$$

This is a standard first-order differential equation with constant coefficients. Solve it and then reverse the change of variable to get the result.

PS: The above requires that $k \neq 0$ everywhere, which is the economically meaningful case.