Analytically tractable Ramsey model: how to solve ODE for optimal trajectories

Question

In Brunner and Strulik (2002) the authors claim, that the solution of \begin{align} \dot c &= \frac{c}{\sigma}(\alpha k^{\alpha-1} - \delta - \rho)\\ \dot k &= k^\alpha - \delta k - c \end{align}

is given by (see eq. 27) \begin{align} c(t) &= \left(1 - \frac{1}{\sigma}\right) k(t)^\alpha\\ k(t) &= \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 &= \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) &= \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]\\ &= \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) &= \frac{1}{\delta\sigma} + \left(z(0) - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t) \\ \Longleftrightarrow \quad k(t)^{1-\alpha} &= \frac{1}{\delta\sigma} + \left(k(0)^{1-\alpha} - \frac{1}{\delta\sigma}\right)\exp((\alpha-1)\delta t)\\ \Longleftrightarrow \quad k(t) &= \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}

Answer 1

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) Multiply 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.