<p>The differential equation</p>
<p><span class="math-container">$$\dot k =  \frac{1}{\sigma} k^\alpha - \delta k$$</span></p>
<p>has the structure of a <strong>Bernoulli equation</strong>. We solve it by the following transformation steps:</p>
<p><strong>1)</strong> Multiply throughout by <span class="math-container">$k^{-\alpha}$</span>:</p>
<p><span class="math-container">$$k^{-\alpha}\dot k =  \frac{1}{\sigma}  - \delta k^{1-\alpha} \tag{1}$$</span></p>
<p><strong>2)</strong> Define the variable
<span class="math-container">$$z \equiv k^{1-\alpha} \implies \dot z = (1-\alpha)k^{-a}\dot k \tag{2}$$</span></p>
<p><strong>3)</strong> Combine to get</p>
<p><span class="math-container">$$(1),(2) \implies \frac {1}{1-\alpha}\dot z = \frac{1}{\sigma} - \delta z $$</span></p>
<p><span class="math-container">$$\implies \dot z + (1-\alpha)\delta z = \frac {1-\alpha}{\sigma}$$</span></p>
<p>This is a standard first-order differential equation with constant coefficients. Solve it and then reverse the change of variable to get the result.</p>
<p><strong>PS:</strong> The above requires that <span class="math-container">$k \neq 0$</span> everywhere, which is the economically meaningful case.</p>

CC BY-SA 4.0
2015-11-23T20:06:54.127Z
2021-12-04T22:36:53.720Z
2021-12-04T22:36:53.720Z
Alecos Papadopoulos
<p>In <a href="http://www.sciencedirect.com/science/article/pii/S0165188900000786" rel="nofollow noreferrer">Brunner and Strulik (2002)</a> 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}</p>

<p>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 <a href="https://economics.stackexchange.com/questions/8257/stochastic-growth-in-continuous-time">thread</a>). 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}</p>

<ul>
<li>How would one proceed?</li>
</ul>

<h2>Solution</h2>

<p>With respect to Alecos answer we may solve the following ODE</p>

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

<p>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}</p>

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

<p>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}</p>

CC BY-SA 3.0
2015-11-23T12:48:42.770Z
I did not read the entire paper but it seems interesting. What I have remarked is that they make an assumption of constant saving rate (page 749). With this assumption, they can take $k$ as a "constant" term for  which allows to find an analytical solution for the differential system. Otherwise, it is not possible.
optimal control
CC BY-SA 3.0
2015-11-23T18:10:53.230Z
The last equation looks like a solvable differential equation. I think dividing everything by $k^{\alpha}$ and doing the following change of variable : $v=k^{1-\alpha}$, might help. This yields a first order differential equation that we know the solution of.
Louis. B
<economic-growth><dynamic-optimization><optimal-control><ode>
Analytically tractable Ramsey model: how to solve ODE for optimal trajectories
clueless
# Analytically tractable Ramsey model: how to solve ODE for optimal trajectories

(Asked by: username_0 on 2015-11-23)

In [Brunner and Strulik (2002)](http://www.sciencedirect.com/science/article/pii/S0165188900000786) 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](https://economics.stackexchange.com/questions/8257/stochastic-growth-in-continuous-time)). 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}

**Tags:** economic-growth, dynamic-optimization, optimal-control, ode
> **Comments:**
> **username_1**: I did not read the entire paper but it seems interesting. What I have remarked is that they make an assumption of constant saving rate (page 749). With this assumption, they can take $k$ as a "constant" term for  which allows to find an analytical solution for the differential system. Otherwise, it is not possible.  
> **username_2**: The last equation looks like a solvable differential equation. I think dividing everything by $k^{\alpha}$ and doing the following change of variable : $v=k^{1-\alpha}$, might help. This yields a first order differential equation that we know the solution of.  

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## 1 answers

### Answer by username_3 on 2015-11-23. Score: 3 | Accepted answer

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.


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