<p>To derive this equation you can use the <a href="https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/core-mathematics/calculus/integrating-factor.html" rel="nofollow noreferrer">integrating factor method</a>. You have the first order ordinary differential equation</p>
<p><span class="math-container">$$\frac{dr}{dt}+n_1r=\frac{s}{b}.$$</span></p>
<p>Integrating factor for this equation is <span class="math-container">$M(t)=e^{\int n_1dt}=e^{n_1t}$</span>. Multiplying both sides of the differential equation by <span class="math-container">$M(t)$</span> yields:</p>
<p><span class="math-container">$$e^{n_1t}\left(\frac{dr}{dt}+n_1r\right)=e^{n_1t}\frac{s}{b}.$$</span></p>
<p>Notice that <span class="math-container">$\frac{d}{dt}\left[ re^{n_1t}\right]=e^{n_1t}\left(\frac{dr}{dt}+n_1r\right)$</span>. Substituting this in the differential equation we get</p>
<p><span class="math-container">$$\frac{d}{dt}\left[ re^{n_1t}\right]=e^{n_1t}\frac{s}{b},$$</span></p>
<p>or</p>
<p><span class="math-container">$$d\left[ re^{n_1t}\right]=e^{n_1t}\frac{s}{b}dt.$$</span></p>
<p>Integrating both sides with respect to the differentials</p>
<p><span class="math-container">$$\int d\left[ re^{n_1t}\right]=\int e^{n_1t}\frac{s}{b}dt,$$</span></p>
<p><span class="math-container">$$re^{n_1t}=e^{n_1t}\frac{s}{n_1b}+C$$</span></p>
<p><span class="math-container">$$r=\frac{s}{n_1b}+Ce^{-n_1t}.$$</span></p>
<p>Evaluating <span class="math-container">$r(t)$</span> at <span class="math-container">$t=0$</span> yields</p>
<p><span class="math-container">$$r_0=\frac{s}{n_1b}+C\implies C=r_0-\frac{s}{n_1b}.$$</span></p>
<p>Hence the solution to the differential equation is</p>
<p><span class="math-container">$$r=\frac{s}{n_1b}+\left(r_0-\frac{s}{n_1b}\right)e^{-n_1t}.$$</span></p>

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2024-08-19T18:18:51.697Z
2024-08-19T22:18:03.753Z
2024-08-19T22:18:03.753Z
manifold
<p>I am attempting to go through the derivations in Robert Solow's 1956 paper 'A Contribution to the Theory of Economic Growth'. On pages 73 to 76 he goes through an example where the production function used in the model is <span class="math-container">$$Y = F(K,L) = \min(\frac{K}{a},\frac{L}{b})$$</span>
On page 74, Solow writes 'Suppose the initial value of the
capital-labor ratio is <span class="math-container">$r_0 &gt; \frac{a}{b}$</span> then <span class="math-container">$\dot{r} = \frac{s}{b} - n_1r$</span>, whose solution is <span class="math-container">$r = (r_0 - \frac{s}{n_1b})e^-{^{n_0}}^t + \frac{s}{n_1b}$</span>'.</p>
<p>I understand why <span class="math-container">$\dot{r} = \frac{s}{b} - n_1r$</span>, but am not sure how this implies that <span class="math-container">$r = (r_0 - \frac{s}{n_1b})e^-{^{n_0}}^t + \frac{s}{n_1b}$</span> ?</p>
<p>Thanks</p>

<economic-growth><solow><model>
Derivation in Robert Solow's "A Contribution to the Theory of Economic Growth"
gguest
# Derivation in Robert Solow's "A Contribution to the Theory of Economic Growth"

(Asked by: username_0 on 2024-08-19)

I am attempting to go through the derivations in Robert Solow's 1956 paper 'A Contribution to the Theory of Economic Growth'. On pages 73 to 76 he goes through an example where the production function used in the model is $$Y = F(K,L) = \min(\frac{K}{a},\frac{L}{b})$$
On page 74, Solow writes 'Suppose the initial value of the
capital-labor ratio is $r\_0 > \frac{a}{b}$ then $\dot{r} = \frac{s}{b} - n\_1r$, whose solution is $r = (r\_0 - \frac{s}{n\_1b})e^-{^{n\_0}}^t + \frac{s}{n\_1b}$'.

I understand why $\dot{r} = \frac{s}{b} - n\_1r$, but am not sure how this implies that $r = (r\_0 - \frac{s}{n\_1b})e^-{^{n\_0}}^t + \frac{s}{n\_1b}$ ?

Thanks

**Tags:** economic-growth, solow, model

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## 1 answers

### Answer by username_1 on 2024-08-19. Score: 4 | Accepted answer

To derive this equation you can use the [integrating factor method](https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/core-mathematics/calculus/integrating-factor.html). You have the first order ordinary differential equation

$$\frac{dr}{dt}+n\_1r=\frac{s}{b}.$$

Integrating factor for this equation is $M(t)=e^{\int n\_1dt}=e^{n\_1t}$. Multiplying both sides of the differential equation by $M(t)$ yields:

$$e^{n\_1t}\left(\frac{dr}{dt}+n\_1r\right)=e^{n\_1t}\frac{s}{b}.$$

Notice that $\frac{d}{dt}\left[ re^{n\_1t}\right]=e^{n\_1t}\left(\frac{dr}{dt}+n\_1r\right)$. Substituting this in the differential equation we get

$$\frac{d}{dt}\left[ re^{n\_1t}\right]=e^{n\_1t}\frac{s}{b},$$

or

$$d\left[ re^{n\_1t}\right]=e^{n\_1t}\frac{s}{b}dt.$$

Integrating both sides with respect to the differentials

$$\int d\left[ re^{n\_1t}\right]=\int e^{n\_1t}\frac{s}{b}dt,$$

$$re^{n\_1t}=e^{n\_1t}\frac{s}{n\_1b}+C$$

$$r=\frac{s}{n\_1b}+Ce^{-n\_1t}.$$

Evaluating $r(t)$ at $t=0$ yields

$$r\_0=\frac{s}{n\_1b}+C\implies C=r\_0-\frac{s}{n\_1b}.$$

Hence the solution to the differential equation is

$$r=\frac{s}{n\_1b}+\left(r\_0-\frac{s}{n\_1b}\right)e^{-n\_1t}.$$


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