Derivation in Robert Solow's "A Contribution to the Theory of Economic Growth"

Question

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

Answer 1

To derive this equation you can use the integrating factor method. 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

$$r=\frac{s}{n_1b}+\left(r_0-\frac{s}{n_1b}\right)e^{-n_1t}.$$