Mathematical methods for economic theory

Martin J. Osborne

8.3 Separable first-order differential equations

Definition
A separable first-order ordinary differential equation is a first-order ordinary differential equation that may be written in the form
x'(t) = f(t)g(x(t))
for functions f and g of a single variable.
Example
The equation
x'(t) = [ex(t)+t/x(t)]√(1 + t2)
is separable because we can write it as
x'(t) = et√(1 + t2) · ex(t)/x(t),
or
x'(t) = f(t)g(x(t))
where f(t) = et√(1 + t2) and g(z) = ez/z.
Example
The equation
x'(t) = F(t) + G(x(t))
is not separable unless either F or G is identically 0: it cannot be written in the form x'(t) = f(t)g(x(t)).
The equation x'(t) = f(t)g(x(t)) may be solved as follows. First look for a solution for which g(x(t)) ≠ 0 for all t. In this case, we may write the equation as
x'(t)/g(x(t)) = f(t).
Thus
t(x'(s)/g(x(s)))ds = ∫tf(s)ds + C.
In this equation, ∫th(s)ds, for any function h, denotes an antiderivative of h evaluated at t. That is, ∫th(s)ds is the value, at t, of a function whose derivative is h.

Now change the variable in the integral on the left: let y = x(s), so that dy = x'(s)ds and the equation becomes

x(t)(1/g(y))dy = ∫tf(t)dt + C.
Assuming we can perform the integration on each side of this equation, we obtain a function of x(t) on the left and a function of t on the right. We may be able to isolate x(t), in which case we have an explicit solution of the equation. If we cannot isolate x(t) then we have an expression that defines x(t) implicitly.

Now consider the possibility of a solution in which g(x(t)) = 0 for some t. If g(x*) = 0 for some x* then x(t) = x* for all t is also a solution, because if g(x(t)) = 0 then x'(t) = 0.

Example
Consider the differential equation
x'(t) = x(tt,
for which we drew a direction field previously. First write the equation as
x'(t)/x(t) = t,
so that
t(x'(s)/x(s))ds = ∫tsds + C.
In the integral on the left, change the variable to y = x(s), to get
x(t)(1/y)dy = ∫tsds + C.
Finally, calculate the integrals, to get
ln x(t) = t2/2 + C.
We can isolate x(t) to obtain
x(t) = Cet2/2 for all t.
The C in this equation is equal to eC from the previous equation. I follow standard practice in using the same letter C to denote the new constant.

This argument is valid only for solutions x(t) with x(t) ≠ 0 for all t. Looking at the original differential equation we see that the function x defined by x(t) = 0 for all t is also a solution.

If we have an initial condition x(t0) = x0 then the value of C is determined by the equation

x0 = Ce(t0)2/2.
The direction field on a previous page shows a solution for the initial condition x(−1) = 1. For this initial condition we get 1 = Ce1/2, so that C = e−1/2.
Example
Consider the differential equation
x'(t) = −2(x(t))2t.
We may write this equation as
x'(t)/(x(t))2 = 2t,
which may be integrated to yield
1/x(t) = t2 + C,
yielding the set of solutions
x(t) = 1/(t2 + C) for all t.
In addition, x(t) = 0 for all t is a solution.

As before, an initial condition determines the exact solution. If x(0) = 1, for example, we need

1 = 1/C
so C = 1. Thus for this initial condition the solution is
x(t) = 1/(t2 + 1).
To take another example, if x(0) = 0, then the solution is x(t) = 0 for all t.
Example
Consider the differential equation
x'(t) = t/((x(t))4 + 1).
Separating the variables yields
((x(t))4 + 1)x'(t) = t.
The value of x4 + 1 is not 0 for any value of x, so all the solutions of the equation are obtained by integrating both sides of this equation, which leads to
(x(t))5/5 + x(t) = t2/2 + C.
In this case we cannot isolate x(t)—we cannot find the solutions explicitly—but know only that they satisfy this equation.
Example (Solow's model of economic growth)
Output is produced from capital, denoted K, and labor, denoted L, according to the production function AK1−aLa, where A is a positive constant and 0 < a < 1. A constant fraction s of output is “saved” (with 0 < s < 1), and used to augment the capital stock. Thus the capital stock changes according to the differential equation
K'(t) = sA(K(t))1−a(L(t))a
and takes the value K0 at t = 0. The labor force is L0 > 0 at t = 0 and grows at a constant rate λ, so that
L'(t)/L(t) = λ.
One way to find a solution of this model is to first solve for L, then substitute the resulting expression into the equation for K'(t) and solve for K.

The equation for L is separable. Integrating both sides yields ln L = λt + C, or L(t) = Ceλt. Given the initial condition, we have C = L0.

Substituting this result into the equation for K'(t) yields

K'(t)  =  sA(K(t))1−a(L0eλt)a
 =  sA(L0)aeaλt(K(t))1−a.
This equation is separable, and may be written as
(K(t))a−1K'(t) = sA(L0)aeaλt.
Integrating both sides, we obtain
(K(t))a/a = sA(L0)aeaλt/aλ + C,
so that
K(t) = (sA(L0)aeaλt/λ + C)1/a.
Given K(0) = K0, we conclude that C = (K0)a − sA(L0)a/λ. Thus
K(t)  =  [sA(L0)a(eaλt − 1)/λ + (K0)a]1/a for all t.
An economically interesting feature of the model is the evolution of the capital-labor ratio K(t)/L(t). We have
K(t)/L(t) = 
[sA(L0)a(eaλt − 1)/λ + (K0)a]1/a
L0eλt
for all t. As t grows without bound, this fraction converges to
(sA/λ)1/a.
A more general version of this model is studied in a later example.