8.3 Separable firstorder differential equations
 Definition

A separable firstorder ordinary differential equation is a firstorder 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) = [e^{x(t)+t}/x(t)]√(1 + t^{2})is separable because we can write it asx'(t) = e^{t}√(1 + t^{2}) · e^{x(t)}/x(t),orx'(t) = f(t)g(x(t))where f(t) = e^{t}√(1 + t^{2}) and g(z) = e^{z}/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)).
Now change the variable in the integral on the left: let y = x(s), so that dy = x'(s)ds and the equation becomes
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(t)·t,for which we drew a direction field previously. First write the equation asx'(t)/x(t) = t,so that∫^{t}(x'(s)/x(s))ds = ∫^{t}sds + C.In the integral on the left, change the variable to y = x(s), to get∫^{x(t)}(1/y)dy = ∫^{t}sds + C.Finally, calculate the integrals, to getln x(t) = t^{2}/2 + C.We can isolate x(t) to obtainx(t) = Ce^{t}^{2}^{/2} for all t.The C in this equation is equal to e^{C} 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(t_{0}) = x_{0} then the value of C is determined by the equation
x_{0} = 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 = Ce^{1/2}, so that C = e^{−1/2}.
 Example

Consider the differential equation
x'(t) = −2(x(t))^{2}t.We may write this equation as−x'(t)/(x(t))^{2} = 2t,which may be integrated to yield1/x(t) = t^{2} + C,yielding the set of solutionsx(t) = 1/(t^{2} + 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/Cso C = 1. Thus for this initial condition the solution isx(t) = 1/(t^{2} + 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 x^{4} + 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) = t^{2}/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 AK^{1−a}L^{a}, 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 K_{0} at t = 0. The labor force is L_{0} > 0 at t = 0 and grows at a constant rate λ, so thatL'(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) = Ce^{λt}. Given the initial condition, we have C = L_{0}.
Substituting this result into the equation for K'(t) yields
K'(t) = sA(K(t))^{1−a}(L_{0}e^{λt})^{a} = sA(L_{0})^{a}e^{aλt}(K(t))^{1−a}. (K(t))^{a−1}K'(t) = sA(L_{0})^{a}e^{aλt}.Integrating both sides, we obtain(K(t))^{a}/a = sA(L_{0})^{a}e^{aλt}/aλ + C,so thatK(t) = (sA(L_{0})^{a}e^{aλt}/λ + C)^{1/a}.Given K(0) = K_{0}, we conclude that C = (K_{0})^{a} − sA(L_{0})^{a}/λ. ThusK(t) = [sA(L_{0})^{a}(e^{aλt} − 1)/λ + (K_{0})^{a}]^{1/a} for all t. K(t)/L(t) = [sA(L_{0})^{a}(e^{aλt} − 1)/λ + (K_{0})^{a}]^{1/a} L_{0}e^{λt} (sA/λ)^{1/a}.A more general version of this model is studied in a later example.