8.2 Firstorder differential equations: existence of a solution
 Definition

A firstorder ordinary differential equation is an ordinary differential equation that may be written in the form
x'(t) = F(t, x(t))for some function F of two variables.
A firstorder differential equation plus a condition of this type—that is, a condition of the form x(t_{0}) = x_{0}, called an initial condition—is called a firstorder initial value problem. Despite the use of the word “initial”, the value of t_{0} in such a problem cannot necessarily be interpreted as a starting point. (Indeed, as I remarked on the previous page, the variable t cannot necessarily be interpreted as time.)
 Definition

A firstorder initial value problem consists of a firstorder ordinary differential equation
x'(t) = F(t, x(t))and an initial conditionx(t_{0}) = x_{0},where t_{0} and x_{0} are numbers.
Existence of a solution
Before trying to find a solution of a firstorder initial value problem, it is useful to know whether a solution exists.A diagram known as a direction field or integral field helps us think about the existence of a solution. We plot t on the horizontal axis and x on the vertical axis, and for each of a set of pairs (t, x) we draw a short line segment through (t, x) with slope F(t, x(t)), which from the equation is equal to x'(t). Such a diagram shows us the rate of change of x for each value of t and x(t).
 Example

The direction field of the equation
x'(t) = x(t)tis shown in the figure below. For example, at every point (t, 0) and every point (0, x) we have x'(t) = 0, so the slope of the line segment through every such point is 0. Similarly, at every point (1, x) we have x'(t) = x, so the slope of the line segment through (1, x) is x for each value of x. (The grid size in the figure is 1/2.)
x'(t)  =  F(t, x(t)) 
x(t_{0})  =  x_{0} 
This construction suggests that any firstorder initial value problem in which the slopes of the line segments in the direction field change continuously as (t, x) changes—that is, in which F is continuous—has a solution. If the partial derivative of F with respect to its second argument is continuous, then in fact the initial value problem has a unique solution.
 Proposition source

If F is a function of two variables that is continuous at (t_{0}, x_{0}) then there exists a number a > 0 and a continuously differentiable function x of a single variable defined on the interval
(t_{0} − a, t_{0} + a) that solves the firstorder initial value problem
x'(t) = F(t, x(t)) x(t_{0}) = x_{0}
 Source hide
 For a proof of existence (the first sentence of the result) see Coddington and Levinson (1955), p. 6. For a proof of uniqueness (the second sentence of the result), see Boyce and DiPrima (1969), pp. 71–78.
 Example

Consider the firstorder initial value problem
x'(t) = (x(t))^{1/2} x(0) = 0. Two solutions of the problem are x(t) = 0 for all t and x(t) = (t/2)^{2} for all t.
Stability of solutions
 Definition
 If for some initial condition a firstorder initial value problem has a solution that is a constant function (independent of t), the value of the constant is an equilibrium or stationary state of the associated differential equation.
 Example

Consider the firstorder initial value problem
x'(t) + x(t) = 2 with x(t_{0}) = x_{0}.The solution of this problem, as we shall see later, isx(t) = (x_{0} − 2)e^{t}^{0}^{−t} + 2.(You can verify that this function is a solution of the problem by calculating x'(t) and substituting it into the differential equation and by checking that x(t_{0}) = x_{0}.)
Thus for x_{0} = 2, the solution of the problem is x(t) = 2 for all t, so that 2 is an equilibrium of the differential equation. For no other value of x_{0} is the solution a constant function, so 2 is the only equilibrium.
 Definition
 If, for all initial conditions, the solution of a firstorder initial value problem converges to an equilibrium of the associated differential equation as t increases without bound, then the equilibrium is globally stable. If, for all values of x_{0} sufficiently close to the equilibrium the solution converges to the equilibrium as t increases without bound, then the equilibrium is locally stable. An equilibrium that is not locally stable is unstable.
 Example

The solution of the initial value problem in the previous example is
x(t) = (x_{0} − 2)e^{t}^{0}^{−t} + 2.For every value of (t_{0}, x_{0}), (x_{0} − 2)e^{t}^{0}^{−t} converges to zero, so the equilibrium x = 2 is globally stable.