First-order first-degree autonomous differential equation

From Calculus
Jump to: navigation, search


Following the convention for autonomous differential equation, we denote the dependent variable by x and independent variable by t.

Form of the differential equation

The differential equation is of the form:

\frac{dx}{dt} = f(x)

Solution method and formula

We convert the differential equation to an integration problem:

\int \frac{dx}{f(x)} = \int dt

and carry out the integrations on both sides. If p is an antiderivative of 1/f, the solution will be:

p(x) = t + C

with C the freely varying parameter over \R: every particular value of C gives a solution. To express x as a function of t, we need to invert p. If we can do so, we'd have:

x = p^{-1}(t + C)

as the general solution.

In addition, there may be stationary solutions. These are solutions that correspond to constant functions x = k that satisfy f(k) = 0.

Related notions


Starting at time zero with value one

Suppose we want to solve the initial value problem for the differential equation:

\frac{dx}{dt} = f(x)

subject to the initial condition that at t = 0, x = 1.

We consider various possibilities for f a function that sends 1 and higher numbers to positive numbers, and make cases based on the growth rate of f:

Nature of f Nature of x in terms of t?
constant function f(x) := c linear function x := 1 + ct
f(x) := x^\gamma, 0 < \gamma < 1 x = O(t^{1/(1 - \gamma)}) (i.e., it grows roughly like a power function of t)
linear function f(x) := mx, m > 0 exponential function x := \exp(mt)
linear times logarithmic, something like x (\ln x + 1) x grows something like a doubly exponential function of t (note: if we wanted growth between exponential and double exponential, we would need something like x(\ln x + 1)^\gamma, 0 < \gamma < 1, and if we wanted triple exponential growth, we would multiply by a double logarithmic term)
f(x) := x^\gamma, \gamma > 1 x grows so fast in terms of t that it reaches \infty in finite time.